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o solve the systems of equations. 618. 4x − 5y = − 50 −x + 2y = 80 619. 1 100 3 100 9 100 x − 3 100 x − 7 100 x − 9 100 y + 1 20 y − 1 100 y − 9 100 z = − 49 z = 13 z = 99 For the following exercises, use Cramer’s Rule to solve the systems of equations. 620. 200x − 300y = 2 400x + 715y = 4 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1341 12 \| ANALYTIC GEOMETRY Figure 12.1 (a) Greek philosopher Aristotle (384–322 BCE) (b) German mathematician and astronomer Johannes Kepler (1571–1630) Chapter Outline 12.1 The Ellipse 12.2 The Hyperbola 12.3 The Parabola 12.4 Rotation of Axes 12.5 Conic Sections in Polar Coordinates Introduction The Greek mathematician Menaechmus (c. 380–c. 320 BCE) is generally credited with discovering the shapes formed by the intersection of a plane and a right circular cone. Depending on how he tilted the plane when it intersected the cone, he formed different shapes at the intersection–beautiful shapes with near-perfect symmetry. It was also said that Aristotle may have had an intuitive understanding of these shapes, as he observed the orbit of the planet to be circular. He presumed that the planets moved in circular orbits around Earth, and for nearly 2000 years this was the commonly held belief. It was not until the Renaissance movement that Johannes Kepler noticed that the orbits of the planet were not circular in nature. His published law of planetary motion in the 1600s changed our view of the solar system forever. He claimed that the sun was at one end of the orbits, and the planets revolved around the sun in an oval-shaped path. 1342 Chapter 12 Analytic Geometry In this chapter, we will investigate the two-dimensional figures that are formed when a right circular cone is intersected by a plane. We will begin by studying each of three figures created in this manner. We will develop defining equations for each figure and then learn how to use these equations to solve a variety of problems. 12.1 \| The Ellipse Learning Objectives In this section, you will: 12.1.1 Write equations of ellipses in standard form. 12.1.2 Graph ellipses centered at the origin. 12.1.3 Graph ellipses not centered at the origin. 12.1.4 Solve applied problems involving ellipses. Figure 12.2 The National Statuary Hall in Washington, D.C. (credit: Greg Palmer, Flickr) Can you imagine standing at one end of a large room and still being able to hear a whisper from a person standing at the other end? The National Statuary Hall in Washington, D.C., shown in Figure 12.2, is such a room.[1] It is an oval-shaped room called a whispering chamber because the shape makes it possible for sound to travel along the walls. In this section, we will investigate the shape of this room and its real-world applications, including how far apart two people in Statuary Hall can stand and still hear each other whisper. Writing Equations of Ellipses in Standard Form A conic section, or conic, is a shape resulting from intersecting a right circular cone with a plane. The angle at which the plane intersects the cone determines the shape, as shown in Figure 12.3. 1. Architect of the Capitol. http://www.aoc.gov. Accessed April 15, 2014. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1343 Figure 12.3 Conic sections can also be described by a set of points in the coordinate plane. Later in this chapter, we will see that the graph of any quadratic equation in two variables is a conic section. The signs of the equations and the coefficients of the variable terms determine the shape. This section focuses on the four variations of the standard form of the equation for the ellipse. An ellipse is the set of all points (x, y) in a plane such that the sum of their distances from two fixed points is a constant. Each fixed point is called a focus (plural: foci). We can draw an ellipse using a piece of cardboard, two thumbtacks, a pencil, and string. Place the thumbtacks in the cardboard to form the foci of the ellipse. Cut a piece of string longer than the distance between the two thumbtacks (the length of the string represents the constant in the definition). Tack each end of the string to the cardboard, and trace a curve with a pencil held taut against the string. The result is an ellipse. See Figure 12.4. Figure 12.4 1344 Chapter 12 Analytic Geometry Every ellipse has two axes of symmetry. The longer axis is called the major axis, and the shorter axis is called the minor axis. Each endpoint of the major axis is the vertex of the ellipse (plural: vertices), and each endpoint of the minor axis is a co-vertex of the ellipse. The center of an ellipse is the midpoint of both the major and minor axes. The axes are perpendicular at the center. The foci always lie on the major axis, and the sum of the distances from the foci to any point on the ellipse (the constant sum) is greater than the distance between the foci. See Figure 12.5. Figure 12.5 In this section, we restrict ellipses to those that are positioned vertically or horizontally in the coordinate plane. That is, the axes will either lie on or be parallel to the x- and y-axes. Later in the chapter, we will see ellipses that are rotated in the coordinate plane. To work with horizontal and vertical ellipses in the coordinate plane, we consider two cases: those that are centered at the origin and those that are centered at a point other than the origin. First we will learn to derive the equations of ellipses, and then we will learn how to write the equations of ellipses in standard form. Later we will use what we learn to draw the graphs. Deriving the Equation of an Ellipse Centered at the Origin To derive the equation of an ellipse centered at the origin, we begin with the foci (−c, 0) and (c, 0). The ellipse is the set of all points (x, y) such that the sum of the distances from (x, y) to the foci is constant, as shown in Figure 12.6. Figure 12.6 If (a, 0) is a vertex of the ellipse, the distance from (−c, 0) to (a, 0) is a − ( − c) = a + c. The distance from (c, 0) to (a, 0) is a − c . The sum of the distances from the foci to the vertex is If (x, y) is a point on the ellipse, then we can define the following variables: (a + c) + (a − c) = 2a This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1345 d1 = the distance from (−c, 0) to (x, y) d2 = the distance from (c, 0) to (x, y) By the definition of an ellipse, d1 + d2 is constant for any point (x, y) on the ellipse. We know that the sum of these distances is 2a for the vertex (a, 0). It follows that d1 + d2 = 2a for any point on the ellipse. We will begin the derivation by applying the distance formula. The rest of the derivation is algebraic. d1 + d2 = (x − ( − c))2 + (y − 0)2 + (x − c)2 + (y − 0)2 = 2a Distance formula (x + c)2 + y2 + (x − c)2 + y2 = 2a (x + c)2 + y2 = 2a − (x − c)2 + y2 ⎡ ⎣2a − (x − c)2 + y2⎤ ⎦ (x + c)2 + y2 = Square both sides. 2 Simplify expressions. Move radical to opposite side. 2 ⎣cx − a2⎤ ⎡ ⎦ x2 + 2cx + c2 + y2 = 4a2 − 4a (x − c)2 + y2 + (x − c)2 + y2 x2 + 2cx + c2 + y2 = 4a2 − 4a (x − c)2 + y2 + x2 − 2cx + c2 + y2 2cx = 4a2 − 4a (x − c)2 + y2 − 2cx 4cx − 4a2 = − 4a (x − c)2 + y2 cx − a2 = − a (x − c)2 + y2 ⎣ (x − c)2 + y2⎤ ⎦ ⎝x2 − 2cx + c2 + y2⎞ ⎠ = a2 ⎡ c2 x2 − 2a2 cx + a4 = a2 ⎛ c2 x2 − 2a2 cx + a4 = a2 x2 − 2a2 cx + a2 c2 + a2 y2 a2 x2 − c2 x2 + a2 y2 = a4 − a2 c2 x2 ⎛ ⎝a2 − c2⎞ ⎝a2 − c2⎞ ⎠ x2 b2 + a2 y2 = a2 b2 a2 b2 x2 b2 a2 b2 + a2 b2 x2 a2 + a2 y2 a2 b2 = y2 b2 = 1 ⎠ + a2 y2 = a2 ⎛ 2 Expand the squares. Expand remaining squares. Combine like terms. Isolate the radical. Divide by 4. Square both sides. Expand the squares. Distribute a2. Rewrite. Factor common terms. Set b2 = a2 − c2. Divide both sides by a2 b2. Simplify. Thus, the standard equation of an ellipse is x2 a2 + y2 b2 = 1. This equation defines an ellipse centered at the origin. If a > b, the ellipse is stretched further in the horizontal direction, and if b > a, direction. the ellipse is stretched further in the vertical Writing Equations of Ellipses Centered at the Origin in Standard Form Standard forms of equations tell us about key features of graphs. Take a moment to recall some of the standard forms of equations we’ve worked with in the past: linear, quadratic, cubic, exponential, logarithmic, and so on. By learning to interpret standard forms of equations, we are bridging the relationship between algebraic and geometric representations of mathematical phenomena. The key features of the ellipse are its center, vertices, co-vertices, foci, and lengths and positions of the major and minor axes. Just as with other equations, we can identify all of these features just by looking at the standard form of the equation. There are four variations of the standard form of the ellipse. These variations are categorized first by the location of the center (the origin or not the origin), and then by the position (horizontal or vertical). Each is presented along with a description of how the parts of the equation relate to the graph. Interpreting these parts allows us to form a mental picture of the ellipse. 1346 Chapter 12 Analytic Geometry Standard Forms of the Equation of an Ellipse with Center (0,0) The standard form of the equation of an ellipse with center (0, 0) and major axis on the x-axis is x2 a2 + y2 b2 = 1 where • a > b • • • • • the length of the major axis is 2a the coordinates of the vertices are (±a, 0) the length of the minor axis is 2b the coordinates of the co-vertices are (0, ± b) the coordinates of the foci are (±c, 0) , where c2 = a2 − b2. See Figure 12.7a The standard form of the equation of an ellipse with center (0, 0) and major axis on the y-axis is x2 b2 + y2 a2 = 1 where • a > b • • • • • the length of the major axis is 2a the coordinates of the vertices are (0, ± a) the length of the minor axis is 2b the coordinates of the co-vertices are (±b, 0) the coordinates of the foci ar
e (0, ± c) , where c2 = a2 − b2. See Figure 12.7b (12.1) (12.2) Note that the vertices, co-vertices, and foci are related by the equation c2 = a2 − b2. When we are given the coordinates of the foci and vertices of an ellipse, we can use this relationship to find the equation of the ellipse in standard form. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1347 Figure 12.7 (a) Horizontal ellipse with center (0, 0) (b) Vertical ellipse with center (0, 0) Given the vertices and foci of an ellipse centered at the origin, write its equation in standard form. 1. Determine whether the major axis lies on the x- or y-axis. a. b. If the given coordinates of the vertices and foci have the form (±a, 0) and ( ± c, 0) respectively, then the major axis is the x-axis. Use the standard form x2 y2 b2 = 1. a2 + If the given coordinates of the vertices and foci have the form (0, ± a) and ( ± c, 0), respectively, then the major axis is the y-axis. Use the standard form x2 y2 a2 = 1. b2 + 2. Use the equation c2 = a2 − b2, along with the given coordinates of the vertices and foci, to solve for b2. 3. Substitute the values for a2 and b2 into the standard form of the equation determined in Step 1. Example 12.1 Writing the Equation of an Ellipse Centered at the Origin in Standard Form What is the standard form equation of the ellipse that has vertices (±8, 0) and foci (±5, 0) ? Solution The foci are on the x-axis, so the major axis is the x-axis. Thus, the equation will have the form The vertices are (±8, 0), so a = 8 and a2 = 64. x2 a2 + y2 b2 = 1 1348 Chapter 12 Analytic Geometry The foci are (±5, 0), so c = 5 and c2 = 25. We know that the vertices and foci are related by the equation c2 = a2 − b2. Solving for b2, we have: c2 = a2 − b2 25 = 64 − b2 b2 = 39 Substitute for c2 and a2. Solve for b2. Now we need only substitute a2 = 64 and b2 = 39 into the standard form of the equation. The equation of the ellipse is x2 64 y2 39 = 1. + 12.1 What is the standard form equation of the ellipse that has vertices (0, ± 4) and foci ⎛ ⎝0, ± 15⎞ ⎠? Can we write the equation of an ellipse centered at the origin given coordinates of just one focus and vertex? Yes. Ellipses are symmetrical, so the coordinates of the vertices of an ellipse centered around the origin will always have the form (±a, 0) or (0, ± a). Similarly, the coordinates of the foci will always have the form (±c, 0) or (0, ± c). Knowing this, we can use a and c from the given points, along with the equation c2 = a2 − b2, find b2. to Writing Equations of Ellipses Not Centered at the Origin Like the graphs of other equations, the graph of an ellipse can be translated. If an ellipse is translated h units horizontally and k units vertically, the center of the ellipse will be (h, k). This translation results in the standard form of the equation we saw previously, with x replaced by (x − h) and y replaced by ⎛ ⎝y − k⎞ ⎠. Standard Forms of the Equation of an Ellipse with Center (h, k) The standard form of the equation of an ellipse with center (h, k) and major axis parallel to the x-axis is (x − h)2 a2 + where • a > b 2 (12.3) ⎛ ⎝y − k⎞ ⎠ b2 = 1 • • • • • the length of the major axis is 2a the coordinates of the vertices are (h ± a, k) the length of the minor axis is 2b the coordinates of the co-vertices are (h, k ± b) the coordinates of the foci are (h ± c, k), where c2 = a2 − b2. See Figure 12.8a The standard form of the equation of an ellipse with center (h, k) and major axis parallel to the y-axis is This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry where • a > b (x − h)2 b2 + 2 1349 (12.4) ⎛ ⎝y − k⎞ ⎠ a2 = 1 • • • • • the length of the major axis is 2a the coordinates of the vertices are (h, k ± a) the length of the minor axis is 2b the coordinates of the co-vertices are (h ± b, k) the coordinates of the foci are (h, k ± c), where c2 = a2 − b2. See Figure 12.8b Just as with ellipses centered at the origin, ellipses that are centered at a point (h, k) have vertices, co-vertices, and foci that are related by the equation c2 = a2 − b2. We can use this relationship along with the midpoint and distance formulas to find the equation of the ellipse in standard form when the vertices and foci are given. Figure 12.8 (a) Horizontal ellipse with center (h, k) (b) Vertical ellipse with center (h, k) 1350 Chapter 12 Analytic Geometry Given the vertices and foci of an ellipse not centered at the origin, write its equation in standard form. 1. Determine whether the major axis is parallel to the x- or y-axis. If the y-coordinates of the given vertices and foci are the same, then the major axis is parallel to the x-axis. Use the standard form (x − h)2 ⎝y − k⎞ 2 ⎛ ⎠ If the x-coordinates of the given vertices and foci are the same, then the major axis is parallel to the y-axis. Use the standard form (x − h)2 ⎝y − k⎞ 2 ⎠ ⎛ + + a2 b2 b2 = 1. a2 = 1. a. b. 2. Identify the center of the ellipse (h, k) using the midpoint formula and the given coordinates for the vertices. 3. Find a2 by solving for the length of the major axis, 2a, which is the distance between the given vertices. 4. Find c2 using h and k, found in Step 2, along with the given coordinates for the foci. 5. Solve for b2 using the equation c2 = a2 − b2. 6. Substitute the values for h, k, a2, and b2 into the standard form of the equation determined in Step 1. Example 12.2 Writing the Equation of an Ellipse Centered at a Point Other Than the Origin What is the standard form equation of the ellipse that has vertices (−2, −8) and (−2, 2) and foci (−2, −7) and (−2, 1) ? Solution The x-coordinates of the vertices and foci are the same, so the major axis is parallel to the y-axis. Thus, the equation of the ellipse will have the form (x − h)2 b2 + ⎛ ⎝y − k⎞ ⎠ 2 a2 = 1 First, we identify the center, (h, k). The center is halfway between the vertices, (−2, − 8) and (−2, 2). Applying the midpoint formula, we have: (h, k) = ⎛ ⎝ −2 + (−2) 2 , −8 + 2 2 ⎞ ⎠ = (−2, −3) Next, we find a2. The length of the major axis, 2a, distance between the y-coordinates of the vertices. is bounded by the vertices. We solve for a by finding the 2a = 2 − (−8) 2a = 10 a = 5 So a2 = 25. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1351 Now we find c2. The foci are given by (h, k ± c). So, (h, k − c) = (−2, −7) and (h, k + c) = (−2, 1). We substitute k = −3 using either of these points to solve for c. k + c = 1 −3 + c = 1 c = 4 So c2 = 16. Next, we solve for b2 using the equation c2 = a2 − b2. c2 = a2 − b2 16 = 25 − b2 b2 = 9 Finally, we substitute the values found for h, k, a2, and b2 into the standard form equation for an ellipse: (x + 2)2 9 + ⎛ ⎠ ⎝y + 3⎞ 25 2 = 1 12.2 What is the standard form equation of the ellipse that has vertices (−3, 3) and (5, 3) and foci ⎝1 − 2 3, 3⎞ ⎛ ⎠ and ⎛ ⎝1 + 2 3, 3⎞ ⎠? Graphing Ellipses Centered at the Origin Just as we can write the equation for an ellipse given its graph, we can graph an ellipse given its equation. To graph ellipses centered at the origin, we use the standard form x2 y2 b2 = 1, a > b for horizontal ellipses and x2 b2 + y2 a2 = 1, a > b for a2 + vertical ellipses. 1352 Chapter 12 Analytic Geometry Given the standard form of an equation for an ellipse centered at (0, 0), sketch the graph. 1. Use the standard forms of the equations of an ellipse to determine the major axis, vertices, co-vertices, and foci. a. b. If the equation is in the form x2 a2 + the major axis is the x-axis y2 b2 = 1, where a > b, then the coordinates of the vertices are (±a, 0) the coordinates of the co-vertices are (0, ± b) the coordinates of the foci are (±c, 0) If the equation is in the form x2 b2 + the major axis is the y-axis y2 a2 = 1, where a > b, then the coordinates of the vertices are (0, ± a) the coordinates of the co-vertices are (±b, 0) the coordinates of the foci are (0, ± c. Solve for c using the equation c2 = a2 − b2. 3. Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse. Example 12.3 Graphing an Ellipse Centered at the Origin Graph the ellipse given by the equation, x2 9 + y2 25 foci. = 1. Identify and label the center, vertices, co-vertices, and Solution First, we determine the position of the major axis. Because 25 > 9, y2 the equation is in the form x2 a2 = 1, where b2 = 9 and a2 = 25. It follows that: b2 + the major axis is on the y-axis. Therefore, • • • • the center of the ellipse is (0, 0) the coordinates of the vertices are (0, ± a) = ⎛ ⎝0, ± 25⎞ ⎠ = (0, ± 5) the coordinates of the co-vertices are (±b, 0) = ⎛ ⎝± 9, 0⎞ ⎠ = (±3, 0) the coordinates of the foci are (0, ± c), where c2 = a2 − b2 Solving for c, we have: This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1353 c = ± a2 − b2 = ± 25 − 9 = ± 16 = ± 4 Therefore, the coordinates of the foci are (0, ± 4). Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse. See Figure 12.9. Figure 12.9 12.3 Graph the ellipse given by the equation x2 36 + y2 4 and foci. = 1. Identify and label the center, vertices, co-vertices, Example 12.4 Graphing an Ellipse Centered at the Origin from an Equation Not in Standard Form Graph the ellipse given by the equation 4x2 + 25y2 = 100. Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci. Solution First, use algebra to rewrite the equation in standard form. 1354 Chapter 12 Analytic Geometry 4x2 + 25y2 = 100 25y2 4x2 100 100 y2 4 = 100 100 x2 25 = 1 + + Next, we determine the position of the major axis. Because 25 > 4, the major axis is on the x-axis. Therefore, the equation is in the form x2 y2 b2 = 1, where a2 = 25 and b2 = 4. It follows that: a2 + • • • • the center of the ellipse is (0, 0) the coordinates of the vertices are (±a, 0) = ⎛ ⎝± 25, 0⎞ ⎠ = (±5, 0) th
e coordinates of the co-vertices are (0, ± b) = ⎛ ⎝0, ± 4⎞ ⎠ = (0, ± 2) the coordinates of the foci are (±c, 0), where c2 = a2 − b2. Solving for c, we have: c = ± a2 − b2 = ± 25 − 4 = ± 21 Therefore the coordinates of the foci are ⎛ ⎝± 21, 0⎞ ⎠. Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse. Figure 12.10 12.4 Graph the ellipse given by the equation 49x2 + 16y2 = 784. Rewrite the equation in standard form. Then identify and label the center, vertices, co-vertices, and foci. Graphing Ellipses Not Centered at the Origin When an ellipse is not centered at the origin, we can still use the standard forms to find the key features of the graph. When the ellipse is centered at some point, (h, k), we use the standard forms (x − h)2 + 2 ⎠ ⎛ ⎝y − k⎞ b2 = 1, a > b for horizontal a2 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1355 ellipses and (x − h)2 b2 + 2 ⎛ ⎠ ⎝y − k⎞ a2 = 1, a > b for vertical ellipses. From these standard equations, we can easily determine the center, vertices, co-vertices, foci, and positions of the major and minor axes. Given the standard form of an equation for an ellipse centered at (h, k), sketch the graph. 1. Use the standard forms of the equations of an ellipse to determine the center, position of the major axis, vertices, co-vertices, and foci. a. b. If the equation is in the form (x − h)2 a2 + 2 ⎛ ⎠ ⎝y − k⎞ b2 = 1, where a > b, then ▪ ▪ ▪ ▪ ▪ the center is (h, k) the major axis is parallel to the x-axis the coordinates of the vertices are (h ± a, k) the coordinates of the co-vertices are (h, k ± b) the coordinates of the foci are (h ± c, k) If the equation is in the form (x − h)2 b2 + 2 ⎛ ⎠ ⎝y − k⎞ a2 = 1, where a > b, then ▪ ▪ ▪ ▪ ▪ the center is (h, k) the major axis is parallel to the y-axis the coordinates of the vertices are (h, k ± a) the coordinates of the co-vertices are (h ± b, k) the coordinates of the foci are (h, k ± c) 2. Solve for c using the equation c2 = a2 − b2. 3. Plot the center, vertices, co-vertices, and foci in the coordinate plane, and draw a smooth curve to form the ellipse. Example 12.5 Graphing an Ellipse Centered at (h, k) Graph the ellipse given by the equation, (x + 2)2 4 + vertices, and foci. ⎠ ⎛ ⎝y − 5⎞ 9 2 = 1. Identify and label the center, vertices, co- Solution First, we determine the position of the major axis. Because 9 > 4, Therefore, the equation is in the form (x − h)2 2 ⎛ the major axis is parallel to the y-axis. + ⎠ ⎝y − k⎞ a2 = 1, where b2 = 4 and a2 = 9. It follows that: b2 • the center of the ellipse is (h, k) = (−2, 5) 1356 Chapter 12 Analytic Geometry • • • the coordinates of (−2, 8) the vertices are (h, k ± a) = ( − 2, 5 ± 9) = ( − 2, 5 ± 3), or (−2, 2) and the coordinates of the co-vertices are (h ± b, k) = ( − 2 ± 4, 5) = ( − 2 ± 2, 5), (0, 5) or (−4, 5) and the coordinates of the foci are (h, k ± c), where c2 = a2 − b2. Solving for c, we have: c = ± a2 − b2 = ± 9 − 4 = ± 5 Therefore, the coordinates of the foci are ⎛ ⎝−2, 5 − 5⎞ ⎠ and ⎛ ⎝−2, 5+ 5⎞ ⎠. Next, we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse. ⎠ ⎛ ⎝y − 2⎞ 20 2 = 1. Identify and label the center, vertices, Figure 12.11 12.5 Graph the ellipse given by the equation (x − 4)2 36 + co-vertices, and foci. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1357 Given the general form of an equation for an ellipse centered at (h, k), express the equation in standard form. 1. Recognize that an ellipse described by an equation in the form ax2 + by2 + cx + dy + e = 0 is in general form. 2. Rearrange the equation by grouping terms that contain the same variable. Move the constant term to the opposite side of the equation. 3. Factor out the coefficients of the x2 and y2 terms in preparation for completing the square. 4. Complete the square for each variable to rewrite the equation in the form of the sum of multiples of two 2 = m3, where m1, m2, and m3 binomials squared set equal to a constant, m1 (x − h)2 + m2 are constants. ⎝y − k⎞ ⎛ ⎠ 5. Divide both sides of the equation by the constant term to express the equation in standard form. Example 12.6 Graphing an Ellipse Centered at (h, k) by First Writing It in Standard Form Graph the ellipse given by the equation 4x2 + 9y2 − 40x + 36y + 100 = 0. Identify and label the center, vertices, co-vertices, and foci. Solution We must begin by rewriting the equation in standard form. 4x2 + 9y2 − 40x + 36y + 100 = 0 Group terms that contain the same variable, and move the constant to the opposite side of the equation. Factor out the coefficients of the squared terms. ⎛ ⎝4x2 − 40x⎞ ⎠ + ⎛ ⎝9y2 + 36y⎞ ⎠ = −100 ⎛ ⎝x2 − 10x⎞ ⎛ ⎝y2 + 4y⎞ ⎠ + 9 ⎠ = −100 4 Complete the square twice. Remember to balance the equation by adding the same constants to each side. ⎛ ⎞ ⎛ ⎞ ⎝x2 − 10x + 25 ⎝y2 + 4y + 4 ⎠ + 9 ⎠ = −100 + 100 + 36 4 Rewrite as perfect squares. 4(x − 5)2 + 9⎛ ⎝y + 2⎞ ⎠ 2 = 36 Divide both sides by the constant term to place the equation in standard form. (x − 5)2 9 + ⎠ ⎛ ⎝y + 2⎞ 4 2 = 1 Now that the equation is in standard form, we can determine the position of the major axis. Because 9 > 4, the major axis is parallel to the x-axis. Therefore, the equation is in the form (x − h)2 2 ⎠ ⎛ + ⎝y − k⎞ b2 = 1, where a2 a2 = 9 and b2 = 4. It follows that: • the center of the ellipse is (h, k) = (5, −2) 1358 Chapter 12 Analytic Geometry • • • the coordinates of the vertices are (h ± a, k) = ⎛ ⎝5 ± 9, −2⎞ ⎠ = (5 ± 3, −2), or (2, −2) and (8, −2) the coordinates of the co-vertices are (h, k ± b) = ⎛ ⎝5, −2 ± 4⎞ ⎠ = (5, −2 ± 2), or (5, −4) and (5, 0) the coordinates of the foci are (h ± c, k), where c2 = a2 − b2. Solving for c, we have: c = ± a2 − b2 = ± 9 − 4 = ± 5 Therefore, the coordinates of the foci are ⎛ ⎝5 − 5, −2⎞ ⎠ and ⎛ ⎝5+ 5, −2⎞ ⎠. Next we plot and label the center, vertices, co-vertices, and foci, and draw a smooth curve to form the ellipse as shown in Figure 12.12. Figure 12.12 Express the equation of the ellipse given in standard form. Identify the center, vertices, co-vertices, and 12.6 foci of the ellipse. 4x2 + y2 − 24x + 2y + 21 = 0 Solving Applied Problems Involving Ellipses Many real-world situations can be represented by ellipses, including orbits of planets, satellites, moons and comets, and shapes of boat keels, rudders, and some airplane wings. A medical device called a lithotripter uses elliptical reflectors to break up kidney stones by generating sound waves. Some buildings, called whispering chambers, are designed with elliptical domes so that a person whispering at one focus can easily be heard by someone standing at the other focus. This occurs because of the acoustic properties of an ellipse. When a sound wave originates at one focus of a whispering chamber, the sound wave will be reflected off the elliptical dome and back to the other focus. See Figure 12.13. In the whisper chamber at the Museum of Science and Industry in Chicago, two people standing at the foci—about 43 feet apart—can hear each other whisper. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1359 Figure 12.13 Sound waves are reflected between foci in an elliptical room, called a whispering chamber. Example 12.7 Locating the Foci of a Whispering Chamber The Statuary Hall in the Capitol Building in Washington, D.C. is a whispering chamber. Its dimensions are 46 feet wide by 96 feet long as shown in Figure 12.14. a. What is the standard form of the equation of the ellipse representing the outline of the room? Hint: assume a horizontal ellipse, and let the center of the room be the point (0, 0). b. If two senators standing at the foci of this room can hear each other whisper, how far apart are the senators? Round to the nearest foot. Figure 12.14 Solution a. We are assuming a horizontal ellipse with center (0, 0), so we need to find an equation of the form y2 x2 b2 = 1, where a > b. We know that the length of the major axis, 2a, is longer than the length a2 + of the minor axis, 2b. So the length of the room, 96, is represented by the major axis, and the width of the room, 46, is represented by the minor axis. ◦ Solving for a, we have 2a = 96, so a = 48, and a2 = 2304. ◦ Solving for b, we have 2b = 46, so b = 23, and b2 = 529. 1360 Chapter 12 Analytic Geometry Therefore, the equation of the ellipse is x2 2304 + y2 529 = 1. b. To find the distance between the senators, we must find the distance between the foci, (±c, 0), where c2 = a2 − b2. Solving for c, we have: c2 = a2 − b2 c2 = 2304 − 529 c = ± 2304 − 529 c = ± 1775 c ≈ ± 42 Substitute using the values found in part (a). Take the square root of both sides. Subtract. Round to the nearest foot. The points (±42, 0) represent the foci. Thus, the distance between the senators is 2(42) = 84 feet. 12.7 Suppose a whispering chamber is 480 feet long and 320 feet wide. a. What is the standard form of the equation of the ellipse representing the room? Hint: assume a horizontal ellipse, and let the center of the room be the point (0, 0). b. If two people are standing at the foci of this room and can hear each other whisper, how far apart are the people? Round to the nearest foot. Access these online resources for additional instruction and practice with ellipses. • Conic Sections: The Ellipse (http://openstaxcollege.org/l/conicellipse) • Graph an Ellipse with Center at the Origin (http://openstaxcollege.org/l/grphellorigin) • Graph an Ellipse with Center Not at the Origin (http://openstaxcollege.org/l/grphellnot) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1361 12.1 EXERCISES Verbal 1. Define an ellipse in terms of its foci. 2. Where must the foci of an ellipse lie? What special case of the ellipse do we have when the 3. major and minor axis are of the same length? For the special case mentioned
above, what would be 4. true about the foci of that ellipse? What can be said about the symmetry of the graph of an 5. ellipse with center at the origin and foci along the y-axis? Algebraic For the following exercises, determine whether the given equations represent ellipses. If yes, write in standard form. 6. 7. 8. 9. 2x2 + y = 4 4x2 + 9y2 = 36 4x2 − y2 = 4 4x2 + 9y2 = 1 10. 4x2 − 8x + 9y2 − 72y + 112 = 0 For the following exercises, write the equation of an ellipse in standard form, and identify the end points of the major and minor axes as well as the foci. (x − 7)2 49 + ⎛ ⎠ ⎝y − 7⎞ 49 2 = 1 19. 20. 21. 22. 23. 24. 25. 26. 4x2 − 8x + 9y2 − 72y + 112 = 0 9x2 − 54x + 9y2 − 54y + 81 = 0 4x2 − 24x + 36y2 − 360y + 864 = 0 4x2 + 24x + 16y2 − 128y + 228 = 0 4x2 + 40x + 25y2 − 100y + 100 = 0 x2 + 2x + 100y2 − 1000y + 2401 = 0 4x2 + 24x + 25y2 + 200y + 336 = 0 9x2 + 72x + 16y2 + 16y + 4 = 0 For the following exercises, find the foci for the given ellipses. 27. 28. 29. 30. 31. (x + 3)2 25 + (x + 1)2 100 + ⎛ ⎠ ⎝y + 1⎞ 36 ⎛ ⎠ ⎝y − 2⎞ 4 2 2 = 1 = 1 x2 + y2 = 1 x2 + 4y2 + 4x + 8y = 1 10x2 + y2 + 200x = 0 x2 4 + y2 49 = 1 x2 100 + y2 64 = 1 x2 + 9y2 = 1 4x2 + 16y2 = 1 (x − 2)2 49 + (x − 2)2 81 + (x + 5)2 4 + ⎠ ⎛ ⎝y − 4⎞ 25 ⎠ ⎛ ⎝y + 1⎞ 16 ⎛ ⎠ ⎝y − 7⎞ 9 11. 12. 13. 14. 15. 16. 17. 18. Graphical For the following exercises, graph the given ellipses, noting center, vertices, and foci x2 25 x2 16 + + y2 36 y2 9 = 1 = 1 4x2 + 9y2 = 1 81x2 + 49y2 = 1 32. 33. 34. 35. 36. 1362 (x − 2)2 64 + ⎛ ⎠ ⎝y − 4⎞ 16 2 = 1 Chapter 12 Analytic Geometry 37. 38. 39. 40. 41. 42. 43. 44. 45. (x + 3)2 9 + ⎛ ⎠ ⎝y − 3⎞ 9 2 = 1 x2 2 + ⎛ ⎠ ⎝y + 1⎞ 5 2 = 1 4x2 − 8x + 16y2 − 32y − 44 = 0 x2 − 8x + 25y2 − 100y + 91 = 0 x2 + 8x + 4y2 − 40y + 112 = 0 64x2 + 128x + 9y2 − 72y − 368 = 0 16x2 + 64x + 4y2 − 8y + 4 = 0 100x2 + 1000x + y2 − 10y + 2425 = 0 53. 4x2 + 16x + 4y2 + 16y + 16 = 0 For the following exercises, use the given information about the graph of each ellipse to determine its equation. Center at the origin, symmetric with respect to the x- 46. and y-axes, focus at (4, 0), and point on graph (0, 3). Center at the origin, symmetric with respect to the x- 47. and y-axes, focus at (0, −2), and point on graph (5, 0). Center at the origin, symmetric with respect to the x48. and y-axes, focus at (3, 0), and major axis is twice as long as minor axis. Center (4, 2) ; vertex (9, 2) ; one focus: ⎛ ⎝4 + 2 6, 2⎞ ⎠ 49. . 50. Center (3, 5) ; vertex (3, 11) ; one focus: ⎛ ⎝3, 5+4 2⎞ ⎠ 51. Center (−3, 4) ; vertex (1, 4) ; one focus: ⎝−3 + 2 3, 4⎞ ⎛ ⎠ For the following exercises, given the graph of the ellipse, determine its equation. 52. This content is available for free at https://cnx.org/content/col11758/1.5 54. 55. Chapter 12 Analytic Geometry 1363 height to the nearest 0.01 foot of the arch at a distance of 4 feet from the center. An arch has the shape of a semi-ellipse. The arch has a 65. height of 12 feet and a span of 40 feet. Find an equation for the ellipse, and use that to find the distance from the center to a point at which the height is 6 feet. Round to the nearest hundredth. A bridge is to be built in the shape of a semi-elliptical 66. arch and is to have a span of 120 feet. The height of the arch at a distance of 40 feet from the center is to be 8 feet. Find the height of the arch at its center. A person in a whispering gallery standing at one focus 67. of the ellipse can whisper and be heard by a person standing at the other focus because all the sound waves that reach the ceiling are reflected to the other person. If a whispering gallery has a length of 120 feet, and the foci are located 30 feet from the center, find the height of the ceiling at the center. A person is standing 8 feet from the nearest wall in a 68. whispering gallery. If that person is at one focus, and the other focus is 80 feet away, what is the length and height at the center of the gallery? 56. Extensions For the following exercises, find the area of the ellipse. The area of an ellipse is given by the formula Area = a ⋅ b ⋅ π. 57. 58. 59. 60. 61. (x − 3)2 9 + (x + 6)2 16 + (x + 1)2 4 + ⎛ ⎠ ⎝y − 3⎞ 16 ⎛ ⎠ ⎝y − 6⎞ 36 ⎛ ⎠ ⎝y − 2 4x2 − 8x + 9y2 − 72y + 112 = 0 9x2 − 54x + 9y2 − 54y + 81 = 0 Real-World Applications Find the equation of the ellipse that will just fit inside a 62. box that is 8 units wide and 4 units high. Find the equation of the ellipse that will just fit inside a 63. box that is four times as wide as it is high. Express in terms of h, the height. An arch has the shape of a semi-ellipse (the top half of 64. an ellipse). The arch has a height of 8 feet and a span of 20 feet. Find an equation for the ellipse, and use that to find the 1364 Chapter 12 Analytic Geometry 12.2 \| The Hyperbola Learning Objectives In this section, you will: 12.2.1 Locate a hyperbola’s vertices and foci. 12.2.2 Write equations of hyperbolas in standard form. 12.2.3 Graph hyperbolas centered at the origin. 12.2.4 Graph hyperbolas not centered at the origin. 12.2.5 Solve applied problems involving hyperbolas. What do paths of comets, supersonic booms, ancient Grecian pillars, and natural draft cooling towers have in common? They can all be modeled by the same type of conic. For instance, when something moves faster than the speed of sound, a shock wave in the form of a cone is created. A portion of a conic is formed when the wave intersects the ground, resulting in a sonic boom. See Figure 12.15. Figure 12.15 A shock wave intersecting the ground forms a portion of a conic and results in a sonic boom. Most people are familiar with the sonic boom created by supersonic aircraft, but humans were breaking the sound barrier long before the first supersonic flight. The crack of a whip occurs because the tip is exceeding the speed of sound. The bullets shot from many firearms also break the sound barrier, although the bang of the gun usually supersedes the sound of the sonic boom. Locating the Vertices and Foci of a Hyperbola In analytic geometry, a hyperbola is a conic section formed by intersecting a right circular cone with a plane at an angle such that both halves of the cone are intersected. This intersection produces two separate unbounded curves that are mirror images of each other. See Figure 12.16. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1365 Figure 12.16 A hyperbola Like the ellipse, the hyperbola can also be defined as a set of points in the coordinate plane. A hyperbola is the set of all points (x, y) in a plane such that the difference of the distances between (x, y) and the foci is a positive constant. Notice that the definition of a hyperbola is very similar to that of an ellipse. The distinction is that the hyperbola is defined in terms of the difference of two distances, whereas the ellipse is defined in terms of the sum of two distances. As with the ellipse, every hyperbola has two axes of symmetry. The transverse axis is a line segment that passes through the center of the hyperbola and has vertices as its endpoints. The foci lie on the line that contains the transverse axis. The conjugate axis is perpendicular to the transverse axis and has the co-vertices as its endpoints. The center of a hyperbola is the midpoint of both the transverse and conjugate axes, where they intersect. Every hyperbola also has two asymptotes that pass through its center. As a hyperbola recedes from the center, its branches approach these asymptotes. The central rectangle of the hyperbola is centered at the origin with sides that pass through each vertex and co-vertex; it is a useful tool for graphing the hyperbola and its asymptotes. To sketch the asymptotes of the hyperbola, simply sketch and extend the diagonals of the central rectangle. See Figure 12.17. Figure 12.17 Key features of the hyperbola 1366 Chapter 12 Analytic Geometry In this section, we will limit our discussion to hyperbolas that are positioned vertically or horizontally in the coordinate plane; the axes will either lie on or be parallel to the x- and y-axes. We will consider two cases: those that are centered at the origin, and those that are centered at a point other than the origin. Deriving the Equation of an Ellipse Centered at the Origin Let (−c, 0) and (c, 0) be the foci of a hyperbola centered at the origin. The hyperbola is the set of all points (x, y) such that the difference of the distances from (x, y) to the foci is constant. See Figure 12.18. Figure 12.18 If (a, 0) is a vertex of the hyperbola, the distance from (−c, 0) to (a, 0) is a − (−c) = a + c. The distance from (c, 0) to (a, 0) is c − a. The sum of the distances from the foci to the vertex is If (x, y) is a point on the hyperbola, we can define the following variables: (a + c) − (c − a) = 2a d2 = the distance from (−c, 0) to (x, y) d1 = the distance from (c, 0) to (x, y) By definition of a hyperbola, d2 − d1 is constant for any point (x, y) on the hyperbola. We know that the difference of these distances is 2a for the vertex (a, 0). It follows that d2 − d1 = 2a for any point on the hyperbola. As with the derivation of the equation of an ellipse, we will begin by applying the distance formula. The rest of the derivation is algebraic. Compare this derivation with the one from the previous section for ellipses. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1367 d2 − d1 = (x − ( − c))2 + (y − 0)2 − (x − c)2 + (y − 0)2 = 2a Distance Formula (x + c)2 + y2 − (x − c)2 + y2 = 2a (x + c)2 + y2 = 2a + (x − c)2 + y2 ⎛ ⎝2a + (x − c)2 + y2⎞ ⎠ (x + c)2 + y2 = Square both sides. 2 Simplify expressions. Move radical to opposite side. x2 + 2cx + c2 + y2 = 4a2 + 4a (x − c)2 + y2 + (x − c)2 + y2 x2 + 2cx + c2 + y2 = 4a2 + 4a (x − c)2 + y2 + x2 − 2cx + c2 + y2 2cx = 4a2 + 4a (x − c)2 + y2 − 2cx 4cx − 4a2 = 4a (x − c)2 + y2 cx − a2 = a (x − c)2 + y2 2 2 ⎝cx − a2⎞ ⎛ ⎠ ⎣ (x − c)2 + y2⎤ ⎦ ⎝x2 − 2cx + c2 + y2⎞ ⎠ = a2 ⎡ c2 x2 − 2a2 cx + a4 = a2 ⎛ c2 x2 − 2a2 cx + a4 = a2 x
2 − 2a2 cx + a2 c2 + a2 y2 a4 + c2 x2 = a2 x2 + a2 c2 + a2 y2 c2 x2 − a2 x2 − a2 y2 = a2 c2 − a4 ⎝c2 − a2⎞ ⎠ − a2 y2 = a2 ⎛ ⎝c2 − a2⎞ x2 ⎛ ⎠ x2 b2 − a2 y2 = a2 b2 a2 b2 x2 b2 a2 b2 − a2 b2 x2 a2 − a2 y2 a2 b2 = y2 b2 = 1 Expand the squares. Expand remaining square. Combine like terms. Isolate the radical. Divide by 4. Square both sides. Expand the squares. Distribute a2 . Combine like terms. Rearrange terms. Factor common terms. Set b2 = c2 − a2. Divide both sides by a2 b2 This equation defines a hyperbola centered at the origin with vertices (±a, 0) and co-vertices (0 ± b). Standard Forms of the Equation of a Hyperbola with Center (0,0) The standard form of the equation of a hyperbola with center (0, 0) and transverse axis on the x-axis is x2 a2 − y2 b2 = 1 (12.5) where • • • • • • • the length of the transverse axis is 2a the coordinates of the vertices are (±a, 0) the length of the conjugate axis is 2b the coordinates of the co-vertices are (0, ± b) the distance between the foci is 2c, where c2 = a2 + b2 the coordinates of the foci are (±c, 0) the equations of the asymptotes are y = ± b ax See Figure 12.19a. The standard form of the equation of a hyperbola with center (0, 0) and transverse axis on the y-axis is Chapter 12 Analytic Geometry (12.6) 1368 where • • • • • • • y2 a2 − x2 b2 = 1 the length of the transverse axis is 2a the coordinates of the vertices are (0, ± a) the length of the conjugate axis is 2b the coordinates of the co-vertices are (±b, 0) the distance between the foci is 2c, where c2 = a2 + b2 the coordinates of the foci are (0, ± c) the equations of the asymptotes are y = ± x a b See Figure 12.19b. Note that the vertices, co-vertices, and foci are related by the equation c2 = a2 + b2. When we are given the equation of a hyperbola, we can use this relationship to identify its vertices and foci. Figure 12.19 (a) Horizontal hyperbola with center (0, 0) (b) Vertical hyperbola with center (0, 0) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1369 Given the equation of a hyperbola in standard form, locate its vertices and foci. 1. Determine whether the transverse axis lies on the x- or y-axis. Notice that a2 is always under the variable with the positive coefficient. So, if you set the other variable equal to zero, you can easily find the intercepts. In the case where the hyperbola is centered at the origin, the intercepts coincide with the vertices. a. b. If the equation has the form x2 a2 − are located at ( ± a, 0), and the foci are located at (±c, 0). y2 b2 = 1, then the transverse axis lies on the x-axis. The vertices If the equation has the form y2 a2 − are located at (0, ± a), and the foci are located at (0, ± c). x2 b2 = 1, then the transverse axis lies on the y-axis. The vertices 2. Solve for a using the equation a = a2. 3. Solve for c using the equation c = a2 + b2. Example 12.8 Locating a Hyperbola’s Vertices and Foci Identify the vertices and foci of the hyperbola with equation y2 49 − x2 32 = 1. Solution The equation has the form y2 a2 − x2 b2 = 1, so the transverse axis lies on the y-axis. The hyperbola is centered at the origin, so the vertices serve as the y-intercepts of the graph. To find the vertices, set x = 0, and solve for y. − 1 = 1 = x2 32 − 02 32 y2 49 y2 49 y2 49 y2 = 49 y = ± 49 = ± 7 1 = The foci are located at (0, ± c). Solving for c, Therefore, the vertices are located at (0, ± 7), and the foci are located at (0, 9). c = a2 + b2 = 49 + 32 = 81 = 9 12.8 Identify the vertices and foci of the hyperbola with equation x2 9 − y2 25 = 1. 1370 Chapter 12 Analytic Geometry Writing Equations of Hyperbolas in Standard Form Just as with ellipses, writing the equation for a hyperbola in standard form allows us to calculate the key features: its center, vertices, co-vertices, foci, asymptotes, and the lengths and positions of the transverse and conjugate axes. Conversely, an equation for a hyperbola can be found given its key features. We begin by finding standard equations for hyperbolas centered at the origin. Then we will turn our attention to finding standard equations for hyperbolas centered at some point other than the origin. Hyperbolas Centered at the Origin Reviewing the standard forms given for hyperbolas centered at (0, 0), we see that the vertices, co-vertices, and foci are related by the equation c2 = a2 + b2. Note that this equation can also be rewritten as b2 = c2 − a2. This relationship is used to write the equation for a hyperbola when given the coordinates of its foci and vertices. Given the vertices and foci of a hyperbola centered at (0, 0), write its equation in standard form. 1. Determine whether the transverse axis lies on the x- or y-axis. a. b. If the given coordinates of the vertices and foci have the form (±a, 0) and (±c, 0), respectively, then the transverse axis is the x-axis. Use the standard form x2 y2 b2 = 1. a2 − If the given coordinates of the vertices and foci have the form (0, ± a) and (0, ± c), respectively, then the transverse axis is the y-axis. Use the standard form y2 a2 − x2 b2 = 1. 2. Find b2 using the equation b2 = c2 − a2. 3. Substitute the values for a2 and b2 into the standard form of the equation determined in Step 1. Example 12.9 Finding the Equation of a Hyperbola Centered at (0,0) Given its Foci and Vertices What is the standard form equation of the hyperbola that has vertices (±6, 0) and foci ⎛ ⎝±2 10, 0⎞ ⎠? Solution The vertices and foci are on the x-axis. Thus, the equation for the hyperbola will have the form x2 a2 − y2 b2 = 1. The vertices are (±6, 0), so a = 6 and a2 = 36. The foci are ⎛ ⎝±2 10, 0⎞ ⎠, so c = 2 10 and c2 = 40. Solving for b2, we have b2 = c2 − a2 b2 = 40 − 36 b2 = 4 Substitute for c2 and a2. Subtract. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1371 Finally, we substitute a2 = 36 and b2 = 4 into the standard form of the equation, x2 a2 − y2 b2 = 1. The equation of the hyperbola is x2 36 − y2 4 = 1, as shown in Figure 12.20. Figure 12.20 12.9 What is the standard form equation of the hyperbola that has vertices (0, ± 2) and foci ⎛ ⎝0, ± 2 5⎞ ⎠? Hyperbolas Not Centered at the Origin Like the graphs for other equations, the graph of a hyperbola can be translated. If a hyperbola is translated h units horizontally and k units vertically, the center of the hyperbola will be (h, k). This translation results in the standard form of the equation we saw previously, with x replaced by (x − h) and y replaced by ⎛ ⎝y − k⎞ ⎠. Standard Forms of the Equation of a Hyperbola with Center (h, k) The standard form of the equation of a hyperbola with center (h, k) and transverse axis parallel to the x-axis is (x − h)2 a2 − 2 ⎛ ⎠ ⎝y − k⎞ b2 = 1 (12.7) where • • • • • • the length of the transverse axis is 2a the coordinates of the vertices are (h ± a, k) the length of the conjugate axis is 2b the coordinates of the co-vertices are (h, k ± b) the distance between the foci is 2c, where c2 = a2 + b2 the coordinates of the foci are (h ± c, k) The asymptotes of the hyperbola coincide with the diagonals of the central rectangle. The length of the rectangle is b 2a and its width is 2b. The slopes of the diagonals are ± a, and each diagonal passes through the center (h, k). 1372 Chapter 12 Analytic Geometry Using the point-slope formula, it is simple to show that the equations of the asymptotes are y = ± Figure 12.21a b a(x − h) + k. See The standard form of the equation of a hyperbola with center (h, k) and transverse axis parallel to the y-axis is ⎛ ⎝y − k⎞ ⎠ 2 a2 − (x − h)2 b2 = 1 (12.8) where • • • • • • the length of the transverse axis is 2a the coordinates of the vertices are (h, k ± a) the length of the conjugate axis is 2b the coordinates of the co-vertices are (h ± b, k) the distance between the foci is 2c, where c2 = a2 + b2 the coordinates of the foci are (h, k ± c) Using the reasoning above, the equations of the asymptotes are y = ± a b(x − h) + k. See Figure 12.21b. Figure 12.21 (a) Horizontal hyperbola with center (h, k) (b) Vertical hyperbola with center (h, k) Like hyperbolas centered at the origin, hyperbolas centered at a point (h, k) have vertices, co-vertices, and foci that are related by the equation c2 = a2 + b2. We can use this relationship along with the midpoint and distance formulas to find the standard equation of a hyperbola when the vertices and foci are given. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1373 Given the vertices and foci of a hyperbola centered at (h, k), write its equation in standard form. 1. Determine whether the transverse axis is parallel to the x- or y-axis. a. If the y-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the x-axis. Use the standard form (x − h)2 ⎝y − k⎞ 2 ⎠ ⎛ a2 − b2 = 1. b. If the x-coordinates of the given vertices and foci are the same, then the transverse axis is parallel to the y-axis. Use the standard form a2 − Identify the center of the hyperbola, (h, k), using the midpoint formula and the given coordinates for the vertices. = 1. ⎛ ⎝y − k⎞ ⎠ 2 (x − h)2 b2 2. 3. Find a2 by solving for the length of the transverse axis, 2a , which is the distance between the given vertices. 4. Find c2 using h and k found in Step 2 along with the given coordinates for the foci. 5. Solve for b2 using the equation b2 = c2 − a2. 6. Substitute the values for h, k, a2, and b2 into the standard form of the equation determined in Step 1. Example 12.10 Finding the Equation of a Hyperbola Centered at (h, k) Given its Foci and Vertices What is the standard form equation of the hyperbola that has vertices at (0, −2) and (6, −2) and foci at (−2, −2) and (8, −2) ? Solution The y-coordinates of the vertices and foci are the same, so the transverse axis is parallel to the x-axis. Thus, the equation of the hyperbola will have the form (x − h)2 a2 − ⎛ ⎝y − k⎞ ⎠ 2 b2 = 1 First, we
identify the center, (h, k). The center is halfway between the vertices (0, −2) and (6, −2). Applying the midpoint formula, we have (h, k2 + (−2) 2 ⎞ ⎠ = (3, −2) Next, we find a2. The length of the transverse axis, 2a, finding the distance between the x-coordinates of the vertices. is bounded by the vertices. So, we can find a2 by 2a = \|0 − 6\| 2a = 6 a = 3 a2 = 9 1374 Chapter 12 Analytic Geometry Now we need to find c2. The coordinates of the foci are (h ± c, k). So (h − c, k) = (−2, −2) and (h + c, k) = (8, −2). We can use the x-coordinate from either of these points to solve for c. Using the point (8, −2), and substituting h = 3 c2 = 25 Next, solve for b2 using the equation b2 = c2 − a2 : b2 = c2 − a2 = 25 − 9 = 16 Finally, substitute the values found for h, k, a2, and b2 into the standard form of the equation. (x − 3)2 9 − (y + 2)2 16 = 1 What is the standard form equation of the hyperbola that has vertices (1, −2) and (1, 8) and foci 12.10 (1, −10) and (1, 16) ? Graphing Hyperbolas Centered at the Origin When we have an equation in standard form for a hyperbola centered at the origin, we can interpret its parts to identify the key features of its graph: the center, vertices, co-vertices, asymptotes, foci, and lengths and positions of the transverse and conjugate axes. To graph hyperbolas centered at the origin, we use the standard form x2 y2 b2 = 1 for horizontal a2 − hyperbolas and the standard form y2 a2 − x2 b2 = 1 for vertical hyperbolas. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1375 Given a standard form equation for a hyperbola centered at (0, 0), sketch the graph. 1. Determine which of the standard forms applies to the given equation. 2. Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the vertices, co-vertices, and foci; and the equations for the asymptotes. a. b. If the equation is in the form x2 a2 − y2 b2 = 1, then ▪ ▪ ▪ ▪ ▪ the transverse axis is on the x-axis the coordinates of the vertices are (±a, 0) the coordinates of the co-vertices are (0, ± b) the coordinates of the foci are (±c, 0) the equations of the asymptotes are y = ± b ax If the equation is in the form y2 x2 b2 = 1, the transverse axis is on the y-axis a2 − then ▪ ▪ ▪ ▪ ▪ the coordinates of the vertices are (0, ± a) the coordinates of the co-vertices are (±b, 0) the coordinates of the foci are (0, ± c) the equations of the asymptotes are y = ± x a b 3. Solve for the coordinates of the foci using the equation c = ± a2 + b2. 4. Plot the vertices, co-vertices, foci, and asymptotes in the coordinate plane, and draw a smooth curve to form the hyperbola. Example 12.11 Graphing a Hyperbola Centered at (0, 0) Given an Equation in Standard Form Graph the hyperbola given by the equation y2 64 − x2 36 asymptotes. = 1. Identify and label the vertices, co-vertices, foci, and Solution The standard form that applies to the given equation is y2 a2 − x2 b2 = 1. Thus, the transverse axis is on the y-axis The coordinates of the vertices are (0, ± a) = ⎛ ⎝0, ± 64⎞ ⎠ = (0, ± 8) The coordinates of the co-vertices are (±b, 0) = ⎛ ⎝± 36, 0⎞ ⎠ = (±6, 0) The coordinates of the foci are (0, ± c), where c = ± a2 + b2. Solving for c, we have 1376 Chapter 12 Analytic Geometry c = ± a2 + b2 = ± 64 + 36 = ± 100 = ± 10 Therefore, the coordinates of the foci are (0, ± 10) The equations of the asymptotes are Plot and label the vertices and co-vertices, and then sketch the central rectangle. Sides of the rectangle are parallel to the axes and pass through the vertices and co-vertices. Sketch and extend the diagonals of the central rectangle to show the asymptotes. The central rectangle and asymptotes provide the framework needed to sketch an accurate graph of the hyperbola. Label the foci and asymptotes, and draw a smooth curve to form the hyperbola, as shown in Figure 12.22. Figure 12.22 12.11 Graph the hyperbola given by the equation x2 144 − y2 81 foci, and asymptotes. = 1. Identify and label the vertices, co-vertices, Graphing Hyperbolas Not Centered at the Origin Graphing hyperbolas centered at a point (h, k) other than the origin is similar to graphing ellipses centered at a point other than the origin. We use the standard forms (x − h)2 a2 − 2 ⎠ ⎛ ⎝y − k⎞ b2 = 1 for horizontal hyperbolas, and ⎛ ⎝y − k⎞ ⎠ 2 a2 − (x − h)2 b2 = 1 for vertical hyperbolas. From these standard form equations we can easily calculate and plot key This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1377 features of the graph: the coordinates of its center, vertices, co-vertices, and foci; the equations of its asymptotes; and the positions of the transverse and conjugate axes. Given a general form for a hyperbola centered at (h, k), sketch the graph. 1. Convert the general form to that standard form. Determine which of the standard forms applies to the given equation. 2. Use the standard form identified in Step 1 to determine the position of the transverse axis; coordinates for the center, vertices, co-vertices, foci; and equations for the asymptotes. a. If the equation is in the form (x − h)2 a2 − 2 ⎛ ⎠ ⎝y − k⎞ b2 = 1, then ▪ ▪ ▪ ▪ ▪ ▪ the transverse axis is parallel to the x-axis the center is (h, k) the coordinates of the vertices are (h ± a, k) the coordinates of the co-vertices are (h, k ± b) the coordinates of the foci are (h ± c, k) the equations of the asymptotes are y = ± b a(x − h) + k b. If the equation is in the form ⎛ ⎝y − k⎞ ⎠ 2 a2 − (x − h)2 b2 = 1, then ▪ ▪ ▪ ▪ ▪ ▪ the transverse axis is parallel to the y-axis the center is (h, k) the coordinates of the vertices are (h, k ± a) the coordinates of the co-vertices are (h ± b, k) the coordinates of the foci are (h, k ± c) the equations of the asymptotes are y = ± a b(x − h) + k 3. Solve for the coordinates of the foci using the equation c = ± a2 + b2. 4. Plot the center, vertices, co-vertices, foci, and asymptotes in the coordinate plane and draw a smooth curve to form the hyperbola. Example 12.12 Graphing a Hyperbola Centered at (h, k) Given an Equation in General Form Graph the hyperbola given by the equation 9x2 − 4y2 − 36x − 40y − 388 = 0. Identify and label the center, vertices, co-vertices, foci, and asymptotes. Solution Start by expressing the equation in standard form. Group terms that contain the same variable, and move the constant to the opposite side of the equation. ⎝9x2 − 36x⎞ ⎛ ⎠ − ⎝4y2 + 40y⎞ ⎛ ⎠ = 388 1378 Chapter 12 Analytic Geometry Factor the leading coefficient of each expression. ⎝x2 − 4x⎞ ⎛ ⎝y2 + 10y⎞ ⎛ ⎠ − 4 ⎠ = 388 9 Complete the square twice. Remember to balance the equation by adding the same constants to each side. ⎞ ⎛ ⎞ ⎛ ⎝y2 + 10y + 25 ⎝x2 − 4x + 4 ⎠ = 388 + 36 − 100 ⎠ − 4 9 Rewrite as perfect squares. Divide both sides by the constant term to place the equation in standard form. 9(x − 2)2 − 4⎛ ⎝y + 5⎞ ⎠ 2 = 324 (x − 2)2 36 − ⎛ ⎠ ⎝y + 5⎞ 81 2 = 1 The standard form that applies to the given equation is (x − h)2 a2 − 2 ⎛ ⎠ ⎝y − k⎞ b2 = 1, where a2 = 36 and b2 = 81, or a = 6 and b = 9. Thus, the transverse axis is parallel to the x-axis. It follows that: • • • • the center of the ellipse is (h, k) = (2, −5) the coordinates of the vertices are (h ± a, k) = (2 ± 6, −5), or (−4, −5) and (8, −5) the coordinates of the co-vertices are (h, k ± b) = (2, − 5 ± 9), or (2, − 14) and (2, 4) the coordinates of the foci are (h ± c, k), where c = ± a2 + b2. Solving for c, we have c = ± 36 + 81 = ± 117 = ± 3 13 Therefore, the coordinates of the foci are ⎛ ⎝2 − 3 13, −5⎞ ⎠ and ⎛ ⎝2 + 3 13, −5⎞ ⎠. The equations of the asymptotes are y = ± b a(x − h) + k = ± 3 2 (x − 2) − 5. Next, we plot and label the center, vertices, co-vertices, foci, and asymptotes and draw smooth curves to form the hyperbola, as shown in Figure 12.23. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1379 Figure 12.23 12.12 Graph the hyperbola given by the standard form of an equation label the center, vertices, co-vertices, foci, and asymptotes. ⎛ ⎠ ⎝y + 4⎞ 100 2 − (x − 3)2 64 = 1. Identify and Solving Applied Problems Involving Hyperbolas As we discussed at the beginning of this section, hyperbolas have real-world applications in many fields, such as astronomy, physics, engineering, and architecture. The design efficiency of hyperbolic cooling towers is particularly interesting. Cooling towers are used to transfer waste heat to the atmosphere and are often touted for their ability to generate power efficiently. Because of their hyperbolic form, these structures are able to withstand extreme winds while requiring less material than any other forms of their size and strength. See Figure 12.24. For example, a 500-foot tower can be made of a reinforced concrete shell only 6 or 8 inches wide! Figure 12.24 Cooling towers at the Drax power station in North Yorkshire, United Kingdom (credit: Les Haines, Flickr) 1380 Chapter 12 Analytic Geometry The first hyperbolic towers were designed in 1914 and were 35 meters high. Today, the tallest cooling towers are in France, standing a remarkable 170 meters tall. In Example 12.13 we will use the design layout of a cooling tower to find a hyperbolic equation that models its sides. Example 12.13 Solving Applied Problems Involving Hyperbolas The design layout of a cooling tower is shown in Figure 12.25. The tower stands 179.6 meters tall. The diameter of the top is 72 meters. At their closest, the sides of the tower are 60 meters apart. Figure 12.25 Project design for a natural draft cooling tower Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola—indicated by the intersection of dashed perpendicular lines in the figure—is the origin of the coordinate plane. Round final values to four decimal places. Solution We are assuming the center of the tower is at the origin, so we can use the standard form of a horizontal hype
rbola centered at the origin: x2 y2 b2 = 1, where the branches of the hyperbola form the sides of the cooling tower. a2 − We must find the values of a2 and b2 to complete the model. First, we find a2. Recall that the length of the transverse axis of a hyperbola is 2a. This length is represented by the distance where the sides are closest, which is given as 65.3 meters. So, 2a = 60. Therefore, a = 30 and a2 = 900. To solve for b2, we need to substitute for x and y in our equation using a known point. To do this, we can use the dimensions of the tower to find some point (x, y) that lies on the hyperbola. We will use the top right corner of the tower to represent that point. Since the y-axis bisects the tower, our x-value can be represented by the radius of the top, or 36 meters. The y-value is represented by the distance from the origin to the top, which is given as 79.6 meters. Therefore, This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1381 x2 a2 − y2 b2 = 1 b2 = = y2 x2 a2 − 1 (79.6)2 (36)2 900 − 1 Standard form of horizontal hyperbola. Isolate b2 Substitute for a2, x, and y ≈ 14400.3636 Round to four decimal places The sides of the tower can be modeled by the hyperbolic equation x2 900 − y2 14400.3636 = 1, or x2 302 − y2 120.00152 = 1 12.13 A design for a cooling tower project is shown in Figure 12.26. Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola—indicated by the intersection of dashed perpendicular lines in the figure—is the origin of the coordinate plane. Round final values to four decimal places. Figure 12.26 Access these online resources for additional instruction and practice with hyperbolas. • Conic Sections: The Hyperbola Part 1 of 2 (http://openstaxcollege.org/l/hyperbola1) • Conic Sections: The Hyperbola Part 2 of 2 (http://openstaxcollege.org/l/hyperbola2) • Graph a Hyperbola with Center at Origin (http://openstaxcollege.org/l/hyperbolaorigin) • Graph a Hyperbola with Center not at Origin (http://openstaxcollege.org/l/hbnotorigin) 1382 Chapter 12 Analytic Geometry 12.2 EXERCISES Verbal 69. Define a hyperbola in terms of its foci. What can we conclude about a hyperbola if 70. asymptotes intersect at the origin? its 71. What must be true of the foci of a hyperbola? If the transverse axis of a hyperbola is vertical, what do 72. we know about the graph? Where must the center of hyperbola be relative to its 73. foci? Algebraic For the the following exercises, determine whether following equations represent hyperbolas. If so, write in standard form. 74. 75. 76. 77. 78. 3y2 + 2x = 6 x2 36 − y2 9 = 1 5y2 + 4x2 = 6x 25x2 − 16y2 = 400 −9x2 + 18x + y2 + 4y − 14 = 0 For the following exercises, write the equation for the hyperbola in standard form if it is not already, and identify the vertices and foci, and write equations of asymptotes. x2 25 − y2 36 = 1 x2 100 − y2 9 = 1 y2 4 − x2 81 = 1 9y2 − 4x2 = 1 (x − 1)2 9 − ⎠ ⎛ ⎝y − 2⎞ 16 2 = 1 ⎛ ⎠ ⎝y − 6⎞ 36 2 − (x + 1)2 16 = 1 79. 80. 81. 82. 83. 84. 85. This content is available for free at https://cnx.org/content/col11758/1.5 (x − 2)2 49 − ⎛ ⎠ ⎝y + 7⎞ 49 2 = 1 86. 87. 88. 89. 90. 91. 92. 93. 4x2 − 8x − 9y2 − 72y + 112 = 0 −9x2 − 54x + 9y2 − 54y + 81 = 0 4x2 − 24x − 36y2 − 360y + 864 = 0 −4x2 + 24x + 16y2 − 128y + 156 = 0 −4x2 + 40x + 25y2 − 100y + 100 = 0 x2 + 2x − 100y2 − 1000y + 2401 = 0 −9x2 + 72x + 16y2 + 16y + 4 = 0 4x2 + 24x − 25y2 + 200y − 464 = 0 For the following exercises, find the equations of the asymptotes for each hyperbola. 94. 95. 96. 97. 98. y2 32 − x2 32 = 1 (x − 3)2 52 − ⎛ ⎝y + 4⎞ ⎠ 2 22 = 1 ⎛ ⎝y − 3⎞ ⎠ 2 32 − (x + 5)2 62 = 1 9x2 − 18x − 16y2 + 32y − 151 = 0 16y2 + 96y − 4x2 + 16x + 112 = 0 Graphical the following exercises, sketch a graph of For hyperbola, labeling vertices and foci. the 99. x2 49 − y2 16 = 1 100. 101. x2 64 y2 9 − y2 4 = 1 − x2 25 = 1 102. 81x2 − 9y2 = 1 Chapter 12 Analytic Geometry 1383 103. 104. 105. 106. 107. 108. 109. 110. 111. 112. ⎛ ⎠ ⎝y + 5⎞ 9 2 − (x − 4)2 25 = 1 (x − 2)2 8 − ⎛ ⎠ ⎝y + 3⎞ 27 2 = 1 ⎛ ⎠ ⎝y − 3⎞ 9 2 − (x − 3)2 9 = 1 −4x2 − 8x + 16y2 − 32y − 52 = 0 x2 − 8x − 25y2 − 100y − 109 = 0 −x2 + 8x + 4y2 − 40y + 88 = 0 64x2 + 128x − 9y2 − 72y − 656 = 0 120. 16x2 + 64x − 4y2 − 8y − 4 = 0 −100x2 + 1000x + y2 − 10y − 2575 = 0 4x2 + 16x − 4y2 + 16y + 16 = 0 For the following exercises, given information about the graph of the hyperbola, find its equation. Vertices at (3, 0) and (−3, 0) and one focus at 113. (5, 0). 121. Vertices at (0, 6) and (0, −6) and one focus at 114. (0, −8). Vertices at (1, 1) and (11, 1) and one focus at 115. (12, 1). 116. Center: (0, 0); vertex: (0, −13); one focus: ⎝0, 313⎞ ⎛ ⎠. 117. Center: (4, 2); vertex: (9, 2); one focus: ⎝4 + 26, 2⎞ ⎛ ⎠. 118. Center: (3, 5); vertex: (3, 11); one focus: ⎝3, 5 + 2 10⎞ ⎛ ⎠. the following exercises, given the graph of For hyperbola, find its equation. the 122. 119. 1384 Chapter 12 Analytic Geometry will hedge The asymptotes y = x and y = − x, and its closest distance to the center fountain is 5 yards. follow the The hedge 130. y = 2x and y = −2x, center fountain is 6 yards. will follow asymptotes the and its closest distance to the 123. Extensions For the following exercises, express the equation for the hyperbola as two functions, with y as a function of x. Express as simply as possible. Use a graphing calculator to sketch the graph of the two functions on the same axes. 124. 125. 126. 127. 128. x2 4 y2 9 − − y2 9 x2 1 = 1 = 1 (x − 2)2 16 − ⎠ ⎛ ⎝y + 3⎞ 25 2 = 1 −4x2 − 16x + y2 − 2y − 19 = 0 4x2 − 24x − y2 − 4y + 16 = 0 Real-World Applications For the following exercises, a hedge is to be constructed in the shape of a hyperbola near a fountain at the center of the yard. Find the equation of the hyperbola and sketch the graph. 131. The hedge will follow the asymptotes y = 1 2 x and y = − 1 2 x, and its closest distance to the center fountain is 10 yards. 132. The hedge will follow the asymptotes y = 2 3 x and y = − 2 3 x, and its closest distance to the center fountain is 12 yards. 133. y = 3 4 The hedge x and y = − 3 4 will follow the asymptotes x, and its closest distance to the center fountain is 20 yards. For the following exercises, assume an object enters our solar system and we want to graph its path on a coordinate system with the sun at the origin and the x-axis as the axis of symmetry for the object's path. Give the equation of the flight path of each object using the given information. The object enters along a path approximated by the 134. line y = x − 2 and passes within 1 au (astronomical unit) of the sun at its closest approach, so that the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line y = − x + 2. The object enters along a path approximated by the 135. line y = 2x − 2 and passes within 0.5 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line y = −2x + 2. The object enters along a path approximated by the 136. line y = 0.5x + 2 and passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line y = −0.5x − 2. 137. line y = 1 3 The object enters along a path approximated by the x − 1 and passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line y = − 1 3 x + 1. 129. 138. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1385 The object It enters along a path approximated by the line y = 3x − 9 and passes within 1 au of the sun at its closest approach, so the sun is one focus of the hyperbola. It then departs the solar system along a path approximated by the line y = −3x + 9. 1386 Chapter 12 Analytic Geometry 12.3 \| The Parabola Learning Objectives In this section, you will: 12.3.1 Graph parabolas with vertices at the origin. 12.3.2 Write equations of parabolas in standard form. 12.3.3 Graph parabolas with vertices not at the origin. 12.3.4 Solve applied problems involving parabolas. Figure 12.27 The Olympic torch concludes its journey around the world when it is used to light the Olympic cauldron during the opening ceremony. (credit: Ken Hackman, U.S. Air Force) Did you know that the Olympic torch is lit several months before the start of the games? The ceremonial method for lighting the flame is the same as in ancient times. The ceremony takes place at the Temple of Hera in Olympia, Greece, and is rooted in Greek mythology, paying tribute to Prometheus, who stole fire from Zeus to give to all humans. One of eleven acting priestesses places the torch at the focus of a parabolic mirror (see Figure 12.27), which focuses light rays from the sun to ignite the flame. Parabolic mirrors (or reflectors) are able to capture energy and focus it to a single point. The advantages of this property are evidenced by the vast list of parabolic objects we use every day: satellite dishes, suspension bridges, telescopes, microphones, spotlights, and car headlights, to name a few. Parabolic reflectors are also used in alternative energy devices, such as solar cookers and water heaters, because they are inexpensive to manufacture and need little maintenance. In this section we will explore the parabola and its uses, including low-cost, energy-efficient solar designs. Graphing Parabolas with Vertices at the Origin In The Ellipse, we saw that an ellipse is formed when a plane cuts through a right circular cone. If the plane is parallel to the edge of the cone, an unbounded curve is formed. This curve is a parabola. See Figure 12.28. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1387 Figure 12.28 Parabola Like the ellipse and hyperbola, the parabola can also be defined
by a set of points in the coordinate plane. A parabola is the set of all points (x, y) in a plane that are the same distance from a fixed line, called the directrix, and a fixed point (the focus) not on the directrix. In Quadratic Functions (https://cnx.org/content/m49346/latest/) , we learned about a parabola’s vertex and axis of symmetry. Now we extend the discussion to include other key features of the parabola. See Figure 12.29. Notice that the axis of symmetry passes through the focus and vertex and is perpendicular to the directrix. The vertex is the midpoint between the directrix and the focus. The line segment that passes through the focus and is parallel to the directrix is called the latus rectum. The endpoints of the latus rectum lie on the curve. By definition, the distance d from the focus to any point P on the parabola is equal to the distance from P to the directrix. Figure 12.29 Key features of the parabola To work with parabolas in the coordinate plane, we consider two cases: those with a vertex at the origin and those with a vertex at a point other than the origin. We begin with the former. 1388 Chapter 12 Analytic Geometry Figure 12.30 Let (x, y) be a point on the parabola with vertex (0, 0), ⎠, and directrix y= −p as shown in Figure 12.30. focus ⎛ The distance d from point (x, y) to point (x, − p) on the directrix is the difference of the y-values: d = y + p. The distance from the focus (0, p) to the point (x, y) is also equal to d and can be expressed using the distance formula. ⎝0, p⎞ d = (x − 0)2 + (y − p)2 = x2 + (y − p)2 Set the two expressions for d equal to each other and solve for y to derive the equation of the parabola. We do this because the distance from (x, y) to ⎛ ⎠ equals the distance from (x, y) to (x, −p). ⎝0, p⎞ We then square both sides of the equation, expand the squared terms, and simplify by combining like terms. x2 + (y − p)2 = y + p x2 + (y − p)2 = (y + p)2 x2 + y2 − 2py + p2 = y2 + 2py + p2 x2 − 2py = 2py x2 = 4py The equations of parabolas with vertex (0, 0) are y2 = 4px when the x-axis is the axis of symmetry and x2 = 4py when the y-axis is the axis of symmetry. These standard forms are given below, along with their general graphs and key features. Standard Forms of Parabolas with Vertex (0, 0) Table 12.1 and Figure 12.31 summarize the standard features of parabolas with a vertex at the origin. Axis of Symmetry Equation Focus Directrix Endpoints of Latus Rectum x-axis y-axis y2 = 4px x2 = 4py Table 12.1 ⎛ ⎝p, 0⎞ ⎠ ⎛ ⎝0, pp, ± 2p⎞ ⎠ ⎛ ⎝±2p, p⎞ ⎠ This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1389 Figure 12.31 (a) When p > 0 and the axis of symmetry is the x-axis, the parabola opens right. (b) When p < 0 and the axis of symmetry is the x-axis, the parabola opens left. (c) When p < 0 and the axis of symmetry is the y-axis, the parabola opens up. (d) When p < 0 and the axis of symmetry is the y-axis, the parabola opens down. The key features of a parabola are its vertex, axis of symmetry, focus, directrix, and latus rectum. See Figure 12.31. When given a standard equation for a parabola centered at the origin, we can easily identify the key features to graph the parabola. A line is said to be tangent to a curve if it intersects the curve at exactly one point. If we sketch lines tangent to the parabola at the endpoints of the latus rectum, these lines intersect on the axis of symmetry, as shown in Figure 12.32. 1390 Chapter 12 Analytic Geometry Figure 12.32 Given a standard form equation for a parabola centered at (0, 0), sketch the graph. 1. Determine which of the standard forms applies to the given equation: y2 = 4px or x2 = 4py. 2. Use the standard form identified in Step 1 to determine the axis of symmetry, focus, equation of the directrix, and endpoints of the latus rectum. a. If the equation is in the form y2 = 4px, then ▪ ▪ the axis of symmetry is the x-axis, y = 0 set 4p equal to the coefficient of x in the given equation to solve for p. If p > 0, parabola opens right. If p < 0, the parabola opens left. the ▪ use p to find the coordinates of the focus, ⎛ ⎝p, 0⎞ ⎠ ▪ use p to find the equation of the directrix, x = − p ▪ use p to find the endpoints of the latus rectum, ⎛ ⎝p, ± 2p⎞ ⎠. Alternately, substitute x = p into the original equation. b. If the equation is in the form x2 = 4py, then ▪ ▪ the axis of symmetry is the y-axis, x = 0 set 4p equal to the coefficient of y in the given equation to solve for p. If p > 0, parabola opens up. If p < 0, the parabola opens down. the ▪ use p to find the coordinates of the focus, ⎛ ⎝0, p⎞ ⎠ ▪ use p to find equation of the directrix, y = − p ▪ use p to find the endpoints of the latus rectum, ⎛ ⎝±2p, p⎞ ⎠ 3. Plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1391 Example 12.14 Graphing a Parabola with Vertex (0, 0) and the x-axis as the Axis of Symmetry Graph y2 = 24x. Identify and label the focus, directrix, and endpoints of the latus rectum. Solution The standard form that applies to the given equation is y2 = 4px. Thus, the axis of symmetry is the x-axis. It follows that: • • • • 24 = 4p, so p = 6. Since p > 0, the parabola opens right the coordinates of the focus are ⎛ ⎝p, 0⎞ ⎠ = (6, 0) the equation of the directrix is x = − p = − 6 the endpoints of the latus rectum have the same x-coordinate at the focus. To find the endpoints, substitute x = 6 into the original equation: (6, ± 12) Next we plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. Figure 12.33 Figure 12.33 12.14 Graph y2 = −16x. Identify and label the focus, directrix, and endpoints of the latus rectum. Example 12.15 Graphing a Parabola with Vertex (0, 0) and the y-axis as the Axis of Symmetry Graph x2 = −6y. Identify and label the focus, directrix, and endpoints of the latus rectum. Solution 1392 Chapter 12 Analytic Geometry The standard form that applies to the given equation is x2 = 4py. Thus, the axis of symmetry is the y-axis. It follows that: • −6 = 4p, so p = − 3 2 . Since p < 0, the parabola opens down. • • • the coordinates of the focus are ⎛ ⎝0, p⎞ ⎠ = ⎛ ⎝0, − 3 2 ⎞ ⎠ the equation of the directrix is y = − p = 3 2 the endpoints of the latus rectum can be found by substituting y = 3 2 into the original equation, ⎛ ⎝±3, − 3 2 ⎞ ⎠ Next we plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. Figure 12.34 12.15 Graph x2 = 8y. Identify and label the focus, directrix, and endpoints of the latus rectum. Writing Equations of Parabolas in Standard Form In the previous examples, we used the standard form equation of a parabola to calculate the locations of its key features. We can also use the calculations in reverse to write an equation for a parabola when given its key features. Given its focus and directrix, write the equation for a parabola in standard form. 1. Determine whether the axis of symmetry is the x- or y-axis. a. b. If the given coordinates of the focus have the form ⎛ ⎝p, 0⎞ ⎠, then the axis of symmetry is the x-axis. Use the standard form y2 = 4px. If the given coordinates of the focus have the form ⎛ ⎝0, p⎞ ⎠, then the axis of symmetry is the y-axis. Use the standard form x2 = 4py. 2. Multiply 4p. 3. Substitute the value from Step 2 into the equation determined in Step 1. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1393 Example 12.16 Writing the Equation of a Parabola in Standard Form Given its Focus and Directrix What is the equation for the parabola with focus ⎛ ⎝− 1 2 ⎞ ⎠ and directrix x = 1 , 0 2 ? Solution The focus has the form ⎛ ⎝p, 0⎞ ⎠, so the equation will have the form y2 = 4px. • Multiplying 4p, we have 4p = 4 ⎛ ⎝− 1 2 ⎞ ⎠ = −2. • Substituting for 4p, we have y2 = 4px = −2x. Therefore, the equation for the parabola is y2 = −2x. 12.16 What is the equation for the parabola with focus ⎛ ⎝0, 7 2 ⎞ ⎠ and directrix y = − 7 2 ? Graphing Parabolas with Vertices Not at the Origin Like other graphs we’ve worked with, the graph of a parabola can be translated. If a parabola is translated h units horizontally and k units vertically, the vertex will be (h, k). This translation results in the standard form of the equation we saw previously with x replaced by (x − h) and y replaced by ⎛ ⎝y − k⎞ ⎠. To graph parabolas with a vertex (h, k) other than the origin, we use the standard form ⎛ that have an axis of symmetry parallel to the x-axis, and (x − h)2 = 4p⎛ parallel to the y-axis. These standard forms are given below, along with their general graphs and key features. 2 = 4p(x − h) for parabolas ⎠ for parabolas that have an axis of symmetry ⎝y − k⎞ ⎝y − k⎞ ⎠ Standard Forms of Parabolas with Vertex (h, k) Table 12.2 and Figure 12.35 summarize the standard features of parabolas with a vertex at a point (h, k). Axis of Symmetry y = k x = h Table 12.2 Equation Focus Directrix Endpoints of Latus Rectum ⎛ ⎝y − k⎞ ⎠ 2 = 4p(x − h) (x − h)2 = 4p⎛ ⎝y − k⎞ ⎠ ⎛ ⎝h + p, k⎞ ⎠ ⎛ ⎝h, k + ph + p, k ± 2p⎞ ⎠ ⎛ ⎝h ± 2p, k + p⎞ ⎠ 1394 Chapter 12 Analytic Geometry Figure 12.35 (a) When p > 0, p > 0, the parabola opens up. (d) When p < 0, the parabola opens right. (b) When p < 0, the parabola opens down. the parabola opens left. (c) When This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1395 Given a standard form equation for a parabola centered at (h, k), sketch the graph. 1. Determine which of (x − h)2 = 4p⎛ ⎝y − k⎞ ⎠. the standard forms applies to the given equation: ⎛ ⎝y − k⎞ ⎠ 2 = 4p(x − h) or 2. Use the standard form identified in Step 1 to determine the vertex, axis of symmetry, focus, equation of the directrix, and endpoints of the latus rectum. a. If the equation is in the form ⎛ ⎝y − k⎞ ⎠ 2 = 4p(x − h), then: ▪ use the given equation to identify h and k for the vertex, (h, k)
▪ use the value of k to determine the axis of symmetry, y = k ▪ set 4p equal to the coefficient of (x − h) in the given equation to solve for p. If p > 0, the parabola opens right. If p < 0, the parabola opens left. ▪ use h, k, and p to find the coordinates of the focus, ⎛ ⎝h + p, k⎞ ⎠ ▪ use h and p to find the equation of the directrix, x = h − p ▪ use h, k, and p to find the endpoints of the latus rectum, ⎛ ⎝h + p, k ± 2p⎞ ⎠ b. If the equation is in the form (x − h)2 = 4p⎛ ⎝y − k⎞ ⎠, then: ▪ use the given equation to identify h and k for the vertex, (h, k) ▪ use the value of h to determine the axis of symmetry, x = h ▪ set 4p equal to the coefficient of ⎛ the parabola opens up. If p < 0, ⎝y − k⎞ the parabola opens down. ⎠ in the given equation to solve for p. If p > 0, ▪ use h, k, and p to find the coordinates of the focus, ⎛ ⎝h, k + p⎞ ⎠ ▪ use k and p to find the equation of the directrix, y = k − p ▪ use h, k, and p to find the endpoints of the latus rectum, ⎛ ⎝h ± 2p, k + p⎞ ⎠ 3. Plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. Example 12.17 Graphing a Parabola with Vertex (h, k) and Axis of Symmetry Parallel to the x-axis Graph ⎛ ⎝y − 1⎞ ⎠ 2 = −16(x + 3). Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum. Solution The standard form that applies to the given equation is ⎛ ⎝y − k⎞ ⎠ 2 = 4p(x − h). Thus, the axis of symmetry is parallel to the x-axis. It follows that: • • the vertex is (h, k) = (−3, 1) the axis of symmetry is y = k = 1 1396 Chapter 12 Analytic Geometry • −16 = 4p, so p = −4. Since p < 0, the parabola opens left. • • • the coordinates of the focus are ⎛ ⎝h + p, k⎞ ⎠ = (−3 + (−4), 1) = (−7, 1) the equation of the directrix is x = h − p = −3 − (−4) = 1 the endpoints of the latus rectum are ⎛ (−7, 9) ⎝h + p, k ± 2p⎞ ⎠ = (−3 + (−4), 1 ± 2(−4)), or (−7, −7) and Next we plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. See Figure 12.36. Figure 12.36 2 = 4(x − 8). Identify and label the vertex, axis of symmetry, focus, directrix, and 12.17 Graph ⎛ ⎝y + 1⎞ ⎠ endpoints of the latus rectum. Example 12.18 Graphing a Parabola from an Equation Given in General Form Graph x2 − 8x − 28y − 208 = 0. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum. Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1397 Start by writing the equation of the parabola in standard form. The standard form that applies to the given equation is (x − h)2 = 4p⎛ ⎠. Thus, the axis of symmetry is parallel to the y-axis. To express the equation of the parabola in this form, we begin by isolating the terms that contain the variable x in order to complete the square. ⎝y − k⎞ x2 − 8x − 28y − 208 = 0 x2 − 8x = 28y + 208 x2 − 8x + 16 = 28y + 208 + 16 (x − 4)2 = 28y + 224 (x − 4)2 = 28(y + 8) (x − 4)2 = 4 ⋅ 7 ⋅ (y + 8) It follows that: • • • • • • the vertex is (h, k) = (4, −8) the axis of symmetry is x = h = 4 since p = 7, p > 0 and so the parabola opens up the coordinates of the focus are ⎛ ⎝h, k + p⎞ ⎠ = (4, −8 + 7) = (4, −1) the equation of the directrix is y = k − p = −8 − 7 = −15 the endpoints of (18, −1) the latus rectum are ⎛ ⎝h ± 2p, k + p⎞ ⎠ = (4 ± 2(7), −8 + 7), or (−10, −1) and Next we plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. See Figure 12.37. Figure 12.37 12.18 Graph (x + 2)2 = −20⎛ ⎝y − 3⎞ ⎠. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum. Solving Applied Problems Involving Parabolas As we mentioned at the beginning of the section, parabolas are used to design many objects we use every day, such as telescopes, suspension bridges, microphones, and radar equipment. Parabolic mirrors, such as the one used to light the Olympic torch, have a very unique reflecting property. When rays of light parallel to the parabola’s axis of symmetry are directed toward any surface of the mirror, the light is reflected directly to the focus. See Figure 12.38. This is why the Olympic torch is ignited when it is held at the focus of the parabolic mirror. 1398 Chapter 12 Analytic Geometry Figure 12.38 Reflecting property of parabolas Parabolic mirrors have the ability to focus the sun’s energy to a single point, raising the temperature hundreds of degrees in a matter of seconds. Thus, parabolic mirrors are featured in many low-cost, energy efficient solar products, such as solar cookers, solar heaters, and even travel-sized fire starters. Example 12.19 Solving Applied Problems Involving Parabolas A cross-section of a design for a travel-sized solar fire starter is shown in Figure 12.39. The sun’s rays reflect off the parabolic mirror toward an object attached to the igniter. Because the igniter is located at the focus of the parabola, the reflected rays cause the object to burn in just seconds. a. Find the equation of the parabola that models the fire starter. Assume that the vertex of the parabolic mirror is the origin of the coordinate plane. b. Use the equation found in part (a) to find the depth of the fire starter. Figure 12.39 Cross-section of a travel-sized solar fire starter Solution a. The vertex of the dish is the origin of the coordinate plane, so the parabola will take the standard form x2 = 4py, where p > 0. The igniter, which is the focus, is 1.7 inches above the vertex of the dish. Thus we have p = 1.7. x2 = 4py x2 = 4(1.7)y x2 = 6.8y Standard form of upward-facing parabola with vertex (0,0) Substitute 1.7 for p. Multiply. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1399 b. The dish extends 4.5 2 = 2.25 inches on either side of the origin. We can substitute 2.25 for x in the equation from part (a) to find the depth of the dish. x2 = 6.8y (2.25)2 = 6.8y y ≈ 0.74 Solve for y. Equation found in part (a). Substitute 2.25 for x. The dish is about 0.74 inches deep. 12.19 Balcony-sized solar cookers have been designed for families living in India. The top of a dish has a diameter of 1600 mm. The sun’s rays reflect off the parabolic mirror toward the “cooker,” which is placed 320 mm from the base. a. Find an equation that models a cross-section of the solar cooker. Assume that the vertex of the parabolic mirror is the origin of the coordinate plane, and that the parabola opens to the right (i.e., has the x-axis as its axis of symmetry). b. Use the equation found in part (a) to find the depth of the cooker. Access these online resources for additional instruction and practice with parabolas. • Conic Sections: The Parabola Part 1 of 2 (http://openstaxcollege.org/l/parabola1) • Conic Sections: The Parabola Part 2 of 2 (http://openstaxcollege.org/l/parabola2) • Parabola with Vertical Axis (http://openstaxcollege.org/l/parabolavertcal) • Parabola with Horizontal Axis (http://openstaxcollege.org/l/parabolahoriz) 1400 Chapter 12 Analytic Geometry 12.3 EXERCISES Verbal 139. Define a parabola in terms of its focus and directrix. 140. If the equation of a parabola is written in standard form and p is positive and the directrix is a vertical line, then what can we conclude about its graph? If the equation of a parabola is written in standard 141. form and p is negative and the directrix is a horizontal line, then what can we conclude about its graph? What is the effect on the graph of a parabola if its 142. equation in standard form has increasing values of p? As the graph of a parabola becomes wider, what will 143. happen to the distance between the focus and directrix? Algebraic For the following exercises, determine whether the given equation is a parabola. If so, rewrite the equation in standard form. 144. y2 = 4 − x2 145. y = 4x2 146. 3x2 − 6y2 = 12 147. 148. ⎛ ⎝y − 3⎞ ⎠ 2 = 8(x − 2) y2 + 12x − 6y − 51 = 0 For the following exercises, rewrite the given equation in standard form, and then determine the vertex (V), focus (F), and directrix (d) of the parabola. 149. x = 8y2 150. y = 1 4 x2 151. y = −4x2 152. x = 1 8 y2 153. x = 36y2 x = 1 36 y2 154. 155. This content is available for free at https://cnx.org/content/col11758/1.5 (x − 1)2 = 4⎛ ⎝y − 1⎞ ⎠ 156. 157. 158. 159. 160. 161. 162. 163. 164. 165. 166. 167. 168. ⎛ ⎝y − 2⎞ ⎠ ⎛ ⎝y − 4⎞ ⎠ 2 = 4 5 (x + 4) 2 = 2(x + 3) (x + 1)2 = 2⎛ ⎝y + 4⎞ ⎠ (x + 4)2 = 24⎛ ⎝y + 1⎞ ⎠ ⎛ ⎝y + 4⎞ ⎠ 2 = 16(x + 4) y2 + 12x − 6y + 21 = 0 x2 − 4x − 24y + 28 = 0 5x2 − 50x − 4y + 113 = 0 y2 − 24x + 4y − 68 = 0 x2 − 4x + 2y − 6 = 0 y2 − 6y + 12x − 3 = 0 3y2 − 4x − 6y + 23 = 0 x2 + 4x + 8y − 4 = 0 Graphical For the following exercises, graph the parabola, labeling the focus and the directrix. 169. x = 1 8 y2 170. y = 36x2 171. y = 1 36 x2 172. y = −9x2 173. 174. 175. ⎛ ⎝y − 2⎞ ⎠ 2 = − 4 3 (x + 2) −5(x + 5)2 = 4⎛ ⎝y + 5⎞ ⎠ −6⎛ ⎝y + 5⎞ ⎠ 2 = 4(x − 4) Chapter 12 Analytic Geometry 1401 176. 177. 178. 179. 180. 181. 182. y2 − 6y − 8x + 1 = 0 x2 + 8x + 4y + 20 = 0 3x2 + 30x − 4y + 95 = 0 y2 − 8x + 10y + 9 = 0 x2 + 4x + 2y + 2 = 0 y2 + 2y − 12x + 61 = 0 −2x2 + 8x − 4y − 24 = 0 For the following exercises, find the equation of the parabola given information about its graph. is (0, 0); directrix is y = 4, focus is 190. is (0, 0); directrix is x = 4, focus is Vertex 183. (0, −4). Vertex 184. (−4, 0). 185. Vertex is (2, 2); directrix is x = 2 − 2, focus is ⎝2 + 2, 2⎞ ⎛ ⎠. 186. Vertex is (−2, 3); directrix is x = − 7 2 , focus is ⎛ ⎝− 1 2 ⎞ , 3 ⎠. 187. Vertex is ⎛ ⎝ 2, − 3⎞ ⎠; directrix is x = 2 2, focus is ⎝0, − 3⎞ ⎛ ⎠. 188. Vertex is (1, 2); directrix is y = 11 3 , focus is 191. ⎛ ⎝1, 1 3 ⎞ ⎠. For the following exercises, determine the equation for the parabola from its graph. 189. 1402 Chapter 12 Analytic Geometry 192. Extensions For the following exercises, the vertex and endpoints of the latus rectum of a parabola are given. Find the equation. 194. V(0, 0),
Endpoints (2, 1), (−2, 1) 195. V(0, 0), Endpoints (−2, 4), (−2, −4) 196. V(1, 2), Endpoints (−5, 5), (7, 5) 197. V(−3, −1), Endpoints (0, 5), (0, −7) 198. ⎛ ⎝5, − 7 V(4, −3), Endpoints 2 ⎞ ⎠, ⎛ ⎝3, − 7 2 ⎞ ⎠ Real-World Applications 199. The mirror in an automobile headlight has a parabolic the focus. On a cross-section with the light bulb at schematic, the parabola is given as the equation of x2 = 4y. At what coordinates should you place the light bulb? 200. If we want to construct the mirror from the previous exercise such that the focus is located at (0, 0.25), what should the equation of the parabola be? 201. A satellite dish is shaped like a paraboloid of revolution. This means that it can be formed by rotating a parabola around its axis of symmetry. The receiver is to be located at the focus. If the dish is 12 feet across at its opening and 4 feet deep at its center, where should the receiver be placed? 202. Consider the satellite dish from the previous exercise. If the dish is 8 feet across at the opening and 2 feet deep, where should we place the receiver? 193. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1403 is A searchlight 203. shaped like a paraboloid of revolution. A light source is located 1 foot from the base the along the axis of symmetry. searchlight is 3 feet across, find the depth. the opening of If 204. If the searchlight from the previous exercise has the light source located 6 inches from the base along the axis of symmetry and the opening is 4 feet, find the depth. An arch is in the shape of a parabola. It has a span of 205. 100 feet and a maximum height of 20 feet. Find the equation of the parabola, and determine the height of the arch 40 feet from the center. If the arch from the previous exercise has a span of 206. 160 feet and a maximum height of 40 feet, find the equation of the parabola, and determine the distance from the center at which the height is 20 feet. An object is projected so as to follow a parabolic path the horizontal 207. given by y = − x2 + 96x, where x is distance traveled in feet and y is the height. Determine the maximum height the object reaches. For the object from the previous exercise, assume the 208. path followed is given by y = −0.5x2 + 80x. Determine how far along the horizontal the object traveled to reach maximum height. 1404 Chapter 12 Analytic Geometry 12.4 \| Rotation of Axes Learning Objectives In this section, you will: 12.4.1 Identify nondegenerate conic sections given their general form equations. 12.4.2 Use rotation of axes formulas. 12.4.3 Write equations of rotated conics in standard form. 12.4.4 Identify conics without rotating axes. As we have seen, conic sections are formed when a plane intersects two right circular cones aligned tip to tip and extending infinitely far in opposite directions, which we also call a cone. The way in which we slice the cone will determine the type of conic section formed at the intersection. A circle is formed by slicing a cone with a plane perpendicular to the axis of symmetry of the cone. An ellipse is formed by slicing a single cone with a slanted plane not perpendicular to the axis of symmetry. A parabola is formed by slicing the plane through the top or bottom of the double-cone, whereas a hyperbola is formed when the plane slices both the top and bottom of the cone. See Figure 12.40. Figure 12.40 The nondegenerate conic sections Ellipses, circles, hyperbolas, and parabolas are sometimes called the nondegenerate conic sections, in contrast to the degenerate conic sections, which are shown in Figure 12.41. A degenerate conic results when a plane intersects the double cone and passes through the apex. Depending on the angle of the plane, three types of degenerate conic sections are possible: a point, a line, or two intersecting lines. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1405 Figure 12.41 Degenerate conic sections Identifying Nondegenerate Conics in General Form In previous sections of this chapter, we have focused on the standard form equations for nondegenerate conic sections. In this section, we will shift our focus to the general form equation, which can be used for any conic. The general form is set equal to zero, and the terms and coefficients are given in a particular order, as shown below. Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 where A, B, and C are not all zero. We can use the values of the coefficients to identify which type conic is represented by a given equation. You may notice that the general form equation has an xy term that we have not seen in any of the standard form equations. As we will discuss later, the xy term rotates the conic whenever B is not equal to zero. 1406 Chapter 12 Analytic Geometry Conic Sections Example ellipse circle 4x2 + 9y2 = 1 4x2 + 4y2 = 1 hyperbola 4x2 − 9y2 = 1 parabola 4x2 = 9y or 4y2 = 9x one line 4x + 9y = 1 intersecting lines (x − 4)⎛ ⎝y + 4⎞ ⎠ = 0 parallel lines (x − 4)(x − 9) = 0 a point 4x2 + 4y2 = 0 no graph 4x2 + 4y2 = − 1 Table 12.3 General Form of Conic Sections A nondegenerate conic section has the general form Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 (12.9) where A, B, and C are not all zero. Table 12.4 summarizes the different conic sections where B = 0, indicates that the conic has not been rotated. and A and C are nonzero real numbers. This This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1407 ellipse Ax2 + Cy2 + Dx + Ey + F = 0, A ≠ C and AC > 0 circle Ax2 + Cy2 + Dx + Ey + F = 0, A = C hyperbola Ax2 − Cy2 + Dx + Ey + F = 0 or − Ax2 + Cy2 + Dx + Ey + F = 0, where A and C are positive parabola Ax2 + Dx + Ey + F = 0 or Cy2 + Dx + Ey + F = 0 Table 12.4 Given the equation of a conic, identify the type of conic. 1. Rewrite the equation in the general form, Ax2 + Bxy + Cy2 + Dx + Ey + F = 0. 2. Identify the values of A and C from the general form. a. b. c. d. If A and C are nonzero, have the same sign, and are not equal to each other, then the graph is an ellipse. If A and C are equal and nonzero and have the same sign, then the graph is a circle. If A and C are nonzero and have opposite signs, then the graph is a hyperbola. If either A or C is zero, then the graph is a parabola. Example 12.20 Identifying a Conic from Its General Form Identify the graph of each of the following nondegenerate conic sections. a. b. c. 4x2 − 9y2 + 36x + 36y − 125 = 0 9y2 + 16x + 36y − 10 = 0 3x2 + 3y2 − 2x − 6y − 4 = 0 d. −25x2 − 4y2 + 100x + 16y + 20 = 0 Solution a. Rewriting the general form, we have A = 4 and C = −9, hyperbola. so we observe that A and C have opposite signs. The graph of this equation is a 1408 Chapter 12 Analytic Geometry b. Rewriting the general form, we have A = 0 and C = 9. We can determine that the equation is a parabola, since A is zero. c. Rewriting the general form, we have A = 3 and C = 3. Because A = C, the graph of this equation is a circle. d. Rewriting the general form, we have A = −25 and C = −4. Because AC > 0 and A ≠ C, the graph of this equation is an ellipse. 12.20 Identify the graph of each of the following nondegenerate conic sections. a. 16y2 − x2 + x − 4y − 9 = 0 b. 16x2 + 4y2 + 16x + 49y − 81 = 0 Finding a New Representation of the Given Equation after Rotating through a Given Angle Until now, we have looked at equations of conic sections without an xy term, which aligns the graphs with the x- and yaxes. When we add an xy term, we are rotating the conic about the origin. If the x- and y-axes are rotated through an angle, then every point on the plane may be thought of as having two representations: (x, y) on the Cartesian plane with say θ, the original x-axis and y-axis, and ⎛ ⎠ on the new plane defined by the new, rotated axes, called the x'-axis and y'-axis. ⎝x′, y′⎞ See Figure 12.42. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1409 Figure 12.42 The graph of the rotated ellipse x2 + y2 – xy – 15 = 0 We will find the relationships between x and y on the Cartesian plane with x′ and y′ on the new rotated plane. See Figure 12.43. Figure 12.43 The Cartesian plane with x- and y-axes and the resulting x′− and y′−axes formed by a rotation by an angle θ. The original coordinate x- and y-axes have unit vectors i and j . The rotated coordinate axes have unit vectors i′ and j′. The angle θ is known as the angle of rotation. See Figure 12.44. We may write the new unit vectors in terms of the original ones. i′ = cos θi + sin θ j j′ = − sin θi + cos θ j 1410 Chapter 12 Analytic Geometry Figure 12.44 Relationship between the old and new coordinate planes. Consider a vector u in the new coordinate plane. It may be represented in terms of its coordinate axes. u = x′ i′ + y′ j′ u = x′(i cos θ + j sin θ) + y′( − i sin θ + j cos θ) u = ix ' cos θ + jx ' sin θ − iy ' sin θ + jy ' cos θ u = ix ' cos θ − iy ' sin θ + jx ' sin θ + jy ' cos θ u = (x ' cos θ − y ' sin θ)i + (x ' sin θ + y ' cos θ) j Substitute. Distribute. Apply commutative property. Factor by grouping. Because u = x′ i′ + y′ j′, we have representations of x and y in terms of the new coordinate system. x = x′ cos θ − y′ sin θ and y = x′ sin θ + y′ cos θ Equations of Rotation If a point (x, y) on the Cartesian plane is represented on a new coordinate plane where the axes of rotation are formed by rotating an angle θ from the positive x-axis, then the coordinates of the point with respect to the new axes are ⎛ ⎝x′, y′⎞ ⎠. We can use the following equations of rotation to define the relationship between (x, y) and ⎛ ⎝x′, y′⎞ ⎠ : and x = x′ cos θ − y′ sin θ y = x′ sin θ + y′ cos θ (12.10) (12.11) Given the equation of a conic, find a new representation after rotating through an angle. 1. Find x and y where x = x′ cos θ − y′ sin θ and y = x′ sin θ + y′ cos θ. 2. Substitute the expression for x and y into in the given equation, then simplify. 3. Write the equations with x
′ and y′ in standard form. Example 12.21 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1411 Finding a New Representation of an Equation after Rotating through a Given Angle Find a new representation of the equation 2x2 − xy + 2y2 − 30 = 0 after rotating through an angle of θ = 45°. Solution Find x and y, where x = x′ cos θ − y′ sin θ and y = x′ sin θ + y′ cos θ. Because θ = 45°, x = x′ cos(45°) − y′ sin(45°) and x = ⎞ ⎠ − y′ ⎛ 1 ⎝ 2 x′ − y′ 2 y = x′ sin(45°) + y′ cos(45°) ⎞ ⎠ + y′ y = ⎛ 1 ⎝ 2 x′ + y′ 2 Substitute x = x′ cosθ − y′ sinθ and y = x′ sin θ + y′ cos θ into 2x2 − xy + 2y2 − 30 = 0. 2 2 ⎛ ⎝ x′ − y′ 2 ⎞ ⎠ − ⎛ ⎝ x′ − y′ 2 ⎛ ⎞ ⎝ ⎠ x′ + y′ 2 ⎛ ⎞ ⎠ + 2 ⎝ x′ + y′ 2 ⎞ ⎠ 2 − 30 = 0 Simplify. 2 (x′ − y′)(x′ − y′) 2 − (x′ − y′)(x′ + y′) 2 (x′ 2 − y′ 2) 2 + 2 (x′ + y′)(x′ + y′) 2 − 30 = 0 FOIL method x′ 2 −2x′ y′ + y′ 2 − + x′ 2 +2x′ y′ + y′ 2 − 30 = 0 Combine like terms. 2x′ 2 + 2y′ 2 − = 30 Combine like terms. 2 ⎛ ⎜ 2x′ 2 + 2y′ 2 − ⎝ 4x′ 2 + 4y′ 2 − (x′ 2 − y′ 2) = 60 4x′ 2 + 4y′ 2 − x′ 2 + y′ 2 = 60 3x′ 2 60 5y′ 2 60 = 60 60 ⎞ ⎟ = 2(30) ⎠ + Multiply both sides by 2. Simplify. Distribute. Set equal to 1. (x′ 2 − y′ 2) 2 (x′ 2 − y′ 2) 2 Write the equations with x′ and y′ in the standard form. This equation is an ellipse. Figure 12.45 shows the graph. x′2 20 + y′2 12 = 1 1412 Chapter 12 Analytic Geometry Figure 12.45 Writing Equations of Rotated Conics in Standard Form Now that we can find the standard form of a conic when we are given an angle of rotation, we will learn how to transform the equation of a conic given in the form Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 into standard form by rotating the axes. To do so, we will rewrite the general form as an equation in the x′ and y′ coordinate system without the x′ y′ term, by rotating the axes by a measure of θ that satisfies cot(2θ) = A − C B (12.12) We have learned already that any conic may be represented by the second degree equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 where A, B, and C are not all zero. However, if B ≠ 0, then we have an xy term that prevents us from rewriting the equation in standard form. To eliminate it, we can rotate the axes by an acute angle θ where cot(2θ) = A − C B . • • • If cot(2θ) > 0, then 2θ is in the first quadrant, and θ is between (0°, 45°). If cot(2θ) < 0, then 2θ is in the second quadrant, and θ is between (45°, 90°). If A = C, then θ = 45°. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1413 Given an equation for a conic in the x′ y′ system, rewrite the equation without the x′ y′ term in terms of x′ and y′, where the x′ and y′ axes are rotations of the standard axes by θ degrees. 1. Find cot(2θ). 2. Find sin θ and cos θ. 3. Substitute sin θ and cos θ into x = x′ cos θ − y′ sin θ and y = x′ sin θ + y′ cos θ. 4. Substitute the expression for x and y into in the given equation, and then simplify. 5. Write the equations with x′ and y′ in the standard form with respect to the rotated axes. Example 12.22 Rewriting an Equation with respect to the x′ and y′ axes without the x′y′ Term Rewrite the equation 8x2 − 12xy + 17y2 = 20 in the x′ y′ system without an x′ y′ term. Solution First, we find cot(2θ). See Figure 12.46. 8x2 − 12xy + 17y2 = 20 ⇒ A = 8, B = − 12 and C = 17 cot(2θ) = cot(2θ) = −9 −12 A − C B = 8 − 17 −12 = 3 4 Figure 12.46 cot(2θ) = 3 4 = adjacent opposite So the hypotenuse is 32 + 42 = h2 9 + 16 = h2 25 = h2 h = 5 Next, we find sin θ and cos θ. 1414 Chapter 12 Analytic Geometry sin θ = 1 − cos(2θ) 2 = 1 − 3 5 2 = sin θ = 1 5 cos θ = 1 + cos(2θ) 2 = 1 + 3 5 2 = cos 10 = 10 = 4 5 Substitute the values of sin θ and cos θ into x = x′ cos θ − y′ sin θ and y = x′ sin θ + y′ cos θ. x = x′ cos θ − y′ sin θ ⎞ ⎞ ⎠ − y′ ⎠ x = x′ ⎛ ⎝ 1 5 ⎛ 2 ⎝ 5 2x′ − y′ 5 and x = y = x′ sin θ + y′ cos θ ⎞ ⎞ ⎠ + y′ ⎠ y = x′ + 2y′ 5 y = 2 Substitute the expressions for x and y into in the given equation, and then simplify. 8 ⎞ ⎠ 2x′ − y′ 5 x′ + 2y′ 2x′ − y′ ⎛ ⎞ ⎞ ⎛ ⎞ ⎛ ⎛ ⎠ + 17 − 12 ⎠ ⎝ ⎝ ⎠ ⎝ ⎝ 5 5 (x′ + 2y′)(x′ + 2y′) (2x′ − y′)(x′ + 2y′) ⎛ ⎞ ⎛ ⎞ ⎠ + 17 ⎠ − 12 ⎝ ⎝ 5 5 ⎛ ⎝2x′ 2 + 3x′ y′ − 2y′ 2⎞ ⎛ ⎝x′ 2 + 4x′ y′ + 4y′ 2⎞ ⎠ + 17 x′ + 2y′ 5 2 = 20 ⎞ ⎠ = 20 ⎛ ⎝ 8 (2x′ − y′)(2x′ − y′) 5 ⎛ ⎝4x′ 2 − 4x′ y′ + y′ 2⎞ ⎠ − 12 ⎠ = 100 8 32x′ 2 − 32x′ y′ + 8y′ 2 − 24x′ 2 − 36x′ y′ + 24y′ 2 + 17x′ 2 + 68x′ y′ + 68y′ 2 = 100 25x′ 2 + 100y′ 2 = 100 25 y′ 2 = 100 100 100 x′ 2 + 100 100 Write the equations with x′ and y′ in the standard form with respect to the new coordinate system. Figure 12.47 shows the graph of the ellipse. x′2 4 + y′2 1 = 1 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1415 Figure 12.47 12.21 Rewrite the 13x2 − 6 3xy + 7y2 = 16 in the x′ y′ system without the x′ y′ term. Example 12.23 Graphing an Equation That Has No x′y′ Terms Graph the following equation relative to the x′ y′ system: x2 + 12xy − 4y2 = 30 Solution First, we find cot(2θ). x2 + 12xy − 4y2 = 20 ⇒ A = 1, B = 12, and C = −4 cot(2θ) = cot(2θ) = A − C B 1 − (−4) 12 cot(2θ) = 5 12 Because cot(2θ) = 5 12 , we can draw a reference triangle as in Figure 12.48. 1416 Chapter 12 Analytic Geometry Figure 12.48 cot(2θ) = 5 12 = adjacent opposite Thus, the hypotenuse is 52 + 122 = h2 25 + 144 = h2 169 = h2 h = 13 Next, we find sin θ and cos θ. We will use half-angle identities. sin θ = 1 − cos(2θ) 2 = cos θ = 1 + cos(2θ) 2 = 1 − 5 13 2 1 + 5 13 2 = = Now we find x and y. 13 13 13 − 5 2 13 + 5 2 13 13 = 8 13 ⋅ 1 2 = 2 13 = 18 13 ⋅ 1 2 = 3 13 x = x′ cos θ − y′ sin θ ⎞ ⎠ − y′ x = x′ ⎛ ⎝ ⎛ ⎝ ⎞ ⎠ 2 13 3 13 3x′ − 2y′ 13 and x = This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1417 y = x′ sin θ + y′ cos θ ⎞ ⎠ + y′ y = x′ ⎛ ⎝ ⎛ ⎝ ⎞ ⎠ 3 13 2 13 2x′ + 3y′ 13 y = Now we substitute x = 3x′ − 2y′ 13 and y = 2x′ + 3y′ 13 into x2 + 12xy − 4y2 = 30. 2 ⎛ ⎝ ⎞ ⎠ + 12 3x′ − 2y′ 13 3x′ − 2y′ 13 ⎛ ⎝ ⎛ ⎝ ⎞ ⎛ ⎞ ⎛ ⎠ ⎝ ⎠ ⎝ ⎞ ⎡ ⎣(3x′ − 2y′)2 + 12(3x′ − 2y′)(2x′ + 3y′) − 4(2x′ + 3y′)2⎤ ⎠ ⎤ ⎡ ⎝4x′ 2 + 12x′ y′ + 9y′ 2⎞ ⎛ ⎝6x′ 2 + 5x′ y′ − 6y′ 2⎞ ⎛ ⎞ ⎣9x′ 2 − 12x′ y′ + 4y′ 2 + 12 ⎦ = 30 ⎠ ⎠ ⎞ ⎡ ⎣9x′ 2 − 12x′ y′ + 4y′ 2 + 72x′ 2 + 60x′ y′ − 72y′ 2 − 16x′ 2 − 48x′ y′ − 36y′ 2⎤ ⎠ 2x′ + 3y′ 13 2x′ + 3y′ 13 ⎞ ⎠ − 4 ⎦ = 30 ⎠ − 4 = 30 1 13 ⎛ ⎝ ⎦ = 30 1 13 ⎛ 1 ⎝ 13 2 Factor. Multiply. Distribute. 65x′ 2 − 104y′ 2 = 390 Multiply. ⎛ ⎝ 1 13 ⎞ ⎣65x′ 2 − 104y′ 2⎤ ⎡ ⎠ ⎦ = 30 Combine like terms. x′ 2 6 − 4y′ 2 15 = 1 Divide by 390. Figure 12.49 shows the graph of the hyperbola x′2 6 − 4y′2 15 = 1. Figure 12.49 Identifying Conics without Rotating Axes Now we have come full circle. How do we identify the type of conic described by an equation? What happens when the axes are rotated? Recall, the general form of a conic is If we apply the rotation formulas to this equation we get the form Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 1418 Chapter 12 Analytic Geometry A′ x′2 + B′ x′ y′ + C′ y′2 + D′ x′ + E′ y′ + F′ = 0 It may be shown that B2 − 4AC = B′2 − 4A′ C′. The expression does not vary after rotation, so we call the expression invariant. The discriminant, B2 − 4AC, is invariant and remains unchanged after rotation. Because the discriminant remains unchanged, observing the discriminant enables us to identify the conic section. Using the Discriminant to Identify a Conic If the equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 is transformed by rotating axes into the equation A′ x′2 + B′ x′ y′ + C′ y′2 + D′ x′ + E′ y′ + F′ = 0, then B2 − 4AC = B′2 − 4A′ C′. The equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 is an ellipse, a parabola, or a hyperbola, or a degenerate case of one of these. If the discriminant, B2 − 4AC, is • < 0, • = 0, • > 0, the conic section is an ellipse the conic section is a parabola the conic section is a hyperbola Example 12.24 Identifying the Conic without Rotating Axes Identify the conic for each of the following without rotating axes. a. b. 5x2 + 2 3xy + 2y2 − 5 = 0 5x2 + 2 3xy + 12y2 − 5 = 0 Solution a. Let’s begin by determining A, B, and C. Now, we find the discriminant. 5 ⏟ A x2 + 2 3 ⏟ B xy + 2 ⏟ C y2 − 5 = 0 B2 − 4AC = ⎛ ⎝2 3⎞ = 4(3) − 40 = 12 − 40 = − 28 < 0 Therefore, 5x2 + 2 3xy + 2y2 − 5 = 0 represents an ellipse. ⎠ 2 − 4(5)(2) b. Again, let’s begin by determining A, B, and C. x2 + 2 3 ⏟ B 5 ⏟ A xy + 12 ⏟ C y2 − 5 = 0 Now, we find the discriminant. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1419 B2 − 4AC = ⎛ = 4(3) − 240 = 12 − 240 = − 228 < 0 Therefore, 5x2 + 2 3xy + 12y2 − 5 = 0 represents an ellipse. ⎝2 3⎞ ⎠ 2 − 4(5)(12) 12.22 Identify the conic for each of the following without rotating axes. a. x2 − 9xy + 3y2 − 12 = 0 b. 10x2 − 9xy + 4y2 − 4 = 0 Access this online resource for additional instruction and practice with conic sections and rotation of axes. • Introduction to Conic Sections (http://openstaxcollege.org/l/introconic) 1420 Chapter 12 Analytic Geometry 12.4 EXERCISES Verbal For the following exercises, find a new representation of the given equation after rotating through the given angle. 209. What effect does the xy term have on the graph of a conic section? 210. If the equation of a conic section is written in the form Ax2 + By2 + Cx + Dy + E = 0 and AB = 0, what can we conclude? If the equation of a conic section is written in the form 211. Ax2 + Bxy + Cy2 + Dx + Ey + F = 0, and B2 − 4AC > 0, what can we conclude? 226. 227. 228. 229. 230. 3x2 + xy + 3y2 − 5 = 0, θ = 45° 4x2 − xy + 4y2 − 2 = 0, θ = 45° 2x2 + 8xy − 1 = 0, θ = 30° −2x2 + 8xy + 1 = 0, θ = 45° 4x2 + 2xy + 4y2 + y + 2 = 0, θ = 45° 212. Given the equation ax2 + 4x + 3y2 − 12 = 0, what can we conclude if a > 0 ? For 213. Ax2 + Bxy + Cy2 + Dx + Ey + F = 0, the equation the value of θ that satisfies cot(2θ) = A − C B gives us what information? Algebraic For the following exercises, determine which conic section is represented based on the given equation. 214. 215. 216. 217. 218. 219. 220. 221. 222. 223. 224. 225. 9x2 + 4y2 + 72x + 36y − 500 = 0 x2 − 10x + 4y − 10 = 0 2x2 − 2y
2 + 4x − 6y − 2 = 0 4x2 − y2 + 8x − 1 = 0 4y2 − 5x + 9y + 1 = 0 2x2 + 3y2 − 8x − 12y + 2 = 0 4x2 + 9xy + 4y2 − 36y − 125 = 0 3x2 + 6xy + 3y2 − 36y − 125 = 0 −3x2 + 3 3xy − 4y2 + 9 = 0 2x2 + 4 3xy + 6y2 − 6x − 3 = 0 −x2 + 4 2xy + 2y2 − 2y + 1 = 0 8x2 + 4 2xy + 4y2 − 10x + 1 = 0 This content is available for free at https://cnx.org/content/col11758/1.5 For the following exercises, determine the angle θ that will eliminate the xy term and write the corresponding equation without the xy term. 231. 232. 233. 234. 235. 236. 237. 238. x2 + 3 3xy + 4y2 + y − 2 = 0 4x2 + 2 3xy + 6y2 + y − 2 = 0 9x2 − 3 3xy + 6y2 + 4y − 3 = 0 −3x2 − 3xy − 2y2 − x = 0 16x2 + 24xy + 9y2 + 6x − 6y + 2 = 0 x2 + 4xy + 4y2 + 3x − 2 = 0 x2 + 4xy + y2 − 2x + 1 = 0 4x2 − 2 3xy + 6y2 − 1 = 0 Graphical For the following exercises, rotate through the given angle based on the given equation. Give the new equation and graph the original and rotated equation. 239. y = − x2, θ = − 45∘ 240. x = y2, θ = 45∘ x2 4 y2 16 + y2 1 + x2 9 = 1, θ = 45∘ = 1, θ = 45∘ 241. 242. 243. Chapter 12 Analytic Geometry 1421 y2 − x2 = 1, θ = 45∘ For the following exercises, determine the value of k based on the given equation. 264. Given 4x2 + kxy + 16y2 + 8x + 24y − 48 = 0, find k for the graph to be a parabola. 265. Given 2x2 + kxy + 12y2 + 10x − 16y + 28 = 0, find k for the graph to be an ellipse. 266. Given 3x2 + kxy + 4y2 − 6x + 20y + 128 = 0, find k for the graph to be a hyperbola. 267. Given kx2 + 8xy + 8y2 − 12x + 16y + 18 = 0, find k for the graph to be a parabola. 268. Given 6x2 + 12xy + ky2 + 16x + 10y + 4 = 0, find k for the graph to be an ellipse. 244. 245. 246. y = x2 2 , θ = 30∘ x = ⎛ ⎝y − 1⎞ ⎠ 2, θ = 30∘ x2 9 + y2 4 = 1, θ = 30∘ For the following exercises, graph the equation relative to the x′ y′ system in which the equation has no x′ y′ term. 247. xy = 9 248. 249. 250. 251. 252. 253. 254. 255. 256. 257. x2 + 10xy + y2 − 6 = 0 x2 − 10xy + y2 − 24 = 0 4x2 − 3 3xy + y2 − 22 = 0 6x2 + 2 3xy + 4y2 − 21 = 0 11x2 + 10 3xy + y2 − 64 = 0 21x2 + 2 3xy + 19y2 − 18 = 0 16x2 + 24xy + 9y2 − 130x + 90y = 0 16x2 + 24xy + 9y2 − 60x + 80y = 0 13x2 − 6 3xy + 7y2 − 16 = 0 4x2 − 4xy + y2 − 8 5x − 16 5y = 0 For the following exercises, determine the angle of rotation in order to eliminate the xy term. Then graph the new set of axes. 258. 259. 260. 261. 262. 263. 6x2 − 5 3xy + y2 + 10x − 12y = 0 6x2 − 5xy + 6y2 + 20x − y = 0 6x2 − 8 3xy + 14y2 + 10x − 3y = 0 4x2 + 6 3xy + 10y2 + 20x − 40y = 0 8x2 + 3xy + 4y2 + 2x − 4 = 0 16x2 + 24xy + 9y2 + 20x − 44y = 0 1422 Chapter 12 Analytic Geometry 12.5 \| Conic Sections in Polar Coordinates Learning Objectives In this section, you will: 12.5.1 Identify a conic in polar form. 12.5.2 Graph the polar equations of conics. 12.5.3 Deﬁne conics in terms of a focus and a directrix. Figure 12.50 Planets orbiting the sun follow elliptical paths. (credit: NASA Blueshift, Flickr) Most of us are familiar with orbital motion, such as the motion of a planet around the sun or an electron around an atomic nucleus. Within the planetary system, orbits of planets, asteroids, and comets around a larger celestial body are often elliptical. Comets, however, may take on a parabolic or hyperbolic orbit instead. And, in reality, the characteristics of the planets’ orbits may vary over time. Each orbit is tied to the location of the celestial body being orbited and the distance and direction of the planet or other object from that body. As a result, we tend to use polar coordinates to represent these orbits. In an elliptical orbit, the periapsis is the point at which the two objects are closest, and the apoapsis is the point at which they are farthest apart. Generally, the velocity of the orbiting body tends to increase as it approaches the periapsis and decrease as it approaches the apoapsis. Some objects reach an escape velocity, which results in an infinite orbit. These bodies exhibit either a parabolic or a hyperbolic orbit about a body; the orbiting body breaks free of the celestial body’s gravitational pull and fires off into space. Each of these orbits can be modeled by a conic section in the polar coordinate system. Identifying a Conic in Polar Form Any conic may be determined by three characteristics: a single focus, a fixed line called the directrix, and the ratio of the distances of each to a point on the graph. Consider the parabola x = 2 + y2 shown in Figure 12.51. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1423 Figure 12.51 In The Parabola, we learned how a parabola is defined by the focus (a fixed point) and the directrix (a fixed line). In this section, we will learn how to define any conic in the polar coordinate system in terms of a fixed point, the focus P(r, θ) at the pole, and a line, the directrix, which is perpendicular to the polar axis. If F is a fixed point, the focus, and D is a fixed line, the directrix, then we can let e be a fixed positive number, called the eccentricity, which we can define as the ratio of the distances from a point on the graph to the focus and the point on the graph to the directrix. Then the set of all points P such that e = PF is a conic. In other words, we can define a conic as PD the set of all points P with the property that the ratio of the distance from P to F to the distance from P to D is equal to the constant e. For a conic with eccentricity e, • • • if 0 ≤ e < 1, the conic is an ellipse if e = 1, if e > 1, the conic is a parabola the conic is an hyperbola With this definition, we may now define a conic in terms of the directrix, x = ± p, Thus, each conic may be written as a polar equation, an equation written in terms of r and θ. the eccentricity e, and the angle θ. The Polar Equation for a Conic For a conic with a focus at the origin, if the directrix is x = ± p, where p is a positive real number, and the eccentricity is a positive real number e, the conic has a polar equation r = ep 1 ± e cos θ For a conic with a focus at the origin, if the directrix is y = ± p, where p eccentricity is a positive real number e, the conic has a polar equation is a positive real number, and the r = ep 1 ± e sin θ 1424 Chapter 12 Analytic Geometry Given the polar equation for a conic, identify the type of conic, the directrix, and the eccentricity. 1. Multiply the numerator and denominator by the reciprocal of the constant in the denominator to rewrite the equation in standard form. 2. Identify the eccentricity e as the coefficient of the trigonometric function in the denominator. 3. Compare e with 1 to determine the shape of the conic. 4. Determine the directrix as x = p if cosine is in the denominator and y = p if sine is in the denominator. Set ep equal to the numerator in standard form to solve for x or y. Example 12.25 Identifying a Conic Given the Polar Form For each of the following equations, identify the conic with focus at the origin, the directrix, and the eccentricity. a. b. c. r = 6 3 + 2 sin θ r = 12 4 + 5 cos θ r = 7 2 − 2 sin θ Solution For each of the three conics, we will rewrite the equation in standard form. Standard form has a 1 as the constant in the denominator. Therefore, in all three parts, the first step will be to multiply the numerator and denominator by the reciprocal of the constant of the original equation, 1 c, where c is that constant. a. Multiply the numerator and denominator by 1 3 ⎞ ⎠ ⎞ ⎠ 6 3 + 2sin Because sin θ is in the denominator, the directrix is y = p. Comparing to standard form, note that e = 2 3 . Therefore, from the numerator, 3 ⎞ ⎠sin sin θ p 2 = ep Since e < 1, the conic is an ellipse. The eccentricity is e = 2 3 and the directrix is y = 3. b. Multiply the numerator and denominator by 1 4 . This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1425 r = 12 4 + 5 cos = 12 ⎛ ⎞ 1 ⎠ ⎝ 4 ⎛ ⎞ ⎞ 1 ⎠cos cos θ Because cos θ is in the denominator, the directrix is x = p. Comparing to standard form, e = 5 4 . Therefore, from the numerator, p 3 = ep 3 = ⎝ ⎝ 5 5 12 = p 5 ⎞ ⎠ p 5 4 Since e > 1, the conic is a hyperbola. The eccentricity is e = 5 4 and the directrix is x = 12 5 = 2.4. c. Multiply the numerator and denominator by 1 2 . r = 7 2 − 2 sin ⎠ sin − sin θ Because sine is in the denominator, the directrix is y = −p. Comparing to standard form, e = 1. Therefore, from the numerator, r = = ep = (1)p = p 7 2 7 2 7 2 Because e = 1, the conic is a parabola. The eccentricity is e = 1 and the directrix is y = − 7 2 = −3.5. 12.23 Identify the conic with focus at the origin, the directrix, and the eccentricity for r = 2 3 − cos θ. Graphing the Polar Equations of Conics When graphing in Cartesian coordinates, each conic section has a unique equation. This is not the case when graphing in polar coordinates. We must use the eccentricity of a conic section to determine which type of curve to graph, and then 1426 Chapter 12 Analytic Geometry determine its specific characteristics. The first step is to rewrite the conic in standard form as we have done in the previous example. In other words, we need to rewrite the equation so that the denominator begins with 1. This enables us to determine e and, therefore, the shape of the curve. The next step is to substitute values for θ and solve for r to plot a few key points. Setting θ equal to 0, π 2 , π, and 3π 2 Example 12.26 provides the vertices so we can create a rough sketch of the graph. Graphing a Parabola in Polar Form Graph r = 5 3 + 3 cos θ. Solution First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 3, which is cos θ r = 5 3 + 3 cos + cos θ Because e = 1, we will graph a parabola with a focus at the origin. The function has a cos θ, and there is an addition sign in the denominator, so the directrix is x = p. = ep = (1)p = p 5 3 5 3 5 3 The directrix is x = 5 3 . Plotting a few key points as in Table 12.5 will enable us to see the vertices. See Figure 12.52 3π 2 r = 5 3 + 3 cos θ
5 6 ≈ 0.83 5 3 ≈ 1.67 undefined 5 3 ≈ 1.67 Table 12.5 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1427 Figure 12.52 Analysis We can check our result with a graphing utility. See Figure 12.53. Figure 12.53 Example 12.27 Graphing a Hyperbola in Polar Form Graph r = 8 2 − 3 sin θ. Solution First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 2, which is 1 2 . 1428 Chapter 12 Analytic Geometry ⎛ ⎝ sin θ r = 8 2 − 3sin sin θ Because e = 3 2 , e > 1, so we will graph a hyperbola with a focus at the origin. The function has a sin θ term and there is a subtraction sign in the denominator, so the directrix is y = −p. 4 = ep ⎞ 3 ⎠ The directrix is y = − 8 3 . Plotting a few key points as in Table 12.6 will enable us to see the vertices. See Figure 12.54. A B π 2 C π D 3π 2 −8 4 8 5 = 1.6 θ r = 8 2 − 3sin θ Table 12.6 0 4 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1429 Figure 12.54 Example 12.28 Graphing an Ellipse in Polar Form Graph r = 10 5 − 4 cos θ. Solution First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 5, which is 1 5 . 1430 Chapter 12 Analytic Geometry 10 ⎛ ⎞ 1 ⎠ ⎝ 5 ⎞ ⎛ ⎞ 1 ⎠cos = 10 5 − 4cos θ = r = 2 1 − 4 5 cos θ Because e = 4 5 , e < 1, so we will graph an ellipse with a focus at the origin. The function has a cos θ, and there is a subtraction sign in the denominator, so the directrix is x = −p. 2 = ep ⎞ 4 ⎠ The directrix is x = − 5 2 . Plotting a few key points as in Table 12.7 will enable us to see the vertices. See Figure 12.55. A 0 10 B C π 2 2 π 10 9 ≈ 1.1 D 3π 2 2 θ r = 10 5 − 4 cos θ Table 12.7 Figure 12.55 Analysis This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1431 We can check our result using a graphing utility. See Figure 12.56. Figure 12.56 r = 10 5 − 4 cos θ window of [–3, 12, 1] by [ – 4, 4, 1], θ min = 0 and θ max = 2π. graphed on a viewing 12.24 Graph r = 2 4 − cos θ. Deﬁning Conics in Terms of a Focus and a Directrix So far we have been using polar equations of conics to describe and graph the curve. Now we will work in reverse; we will use information about the origin, eccentricity, and directrix to determine the polar equation. Given the focus, eccentricity, and directrix of a conic, determine the polar equation. 1. Determine whether the directrix is horizontal or vertical. If the directrix is given in terms of y, we use the general polar form in terms of sine. If the directrix is given in terms of x, we use the general polar form in terms of cosine. 2. Determine the sign in the denominator. If p < 0, use subtraction. If p > 0, use addition. 3. Write the coefficient of the trigonometric function as the given eccentricity. 4. Write the absolute value of p in the numerator, and simplify the equation. Example 12.29 Finding the Polar Form of a Vertical Conic Given a Focus at the Origin and the Eccentricity and Directrix Find the polar form of the conic given a focus at the origin, e = 3 and directrix y = − 2. 1432 Chapter 12 Analytic Geometry Solution The directrix is y = −p, Because y = −2, –2 < 0, of so we know the trigonometric function in the denominator is sine. so we know there is a subtraction sign in the denominator. We use the standard form and e = 3 and \|−2\| = 2 = p. Therefore, r = ep 1 − e sin θ r = r = (3)(2) 1 − 3 sin θ 6 1 − 3 sin θ Example 12.30 Finding the Polar Form of a Horizontal Conic Given a Focus at the Origin and the Eccentricity and Directrix Find the polar form of a conic given a focus at the origin, e = 3 5 , and directrix x = 4. Solution Because the directrix is x = p, we know the function in the denominator is cosine. Because x = 4, 4 > 0, so we know there is an addition sign in the denominator. We use the standard form of r = ep 1 + e cos θ and \|4\| = 4 = p. and e = 3 5 Therefore, This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1433 r = r = r = r = ⎞ ⎠(4) cos θ ⎛ 3 ⎝ 5 1 + 3 5 12 5 1 + 3 cos θ 5 cos θ 12 5 ⎞ ⎠ + 3 5 12 5 cos = 12 5 5 5 + 3 cos θ 12 5 + 3 cos θ r = 12.25 Find the polar form of the conic given a focus at the origin, e = 1, and directrix x = −1. Example 12.31 Converting a Conic in Polar Form to Rectangular Form Convert the conic r = 1 5 − 5sin θ to rectangular form. Solution We will rearrange the formula to use the identities r = x2 + y2, x = r cos θ, and y = r sin θ. 1 5 − 5 sin θ 1 5 − 5 sin θ ⋅ (5 − 5 sin θ) Eliminate the fraction. r = r ⋅ (5 − 5 sin θ) = 5r − 5r sin θ = 1 5r = 1 + 5r sin θ 25r 2 = (1 + 5r sin θ)2 25(x2 + y2) = (1 + 5y)2 25x2 + 25y2 = 1 + 10y + 25y2 25x2 − 10y = 1 Distribute. Isolate 5r. Square both sides. Substitute r = x2 + y2 and y = r sin θ. Distribute and use FOIL. Rearrange terms and set equal to 1. 12.26 Convert the conic r = 2 1 + 2 cos θ to rectangular form. 1434 Chapter 12 Analytic Geometry Access these online resources for additional instruction and practice with conics in polar coordinates. • Polar Equations of Conic Sections (http://openstaxcollege.org/l/determineconic) • Graphing Polar Equations of Conics - 1 (http://openstaxcollege.org/l/graphconic1) • Graphing Polar Equations of Conics - 2 (http://openstaxcollege.org/l/graphconic2) this website (http://openstaxcollege.org/l/PreCalcLPC10) Visit Learningpod. for additional practice questions from This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1435 12.5 EXERCISES Verbal Explain how eccentricity determines which conic 269. section is given. If a conic section is written as a polar equation, what 270. must be true of the denominator? 271. If a conic section is written as a polar equation, and the denominator involves sin θ, what conclusion can be drawn about the directrix? 272. If the directrix of a conic section is perpendicular to the polar axis, what do we know about the equation of the graph? What do we know about the focus/foci of a conic 273. section if it is written as a polar equation? Algebraic For the following exercises, identify the conic with a focus at the origin, and then give the directrix and eccentricity. 274. 275. 276. 277. 278. 279. 280. 281. r = 6 1 − 2 cos θ r = 3 4 − 4 sin θ r = 8 4 − 3 cos θ r = 5 1 + 2 sin θ r = 16 4 + 3 cos θ r = 3 10 + 10 cos θ r = 2 1 − cos θ r = 4 7 + 2 cos θ 282. r(1 − cos θ) = 3 283. r(3 + 5sin θ) = 11 284. r(4 − 5sin θ) = 1 285. r(7 + 8cos θ) = 7 For the following exercises, convert the polar equation of a conic section to a rectangular equation. 286. 287. 288. 289. 290. 291. 292. 293. r = 4 1 + 3 sin θ r = 2 5 − 3 sin θ r = 8 3 − 2 cos θ r = 3 2 + 5 cos θ r = 4 2 + 2 sin θ r = 3 8 − 8 cos θ r = 2 6 + 7 cos θ r = 5 5 − 11 sin θ 294. r(5 + 2 cos θ) = 6 295. r(2 − cos θ) = 1 296. r(2.5 − 2.5 sin θ) = 5 297. r = 6sec θ −2 + 3 sec θ 298. r = 6csc θ 3 + 2 csc θ For the following exercises, graph the given conic section. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse, label the vertices and foci. If it is a hyperbola, label the vertices and foci. 299. 300. 301. 302. 303. 304. r = 5 2 + cos θ r = 2 3 + 3 sin θ r = 10 5 − 4 sin θ r = 3 1 + 2 cos θ r = 8 4 − 5 cos θ r = 3 4 − 4 cos θ 1436 305. 306. r = 2 1 − sin θ r = 6 3 + 2 sin θ 307. r(1 + cos θ) = 5 308. r(3 − 4sin θ) = 9 309. r(3 − 2sin θ) = 6 310. r(6 − 4cos θ) = 5 For the following exercises, find the polar equation of the conic with focus at the origin and the given eccentricity and directrix. Chapter 12 Analytic Geometry 324. xy = 2 325. x2 + xy + y2 = 4 326. 327. 2x2 + 4xy + 2y2 = 9 16x2 + 24xy + 9y2 = 4 328. 2xy + y = 1 311. Directrix: x = 4; e = 1 5 312. Directrix: x = − 4; e = 5 313. Directrix: y = 2; e = 2 314. Directrix: y = − 2; e = 1 2 315. Directrix: x = 1; e = 1 316. Directrix: x = − 1; e = 1 317. 318. 319. 320. 321. Directrix Directrix Directrix: y = 4; e = 3 2 Directrix: x = −2; e = 8 3 Directrix: x = −5; e = 3 4 322. Directrix: y = 2; e = 2.5 323. Directrix: x = −3; e = 1 3 Extensions Recall from Rotation of Axes that equations of conics with an xy term have rotated graphs. For the following exercises, express each equation in polar form with r as a function of θ. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1437 CHAPTER 12 REVIEW KEY TERMS angle of rotation an acute angle formed by a set of axes rotated from the Cartesian plane where, if cot(2θ) > 0, then θ is between (0°, 45°); if cot(2θ) < 0, then θ is between (45°, 90°); and if cot(2θ) = 0, then θ = 45° center of a hyperbola the midpoint of both the transverse and conjugate axes of a hyperbola center of an ellipse the midpoint of both the major and minor axes conic section any shape resulting from the intersection of a right circular cone with a plane conjugate axis the axis of a hyperbola that is perpendicular to the transverse axis and has the co-vertices as its endpoints degenerate conic sections any of the possible shapes formed when a plane intersects a double cone through the apex. Types of degenerate conic sections include a point, a line, and intersecting lines. directrix a line perpendicular to the axis of symmetry of a parabola; a line such that the ratio of the distance between the points on the conic and the focus to the distance to the directrix is constant the ratio of the distances from a point P on the graph to the focus F and to the directrix D represented by eccentricity e = PF PD, where e is a positive real number ellipse the set of all points (x, y) in a plane such that the sum of their distances from two fixed points is a constant foci plural of focus focus (of a parabola) a fixed point in the interior of a parabola that lies on the axis of symmetry focus (of an ellipse) one of the two fixed points on the major axis of an ellipse such that the sum of the distances from these points to an
y point (x, y) on the ellipse is a constant hyperbola the set of all points (x, y) in a plane such that the difference of the distances between (x, y) and the foci is a positive constant latus rectum parabola the line segment that passes through the focus of a parabola parallel to the directrix, with endpoints on the major axis the longer of the two axes of an ellipse minor axis the shorter of the two axes of an ellipse nondegenerate conic section a shape formed by the intersection of a plane with a double right cone such that the plane does not pass through the apex; nondegenerate conics include circles, ellipses, hyperbolas, and parabolas parabola the set of all points (x, y) in a plane that are the same distance from a fixed line, called the directrix, and a fixed point (the focus) not on the directrix polar equation an equation of a curve in polar coordinates r and θ transverse axis the axis of a hyperbola that includes the foci and has the vertices as its endpoints KEY EQUATIONS 1438 Chapter 12 Analytic Geometry Horizontal ellipse, center at origin Vertical ellipse, center at origin Horizontal ellipse, center (h, k) Vertical ellipse, center (h, k) x2 a2 + y2 b2 = 1, a > b x2 b2 + y2 a2 = 1, a > b (x − h)2 a2 + (x − h)2 b2 + ⎛ ⎝y − k⎞ ⎠ 2 b2 = 1, a > b ⎛ ⎝y − k⎞ ⎠ 2 a2 = 1, a > b Hyperbola, center at origin, transverse axis on x-axis Hyperbola, center at origin, transverse axis on y-axis x2 a2 − y2 b2 = 1 y2 a2 − x2 b2 = 1 Hyperbola, center at (h, k), transverse axis parallel to x-axis (x − h)2 a2 − ⎛ ⎝y − k⎞ ⎠ 2 b2 = 1 Hyperbola, center at (h, k), transverse axis parallel to y-axis ⎛ ⎝y − k⎞ ⎠ 2 a2 − (x − h)2 b2 = 1 Parabola, vertex at origin, axis of symmetry on x-axis Parabola, vertex at origin, axis of symmetry on y-axis Parabola, vertex at (h, k), axis of symmetry on x-axis Parabola, vertex at (h, k), axis of symmetry on y-axis y2 = 4px x2 = 4py ⎛ ⎝y − k⎞ ⎠ 2 = 4p(x − h) (x − h)2 = 4p⎛ ⎝y − k⎞ ⎠ General Form equation of a conic section Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 Rotation of a conic section Angle of rotation x = x′ cos θ − y′ sin θ y = x′ sin θ + y′ cos θ θ, where cot(2θ) = A − C B This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 12 Analytic Geometry 1439 KEY CONCEPTS 12.1 The Ellipse • An ellipse is the set of all points (x, y) in a plane such that the sum of their distances from two fixed points is a constant. Each fixed point is called a focus (plural: foci). • When given the coordinates of the foci and vertices of an ellipse, we can write the equation of the ellipse in standard form. See Example 12.1 and Example 12.2. • When given an equation for an ellipse centered at the origin in standard form, we can identify its vertices, covertices, foci, and the lengths and positions of the major and minor axes in order to graph the ellipse. See Example 12.3 and Example 12.4. • When given the equation for an ellipse centered at some point other than the origin, we can identify its key features and graph the ellipse. See Example 12.5 and Example 12.6. • Real-world situations can be modeled using the standard equations of ellipses and then evaluated to find key features, such as lengths of axes and distance between foci. See Example 12.7. 12.2 The Hyperbola • A hyperbola is the set of all points (x, y) in a plane such that the difference of the distances between (x, y) and the foci is a positive constant. • The standard form of a hyperbola can be used to locate its vertices and foci. See Example 12.8. • When given the coordinates of the foci and vertices of a hyperbola, we can write the equation of the hyperbola in standard form. See Example 12.9 and Example 12.10. • When given an equation for a hyperbola, we can identify its vertices, co-vertices, foci, asymptotes, and lengths and positions of the transverse and conjugate axes in order to graph the hyperbola. See Example 12.11 and Example 12.12. • Real-world situations can be modeled using the standard equations of hyperbolas. For instance, given the dimensions of a natural draft cooling tower, we can find a hyperbolic equation that models its sides. See Example 12.13. 12.3 The Parabola • A parabola is the set of all points (x, y) in a plane that are the same distance from a fixed line, called the directrix, and a fixed point (the focus) not on the directrix. • The standard form of a parabola with vertex (0, 0) and the x-axis as its axis of symmetry can be used to graph the parabola. If p > 0, the parabola opens right. If p < 0, the parabola opens left. See Example 12.14. • The standard form of a parabola with vertex (0, 0) and the y-axis as its axis of symmetry can be used to graph the parabola. If p > 0, the parabola opens up. If p < 0, the parabola opens down. See Example 12.15. • When given the focus and directrix of a parabola, we can write its equation in standard form. See Example 12.16. • The standard form of a parabola with vertex (h, k) and axis of symmetry parallel to the x-axis can be used to graph the parabola. If p > 0, the parabola opens right. If p < 0, the parabola opens left. See Example 12.17. • The standard form of a parabola with vertex (h, k) and axis of symmetry parallel to the y-axis can be used to graph the parabola. If p > 0, the parabola opens up. If p < 0, the parabola opens down. See Example 12.18. • Real-world situations can be modeled using the standard equations of parabolas. For instance, given the diameter and focus of a cross-section of a parabolic reflector, we can find an equation that models its sides. See Example 12.19. 1440 Chapter 12 Analytic Geometry 12.4 Rotation of Axes • Four basic shapes can result from the intersection of a plane with a pair of right circular cones connected tail to tail. They include an ellipse, a circle, a hyperbola, and a parabola. • A nondegenerate conic section has the general form Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 where A, B and C are not all zero. The values of A, B, and C determine the type of conic. See Example 12.20. • Equations of conic sections with an xy term have been rotated about the origin. See Example 12.21. • The general form can be transformed into an equation in the x′ and y′ coordinate system without the x′ y′ term. See Example 12.22 and Example 12.23. • An expression is described as invariant if it remains unchanged after rotating. Because the discriminant is invariant, observing it enables us to identify the conic section. See Example 12.24. 12.5 Conic Sections in Polar Coordinates • Any conic may be determined by a single focus, the corresponding eccentricity, and the directrix. We can also define a conic in terms of a fixed point, the focus P(r, θ) at the pole, and a line, the directrix, which is perpendicular to the polar axis. • A conic is the set of all points e = PF PD, where eccentricity e is a positive real number. Each conic may be written in terms of its polar equation. See Example 12.25. • The polar equations of conics can be graphed. See Example 12.26, Example 12.27, and Example 12.28. • Conics can be defined in terms of a focus, a directrix, and eccentricity. See Example 12.29 and Example 12.30. • We can use the identities r = x2 + y2, x = r cos θ, and y = r sin θ to convert the equation for a conic from polar to rectangular form. See Example 12.31. CHAPTER 12 REVIEW EXERCISES The Ellipse For the following exercises, write the equation of the ellipse in standard form. Then identify the center, vertices, and foci. 329. x2 25 + y2 64 = 1 330. (x − 2)2 100 + ⎠ ⎛ ⎝y + 3⎞ 36 2 = 1 331. 9x2 + y2 + 54x − 4y + 76 = 0 332. 9x2 + 36y2 − 36x + 72y + 36 = 0 For the following exercises, graph the ellipse, noting center, vertices, and foci. 333. x2 36 + y2 9 = 1 This content is available for free at https://cnx.org/content/col11758/1.5 334. (x − 4)2 25 + ⎛ ⎠ ⎝y + 3⎞ 49 2 = 1 335. 4x2 + y2 + 16x + 4y − 44 = 0 336. 2x2 + 3y2 − 20x + 12y + 38 = 0 For the following exercises, use the given information to find the equation for the ellipse. 337. Center at (0, 0), focus at (3, 0), vertex at (−5, 0) Center at (2, −2), vertex at (7, −2), focus at 338. (4, −2) 339. A whispering gallery is to be constructed such that the foci are located 35 feet from the center. If the length of the gallery is to be 100 feet, what should the height of the ceiling be? Chapter 12 Analytic Geometry 1441 The Hyperbola For the following exercises, write the equation of the hyperbola in standard form. Then give the center, vertices, and foci. For the following exercises, graph the parabola, labeling vertex, focus, and directrix. 354. x2 + 4y = 0 340. x2 81 − y2 9 = 1 355. ⎛ ⎝y − 1⎞ ⎠ 2 = 1 2 (x + 3) 341. ⎛ ⎠ ⎝y + 1⎞ 16 2 − (x − 4)2 36 = 1 342. 9y2 − 4x2 + 54y − 16x + 29 = 0 343. 3x2 − y2 − 12x − 6y − 9 = 0 For the following exercises, graph the hyperbola, labeling vertices and foci. 344. x2 9 − y2 16 = 1 345. ⎛ ⎠ ⎝y − 1⎞ 49 2 − (x + 1)2 4 = 1 356. x2 − 8x − 10y + 46 = 0 357. 2y2 + 12y + 6x + 15 = 0 For the following exercises, write the equation of the parabola using the given information. 358. Focus at (−4, 0); directrix is x = 4 359. Focus at ⎛ ⎝2, 9 8 ⎞ ⎠; directrix is y = 7 8 360. A cable TV receiving dish is the shape of a paraboloid of revolution. Find the location of the receiver, which is placed at the focus, if the dish is 5 feet across at its opening and 1.5 feet deep. 346. x2 − 4y2 + 6x + 32y − 91 = 0 Rotation of Axes 347. 2y2 − x2 − 12y − 6 = 0 For the following exercises, find the equation of the hyperbola. 348. Center at (0, 0), vertex at (0, 4), focus at (0, −6) 349. Foci at (3, 7) and (7, 7), vertex at (6, 7) The Parabola For the following exercises, write the equation of the parabola in standard form. Then give the vertex, focus, and directrix. 350. y2 = 12x 351. (x + 2)2 = 1 2 ⎛ ⎝y − 1⎞ ⎠ For the following exercises, determine which of the conic sections is represented. 361. 16x2 + 24xy + 9y2 + 24x − 60y − 60 = 0 362. 4x2 + 14xy + 5y2 + 18x − 6y + 30 = 0 363. 4x2 + xy + 2y2 + 8x − 26y + 9 = 0 For the following exercises, determine the angle θ that
will corresponding eliminate and write the xy term, the equation without the xy term. 364. x2 + 4xy − 2y2 − 6 = 0 365. x2 − xy + y2 − 6 = 0 For the following exercises, graph the equation relative to the x′ y′ system in which the equation has no x′ y′ term. 352. y2 − 6y − 6x − 3 = 0 366. 9x2 − 24xy + 16y2 − 80x − 60y + 100 = 0 353. x2 + 10x − y + 23 = 0 367. x2 − xy + y2 − 2 = 0 1442 Chapter 12 Analytic Geometry 368. 6x2 + 24xy − y2 − 12x + 26y + 11 = 0 Conic Sections in Polar Coordinates For the following exercises, given the polar equation of the conic with focus at the origin, identify the eccentricity and directrix. 369. r = 10 1 − 5 cos θ 370. r = 6 3 + 2 cos θ 371. r = 1 4 + 3 sin θ 372. r = 3 5 − 5 sin θ For the following exercises, graph the conic given in polar form. If it the vertex, focus, and directrix. If it is an ellipse or a hyperbola, label the vertices and foci. is a parabola, label 373. r = 3 1 − sin θ 374. r = 8 4 + 3 sin θ 375. r = 10 4 + 5 cos θ 376. r = 9 3 − 6 cos θ For the following exercises, given information about the graph of a conic with focus at the origin, find the equation in polar form. 377. Directrix is x = 3 and eccentricity e = 1 378. Directrix is y = −2 and eccentricity e = 4 CHAPTER 12 PRACTICE TEST For the following exercises, write the equation in standard form and state the center, vertices, and foci. 381. (x − 3)2 64 + ⎠ ⎛ ⎝y − 2⎞ 36 2 = 1 379. x2 9 + y2 4 = 1 380. 9y2 + 16x2 − 36y + 32x − 92 = 0 For the following exercises, sketch the graph, identifying the center, vertices, and foci. This content is available for free at https://cnx.org/content/col11758/1.5 382. 2x2 + y2 + 8x − 6y − 7 = 0 383. Write the standard form equation of an ellipse with a center at (1, 2), vertex at (7, 2), and focus at (4, 2). Chapter 12 Analytic Geometry 1443 384. A whispering gallery is to be constructed with a length of 150 feet. If the foci are to be located 20 feet away from the wall, how high should the ceiling be? 396. 3x2 − 2xy + 3y2 = 4 397. x2 + 4xy + 4y2 + 6x − 8y = 0 For the following exercises, write the equation of the hyperbola in standard form, and give the center, vertices, foci, and asymptotes. 385. x2 49 − y2 81 = 1 For the following exercises, rewrite in the x′ y′ system without the x′ y′ term, and graph the rotated graph. 398. 11x2 + 10 3xy + y2 = 4 386. 16y2 − 9x2 + 128y + 112 = 0 399. 16x2 + 24xy + 9y2 − 125x = 0 For the following exercises, graph the hyperbola, noting its center, vertices, and foci. State the equations of the asymptotes. For the following exercises, identify the conic with focus at the origin, and then give the directrix and eccentricity. 387. (x − 3)2 25 − ⎛ ⎠ ⎝y + 3⎞ 1 2 = 1 388. y2 − x2 + 4y − 4x − 18 = 0 400. r = 3 2 − sin θ 401. r = 5 4 + 6 cos θ 389. Write the standard form equation of a hyperbola with foci at (1, 0) and (1, 6), and a vertex at (1, 2). For the following exercises, graph the given conic section. If it is a parabola, label vertex, focus, and directrix. If it is an ellipse or a hyperbola, label vertices and foci. For the following exercises, write the equation of the parabola in standard form, and give the vertex, focus, and equation of the directrix. 390. y2 + 10x = 0 391. 3x2 − 12x − y + 11 = 0 For the following exercises, graph the parabola, labeling the vertex, focus, and directrix. 402. r = 12 4 − 8 sin θ 403. r = 2 4 + 4 sin θ 404. Find a polar equation of the conic with focus at the origin, eccentricity of e = 2, and directrix: x = 3. 392. (x − 1)2 = −4⎛ ⎝y + 3⎞ ⎠ 393. y2 + 8x − 8y + 40 = 0 394. Write the equation of a parabola with a focus at (2, 3) and directrix y = −1. 395. A searchlight is shaped like a paraboloid of revolution. If the light source is located 1.5 feet from the base along the axis of symmetry, and the depth of the searchlight is 3 feet, what should the width of the opening be? For the following exercises, determine which conic section is represented by the given equation, and then determine the angle θ that will eliminate the xy term. 1444 Chapter 12 Analytic Geometry This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1445 13 \| SEQUENCES, PROBABILITY, AND COUNTING THEORY Figure 13.1 (credit: Robert S. Donovan, Flickr.) Chapter Outline 13.1 Sequences and Their Notations 13.2 Arithmetic Sequences 13.3 Geometric Sequences 13.4 Series and Their Notations 13.5 Counting Principles 13.6 Binomial Theorem 13.7 Probability Introduction A lottery winner has some big decisions to make regarding what to do with the winnings. Buy a villa in Saint Barthélemy? A luxury convertible? A cruise around the world? The likelihood of winning the lottery is slim, but we all love to fantasize about what we could buy with the winnings. One of the first things a lottery winner has to decide is whether to take the winnings in the form of a lump sum or as a series of regular payments, called an annuity, over the next 30 years or so. 1446 Chapter 13 Sequences, Probability, and Counting Theory This decision is often based on many factors, such as tax implications, interest rates, and investment strategies. There are also personal reasons to consider when making the choice, and one can make many arguments for either decision. However, most lottery winners opt for the lump sum. In this chapter, we will explore the mathematics behind situations such as these. We will take an in-depth look at annuities. We will also look at the branch of mathematics that would allow us to calculate the number of ways to choose lottery numbers and the probability of winning. 13.1 \| Sequences and Their Notations Learning Objectives In this section, you will: 13.1.1 Write the terms of a sequence defined by an explicit formula. 13.1.2 Write the terms of a sequence defined by a recursive formula. 13.1.3 Use factorial notation. A video game company launches an exciting new advertising campaign. They predict the number of online visits to their website, or hits, will double each day. The model they are using shows 2 hits the first day, 4 hits the second day, 8 hits the third day, and so on. See Table 13.1. Day Hits 1 2 2 4 3 8 4 5 … 16 32 … Table 13.1 If their model continues, how many hits will there be at the end of the month? To answer this question, we’ll first need to know how to determine a list of numbers written in a specific order. In this section, we will explore these kinds of ordered lists. Writing the Terms of a Sequence Defined by an Explicit Formula One way to describe an ordered list of numbers is as a sequence. A sequence is a function whose domain is a subset of the counting numbers. The sequence established by the number of hits on the website is {2, 4, 8, 16, 32, … }. The ellipsis (…) indicates that the sequence continues indefinitely. Each number in the sequence is called a term. The first five terms of this sequence are 2, 4, 8, 16, and 32. Listing all of the terms for a sequence can be cumbersome. For example, finding the number of hits on the website at the end of the month would require listing out as many as 31 terms. A more efficient way to determine a specific term is by writing a formula to define the sequence. One type of formula is an explicit formula, which defines the terms of a sequence using their position in the sequence. Explicit formulas are helpful if we want to find a specific term of a sequence without finding all of the previous terms. We can use the formula to find the nth term of the sequence, where n is any positive number. In our example, each number in the sequence is double the previous number, so we can use powers of 2 to write a formula for the nth term. The first term of the sequence is 21 = 2, the second term is 22 = 4, the third term is 23 = 8, and so on. The nth term of the sequence can be found by raising 2 to the nth power. An explicit formula for a sequence is named by a lower case letter a, b, c... with the subscript n. The explicit formula for this sequence is This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1447 an = 2 n . Now that we have a formula for the nth term of the sequence, we can answer the question posed at the beginning of this section. We were asked to find the number of hits at the end of the month, which we will take to be 31 days. To find the number of hits on the last day of the month, we need to find the 31st term of the sequence. We will substitute 31 for n in the formula. a31 = 231 = 2,147,483,648 If the doubling trend continues, the company will get 2,147,483,648 hits on the last day of the month. That is over 2.1 billion hits! The huge number is probably a little unrealistic because it does not take consumer interest and competition into account. It does, however, give the company a starting point from which to consider business decisions. Another way to represent the sequence is by using a table. The first five terms of the sequence and the nth term of the sequence are shown in Table 13.2. n nth term of the sequence, an Table 13.2 1 2 2 4 3 8 4 5 16 32 n n 2 Graphing provides a visual representation of the sequence as a set of distinct points. We can see from the graph in Figure 13.2 that the number of hits is rising at an exponential rate. This particular sequence forms an exponential function. Figure 13.2 Lastly, we can write this particular sequence as ⎧ ⎨2, 4, 8, 16, 32, … , 2 ⎩ n , … ⎫ ⎬. ⎭ A sequence that continues indefinitely is called an infinite sequence. The domain of an infinite sequence is the set of counting numbers. If we consider only the first 10 terms of the sequence, we could write This sequence is called a finite sequence because it does not continue indefinitely. ⎧ ⎨2, 4, 8, 16, 32, … , 2 ⎩ n , … , 1024 ⎬. ⎭ ⎫ 1448 Chapter 13 Sequences, Probability, and Counting Theory Sequence A sequence is a function whose domain is the set of positive integers. A finite sequence is a sequence whose domain consists of
only the first n positive integers. The numbers in a sequence are called terms. The variable a with a number subscript is used to represent the terms in a sequence and to indicate the position of the term in the sequence. a1, a2, a3, … , an, … the first term of the sequence, a2 We call a1 so on. The term an is called the nth term of the sequence, or the general term of the sequence. An explicit formula defines the nth term of a sequence using the position of the term. A sequence that continues indefinitely is an infinite sequence. the second term of the sequence, a3 the third term of the sequence, and Does a sequence always have to begin with a1 ? No. In certain problems, it may be useful to define the initial term as a0 instead of a1. In these problems, the domain of the function includes 0. Given an explicit formula, write the first n terms of a sequence. 1. Substitute each value of n into the formula. Begin with n = 1 to find the first term, a1. 2. To find the second term, a2, use n = 2. 3. Continue in the same manner until you have identified all n terms. Example 13.1 Writing the Terms of a Sequence Defined by an Explicit Formula Write the first five terms of the sequence defined by the explicit formula an = − 3n + 8. Solution Substitute n = 1 into the formula. Repeat with values 2 through 5 for n a1 = − 3(1) + 8 = 5 a2 = − 3(2) + 8 = 2 a3 = − 3(3) + 8 = − 1 a4 = − 3(4) + 8 = − 4 a5 = − 3(5) + 8 = − 7 The first five terms are {5, 2, −1, −4, −7}. Analysis The sequence values can be listed in a table. A table, such as Table 13.2, is a convenient way to input the function into a graphing utility. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1449 n an 1 5 2 2 3 4 5 –1 –4 –7 Table 13.2 A graph can be made from this table of values. From the graph in Figure 13.3, we can see that this sequence represents a linear function, but notice the graph is not continuous because the domain is over the positive integers only. Figure 13.3 13.1 Write the first five terms of the sequence defined by the explicit formula tn = 5n − 4. Investigating Alternating Sequences Sometimes sequences have terms that are alternate. In fact, the terms may actually alternate in sign. The steps to finding terms of the sequence are the same as if the signs did not alternate. However, the resulting terms will not show increase or decrease as n increases. Let’s take a look at the following sequence. Notice the first term is greater than the second term, the second term is less than the third term, and the third term is greater than the fourth term. This trend continues forever. Do not rearrange the terms in numerical order to interpret the sequence. {2, −4, 6, −8} 1450 Chapter 13 Sequences, Probability, and Counting Theory Given an explicit formula with alternating terms, write the first n terms of a sequence. 1. Substitute each value of n into the formula. Begin with n = 1 to find the first term, a1. The sign of the term is given by the (−1) n in the explicit formula. 2. To find the second term, a2, use n = 2. 3. Continue in the same manner until you have identified all n terms. Example 13.2 Writing the Terms of an Alternating Sequence Defined by an Explicit Formula Write the first five terms of the sequence. an = Solution Substitute n = 1, n = 2, and so on in the formula. ( − 1) n n2 n + 1 n = 1 a1 = n = 2 a2 = n = 3 a3 = n = 4 a4 = n = 5 a5 = ( − 1)1 22 1 + 1 ( − 1)2 22 2 + 1 ( − 1)3 32 3 + 1 ( − 1)4 42 4 + 1 ( − 1)5 52 = 16 5 = − 25 6 , 4 3 ,−9 4 , 16 5 ,−25 6 ⎫ ⎬. ⎭ The first five terms are ⎧ ⎨− 1 2 ⎩ Analysis The graph of this function, shown in Figure 13.4, looks different from the ones we have seen previously in this section because the terms of the sequence alternate between positive and negative values. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1451 Figure 13.4 In Example 13.2, does the (–1) to the power of n account for the oscillations of signs? Yes, the power might be n, n + 1, n − 1, and so on, but any odd powers will result in a negative term, and any even power will result in a positive term. 13.2 Write the first five terms of the sequence: an = 4n ( − 2) n Investigating Piecewise Explicit Formulas We’ve learned that sequences are functions whose domain is over the positive integers. This is true for other types of functions, including some piecewise functions. Recall that a piecewise function is a function defined by multiple subsections. A different formula might represent each individual subsection. Given an explicit formula for a piecewise function, write the first n terms of a sequence 1. Identify the formula to which n = 1 applies. 2. To find the first term, a1, use n = 1 in the appropriate formula. 3. Identify the formula to which n = 2 applies. 4. To find the second term, a2, use n = 2 in the appropriate formula. 5. Continue in the same manner until you have identified all n terms. Example 13.3 Writing the Terms of a Sequence Defined by a Piecewise Explicit Formula 1452 Chapter 13 Sequences, Probability, and Counting Theory Write the first six terms of the sequence. an = ⎧ n2 if n is not divisible by 3 ⎨ n if n is divisible by 3 ⎩ 3 Solution Substitute n = 1, n = 2, and so on in the appropriate formula. Use n2 when n is not a multiple of 3. Use n 3 when n is a multiple of 3. a1 = 12 = 1 a2 = 22 = 4 a3 = 3 = 1 3 a4 = 42 = 16 a5 = 52 = 25 a6 = 6 3 = 2 1 is not a multiple of 3. Use n2. 2 is not a multiple of 3. Use n2. 3 is a multiple of 3. Use n 3 . 4 is not a multiple of 3. Use n2. 5 is not a multiple of 3. Use n2. 6 is a multiple of 3. Use n 3 . The first six terms are {1, 4, 1, 16, 25, 2}. Analysis Every third point on the graph shown in Figure 13.5 stands out from the two nearby points. This occurs because the sequence was defined by a piecewise function. Figure 13.5 13.3 Write the first six terms of the sequence. an = ⎧ 2n3 if n is odd ⎨ 5n if n is even ⎩ 2 Finding an Explicit Formula Thus far, we have been given the explicit formula and asked to find a number of terms of the sequence. Sometimes, the explicit formula for the nth term of a sequence is not given. Instead, we are given several terms from the sequence. When this happens, we can work in reverse to find an explicit formula from the first few terms of a sequence. The key to finding an explicit formula is to look for a pattern in the terms. Keep in mind that the pattern may involve alternating terms, formulas for numerators, formulas for denominators, exponents, or bases. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1453 Given the first few terms of a sequence, find an explicit formula for the sequence. 1. Look for a pattern among the terms. 2. If the terms are fractions, look for a separate pattern among the numerators and denominators. 3. Look for a pattern among the signs of the terms. 4. Write a formula for an in terms of n. Test your formula for n = 1, n = 2, and n = 3. Example 13.4 Writing an Explicit Formula for the nth Term of a Sequence Write an explicit formula for the nth term of each sequence. a. b. c. ⎧ ⎨− 2 11 ⎩ , 3 13 , − 4 15 , 5 17 , − 6 19 , … ⎫ ⎬ ⎭ ⎧ ⎨ − 2 25 ⎩ , − 2 125 , − 2 625 , − 2 3,125 , − 2 15,625 , … ⎫ ⎬ ⎭ ⎧ ⎩ ⎨e4 ,e5 ,e6 ,e7 ,e8 , … ⎫ ⎬ ⎭ Solution Look for the pattern in each sequence. a. The terms alternate between positive and negative. We can use ( − 1) n to make the terms alternate. The numerator can be represented by n + 1. The denominator can be represented by 2n + 9. an = n ( − 1) (n + 1) 2n + 9 b. The terms are all negative. So we know that the fraction is negative, the numerator is 2, and the denominator can be represented by n + 1. 5 an = − 2 n + 1 5 c. The terms are powers of e. For n = 1, the first term is e4 so the exponent must be n + 3. an = en + 3 13.4 Write an explicit formula for the nth term of the sequence. {9, − 81, 729, − 6,561, 59,049, …} 13.5 Write an explicit formula for the nth term of the sequence. ⎧ ⎨− 3 4 ⎩ , − 9 8 , − 27 12 , − 81 16 , − 243 20 ⎫ , ... ⎬ ⎭ 1454 Chapter 13 Sequences, Probability, and Counting Theory 13.6 Write an explicit formula for the nth term of the sequence. ⎧ ⎨ 1 e2, ⎩ 1 e, 1, e, e2, ... ⎫ ⎬ ⎭ Writing the Terms of a Sequence Defined by a Recursive Formula Sequences occur naturally in the growth patterns of nautilus shells, pinecones, tree branches, and many other natural structures. We may see the sequence in the leaf or branch arrangement, the number of petals of a flower, or the pattern of the chambers in a nautilus shell. Their growth follows the Fibonacci sequence, a famous sequence in which each term can be found by adding the preceding two terms. The numbers in the sequence are 1, 1, 2, 3, 5, 8, 13, 21, 34,…. Other examples from the natural world that exhibit the Fibonacci sequence are the Calla Lily, which has just one petal, the Black-Eyed Susan with 13 petals, and different varieties of daisies that may have 21 or 34 petals. Each term of the Fibonacci sequence depends on the terms that come before it. The Fibonacci sequence cannot easily be written using an explicit formula. Instead, we describe the sequence using a recursive formula, a formula that defines the terms of a sequence using previous terms. A recursive formula always has two parts: the value of an initial term (or terms), and an equation defining an in terms of preceding terms. For example, suppose we know the following: We can find the subsequent terms of the sequence using the first term. a1 = 3 an = 2an − 1 − 1, for n ≥ 2 a1 = 3 a2 = 2a1 − 1 = 2(3) − 1 = 5 a3 = 2a2 − 1 = 2(5) − 1 = 9 a4 = 2a3 − 1 = 2(9) − 1 = 17 So the first four terms of the sequence are {3, 5, 9, 17} . The recursive formula for the Fibonacci sequence states the first two terms and defines each successive term as the sum of the preceding two terms. a1 = 1 a2 = 1 an = an − 1 + an − 2, for
n ≥ 3 To find the tenth term of the sequence, for example, we would need to add the eighth and ninth terms. We were told previously that the eighth and ninth terms are 21 and 34, so a10 = a9 + a8 = 34 + 21 = 55 Recursive Formula A recursive formula is a formula that defines each term of a sequence using preceding term(s). Recursive formulas must always state the initial term, or terms, of the sequence. Must the first two terms always be given in a recursive formula? No. The Fibonacci sequence defines each term using the two preceding terms, but many recursive formulas define each term using only one preceding term. These sequences need only the first term to be defined. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1455 Given a recursive formula with only the first term provided, write the first n terms of a sequence. 1. Identify the initial term, a1, which is given as part of the formula. This is the first term. 2. To find the second term, a2, substitute the initial term into the formula for an − 1. Solve. 3. To find the third term, a3, substitute the second term into the formula. Solve. 4. Repeat until you have solved for the nth term. Example 13.5 Writing the Terms of a Sequence Defined by a Recursive Formula Write the first five terms of the sequence defined by the recursive formula. a1 = 9 an = 3an − 1 − 20, for n ≥ 2 Solution The first term is given in the formula. For each subsequent term, we replace an − 1 preceding term. with the value of the a1 = 9 a2 = 3a1 − 20 = 3(9) − 20 = 27 − 20 = 7 a3 = 3a2 − 20 = 3(7) − 20 = 21 − 20 = 1 a4 = 3a3 − 20 = 3(1) − 20 = 3 − 20 = − 17 a5 = 3a4 − 20 = 3( − 17) − 20 = − 51 − 20 = − 71 The first five terms are {9, 7, 1, – 17, – 71}. See Figure 13.6. Figure 13.6 13.7 Write the first five terms of the sequence defined by the recursive formula. a1 = 2 an = 2an − 1 + 1, for n ≥ 2 1456 Chapter 13 Sequences, Probability, and Counting Theory Given a recursive formula with two initial terms, write the first n terms of a sequence. 1. 2. Identify the initial term, a1, which is given as part of the formula. Identify the second term, a2, which is given as part of the formula. 3. To find the third term, substitute the initial term and the second term into the formula. Evaluate. 4. Repeat until you have evaluated the nth term. Example 13.6 Writing the Terms of a Sequence Defined by a Recursive Formula Write the first six terms of the sequence defined by the recursive formula. a1 = 1 a2 = 2 an = 3an − 1 + 4an − 2 , for n ≥ 3 Solution The first two terms are given. For each subsequent term, we replace an − 1 preceding terms. and an − 2 with the values of the two a3 = 3a2 + 4a1 = 3(2) + 4(1) = 10 a4 = 3a3 + 4a2 = 3(10) + 4(2) = 38 a5 = 3a4 + 4a3 = 3(38) + 4(10) = 154 a6 = 3a5 + 4a4 = 3(154) + 4(38) = 614 The first six terms are {1,2,10,38,154,614}. See Figure 13.7. Figure 13.7 13.8 Write the first 8 terms of the sequence defined by the recursive formula. a1 = 0 a2 = 1 a3 = 1 an = an − 1 an − 2 + an − 3 , for n ≥ 4 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory Using Factorial Notation The formulas for some sequences include products of consecutive positive integers. n factorial, written as n !, product of the positive integers from 1 to n. For example = 24 = 120 1457 is the An example of formula containing a factorial is an = (n + 1) !. The sixth term of the sequence can be found by substituting 6 for n. The factorial of any whole number n is n(n − 1) ! We can therefore also think of 5 ! as 5 ⋅ 4 !. a6 = (6 + 1 = 5040 Factorial n factorial is a mathematical operation that can be defined using a recursive formula. The factorial of n, denoted n !, is defined for a positive integer n as(n − 1)(n − 2) ⋯ (2)(1), for n ≥ 2 (13.1) The special case 0 ! is defined as 0 ! = 1. Can factorials always be found using a calculator? No. Factorials get large very quickly—faster than even exponential functions! When the output gets too large for the calculator, it will not be able to calculate the factorial. Example 13.7 Writing the Terms of a Sequence Using Factorials Write the first five terms of the sequence defined by the explicit formula an = 5n (n + 2) ! . Solution Substitute n = 1, n = 2, and so on in the formula a1 = a2 = a3 = a4 = a5 = 5(1) (1 + 2) ! 5(2) (2 + 2) ! 5(3) (3 + 2) ! 5(4) (4 + 2) ! 5(5) (5 + 2) ! = 5 3 ! = 10 4 ! = 15 5 ! = 20 6 ! = 25 10 4 · 3 · 2 · 1 15 5 · 4 · 3 · 2 · 1 20 12 = 1 8 = 1 36 = 5 25 ,008 1458 Chapter 13 Sequences, Probability, and Counting Theory The first five terms are ⎧ ⎨5 6 ⎩ , 5 12 , 1 8 , 1 36 , 5 1,008 ⎫ ⎬. ⎭ Analysis Figure 13.8 shows the graph of the sequence. Notice that, since factorials grow very quickly, the presence of the factorial term in the denominator results in the denominator becoming much larger than the numerator as n increases. This means the quotient gets smaller and, as the plot of the terms shows, the terms are decreasing and nearing zero. Figure 13.8 13.9 Write the first five terms of the sequence defined by the explicit formula an = (n + 1) ! 2n . Access this online resource for additional instruction and practice with sequences. • Finding Terms in a Sequence (http://openstaxcollege.org/l/findingterms) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1459 13.1 EXERCISES Verbal 1. Discuss the meaning of a sequence. If a finite sequence is defined by a formula, what is its domain? What about an infinite sequence? 2. Describe three ways that a sequence can be defined. 3. Is the ordered set of even numbers an infinite sequence? What about the ordered set of odd numbers? Explain why or why not. 4. What happens to the terms an of a sequence when there is a negative factor in the formula that is raised to a power that includes n ? What is the term used to describe this phenomenon? What is a factorial, and how is it denoted? Use an 5. example to illustrate how factorial notation can be beneficial. Algebraic For the following exercises, write the first four terms of the sequence. 6. 7. 8. 9. 10. 11. 12. 13. an = 2 n − 2 an = − 16 n + 1 an = − (−5) n − 1 n an = 2 n3 an = 2n + 1 n3 an = 1.25 ⋅ (−4) n − 1 an = − 4 ⋅ (−6) n − 1 an = n2 2n + 1 14. an = (−10) n + 1 15. an = − ⎛ ⎜ 4 ⋅ ( − 5) ⎝ 5 n − 1 ⎞ ⎟ ⎠ For the following exercises, write the first eight terms of the piecewise sequence. 16. an = ⎧ ⎨ ⎩ n ( − 2) n − 1 (3) − 2 if n is even if n is odd 17. 18. 19. 20. an = ⎧ ⎨ ⎩ if n ≤ 5 n2 2n + 1 n2 − 5 if n >5 an = ⎧ (2n + 1)2 if n is divisible by 4 ⎨ 2 ⎩ n if n is not divisible by 4 an = an = ⎧ ⎨ ⎩ −0.6 ⋅ 5 n − 1 2.5 ⋅ ( − 2) if n is prime or 1 n − 1 if n is composite ⎧ ⎨ ⎩ 4(n2 − 2) if n ≤ 3 or n > 6 n2 − 2 4 if 3 < n ≤ 6 For the following exercises, write an explicit formula for each sequence. 21. 22. 23. 24. 25. 4, 7, 12, 19, 28, … −4, 2, − 10, 14, − 34, … 1, 1, 4 3 , 2, 16 5 , … 1 + e2, 1 − e2 0, 1 − e1 1 + e3, 1 − e3 1 + e4, 1 − e4 1 + e5, … 1 16 , … For the following exercises, write the first five terms of the sequence. 26. a1 = 9, an = an − 1 + n 27. a1 = 3, an = (−3)an − 1 28. 29. 30. a1 = − 4, an = a1 = − 1, an = an − 1 + 2n an − 1 − 1 n − 1 (−3) an − 1 − 2 a1 = − 30, an = ⎛ ⎝2 + an − 1 n ⎛ ⎞ ⎠ ⎝ 1 2 ⎞ ⎠ For the following exercises, write the first eight terms of the sequence. 1460 31. 32. 33. Chapter 13 Sequences, Probability, and Counting Theory 4 + n 2n if n is even ⎧ ⎨ ⎩ 3 + n if n is odd a1 = 1 24 , a2 = 1, an = ⎛ ⎝2an − 2 ⎛ ⎞ ⎠ ⎝3an − 1 ⎞ ⎠ an = a1 = − 1, a2 = 5, an = an − 2 ⎛ ⎝3 − an − 1 ⎞ ⎠ 49. a1 = 2, a2 = 10, an = 2⎛ ⎝an − 1 + 2⎞ an − 2 ⎠ a1 = 2, an = ⎛ ⎝−an − 1 + 1⎞ ⎠ 2 50. an = 1, an = an − 1 + 8 For the following exercises, write a recursive formula for each sequence. 51. an = (n + 1) ! (n − 1) ! 34. 35. 36. 37. 38. −2.5, − 5, − 10, − 20, − 40, … −8, − 6, − 3, 1, 6, … 2, 4, 12, 48, 240, … 35, 38, 41, 44, 47, … 15, 3, 3 5 , 3 25 , 3 125 , ⋯ For the following exercises, evaluate the factorial. 39. 6 ! 40. 41. 42. ⎛ ⎝ 12 6 ⎞ ⎠! 12 ! 6 ! 100 ! 99 ! For the following exercises, write the first four terms of the sequence. 43. an = n ! n2 44. an = 3 ⋅ n ! 4 ⋅ n ! 45. an = n ! n2 − n − 1 46. an = 100 ⋅ n n(n − 1) ! Graphical For the following exercises, graph the first five terms of the indicated sequence an = (−1) n n + n 47. 48. This content is available for free at https://cnx.org/content/col11758/1.5 For the following exercises, write an explicit formula for the sequence using the first five points shown on the graph. 52. 53. 54. For the following exercises, write a recursive formula for the sequence using the first five points shown on the graph. Chapter 13 Sequences, Probability, and Counting Theory 1461 55. 56. Technology Follow these steps recursively using a graphing calculator: to evaluate a sequence defined • On the home screen, key in the value for the initial term a1 and press [ENTER]. • Enter the recursive formula by keying in all numerical values given in the formula, along with the key strokes [2ND] ANS for the previous term an − 1. Press [ENTER]. • Continue pressing [ENTER] to calculate the values for each successive term. For the following exercises, use the steps above to find the indicated term or terms for the sequence. 57. Find a1 = 87 111 give fractional results. the first , an = 4 3 five an − 1 + 12 37 of terms the sequence . Use the >Frac feature to 59. Find a1 = 2, the first five terms of the sequence an = 2 [(an − 1) − 1] + 1. 60. Find the a1 = 8, an = ⎛ ten first ⎝an − 1 + 1⎞ ⎠! an − 1 ! . terms of the sequence Find 61. a1 = 2, an = nan − 1 the tenth term of the sequence Follow these steps to evaluate a finite sequence defined by an explicit formula. Using a TI-84, do the following. • In the home screen, press [2ND] LIST. • Scroll over to OPS and choose “seq(” from the dropdown list. Press [ENTER]. • • • • In the line headed “Expr:” type in the explicit formula, us
ing the [X,T, θ, n] button for n In the line headed “Variable:” type in the variable used on the previous step. In the line headed “start:” key in the value of n that begins the sequence. In the line headed “end:” key in the value of n that ends the sequence. • Press [ENTER] 3 times to return to the home screen. You will see the sequence syntax on the screen. Press [ENTER] to see the list of terms for the finite sequence defined. Use the right arrow key to scroll through the list of terms. Using a TI-83, do the following. • In the home screen, press [2ND] LIST. • Scroll over to OPS and choose “seq(” from the dropdown list. Press [ENTER]. • Enter the items in the order “Expr”, “Variable”, “start”, “end” separated by commas. See the instructions above for the description of each item. • Press [ENTER] to see the list of terms for the finite sequence defined. Use the right arrow key to scroll through the list of terms. For the following exercises, use the steps above to find the indicated terms for the sequence. Round to the nearest thousandth when necessary. first List 62. an = − 28 9 the n + 5 3 . five terms of the sequence 63. List six the first terms n3 − 3.5n2 + 4.1n − 1.5 2.4n . of the sequence 58. Find a1 = 625, 15th the an = 0.8an − 1 + 18. term of the sequence an = 64. 1462 List an = five terms of the sequence Chapter 13 Sequences, Probability, and Counting Theory the first 15n ⋅ (−2) n − 1 47 65. List an = 5.7 n the four first + 0.275(n − 1) ! terms of the sequence 66. List the first six terms of the sequence an = n ! n . Extensions Consider the sequence defined by an = − 6 − 8n. Is 67. an = − 421 a term in the sequence? Verify the result. 68. What term in the sequence an = n2 + 4n + 4 2(n + 2) has the value 41 ? Verify the result. a Find recursive sequence 69. 1, 0, − 1, − 1, 0, 1, 1, 0, − 1, − 1, 0, 1, 1, ... . (Hint: find a pattern for an based on the first two terms.) formula the for 70. Calculate the first eight terms of the sequences and bn = n3 + 3n2 + 2n, and then make the relationship between these two an = (n + 2) ! (n − 1) ! a conjecture about sequences. 71. Prove the conjecture made in the preceding exercise. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1463 13.2 \| Arithmetic Sequences Learning Objectives In this section, you will: 13.2.1 Find the common difference for an arithmetic sequence. 13.2.2 Write terms of an arithmetic sequence. 13.2.3 Use a recursive formula for an arithmetic sequence. 13.2.4 Use an explicit formula for an arithmetic sequence. Companies often make large purchases, such as computers and vehicles, for business use. The book-value of these supplies decreases each year for tax purposes. This decrease in value is called depreciation. One method of calculating depreciation is straight-line depreciation, in which the value of the asset decreases by the same amount each year. As an example, consider a woman who starts a small contracting business. She purchases a new truck for $25,000. After five years, she estimates that she will be able to sell the truck for $8,000. The loss in value of the truck will therefore be $17,000, which is $3,400 per year for five years. The truck will be worth $21,600 after the first year; $18,200 after two years; $14,800 after three years; $11,400 after four years; and $8,000 at the end of five years. In this section, we will consider specific kinds of sequences that will allow us to calculate depreciation, such as the truck’s value. Finding Common Differences The values of the truck in the example are said to form an arithmetic sequence because they change by a constant amount each year. Each term increases or decreases by the same constant value called the common difference of the sequence. For this sequence, the common difference is –3,400. The sequence below is another example of an arithmetic sequence. In this case, the constant difference is 3. You can choose any term of the sequence, and add 3 to find the subsequent term. Arithmetic Sequence An arithmetic sequence is a sequence that has the property that the difference between any two consecutive terms is a constant. This constant is called the common difference. If a1 is the first term of an arithmetic sequence and d is the common difference, the sequence will be: ⎧ ⎨an ⎩ ⎫ ⎬ = ⎭ ⎧ ⎨a1, a1 + d, a1 + 2d, a1 + 3d, ... ⎩ ⎫ ⎬ ⎭ Example 13.8 Finding Common Differences Is each sequence arithmetic? If so, find the common difference. a. b. {1, 2, 4, 8, 16, ...} { − 3, 1, 5, 9, 13, ...} 1464 Chapter 13 Sequences, Probability, and Counting Theory Solution Subtract each term from the subsequent term to determine whether a common difference exists. a. The sequence is not arithmetic because there is no common difference. b. The sequence is arithmetic because there is a common difference. The common difference is 4. Analysis The graph of each of these sequences is shown in Figure 13.9. We can see from the graphs that, although both sequences show growth, a is not linear whereas b is linear. Arithmetic sequences have a constant rate of change so their graphs will always be points on a line. Figure 13.9 If we are told that a sequence is arithmetic, do we have to subtract every term from the following term to find the common difference? No. If we know that the sequence is arithmetic, we can choose any one term in the sequence, and subtract it from the subsequent term to find the common difference. 13.10 Is the given sequence arithmetic? If so, find the common difference. {18, 16, 14, 12, 10, … } 13.11 Is the given sequence arithmetic? If so, find the common difference. {1, 3, 6, 10, 15, … } Writing Terms of Arithmetic Sequences Now that we can recognize an arithmetic sequence, we will find the terms if we are given the first term and the common difference. The terms can be found by beginning with the first term and adding the common difference repeatedly. In addition, any term can also be found by plugging in the values of n and d into formula below. an = a1 + (n − 1)d This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1465 Given the first term and the common difference of an arithmetic sequence, find the first several terms. 1. Add the common difference to the first term to find the second term. 2. Add the common difference to the second term to find the third term. 3. Continue until all of the desired terms are identified. 4. Write the terms separated by commas within brackets. Example 13.9 Writing Terms of Arithmetic Sequences Write the first five terms of the arithmetic sequence with a1 = 17 and d = − 3 . Solution Adding − 3 is the same as subtracting 3. Beginning with the first term, subtract 3 from each term to find the next term. The first five terms are {17, 14, 11, 8, 5} Analysis As expected, the graph of the sequence consists of points on a line as shown in Figure 13.10. Figure 13.10 13.12 List the first five terms of the arithmetic sequence with a1 = 1 and d = 5 . Given any the first term and any other term in an arithmetic sequence, find a given term. 1. Substitute the values given for a1, an, n into the formula an = a1 + (n − 1)d to solve for d. 2. Find a given term by substituting the appropriate values for a1, n, and d into the formula an = a1 + (n − 1)d. Example 13.10 Writing Terms of Arithmetic Sequences 1466 Chapter 13 Sequences, Probability, and Counting Theory Given a1 = 8 and a4 = 14 , find a5 . Solution The sequence can be written in terms of the initial term 8 and the common difference d . {8, 8 + d, 8 + 2d, 8 + 3d} We know the fourth term equals 14; we know the fourth term has the form a1 + 3d = 8 + 3d . We can find the common difference d . an = a1 + (n − 1)d a4 = a1 + 3d a4 = 8 + 3d 14 = 8 + 3d d = 2 Write the fourth term of the sequence in terms of a1 and d. Substitute 14 for a4. Solve for the common diffe ence . Find the fifth term by adding the common difference to the fourth term. a5 = a4 + 2 = 16 Analysis Notice that the common difference is added to the first term once to find the second term, twice to find the third term, three times to find the fourth term, and so on. The tenth term could be found by adding the common difference to the first term nine times or by using the equation an = a1 + (n − 1)d. 13.13 Given a3 = 7 and a5 = 17 , find a2 . Using Recursive Formulas for Arithmetic Sequences Some arithmetic sequences are defined in terms of the previous term using a recursive formula. The formula provides an algebraic rule for determining the terms of the sequence. A recursive formula allows us to find any term of an arithmetic sequence using a function of the preceding term. Each term is the sum of the previous term and the common difference. For example, if the common difference is 5, then each term is the previous term plus 5. As with any recursive formula, the first term must be given. an = an − 1 + d n ≥ 2 Recursive Formula for an Arithmetic Sequence The recursive formula for an arithmetic sequence with common difference d is: an = an − 1 + d n ≥ 2 (13.2) Given an arithmetic sequence, write its recursive formula. 1. Subtract any term from the subsequent term to find the common difference. 2. State the initial term and substitute the common difference into the recursive formula for arithmetic sequences. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1467 Example 13.11 Writing a Recursive Formula for an Arithmetic Sequence Write a recursive formula for the arithmetic sequence. { − 18, − 7, 4, 15, 26, …} Solution The first term is given as −18 . The common difference can be found by subtracting the first term from the second term. Substitute the initial term and the common difference into the recursive formula for arithmetic sequences. d = −7 − (−18) = 11 a1 = − 18 an = an − 1 + 11, for n ≥ 2 Analysis We see that the common
difference is the slope of the line formed when we graph the terms of the sequence, as shown in Figure 13.11. The growth pattern of the sequence shows the constant difference of 11 units. Figure 13.11 Do we have to subtract the first term from the second term to find the common difference? No. We can subtract any term in the sequence from the subsequent term. It is, however, most common to subtract the first term from the second term because it is often the easiest method of finding the common difference. 13.14 Write a recursive formula for the arithmetic sequence. {25, 37, 49, 61, …} Using Explicit Formulas for Arithmetic Sequences We can think of an arithmetic sequence as a function on the domain of the natural numbers; it is a linear function because it has a constant rate of change. The common difference is the constant rate of change, or the slope of the function. We can construct the linear function if we know the slope and the vertical intercept. To find the y-intercept of the function, we can subtract the common difference from the first term of the sequence. Consider the following sequence. an = a1 + d(n − 1) 1468 Chapter 13 Sequences, Probability, and Counting Theory The common difference is −50 , so the sequence represents a linear function with a slope of −50 . To find the y intercept, we subtract −50 from 200 : 200 − ( − 50) = 200 + 50 = 250 . You can also find the y -intercept by graphing the function and determining where a line that connects the points would intersect the vertical axis. The graph is shown in Figure 13.12. Figure 13.12 Recall the slope-intercept form of a line is y = mx + b. When dealing with sequences, we use an in place of y and n in place of x. If we know the slope and vertical intercept of the function, we can substitute them for m and b in the slopeintercept form of a line. Substituting − 50 for the slope and 250 for the vertical intercept, we get the following equation: an = − 50n + 250 We do not need to find the vertical intercept to write an explicit formula for an arithmetic sequence. Another explicit formula for this sequence is an = 200 − 50(n − 1) , which simplifies to an = − 50n + 250. Explicit Formula for an Arithmetic Sequence An explicit formula for the nth term of an arithmetic sequence is given by an = a1 + d(n − 1) (13.3) Given the first several terms for an arithmetic sequence, write an explicit formula. 1. Find the common difference, a2 − a1. 2. Substitute the common difference and the first term into an = a1 + d(n − 1). Example 13.12 Writing the nth Term Explicit Formula for an Arithmetic Sequence Write an explicit formula for the arithmetic sequence. {2, 12, 22, 32, 42, …} Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1469 The common difference can be found by subtracting the first term from the second term. d = a2 − a1 = 12 − 2 = 10 The common difference is 10. Substitute the common difference and the first term of the sequence into the formula and simplify. an = 2 + 10(n − 1) an = 10n − 8 Analysis The graph of this sequence, represented in Figure 13.13, shows a slope of 10 and a vertical intercept of −8 . Figure 13.13 13.15 Write an explicit formula for the following arithmetic sequence. {50, 47, 44, 41, … } Finding the Number of Terms in a Finite Arithmetic Sequence Explicit formulas can be used to determine the number of terms in a finite arithmetic sequence. We need to find the common difference, and then determine how many times the common difference must be added to the first term to obtain the final term of the sequence. Given the first three terms and the last term of a finite arithmetic sequence, find the total number of terms. 1. Find the common difference d. 2. Substitute the common difference and the first term into an = a1 + d(n – 1). 3. Substitute the last term for an and solve for n. Example 13.13 Finding the Number of Terms in a Finite Arithmetic Sequence Find the number of terms in the finite arithmetic sequence. {8, 1, –6, ..., –41} 1470 Chapter 13 Sequences, Probability, and Counting Theory Solution The common difference can be found by subtracting the first term from the second term. 1 − 8 = − 7 The common difference is −7 . Substitute the common difference and the initial term of the sequence into the nth term formula and simplify. an = a1 + d(n − 1) an = 8 + − 7(n − 1) an = 15 − 7n −41 = 15 − 7n 8 = n Substitute −41 for an and solve for n There are eight terms in the sequence. 13.16 Find the number of terms in the finite arithmetic sequence. {6, 11, 16, ..., 56} Solving Application Problems with Arithmetic Sequences In many application problems, it often makes sense to use an initial term of a0 the explicit formula slightly to account for the difference in initial terms. We use the following formula: instead of a1. In these problems, we alter an = a0 + dn Example 13.14 Solving Application Problems with Arithmetic Sequences A five-year old child receives an allowance of $1 each week. His parents promise him an annual increase of $2 per week. a. Write a formula for the child’s weekly allowance in a given year. b. What will the child’s allowance be when he is 16 years old? Solution a. The situation can be modeled by an arithmetic sequence with an initial term of 1 and a common difference of 2. Let A be the amount of the allowance and n be the number of years after age 5. Using the altered explicit formula for an arithmetic sequence we get: An = 1 + 2n b. We can find the number of years since age 5 by subtracting. 16 − 5 = 11 We are looking for the child’s allowance after 11 years. Substitute 11 into the formula to find the child’s allowance at age 16. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1471 The child’s allowance at age 16 will be $23 per week. A11 = 1 + 2(11) = 23 13.17 A woman decides to go for a 10-minute run every day this week and plans to increase the time of her daily run by 4 minutes each week. Write a formula for the time of her run after n weeks. How long will her daily run be 8 weeks from today? Access this online resource for additional instruction and practice with arithmetic sequences. • Arithmetic Sequences (http://openstaxcollege.org/l/arithmeticseq) 1472 Chapter 13 Sequences, Probability, and Counting Theory 13.2 EXERCISES Verbal 72. What is an arithmetic sequence? How is the common difference of an arithmetic 73. sequence found? How do we determine whether a sequence is 74. arithmetic? 75. What are the main differences between using a recursive formula and using an explicit formula to describe an arithmetic sequence? Describe 76. and how linear sequences are similar. How are they different? functions arithmetic Algebraic For the following exercises, find the common difference for the arithmetic sequence provided. 77. 78. {5, 11, 17, 23, 29, ...} ⎧ ⎨0, 1 2 ⎩ , 1, 3 2 ⎫ , 2, ... ⎬ ⎭ the following exercises, determine whether For sequence is arithmetic. If so find the common difference. the 79. 80. {11.4, 9.3, 7.2, 5.1, 3, ...} {4, 16, 64, 256, 1024, ...} For the following exercises, write the first five terms of the arithmetic sequence given the first term and common difference. 81. a1 = −25 , d = −9 82. a1 = 0 , d = 2 3 For the following exercises, write the first five terms of the arithmetic series given two terms. 83. a1 = 17, a7 = − 31 84. a13 = − 60, a33 = − 160 For the following exercises, find the specified term for the arithmetic sequence given the first term and common difference. First term is 3, common difference is 4, find the 5th 85. term. 86. This content is available for free at https://cnx.org/content/col11758/1.5 First term is 4, common difference is 5, find the 4th term. First term is 5, common difference is 6, find the 8th 87. term. First term is 6, common difference is 7, find the 6th 88. term. First term is 7, common difference is 8, find the 7th 89. term. For the following exercises, find the first term given two terms from an arithmetic sequence. Find the first term or a1 90. a6 = 12 and a14 = 28. Find the first term or a1 91. a7 = 21 and a15 = 42. Find the first term or a1 92. a8 = 40 and a23 = 115. Find the first term or a1 93. a9 = 54 and a17 = 102. Find the first term or a1 94. a11 = 11 and a21 = 16. of an arithmetic sequence if of an arithmetic sequence if of an arithmetic sequence if of an arithmetic sequence if of an arithmetic sequence if For the following exercises, find the specified term given two terms from an arithmetic sequence. 95. a1 = 33 and a7 = − 15. Find a4. 96. a3 = − 17.1 and a10 = − 15.7. Find a21. For the following exercises, use the recursive formula to write the first five terms of the arithmetic sequence. 97. 98. a1 = 39; an = an − 1 − 3 a1 = − 19; an = an − 1 − 1.4 For the following exercises, write a recursive formula for each arithmetic sequence. 99. an = {40, 60, 80, ...} 100. 101. an = ⎧ ⎨17, 26, 35, ...⎫ ⎬ ⎭ ⎩ an = ⎧ ⎨ − 1, 2, 5, ...⎫ ⎬ ⎭ ⎩ Chapter 13 Sequences, Probability, and Counting Theory 1473 102. an = {12, 17, 22, ...} 123. an = ⎧ ⎨−5, − 10 3 ⎩ , − 5 3 , … ⎫ ⎬ ⎭ 103. an = ⎧ ⎨ − 15, − 7, 1, ...⎫ ⎬ ⎭ ⎩ 104. an = {8.9, 10.3, 11.7, ...} For the following exercises, find the number of terms in the given finite arithmetic sequence. 105. 106. an = ⎧ ⎨ − 0.52, − 1.02, − 1.52, ...⎫ ⎬ ⎭ ⎩ an = ⎧ ⎨1 5 ⎩ , 9 20 , 7 10 ⎫ , ... ⎬ ⎭ 107. an = ⎧ ⎨− , ... ⎬ ⎭ 108. an = ⎧ ⎨1 6 ⎩ , − 11 12 ⎫ , − 2, ... ⎬ ⎭ 124. 125. 126. an = ⎧ ⎨3, − 4, − 11, ..., − 60⎫ ⎬ ⎭ ⎩ an = ⎧ ⎨1.2, 1.4, 1.6, ..., 3.8⎫ ⎬ ⎭ ⎩ an = ⎧ ⎨1 2 ⎩ , 2, 7 2 , ..., 8 ⎫ ⎬ ⎭ Graphical For the following exercises, determine whether the graph shown represents an arithmetic sequence. For the following exercises, write a recursive formula for the given arithmetic sequence, and then find the specified term. 127. 109. an = {7, 4, 1, ...}; Find the 17th term. 110. an = {4, 11, 18, ...}; Find the 14th term. 111. an = ⎧ ⎨2, 6, 10, ...⎫ ⎭ ⎩
⎬; Find the 12th term. For the following exercises, use the explicit formula to write the first five terms of the arithmetic sequence. 112. an = 24 − 4n 113. an = 1 2 n − 1 2 For the following exercises, write an explicit formula for each arithmetic sequence. 114. 115. 116. an = ⎧ ⎨3, 5, 7, ...⎫ ⎬ ⎭ ⎩ an = ⎧ ⎨32, 24, 16, ...⎫ ⎬ ⎭ ⎩ an = ⎧ ⎨ − 5, 95, 195, ...⎫ ⎬ ⎭ ⎩ 117. an = {−17, −217, −417,...} 118. 119. 120. 121. an = ⎧ ⎨1.8, 3.6, 5.4, ...⎫ ⎬ ⎭ ⎩ an = ⎧ ⎨−18.1, −16.2, −14.3, ...⎫ ⎬ ⎭ ⎩ an = ⎧ ⎨15.8, 18.5, 21.2, ...⎫ ⎬ ⎭ ⎩ an = ⎧ ⎨1 3 ⎩ , − 4 3 , −3, ... ⎫ ⎬ ⎭ 122. an = ⎧ ⎨0, 1 3 ⎩ ⎫ , ... ⎬ ⎭ , 2 3 128. 1474 Chapter 13 Sequences, Probability, and Counting Theory ◦ Set TblStart = 1 ◦ Set ΔTbl = 1 ◦ Set Indpnt: Auto and Depend: Auto • Press [2ND] then [GRAPH] to go to the TABLE What are the first seven terms shown in the column 132. with the heading u(n)? Use the scroll-down arrow to scroll to n = 50. What 133. value is given for u(n)? Press [WINDOW]. 134. nMin = 1, nMax = 5, xMin = 0, xMax = 6, yMin = − 1, and yMax = 14. Then press sequence as it appears on the graphing calculator. [GRAPH]. Graph Set the For the following exercises, follow the steps given above to work with the arithmetic sequence an = 1 n + 5 using a 2 graphing calculator. What are the first seven terms shown in the column 135. with the heading u(n) in the TABLE feature? 136. Graph the sequence as it appears on the graphing calculator. Be sure to adjust the WINDOW settings as needed. Extensions Give two examples of arithmetic sequences whose 4th 137. terms are 9. Give two examples of arithmetic sequences whose 138. 10th terms are 206. For the following exercises, use the information provided to graph the first 5 terms of the arithmetic sequence. 129. a1 = 0, d = 4 130. a1 = 9; an = an − 1 − 10 131. an = − 12 + 5n Technology For the following exercises, follow the steps to work with the arithmetic sequence an = 3n − 2 using a graphing calculator: Find the 5th term of 139. {9b, 5b, b, … }. the arithmetic sequence • Press [MODE] ◦ Select SEQ in the fourth line ◦ Select DOT in the fifth line ◦ Press [ENTER] • Press [Y=] ◦ ◦ ◦ nMin is the first counting number for the sequence. Set nMin = 1 u(n) is the pattern for the sequence. Set u(n) = 3n − 2 u(nMin) is the sequence. Set u(nMin) = 1 first number in the • Press [2ND] then [WINDOW] to go to TBLSET This content is available for free at https://cnx.org/content/col11758/1.5 Find the 11th term of 140. {3a − 2b, a + 2b, − a + 6b … }. the arithmetic sequence At 141. {5.4, 14.5, 23.6, ...} exceed 151? which term does the sequence 142. At which term does the sequence ⎧ ⎨17 3 ⎩ , 31 6 , 14 3 ⎫ , ... ⎬ ⎭ begin to have negative values? For which terms does the finite arithmetic sequence 143. ⎧ ⎨5 2 ⎩ , 19 8 , 9 4 , ..., 1 8 ⎫ ⎬ have integer values? ⎭ Write an arithmetic sequence using a recursive 144. formula. Show the first 4 terms, and then find the 31st term. Chapter 13 Sequences, Probability, and Counting Theory 1475 145. Write an arithmetic sequence using an explicit formula. Show the first 4 terms, and then find the 28th term. 1476 Chapter 13 Sequences, Probability, and Counting Theory 13.3 \| Geometric Sequences Learning Objectives In this section, you will: 13.3.1 Find the common ratio for a geometric sequence. 13.3.2 List the terms of a geometric sequence. 13.3.3 Use a recursive formula for a geometric sequence. 13.3.4 Use an explicit formula for a geometric sequence. Many jobs offer an annual cost-of-living increase to keep salaries consistent with inflation. Suppose, for example, a recent college graduate finds a position as a sales manager earning an annual salary of $26,000. He is promised a 2% cost of living increase each year. His annual salary in any given year can be found by multiplying his salary from the previous year by 102%. His salary will be $26,520 after one year; $27,050.40 after two years; $27,591.41 after three years; and so on. When a salary increases by a constant rate each year, the salary grows by a constant factor. In this section, we will review sequences that grow in this way. Finding Common Ratios The yearly salary values described form a geometric sequence because they change by a constant factor each year. Each term of a geometric sequence increases or decreases by a constant factor called the common ratio. The sequence below is an example of a geometric sequence because each term increases by a constant factor of 6. Multiplying any term of the sequence by the common ratio 6 generates the subsequent term. Definition of a Geometric Sequence A geometric sequence is one in which any term divided by the previous term is a constant. This constant is called the common ratio of the sequence. The common ratio can be found by dividing any term in the sequence by the previous term. If a1 is the initial term of a geometric sequence and r is the common ratio, the sequence will be ⎧ ⎨a1, ⎩ a1 r, a1 r 2, a1 r 3, ... ⎫ ⎬. ⎭ Given a set of numbers, determine if they represent a geometric sequence. 1. Divide each term by the previous term. 2. Compare the quotients. If they are the same, a common ratio exists and the sequence is geometric. Example 13.15 Finding Common Ratios Is the sequence geometric? If so, find the common ratio. a. b. 1, 2, 4, 8, 16, ... 48, 12, 4, 2, ... This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1477 Solution Divide each term by the previous term to determine whether a common ratio exists. a. b. = 2 2 1 The sequence is geometric because there is a common ratio. The common ratio is 2. 16 12 48 The sequence is not geometric because there is not a common ratio. = 1 3 = 1 2 4 12 2 4 Analysis The graph of each sequence is shown in Figure 13.14. It seems from the graphs that both (a) and (b) appear have the form of the graph of an exponential function in this viewing window. However, we know that (a) is geometric and so this interpretation holds, but (b) is not. Figure 13.14 If you are told that a sequence is geometric, do you have to divide every term by the previous term to find the common ratio? No. If you know that the sequence is geometric, you can choose any one term in the sequence and divide it by the previous term to find the common ratio. 13.18 Is the sequence geometric? If so, find the common ratio. 5, 10, 15, 20, ... 13.19 Is the sequence geometric? If so, find the common ratio. 100, 20, 4, 4 5 , ... Writing Terms of Geometric Sequences Now that we can identify a geometric sequence, we will learn how to find the terms of a geometric sequence if we are given the first term and the common ratio. The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. For instance, if the first term of a geometric sequence is a1 = − 2 and the common ratio is r = 4, we can find subsequent terms by multiplying −2 ⋅ 4 to get −8 then multiplying the result −8 ⋅ 4 to get −32 and so on. 1478 Chapter 13 Sequences, Probability, and Counting Theory a1 = − 2 a2 = ( − 2 ⋅ 4) = − 8 a3 = ( − 8 ⋅ 4) = − 32 a4 = ( − 32 ⋅ 4) − 128 The first four terms are {–2, –8, –32, –128}. Given the first term and the common factor, find the first four terms of a geometric sequence. 1. Multiply the initial term, a1, by the common ratio to find the next term, a2. 2. Repeat the process, using an = a2 to find a3 and then a3 to find a4, until all four terms have been identified. 3. Write the terms separated by commons within brackets. Example 13.16 Writing the Terms of a Geometric Sequence List the first four terms of the geometric sequence with a1 = 5 and r = –2. Solution Multiply a1 by −2 to find a2. Repeat the process, using a2 to find a3, and so on. a1 = 5 a2 = − 2a1 = − 10 a3 = − 2a2 = 20 a4 = − 2a3 = − 40 The first four terms are {5, –10, 20, –40}. 13.20 List the first five terms of the geometric sequence with a1 = 18 and r = 1 3 . Using Recursive Formulas for Geometric Sequences A recursive formula allows us to find any term of a geometric sequence by using the previous term. Each term is the product of the common ratio and the previous term. For example, suppose the common ratio is 9. Then each term is nine times the previous term. As with any recursive formula, the initial term must be given. Recursive Formula for a Geometric Sequence The recursive formula for a geometric sequence with common ratio r and first term a1 is an = ran − 1, n ≥ 2 (13.4) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1479 Given the first several terms of a geometric sequence, write its recursive formula. 1. State the initial term. 2. Find the common ratio by dividing any term by the preceding term. 3. Substitute the common ratio into the recursive formula for a geometric sequence. Example 13.17 Using Recursive Formulas for Geometric Sequences Write a recursive formula for the following geometric sequence. {6, 9, 13.5, 20.25, ...} Solution The first term is given as 6. The common ratio can be found by dividing the second term by the first term. Substitute the common ratio into the recursive formula for geometric sequences and define a1. r = 9 6 = 1.5 an = ran − 1 an = 1.5an − 1 for n ≥ 2 a1 = 6 Analysis The sequence of data points follows an exponential pattern. The common ratio is also the base of an exponential function as shown in Figure 13.15 Figure 13.15 Do we have to divide the second term by the first term to find the common ratio? No. We can divide any term in the sequence by the previous term. It is, however, most common to divide the second term by the first term because it is often the easiest method of finding the common ratio. 13.21 Write a recursive formula for the following geometric sequence. ⎧ ⎨2, 4 3 ⎩ , 8 9 , 16 27 , ... ⎫ ⎬ ⎭ Using Explicit Formulas for Geometric Sequences Because a geometric sequence is an exponential function whose domain is the set of positive integer
s, and the common ratio is the base of the function, we can write explicit formulas that allow us to find particular terms. 1480 Chapter 13 Sequences, Probability, and Counting Theory an = a1 r n − 1 Let’s take a look at the sequence {18, 36, 72, 144, 288, ...}. This is a geometric sequence with a common ratio of 2 and an exponential function with a base of 2. An explicit formula for this sequence is The graph of the sequence is shown in Figure 13.16. an = 18 · 2 n − 1 Figure 13.16 Explicit Formula for a Geometric Sequence The nth term of a geometric sequence is given by the explicit formula: an = a1 r n − 1 (13.5) Example 13.18 Writing Terms of Geometric Sequences Using the Explicit Formula Given a geometric sequence with a1 = 3 and a4 = 24, find a2. Solution The sequence can be written in terms of the initial term and the common ratio r. Find the common ratio using the given fourth term. 3, 3r, 3r 2, 3r 3, ... This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1481 an = a1 r n − 1 a4 = 3r 3 24 = 3r 3 8 = r 3 r = 2 Write the fourth term of sequence in terms of α1 and r Substitute 24 for a4 Divide Solve for the common ratio Find the second term by multiplying the first term by the common ratio. a2 = 2a1 = 2(3) = 6 Analysis The common ratio is multiplied by the first term once to find the second term, twice to find the third term, three times to find the fourth term, and so on. The tenth term could be found by multiplying the first term by the common ratio nine times or by multiplying by the common ratio raised to the ninth power. 13.22 Given a geometric sequence with a2 = 4 and a3 = 32 , find a6. Example 13.19 Writing an Explicit Formula for the nth Term of a Geometric Sequence Write an explicit formula for the nth term of the following geometric sequence. {2, 10, 50, 250, ...} Solution The first term is 2. The common ratio can be found by dividing the second term by the first term. The common ratio is 5. Substitute the common ratio and the first term of the sequence into the formula. 10 2 = 5 The graph of this sequence in Figure 13.17 shows an exponential pattern. an = a1 r (n − 1) n − 1 an = 2 ⋅ 5 1482 Chapter 13 Sequences, Probability, and Counting Theory Figure 13.17 13.23 Write an explicit formula for the following geometric sequence. {–1, 3, –9, 27, ...} Solving Application Problems with Geometric Sequences In real-world scenarios involving arithmetic sequences, we may need to use an initial term of a0 problems, we can alter the explicit formula slightly by using the following formula: instead of a1. In these an = a0 r n Example 13.20 Solving Application Problems with Geometric Sequences In 2013, the number of students in a small school is 284. It is estimated that the student population will increase by 4% each year. a. Write a formula for the student population. b. Estimate the student population in 2020. Solution a. The situation can be modeled by a geometric sequence with an initial term of 284. The student population will be 104% of the prior year, so the common ratio is 1.04. Let P be the student population and n be the number of years after 2013. Using the explicit formula for a geometric sequence we get b. We can find the number of years since 2013 by subtracting. Pn = 284 ⋅ 1.04 n 2020 − 2013 = 7 We are looking for the population after 7 years. We can substitute 7 for n to estimate the population in 2020. P7 = 284 ⋅ 1.047 ≈ 374 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1483 The student population will be about 374 in 2020. A business starts a new website. Initially the number of hits is 293 due to the curiosity factor. The 13.24 business estimates the number of hits will increase by 2.6% per week. a. Write a formula for the number of hits. b. Estimate the number of hits in 5 weeks. Access these online resources for additional instruction and practice with geometric sequences. • Geometric Sequences (http://openstaxcollege.org/l/geometricseq) • Determine the Type of Sequence (http://openstaxcollege.org/l/sequencetype) • Find the Formula for a Sequence (http://openstaxcollege.org/l/sequenceformula) 1484 Chapter 13 Sequences, Probability, and Counting Theory 13.3 EXERCISES Verbal 146. What is a geometric sequence? How is the common ratio of a geometric sequence 147. found? What is the procedure for determining whether a 148. sequence is geometric? What 149. sequence and a geometric sequence? is the difference between an arithmetic For the following exercises, find the specified term for the geometric sequence, given the first term and common ratio. The first term is 2, and the common ratio is 3. Find 163. the 5th term. 164. The first term is 16 and the common ratio is − 1 3 . Find the 4th term. For the following exercises, find the specified term for the geometric sequence, given the first four terms. Describe how exponential functions and geometric 150. sequences are similar. How are they different? 165. an = {−1, 2, − 4, 8, ...}. Find a12. Algebraic For the following exercises, find the common ratio for the geometric sequence. 151. 152. 153. 1, 3, 9, 27, 81, ... −0.125, 0.25, − 0.5, 1, − 2, ... −2 32 , − 1 128 , ... 166. an = ⎧ ⎨−2 27 ⎫ , ... ⎭ ⎬. Find a7. For the following exercises, write the first five terms of the geometric sequence. 167. a1 = − 486, an = − 1 3 an − 1 168. a1 = 7, an = 0.2an − 1 the following exercises, determine whether For sequence is geometric. If so, find the common ratio. the For the following exercises, write a recursive formula for each geometric sequence. 154. 155. 156. 157. 158. −6, − 12, − 24, − 48, − 96, ... 5, 5.2, 5.4, 5.6, 5.8, ... −1 16 , ... 6, 8, 11, 15, 20, ... 0.8, 4, 20, 100, 500, ... For the following exercises, write the first five terms of the geometric sequence, given the first term and common ratio. 159. a1 = 8, r = 0.3 160. a1 = 5, r = 1 5 169. an = {−1, 5, − 25, 125, ...} 170. an = {−32, − 16, − 8, − 4, ...} 171. an = {14, 56, 224, 896, ...} 172. an = {10, − 3, 0.9, − 0.27, ...} 173. an = {0.61, 1.83, 5.49, 16.47, ...} 174. an = ⎧ ⎨3 5 ⎩ , 1 10 , 1 60 , 1 360 ⎫ , ... ⎬ ⎭ 175. an = ⎧ ⎨−2, 4 3 ⎩ , − 8 9 , 16 27 ⎫ , ... ⎬ ⎭ 176. an = ⎧ ⎨ 1 512 ⎩ , − 1 128 , 1 32 , − 1 8 ⎫ , ... ⎬ ⎭ For the following exercises, write the first five terms of the geometric sequence, given any two terms. For the following exercises, write the first five terms of the geometric sequence. 161. a7 = 64, a10 = 512 162. a6 = 25, a8 = 6.25 177. an = − 4 ⋅ 5 n − 1 178. an = 12 ⋅ n − 1 ⎛ ⎝− 1 2 ⎞ ⎠ This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1485 For the following exercises, write an explicit formula for each geometric sequence. 179. an = {−2, − 4, − 8, − 16, ...} 180. an = {1, 3, 9, 27, ...} 181. an = {−4, − 12, − 36, − 108, ...} 182. an = {0.8, − 4, 20, − 100, ...} 183. 184. an = ⎧ ⎨ − 1.25, − 5, − 20, − 80, ...⎫ ⎬ ⎭ ⎩ an = ⎧ ⎨−1, − 4 5 ⎩ , − 16 25 , − 64 125 ⎫ , ... ⎬ ⎭ 185. an = 186. an = ⎧ ⎨2, 1 3 ⎩ , 1 18 , 1 108 ⎫ , ... ⎬ ⎭ 192. ⎧ ⎨3, − 1, 1 3 ⎩ , − 1 9 ⎫ , ... ⎬ ⎭ For the following exercises, find the specified term for the geometric sequence given. 187. Let a1 = 4, an = − 3an − 1. Find a8. 188. Let an = − n − 1 ⎛ ⎝− 1 3 ⎞ ⎠ . Find a12. For the following exercises, find the number of terms in the given finite geometric sequence. 189. an = {−1, 3, − 9, ..., 2187} ⎧ ⎨2, 1, 1 2 ⎩ , ..., 1 1024 ⎫ ⎬ ⎭ 190. an = Graphical For the following exercises, determine whether the graph shown represents a geometric sequence. 191. For the following exercises, use the information provided to graph the first five terms of the geometric sequence. 193. a1 = 1, r = 1 2 194. a1 = 3, an = 2an − 1 195. an = 27 ⋅ 0.3 n − 1 Extensions 196. 1486 Chapter 13 Sequences, Probability, and Counting Theory Use recursive formulas to give two examples of geometric sequences whose 3rd terms are 200. Use explicit 197. geometric sequences whose 7th terms are 1024. formulas to give two examples of Find the 5th term of 198. {b, 4b, 16b, ...}. the geometric sequence Find the 7th term of 199. {64a( − b), 32a( − 3b), 16a( − 9b), ...}. the geometric sequence At 200. {10, 12, 14.4, 17.28, ...} exceed 100 ? which does term the sequence 201. ⎧ ⎨ 1 2187 ⎩ At , 1 729 which , 1 243 , 1 81 does term ⎫ ⎬ begin to have integer values? ⎭ the sequence ... 202. For which term does n − 1 ⎛ a n = − 36 ⎝ ⎞ ⎠ 2 3 first have a non-integer value? the geometric sequence 203. Use the recursive formula to write a geometric sequence whose common ratio is an integer. Show the first four terms, and then find the 10th term. Use the explicit 204. formula to write a geometric sequence whose common ratio is a decimal number between 0 and 1. Show the first 4 terms, and then find the 8th term. Is it possible for a sequence to be both arithmetic and 205. geometric? If so, give an example. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1487 13.4 \| Series and Their Notations Learning Objectives 13.4.1 Use summation notation. 13.4.2 Use the formula for the sum of the ﬁrst n terms of an arithmetic series. 13.4.3 Use the formula for the sum of the ﬁrst n terms of a geometric series. 13.4.4 Use the formula for the sum of an inﬁnite geometric series. 13.4.5 Solve annuity problems. A couple decides to start a college fund for their daughter. They plan to invest $50 in the fund each month. The fund pays 6% annual interest, compounded monthly. How much money will they have saved when their daughter is ready to start college in 6 years? In this section, we will learn how to answer this question. To do so, we need to consider the amount of money invested and the amount of interest earned. Using Summation Notation To find the total amount of money in the college fund and the sum of the amounts deposited, we need to add the amounts
deposited each month and the amounts earned monthly. The sum of the terms of a sequence is called a series. Consider, for example, the following series. The nth partial sum of a series is the sum of a finite number of consecutive terms beginning with the first term. The notation Sn represents the partial sum. 3 + 7 + 11 + 15 + 19 + ... S1 = 3 S2 = 3 + 7 = 10 S3 = 3 + 7 + 11 = 21 S4 = 3 + 7 + 11 + 15 = 36 Summation notation is used to represent series. Summation notation is often known as sigma notation because it uses the Greek capital letter sigma, Σ, to represent the sum. Summation notation includes an explicit formula and specifies the first and last terms in the series. An explicit formula for each term of the series is given to the right of the sigma. A variable called the index of summation is written below the sigma. The index of summation is set equal to the lower limit of summation, which is the number used to generate the first term in the series. The number above the sigma, called the upper limit of summation, is the number used to generate the last term in a series. If we interpret the given notation, we see that it asks us to find the sum of the terms in the series ak = 2k for k = 1 through k = 5. We can begin by substituting the terms for k and listing out the terms of this series. a1 = 2(1) = 2 a2 = 2(2) = 4 a3 = 2(3) = 6 a4 = 2(4) = 8 a5 = 2(5) = 10 We can find the sum of the series by adding the terms: 5 ∑ k = 1 2k = 2 + 4 + 6 + 8 + 10 = 30 1488 Chapter 13 Sequences, Probability, and Counting Theory Summation Notation The sum of the first n terms of a series can be expressed in summation notation as follows: n ∑ k = 1 ak This notation tells us to find the sum of ak from k = 1 to k = n. k is called the index of summation, 1 is the lower limit of summation, and n is the upper limit of summation. Does the lower limit of summation have to be 1? No. The lower limit of summation can be any number, but 1 is frequently used. We will look at examples with lower limits of summation other than 1. Given summation notation for a series, evaluate the value. 1. 2. Identify the lower limit of summation. Identify the upper limit of summation. 3. Substitute each value of k from the lower limit to the upper limit into the formula. 4. Add to find the sum. Example 13.21 Using Summation Notation 7 Evaluate ∑ k = 3 k 2. Solution According to the notation, the lower limit of summation is 3 and the upper limit is 7. So we need to find the sum of k 2 from k = 3 to k = 7. We find the terms of the series by substituting k = 3,4,5,6, and 7 into the function k 2. We add the terms to find the sum. 7 ∑ k = 3 k 2 = 32 + 42 + 52 + 62 + 72 = 9 + 16 + 25 + 36 + 49 = 135 13.25 5 Evaluate ∑ k = 2 (3k – 1). Using the Formula for Arithmetic Series Just as we studied special types of sequences, we will look at special types of series. Recall that an arithmetic sequence is a sequence in which the difference between any two consecutive terms is the common difference, d. The sum of the terms of an arithmetic sequence is called an arithmetic series. We can write the sum of the first n terms of an arithmetic series as: This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1489 We can also reverse the order of the terms and write the sum as Sn = a1 + (a1 + d) + (a1 + 2d) + ... + (an – d) + an. Sn = an + (an – d) + (an – 2d) + ... + (a1 + d) + a1. If we add these two expressions for the sum of the first n terms of an arithmetic series, we can derive a formula for the sum of the first n terms of any arithmetic series. Sn = a1 + (a1 + d) + (a1 + 2d) + ... + (an – d) + an + Sn = an + (an – d) + (an – 2d) + ... + (a1 + d) + a1 2Sn = (a1 + an) + (a1 + an) + ... + (a1 + an) Because there are n terms in the series, we can simplify this sum to 2Sn = n(a1 + an). We divide by 2 to find the formula for the sum of the first n terms of an arithmetic series. Sn = n(a1 + an) 2 Formula for the Sum of the First n Terms of an Arithmetic Series An arithmetic series is the sum of the terms of an arithmetic sequence. The formula for the sum of the first n terms of an arithmetic sequence is Sn = n(a1 + an) 2 (13.6) Given terms of an arithmetic series, find the sum of the first n terms. 1. Identify a1 and an. 2. Determine n. 3. Substitute values for a1 , an , and n into the formula Sn = n(a1 + an) 2 . 4. Simplify to find Sn. Example 13.22 Finding the First n Terms of an Arithmetic Series Find the sum of each arithmetic series. a. b. 5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 + 32 20 + 15 + 10 +…+ −50 12 c. ∑ k = 1 3k − 8 Solution a. We are given a1 = 5 and an = 32. 1490 Chapter 13 Sequences, Probability, and Counting Theory Count the number of terms in the sequence to find n = 10. Substitute values for a1, an , and n into the formula and simplify. b. We are given a1 = 20 and an = − 50. S10 = Sn = n(a1 + an) 2 10(5 + 32) 2 = 185 Use the formula for the general term of an arithmetic sequence to find n. an = a1 + (n − 1)d −50 = 20 + (n − 1)( − 5) −70 = (n − 1)( − 5) 14 = n − 1 15 = n Substitute values for a1, an , n into the formula and simplify. Sn = n(a1 + an) 2 S15 = 15(20 − 50) 2 = − 225 c. To find a1, substitute k = 1 into the given explicit formula. ak = 3k − 8 a1 = 3(1) − 8 = − 5 We are given that n = 12. To find a12, substitute k = 12 into the given explicit formula. Substitute values for a1, an, and n into the formula and simplify. ak = 3k − 8 a12 = 3(12) − 8 = 28 Sn = n(a1 + an) 2 S12 = 12( − 5 + 28) 2 = 138 Use the formula to find the sum of each arithmetic series. 13.26 1.4 + 1.6 + 1.8 + 2.0 + 2.2 + 2.4 + 2.6 + 2.8 + 3.0 + 3.2 + 3.4 13.27 13 + 21 + 29 + … + 69 13.28 10 ∑ k = 1 5 − 6k This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1491 Example 13.23 Solving Application Problems with Arithmetic Series On the Sunday after a minor surgery, a woman is able to walk a half-mile. Each Sunday, she walks an additional quarter-mile. After 8 weeks, what will be the total number of miles she has walked? Solution This problem can be modeled by an arithmetic series with a1 = 1 2 number of miles walked after 8 weeks, so we know that n = 8, and we are looking for S8. To find a8, we can use the explicit formula for an arithmetic sequence. . We are looking for the total and d = 1 4 an = a1 + d(n − 1) a8 = 1 + 1 2 4 (8 − 1) = 9 4 We can now use the formula for arithmetic series. Sn = S8 = She will have walked a total of 11 miles. n(a1 + an) 2 2 + 9 4) 2 8(1 = 11 A man earns $100 in the first week of June. Each week, he earns $12.50 more than the previous week. 13.29 After 12 weeks, how much has he earned? Using the Formula for Geometric Series Just as the sum of the terms of an arithmetic sequence is called an arithmetic series, the sum of the terms in a geometric sequence is called a geometric series. Recall that a geometric sequence is a sequence in which the ratio of any two consecutive terms is the common ratio, r. We can write the sum of the first n terms of a geometric series as Sn = a1 + ra1 + r 2 a1 + ... + r n – 1 a1. Just as with arithmetic series, we can do some algebraic manipulation to derive a formula for the sum of the first n terms of a geometric series. We will begin by multiplying both sides of the equation by r. Next, we subtract this equation from the original equation. rSn = ra1 + r 2 a1 + r 3 a1 + ... + r n a1 Sn = a1 + ra1 + r 2 a1 + ... + r n – 1 a1 −rSn = − (ra1 + r 2 a1 + r 3 a1 + ... + r n a1) (1 − r)Sn = a1 − r n a1 Notice that when we subtract, all but the first term of the top equation and the last term of the bottom equation cancel out. To obtain a formula for Sn, divide both sides by (1 − r). Sn = a1( 1492 Chapter 13 Sequences, Probability, and Counting Theory Formula for the Sum of the First n Terms of a Geometric Series A geometric series is the sum of the terms in a geometric sequence. The formula for the sum of the first n terms of a geometric sequence is represented as Sn = a1(13.7) Given a geometric series, find the sum of the first n terms. 1. Identify a1, r, and n. 2. Substitute values for a1, r, and n into the formula Sn = a1(. Simplify to find Sn. Example 13.24 Finding the First n Terms of a Geometric Series Use the formula to find the indicated partial sum of each geometric series. a. S11 for the series 8 + -4 + 2 + … k = 1 b. ∑ 6 3 ⋅ 2 k Solution a. a1 = 8, and we are given that n = 11. We can find r by dividing the second term of the series by the first. Substitute values for a1, r, r = −4 8 and n into the formula and simplify. = − 1 2 Sn = S11 = ) a1 (1 − r n 1 − r ⎛ ⎛ ⎝− 1 ⎝1 − 2 ⎛ ⎝− 1 1 − 2 8 ⎞ ⎠ ⎞ ⎠ 11⎞ ⎠ ≈ 5.336 b. Find a1 by substituting k = 1 into the given explicit formula. a1 = 3 ⋅ 21 = 6 We can see from the given explicit formula that r = 2. The upper limit of summation is 6, so n = 6. Substitute values for a1, r, and n into the formula, and simplify. Sn = S6 = a1(1 − r n 1 − r 6(1 − 26) 1 − 2 ) = 378 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1493 Use the formula to find the indicated partial sum of each geometric series. 13.30 S20 for the series 1,000 + 500 + 250 + … 13.31 8 ∑ k = 1 k 3 Example 13.25 Solving an Application Problem with a Geometric Series At a new job, an employee’s starting salary is $26,750. He receives a 1.6% annual raise. Find his total earnings at the end of 5 years. Solution The problem can be represented by a geometric series with a1 = 26, 750; n = 5; and r = 1.016. Substitute values for a1 , r, and n into the formula and simplify to find the total amount earned at the end of 5 years. Sn = S5 = a1(1 − r n 1 − r ) 26,750(1 − 1.0165) 1 − 1.016 ≈ 138,099.03 He will have earned a total of $138,099.03 by the end of 5 years. At a new job, an employee’s starting salary is $32,100. She receives a 2% annual raise. How much will 13.32 she have earned by the end of
8 years? Using the Formula for the Sum of an Infinite Geometric Series Thus far, we have looked only at finite series. Sometimes, however, we are interested in the sum of the terms of an infinite sequence rather than the sum of only the first n terms. An infinite series is the sum of the terms of an infinite sequence. An example of an infinite series is 2 + 4 + 6 + 8 + ... ∞ This series can also be written in summation notation as ∑ k = 1 2k, where the upper limit of summation is infinity. Because the terms are not tending to zero, the sum of the series increases without bound as we add more terms. Therefore, the sum of this infinite series is not defined. When the sum is not a real number, we say the series diverges. Determining Whether the Sum of an Infinite Geometric Series is Defined If the terms of an infinite geometric series approach 0, the sum of an infinite geometric series can be defined. The terms in this series approach 0: 1 + 0.2 + 0.04 + 0.008 + 0.0016 + ... The common ratio r = 0.2. As n gets very large, the values of r n get very small and approach 0. Each successive term affects the sum less than the preceding term. As each succeeding term gets closer to 0, the sum of the terms approaches a finite value. The terms of any infinite geometric series with −1 < r < 1 approach 0; the sum of a geometric series is defined when −1 < r < 1. 1494 Chapter 13 Sequences, Probability, and Counting Theory Determining Whether the Sum of an Infinite Geometric Series is Defined The sum of an infinite series is defined if the series is geometric and −1 < r < 1. Given the first several terms of an infinite series, determine if the sum of the series exists. 1. Find the ratio of the second term to the first term. 2. Find the ratio of the third term to the second term. 3. Continue this process to ensure the ratio of a term to the preceding term is constant throughout. If so, the series is geometric. 4. If a common ratio, r, was found in step 3, check to see if −1 < r < 1 . If so, the sum is defined. If not, the sum is not defined. Example 13.26 Determining Whether the Sum of an Infinite Series is Defined Determine whether the sum of each infinite series is defined. a. b. 12 + + ... ∞ c. ∑ k = 1 k 27 ⋅ (1 3 ) ∞ d. ∑ k = 1 5k Solution a. The ratio of the second term to the first is 2 3 second, 1 2 . The series is not geometric. , which is not the same as the ratio of the third term to the b. The ratio of the second term to the first is the same as the ratio of the third term to the second. The series is geometric with a common ratio of 2 3 . The sum of the infinite series is defined. c. The given formula is exponential with a base of 1 3 ; the series is geometric with a common ratio of 1 3 . The sum of the infinite series is defined. d. The given formula is not exponential; the series is not geometric because the terms are increasing, and so cannot yield a finite sum. Determine whether the sum of the infinite series is defined. 13.33 + ... This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1495 13.34 24 + (−12) + 6 + (−3) + ... 13.35 ∞ ∑ k = 1 15 ⋅ ( – 0.3) k Finding Sums of Infinite Series When the sum of an infinite geometric series exists, we can calculate the sum. The formula for the sum of an infinite series is related to the formula for the sum of the first n terms of a geometric series. We will examine an infinite series with r = 1 2 . What happens to r n as n increases? Sn = a1( 16 The value of r n decreases rapidly. What happens for greater values of n ? 10 20 30 (1 2 ) (1 2 ) (1 2 ) = 1 1,024 = = 1 1,048,576 1 1,073,741,824 As n gets very large, r n gets very small. We say that, as n increases without bound, r n approaches 0. As r n approaches 0, 1 − r n approaches 1. When this happens, the numerator approaches a1. This give us a formula for the sum of an infinite geometric series. Formula for the Sum of an Infinite Geometric Series The formula for the sum of an infinite geometric series with −1 < r < 1 is S = a1 1 − r (13.8) Given an infinite geometric series, find its sum. 1. Identify a1 and r. 2. Confirm that – 1 < r < 1. 3. Substitute values for a1 and r into the formula, S = a1 1 − r. 4. Simplify to find S. 1496 Chapter 13 Sequences, Probability, and Counting Theory Example 13.27 Finding the Sum of an Infinite Geometric Series Find the sum, if it exists, for the following: a. b. 10 + 9 + 8 + 7 + … 248.6 + 99.44 + 39.776 + … ∞ c. ∑ k = 1 k – 1 4,374 ⋅ ( – 1 3 ) ∞ d4 3 ) Solution a. There is not a constant ratio; the series is not geometric. b. There is a constant ratio; the series is geometric. a1 = 248.6 and r = 99.44 248.6 = 0.4, Substitute a1 = 248.6 and r = 0.4 into the formula and simplify to find the sum: so the sum exists. S = a1 1 − r S = 248.6 1 − 0.4 ¯ = 414. 3 c. The formula is exponential, so the series is geometric with r = – 1 3 . Find a1 by substituting k = 1 into the given explicit formula: a1 = 4,374 ⋅ ( – 1 3 ) 1 – 1 = 4,374 Substitute a1 = 4,374 and r = − 1 3 into the formula, and simplify to find the sum: S = a1 1 − r S = 4,374 1 − ( − 1 3) = 3,280.5 d. The formula is exponential, so the series is geometric, but r > 1. The sum does not exist. Example 13.28 Finding an Equivalent Fraction for a Repeating Decimal ¯ Find an equivalent fraction for the repeating decimal 0. 3 Solution ¯ We notice the repeating decimal 0. 3 = 0.333... so we can rewrite the repeating decimal as a sum of terms. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1497 Looking for a pattern, we rewrite the sum, noticing that we see the first term multiplied to 0.1 in the second term, and the second term multiplied to 0.1 in the third term. ¯ 0. 3 = 0.3 + 0.03 + 0.003 + ... Notice the pattern; we multiply each consecutive term by a common ratio of 0.1 starting with the first term of 0.3. So, substituting into our formula for an infinite geometric sum, we have Sn = a1 1 − r = 0.3 1 − 0.1 = 0.3 0.9 = 1 3 . Find the sum, if it exists. 13.36 2 + 2 3 + 2 9 + ... 13.37 ∞ ∑ k = 1 0.76k + 1 13.38 ∞ ∑ k = 1 k ⎛ ⎝− 3 8 ⎞ ⎠ Solving Annuity Problems At the beginning of the section, we looked at a problem in which a couple invested a set amount of money each month into a college fund for six years. An annuity is an investment in which the purchaser makes a sequence of periodic, equal payments. To find the amount of an annuity, we need to find the sum of all the payments and the interest earned. In the example, the couple invests $50 each month. This is the value of the initial deposit. The account paid 6% annual interest, compounded monthly. To find the interest rate per payment period, we need to divide the 6% annual percentage interest (APR) rate by 12. So the monthly interest rate is 0.5%. We can multiply the amount in the account each month by 100.5% to find the value of the account after interest has been added. We can find the value of the annuity right after the last deposit by using a geometric series with a1 = 50 and r = 100.5% = 1.005. After the first deposit, the value of the annuity will be $50. Let us see if we can determine the amount in the college fund and the interest earned. We can find the value of the annuity after n deposits using the formula for the sum of the first n terms of a geometric series. In 6 years, there are 72 months, so n = 72. We can substitute a1 = 50, and n = 72 into the formula, and simplify to find the value of the annuity after 6 years. r = 1.005, S72 = 50(1 − 1.00572) 1 − 1.005 ≈ 4,320.44 After the last deposit, the couple will have a total of $4,320.44 in the account. Notice, the couple made 72 payments of $50 each for a total of 72(50) = $3,600. This means that because of the annuity, the couple earned $720.44 interest in their college fund. 1498 Chapter 13 Sequences, Probability, and Counting Theory Given an initial deposit and an interest rate, find the value of an annuity. 1. Determine a1 , the value of the initial deposit. 2. Determine n, the number of deposits. 3. Determine r. a. Divide the annual interest rate by the number of times per year that interest is compounded. b. Add 1 to this amount to find r. 4. Substitute values for a1 , r, and n into the formula for the sum of the first n terms of a geometric series, Sn = a1(. Simplify to find Sn, the value of the annuity after n deposits. Example 13.29 Solving an Annuity Problem A deposit of $100 is placed into a college fund at the beginning of every month for 10 years. The fund earns 9% annual interest, compounded monthly, and paid at the end of the month. How much is in the account right after the last deposit? Solution The value of the initial deposit is $100, so a1 = 100. A total of 120 monthly deposits are made in the 10 years, so n = 120. To find r, divide the annual interest rate by 12 to find the monthly interest rate and add 1 to represent the new monthly deposit. r = 1 + 0.09 12 Substitute a1 = 100, r = 1.0075, and n = 120 into the formula for the sum of the first n terms of a geometric series, and simplify to find the value of the annuity. = 1.0075 S120 = 100(1 − 1.0075120) 1 − 1.0075 ≈ 19,351.43 So the account has $19,351.43 after the last deposit is made. 13.39 At the beginning of each month, $200 is deposited into a retirement fund. The fund earns 6% annual interest, compounded monthly, and paid into the account at the end of the month. How much is in the account if deposits are made for 10 years? Access these online resources for additional instruction and practice with series. • Arithmetic Series (http://openstaxcollege.org/l/arithmeticser) • Geometric Series (http://openstaxcollege.org/l/geometricser) • Summation Notation (http://openstaxcollege.org/l/sumnotation) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1499 13.4 EXERCISES Verbal 206. What is an nth partial sum? For th
e following exercises, use the formula for the sum of the first n terms of each geometric sequence, and then state the indicated sum. What 207. sequence and an arithmetic series? is the difference between an arithmetic 208. What is a geometric series? How is finding the sum of an infinite geometric series 209. different from finding the nth partial sum? 210. What is an annuity? Algebraic For the following exercises, express each description of a sum using summation notation. 211. The sum of terms m2 + 3m from m = 1 to m = 5 212. The sum from of n = 0 to n = 4 of 5n 213. The sum of 6k − 5 from k = − 2 to k = 1 The sum that results from adding the number 4 five 214. times For the following exercises, express each arithmetic sum using summation notation. 224. 225. 226 11 ∑ a = 1 n − 1 5 ⋅ 2 64 ⋅ 0.2 a − 1 For the following exercises, determine whether the infinite series has a sum. If so, write the formula for the sum. If not, state the reason. 227. 228. 229. 230. 12 + 18 + 24 + 30 + ... 2 + 1.6 + 1.28 + 1.024 + ... ∞ ∑ − ⎛ ⎝− 1 2 ⎞ ⎠ 215. 216. 217. 5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 + 50 Graphical 10 + 18 + 26 + … + 162 For the following exercises, use the following scenario. Javier makes monthly deposits into a savings account. He opened the account with an initial deposit of $50. Each month thereafter he increased the previous deposit amount by $20. For the following exercises, use the formula for the sum of the first n terms of each arithmetic sequence. Graph the arithmetic sequence showing one year of 231. Javier’s deposits. 218. 219. 220 Graph the arithmetic series showing the monthly 232. sums of one year of Javier’s deposits. 19 + 25 + 31 + … + 73 For the following exercises, use the geometric series 3.2 + 3.4 + 3.6 + … + 5. For the following exercises, express each geometric sum using summation notation. 233. Graph the first 7 partial sums of the series. 221. 222. 223. 1 + 3 + 9 + 27 + 81 + 243 + 729 + 2187 234. What number does Sn seem to be approaching in the 8 + 4 + 2 + … + 0.125 − 1 6 + 1 12 − 1 24 + … + 1 768 graph? Find the sum to explain why this makes sense. Numeric For the following exercises, find the indicated sum. 235. 1500 14 ∑ a = 1 a 236. 237. 238. 6 ∑ n = 1 n(n − 2) 17 ∑ For the following exercises, use the formula for the sum of the first n terms of an arithmetic series to find the sum. 239. 240. 241. 242. −1.7 + − 0.4 + 0.9 + 2.2 + 3.5 + 4.8 6 + 15 2 + 9 + 21 2 + 12 + 27 2 + 15 −1 + 3 + 7 + ... + 31 11 ∑ For the following exercises, use the formula for the sum of the first n terms of a geometric series to find the partial sum. 243. 244. 245. 246. S6 S7 for the series −2 − 10 − 50 − 250... for the series 0.4 − 2 + 10 − 50... 9 ∑ k = 1 10 ∑ 2 ⋅ ⎞ ⎠ ⎛ ⎝ 1 2 For the following exercises, find the sum of the infinite geometric series. 4 + 2 + 1 + 1 2 ... −1 − 1 4 − 1 16 − 1 64 ... 247. 248. 249. 250. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory ∞ ∑ n = 1 4.6 ⋅ 0.5 n − 1 For the following exercises, determine the value of the annuity for the indicated monthly deposit amount, the number of deposits, and the interest rate. 251. Deposit amount: $50; rate: 5%, compounded monthly total deposits: 60; interest Deposit amount: $150; total deposits: 24; interest 252. rate: 3%, compounded monthly Deposit amount: $450; total deposits: 60; interest 253. rate: 4.5%, compounded quarterly Deposit amount: $100; total deposits: 120; interest 254. rate: 10%, compounded semi-annually Extensions 255. The sum of terms 50 − k 2 from k = x through 7 is 115. What is x? 256. Write an explicit formula for ak such that 6 ∑ k = 0 257. n ∑ k = 1 ak = 189. Assume this is an arithmetic series. Find the smallest value of n such that (3k – 5) > 100. How many terms must be added before the series 258. −1 − 3 − 5 − 7.... has a sum less than −75 ? 259. Write 0.65 as an infinite geometric series using summation notation. Then use the formula for finding the sum of an infinite geometric series to convert 0.65 to a fraction. 260. The sum of an infinite geometric series is five times the value of the first term. What is the common ratio of the series? 261. To get the best loan rates available, the Riches want to save enough money to place 20% down on a $160,000 home. They plan to make monthly deposits of $125 in an investment account interest compounded semi-annually. Will the Riches have enough for a 20% down payment after five years of saving? How much money will they have saved? that offers 8.5% annual 262. Chapter 13 Sequences, Probability, and Counting Theory 1501 Karl has two years to save $10, 000 to buy a used car when he graduates. To the nearest dollar, what would his monthly deposits need to be if he invests in an account offering a 4.2% annual rate that compounds monthly? interest Real-World Applications Keisha devised a week-long study plan to prepare for 263. finals. On the first day, she plans to study for 1 hour, and each successive day she will increase her study time by 30 minutes. How many hours will Keisha have studied after one week? A boulder rolled down a mountain, traveling 6 feet in 264. the first second. Each successive second, its distance increased by 8 feet. How far did the boulder travel after 10 seconds? A scientist places 50 cells in a petri dish. Every hour, 265. the population increases by 1.5%. What will the cell count be after 1 day? A pendulum travels a distance of 3 feet on its first 266. swing. On each successive swing, it travels 3 4 the distance of the previous swing. What is the total distance traveled by the pendulum when it stops swinging? Rachael deposits $1,500 into a retirement fund each 267. year. The fund earns 8.2% annual interest, compounded monthly. If she opened her account when she was 19 years old, how much will she have by the time she is 55? How much of that amount will be interest earned? 1502 Chapter 13 Sequences, Probability, and Counting Theory 13.5 \| Counting Principles Learning Objectives In this section, you will: 13.5.1 Solve counting problems using the Addition Principle. 13.5.2 Solve counting problems using the Multiplication Principle. 13.5.3 Solve counting problems using permutations involving n distinct objects. 13.5.4 Solve counting problems using combinations. 13.5.5 Find the number of subsets of a given set. 13.5.6 Solve counting problems using permutations involving n non-distinct objects. A new company sells customizable cases for tablets and smartphones. Each case comes in a variety of colors and can be personalized for an additional fee with images or a monogram. A customer can choose not to personalize or could choose to have one, two, or three images or a monogram. The customer can choose the order of the images and the letters in the monogram. The company is working with an agency to develop a marketing campaign with a focus on the huge number of options they offer. Counting the possibilities is challenging! We encounter a wide variety of counting problems every day. There is a branch of mathematics devoted to the study of counting problems such as this one. Other applications of counting include secure passwords, horse racing outcomes, and college scheduling choices. We will examine this type of mathematics in this section. Using the Addition Principle The company that sells customizable cases offers cases for tablets and smartphones. There are 3 supported tablet models and 5 supported smartphone models. The Addition Principle tells us that we can add the number of tablet options to the number of smartphone options to find the total number of options. By the Addition Principle, there are 8 total options, as we can see in Figure 13.18. Figure 13.18 The Addition Principle According to the Addition Principle, if one event can occur in m ways and a second event with no common outcomes can occur in n ways, then the first or second event can occur in m + n ways. Example 13.30 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1503 Using the Addition Principle There are 2 vegetarian entrée options and 5 meat entrée options on a dinner menu. What is the total number of entrée options? Solution We can add the number of vegetarian options to the number of meat options to find the total number of entrée options. There are 7 total options. A student is shopping for a new computer. He is deciding among 3 desktop computers and 4 laptop 13.40 computers. What is the total number of computer options? Using the Multiplication Principle The Multiplication Principle applies when we are making more than one selection. Suppose we are choosing an appetizer, an entrée, and a dessert. If there are 2 appetizer options, 3 entrée options, and 2 dessert options on a fixed-price dinner menu, there are a total of 12 possible choices of one each as shown in the tree diagram in Figure 13.19. Figure 13.19 The possible choices are: 1. 2. 3. 4. 5. 6. 7. 8. 9. soup, chicken, cake soup, chicken, pudding soup, fish, cake soup, fish, pudding soup, steak, cake soup, steak, pudding salad, chicken, cake salad, chicken, pudding salad, fish, cake 10. salad, fish, pudding 1504 Chapter 13 Sequences, Probability, and Counting Theory 11. salad, steak, cake 12. salad, steak, pudding We can also find the total number of possible dinners by multiplying. We could also conclude that there are 12 possible dinner choices simply by applying the Multiplication Principle. # of appetizer options × # of entree options × # of dessert options 2 × 3 × 2 = 12 The Multiplication Principle According to the Multiplication Principle, if one event can occur in m ways and a second event can occur in n ways after the first event has occurred, then the two events can occur in m×n ways. This is also known as the Fundamental Counting Principle. Example 13.31 Using the Multiplication Principle Diane packed 2 skirts, 4 blouses, and a sweater for her business t
rip. She will need to choose a skirt and a blouse for each outfit and decide whether to wear the sweater. Use the Multiplication Principle to find the total number of possible outfits. Solution To find the total number of outfits, find the product of the number of skirt options, the number of blouse options, and the number of sweater options. There are 16 possible outfits. 13.41 A restaurant offers a breakfast special that includes a breakfast sandwich, a side dish, and a beverage. There are 3 types of breakfast sandwiches, 4 side dish options, and 5 beverage choices. Find the total number of possible breakfast specials. Finding the Number of Permutations of n Distinct Objects The Multiplication Principle can be used to solve a variety of problem types. One type of problem involves placing objects in order. We arrange letters into words and digits into numbers, line up for photographs, decorate rooms, and more. An ordering of objects is called a permutation. Finding the Number of Permutations of n Distinct Objects Using the Multiplication Principle To solve permutation problems, it is often helpful to draw line segments for each option. That enables us to determine the number of each option so we can multiply. For instance, suppose we have four paintings, and we want to find the number of ways we can hang three of the paintings in order on the wall. We can draw three lines to represent the three places on the wall. There are four options for the first place, so we write a 4 on the first line. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1505 After the first place has been filled, there are three options for the second place so we write a 3 on the second line. After the second place has been filled, there are two options for the third place so we write a 2 on the third line. Finally, we find the product. There are 24 possible permutations of the paintings. Given n distinct options, determine how many permutations there are. 1. Determine how many options there are for the first situation. 2. Determine how many options are left for the second situation. 3. Continue until all of the spots are filled. 4. Multiply the numbers together. Example 13.32 Finding the Number of Permutations Using the Multiplication Principle At a swimming competition, nine swimmers compete in a race. a. How many ways can they place first, second, and third? b. How many ways can they place first, second, and third if a swimmer named Ariel wins first place? (Assume there is only one contestant named Ariel.) c. How many ways can all nine swimmers line up for a photo? Solution a. Draw lines for each place. There are 9 options for first place. Once someone has won first place, there are 8 remaining options for second place. Once first and second place have been won, there are 7 remaining options for third place. Multiply to find that there are 504 ways for the swimmers to place. b. Draw lines for describing each place. 1506 Chapter 13 Sequences, Probability, and Counting Theory We know Ariel must win first place, so there is only 1 option for first place. There are 8 remaining options for second place, and then 7 remaining options for third place. Multiply to find that there are 56 ways for the swimmers to place if Ariel wins first. c. Draw lines for describing each place in the photo. There are 9 choices for the first spot, then 8 for the second, 7 for the third, 6 for the fourth, and so on until only 1 person remains for the last spot. There are 362,880 possible permutations for the swimmers to line up. Analysis Note that in part c, we found there were 9! ways for 9 people to line up. The number of permutations of n distinct objects can always be found by n !. A family of five is having portraits taken. Use the Multiplication Principle to find the following. 13.42 How many ways can the family line up for the portrait? 13.43 How many ways can the photographer line up 3 family members? 13.44 How many ways can the family line up for the portrait if the parents are required to stand on each end? Finding the Number of Permutations of n Distinct Objects Using a Formula For some permutation problems, it is inconvenient to use the Multiplication Principle because there are so many numbers to multiply. Fortunately, we can solve these problems using a formula. Before we learn the formula, let’s look at two common notations for permutations. If we have a set of n objects and we want to choose r objects from the set in order, we write P(n, r). Another way to write this is nPr, a notation commonly seen on computers and calculators. To calculate P(n, r), we begin by finding n !, the number of ways to line up all n objects. We then divide by (n − r) ! to cancel out the (n − r) items that we do not wish to line up. Let’s see how this works with a simple example. Imagine a club of six people. They need to elect a president, a vice president, and a treasurer. Six people can be elected president, any one of the five remaining people can be elected vice president, and any of the remaining four people could be elected treasurer. The number of ways this may be done is 6×5×4 = 120. Using factorials, we get the same result. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1507 There are 120 ways to select 3 officers in order from a club with 6 members. We refer to this as a permutation of 6 taken 3 at a time. The general formula is as follows = 120 P(n, r) = n ! (n − r) ! Note that the formula stills works if we are choosing all n objects and placing them in order. In that case we would be dividing by (n − n) ! or 0 !, which we said earlier is equal to 1. So the number of permutations of n objects taken n at a time is n ! 1 or just n !. Formula for Permutations of n Distinct Objects Given n distinct objects, the number of ways to select r objects from the set in order is P(n, r) = n ! (n − r) ! (13.9) Given a word problem, evaluate the possible permutations. 1. 2. Identify n from the given information. Identify r from the given information. 3. Replace n and r in the formula with the given values. 4. Evaluate. Example 13.33 Finding the Number of Permutations Using the Formula A professor is creating an exam of 9 questions from a test bank of 12 questions. How many ways can she select and arrange the questions? Solution Substitute n = 12 and r = 9 into the permutation formula and simplify. P(n, r) = P(12, 9) = n ! (n − r) ! 12 ! (12 − 9) ! = 12 ! 3 ! = 79,833,600 There are 79,833,600 possible permutations of exam questions! Analysis We can also use a calculator to find permutations. For this problem, we would enter 15, press the n Pr function, enter 12, and then press the equal sign. The n Pr function may be located under the MATH menu with probability commands. Could we have solved Example 13.33 using the Multiplication Principle? Yes. We could have multiplied 15 ⋅ 14 ⋅ 13 ⋅ 12 ⋅ 11 ⋅ 10 ⋅ to find the same answer. 1508 Chapter 13 Sequences, Probability, and Counting Theory A play has a cast of 7 actors preparing to make their curtain call. Use the permutation formula to find the following. 13.45 How many ways can the 7 actors line up? 13.46 How many ways can 5 of the 7 actors be chosen to line up? Find the Number of Combinations Using the Formula So far, we have looked at problems asking us to put objects in order. There are many problems in which we want to select a few objects from a group of objects, but we do not care about the order. When we are selecting objects and the order does not matter, we are dealing with combinations. A selection of r objects from a set of n objects where the order does not matter can be written as C(n, r). Just as with permutations, C(n, r) can also be written as n Cr. In this case, the general formula is as follows. C(n, r) = n ! r !(n − r) ! An earlier problem considered choosing 3 of 4 possible paintings to hang on a wall. We found that there were 24 ways to select 3 of the 4 paintings in order. But what if we did not care about the order? We would expect a smaller number because selecting paintings 1, 2, 3 would be the same as selecting paintings 2, 3, 1. To find the number of ways to select 3 of the 4 paintings, disregarding the order of the paintings, divide the number of permutations by the number of ways to order 3 paintings. There are ways to order 3 paintings. There are 24 , or 4 ways to select 3 of the 4 paintings. 6 This number makes sense because every time we are selecting 3 paintings, we are not selecting 1 painting. There are 4 paintings we could choose not to select, so there are 4 ways to select 3 of the 4 paintings. Formula for Combinations of n Distinct Objects Given n distinct objects, the number of ways to select r objects from the set is C(n, r) = n ! r !(n − r) ! (13.10) Given a number of options, determine the possible number of combinations. 1. 2. Identify n from the given information. Identify r from the given information. 3. Replace n and r in the formula with the given values. 4. Evaluate. Example 13.34 Finding the Number of Combinations Using the Formula A fast food restaurant offers five side dish options. Your meal comes with two side dishes. a. How many ways can you select your side dishes? b. How many ways can you select 3 side dishes? This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1509 Solution a. We want to choose 2 side dishes from 5 options. b. We want to choose 3 side dishes from 5 options. C(5, 2) = 5 ! 2 !(5 − 2) ! = 10 C(5, 3) = 5 ! 3 !(5 − 3) ! = 10 Analysis We can also use a graphing calculator to find combinations. Enter 5, then press n Cr, enter 3, and then press the equal sign. The n Cr, function may be located under the MATH menu with probability commands. Is it a coincidence that parts (a) and (b) in Example 13.34 have the same answers? No. W
hen we choose r objects from n objects, we are not choosing (n – r) objects. Therefore, C(n, r) = C(n, n – r). An ice cream shop offers 10 flavors of ice cream. How many ways are there to choose 3 flavors for a 13.47 banana split? Finding the Number of Subsets of a Set We have looked only at combination problems in which we chose exactly r objects. In some problems, we want to consider choosing every possible number of objects. Consider, for example, a pizza restaurant that offers 5 toppings. Any number of toppings can be ordered. How many different pizzas are possible? To answer this question, we need to consider pizzas with any number of toppings. There is C(5, 0) = 1 way to order a pizza with no toppings. There are C(5, 1) = 5 ways to order a pizza with exactly one topping. If we continue this process, we get C(5, 0) + C(5, 1) + C(5, 2) + C(5, 3) + C(5, 4) + C(5, 5) = 32 There are 32 possible pizzas. This result is equal to 25. We are presented with a sequence of choices. For each of the n objects we have two choices: include it in the subset or not. So for the whole subset we have made n choices, each with two options. So there are a total of 2 · 2 · 2 · … · 2 possible resulting subsets, all the way from the empty subset, which we obtain when we say “no” each time, to the original set itself, which we obtain when we say “yes” each time. Formula for the Number of Subsets of a Set A set containing n distinct objects has 2 n subsets. Example 13.35 Finding the Number of Subsets of a Set 1510 Chapter 13 Sequences, Probability, and Counting Theory A restaurant offers butter, cheese, chives, and sour cream as toppings for a baked potato. How many different ways are there to order a potato? Solution We are looking for the number of subsets of a set with 4 objects. Substitute n = 4 into the formula. There are 16 possible ways to order a potato. n 2 = 24 = 16 A sundae bar at a wedding has 6 toppings to choose from. Any number of toppings can be chosen. How 13.48 many different sundaes are possible? Finding the Number of Permutations of n Non-Distinct Objects We have studied permutations where all of the objects involved were distinct. What happens if some of the objects are indistinguishable? For example, suppose there is a sheet of 12 stickers. If all of the stickers were distinct, there would be 12 ! ways to order the stickers. However, 4 of the stickers are identical stars, and 3 are identical moons. Because all of the objects are not distinct, many of the 12 ! permutations we counted are duplicates. The general formula for this situation is as follows. n ! r1 !r2 ! … rk ! In this example, we need to divide by the number of ways to order the 4 stars and the ways to order the 3 moons to find the number of unique permutations of the stickers. There are 4 ! ways to order the stars and 3 ! ways to order the moon. There are 3,326,400 ways to order the sheet of stickers. 12 ! 4 !3 ! = 3,326,400 Formula for Finding the Number of Permutations of n Non-Distinct Objects If there are n elements in a set and r1 are alike, r2 are alike, r3 are alike, and so on through rk, the number of permutations can be found by n ! r1 !r2 ! … rk ! (13.11) Example 13.36 Finding the Number of Permutations of n Non-Distinct Objects Find the number of rearrangements of the letters in the word DISTINCT. Solution There are 8 letters. Both I and T are repeated 2 times. Substitute n = 8, formula. r1 = 2, and r2 = 2 into the This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1511 There are 10,080 arrangements. 8 ! 2 !2 ! = 10,080 13.49 Find the number of rearrangements of the letters in the word CARRIER. Access these online resources for additional instruction and practice with combinations and permutations. • Combinations (http://openstaxcollege.org/l/combinations) • Permutations (http://openstaxcollege.org/l/permutations) 1512 Chapter 13 Sequences, Probability, and Counting Theory 13.5 EXERCISES Verbal For the following exercises, assume that there are n ways an event A can happen, m ways an event B can happen, and that A and B are non-overlapping. Use the Addition Principle of counting to explain how 268. many ways event A or B can occur. Use the Multiplication Principle of counting to 269. explain how many ways event A and B can occur. Answer the following questions. How many two-letter strings—the first letter from A 279. and the second letter from B — can be formed from the sets A = {b, c, d} and B = {a, e, i, o, u} ? How many ways are there to construct a string of 3 280. digits if numbers can be repeated? How many ways are there to construct a string of 3 281. digits if numbers cannot be repeated? For the following exercises, compute the value of the expression. to apply 270. When given two separate events, how do we know or the whether Multiplication possible outcomes? What conjunctions may help to determine which operations to use? the Addition Principle calculating Principle when 271. Describe how the permutation of n objects differs from the permutation of choosing r objects from a set of n objects. Include how each is calculated. 272. What is the term for the arrangement that selects r objects from a set of n objects when the order of the r objects is not is the formula for calculating the number of possible outcomes for this type of arrangement? important? What Numeric For the following exercises, determine whether to use the Addition Principle or the Multiplication Principle. Then perform the calculations. Let 273. the set A = { − 5, − 3, − 1, 2, 3, 4, 5, 6}. How many ways are there to choose a negative or an even number from A? the Let 274. B = { − 23, − 16, − 7, − 2, 20, 36, 48, 72}. many ways are there to choose a positive or an odd number from A ? set How How many ways are there to pick a red ace or a club 275. from a standard card playing deck? How many ways are there to pick a paint color from 5 276. shades of green, 4 shades of blue, or 7 shades of yellow? How many outcomes are possible from tossing a pair 277. of coins? How many outcomes are possible from tossing a coin 278. and rolling a 6-sided die? This content is available for free at https://cnx.org/content/col11758/1.5 282. P(5, 2) 283. P(8, 4) 284. P(3, 3) 285. P(9, 6) 286. P(11, 5) 287. C(8, 5) 288. C(12, 4) 289. C(26, 3) 290. C(7, 6) 291. C(10, 3) For the following exercises, find the number of subsets in each given set. 292. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} 293. {a, b, c, … , z} A set containing 5 distinct numbers, 4 distinct letters, 294. and 3 distinct symbols 295. The set of even numbers from 2 to 28 The set of two-digit numbers between 1 and 100 296. containing the digit 0 For the following exercises, find the distinct number of arrangements. 297. The letters in the word “juggernaut” 298. The letters in the word “academia” 299. Chapter 13 Sequences, Probability, and Counting Theory 1513 The letters in the word “academia” that begin and end in “a” 300. The symbols in the string #,#,#,@,@,$,$,$,%,%,%,% The symbols in the string #,#,#,@,@,$,$,$,%,%,%,% 301. that begin and end with “%” Extensions 302. The set, S consists of 900,000,000 whole numbers, each being the same number of digits long. How many digits long is a number from S ? (Hint: use the fact that a whole number cannot start with the digit 0.) subsets The number of 5-element from a set 303. containing n elements is equal to the number of 6-element subsets from the same set. What is the value of n ? (Hint: the order in which the elements for the subsets are chosen is not important.) 304. Can C(n, r) ever equal P(n, r) ? Explain. Suppose a set A has 2,048 subsets. How many 305. distinct objects are contained in A ? Hector wants to place billboard advertisements throughout the county for his new business. How many ways can Hector choose 15 neighborhoods to advertise in if there are 30 neighborhoods in the county? An art store has 4 brands of paint pens in 12 different 312. colors and 3 types of ink. How many paint pens are there to choose from? How many ways can a committee of 3 freshmen and 4 313. juniors be formed from a group of 8 freshmen and 11 juniors? How many ways can a baseball coach arrange the 314. order of 9 batters if there are 15 players on the team? A conductor needs 5 cellists and 5 violinists to play at 315. a diplomatic event. To do this, he ranks the orchestra’s 10 cellists and 16 violinists in order of musical proficiency. What is the ratio of the total cellist rankings possible to the total violinist rankings possible? A motorcycle shop has 10 choppers, 6 bobbers, and 5 316. café racers—different types of vintage motorcycles. How many ways can the shop choose 3 choppers, 5 bobbers, and 2 café racers for a weekend showcase? 306. How many arrangements can be made from the letters of the word “mountains” if all the vowels must form a string? A skateboard shop stocks 10 types of board decks, 3 317. types of trucks, and 4 types of wheels. How many different skateboards can be constructed? Just-For-Kicks Sneaker Company offers an online 318. customizing service. How many ways are there to design a custom pair of Just-For-Kicks sneakers if a customer can choose from a basic shoe up to 11 customizable options? A car wash offers the following optional services to 319. the basic wash: clear coat wax, triple foam polish, undercarriage wash, rust inhibitor, wheel brightener, air freshener, and interior shampoo. How many washes are possible if any number of options can be added to the basic wash? Susan bought 20 plants to arrange along the border of 320. her garden. How many distinct arrangements can she make if the plants are comprised of 6 tulips, 6 roses, and 8 daisies? How many unique ways can a string of Christmas 321. lights be arranged from 9 red, 10 green, 6 white, and 12 gold color bulbs? Real-World Applications A family consisting of 2 parents and 3 children is to 307. pose for a picture with 2 family members in
the front and 3 in the back. a. How many arrangements are possible with no restrictions? b. How many arrangements are possible if the parents must sit in the front? c. How many arrangements are possible if the parents must be next to each other? A cell phone company offers 6 different voice 308. packages and 8 different data packages. Of those, 3 packages include both voice and data. How many ways are there to choose either voice or data, but not both? In horse racing, a “trifecta” occurs when a bettor wins 309. by selecting the first three finishers in the exact order (1st place, 2nd place, and 3rd place). How many different trifectas are possible if there are 14 horses in a race? A wholesale T-shirt company offers sizes small, 310. medium, large, and extra-large in organic or non-organic cotton and colors white, black, gray, blue, and red. How many different T-shirts are there to choose from? 311. 1514 Chapter 13 Sequences, Probability, and Counting Theory 13.6 \| Binomial Theorem Learning Objectives In this section, you will: 13.6.1 Apply the Binomial Theorem. A polynomial with two terms is called a binomial. We have already learned to multiply binomials and to raise binomials to powers, but raising a binomial to a high power can be tedious and time-consuming. In this section, we will discuss a shortcut that will allow us to find (x + y) n without multiplying the binomial by itself n times. Identifying Binomial Coefficients In Counting Principles, we studied combinations. In the shortcut to finding (x + y) n to find the coefficients that will appear in the expansion of the binomial. In this case, we use the notation ⎛ r ⎝ , we will need to use combinations ⎞ ⎠ instead of n C(n, r), but it can be calculated in the same way. So n () ⎛ r ⎝ ⎞ ⎠ = C(n, r) = n ! r !(n − r) ! n The combination ⎛ r ⎝ ⎞ ⎠ is called a binomial coefficient. An example of a binomial coefficient is ⎛ 5 ⎝ 2 ⎞ ⎠ = C(5, 2) = 10. Binomial Coefficients If n and r are integers greater than or equal to 0 with n ≥ r, then the binomial coefficient is n ⎛ r ⎝ ⎞ ⎠ = C(n, r) = n ! r !(n − r) ! Is a binomial coefficient always a whole number? Yes. Just as the number of combinations must always be a whole number, a binomial coefficient will always be a whole number. Example 13.37 Finding Binomial Coefficients Find each binomial coefficient. a. b. c Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1515 Use the formula to calculate each binomial coefficient. You can also use the nCr function on your calculator. n ⎛ r ⎝ ⎞ ⎠ = C(n, r) = n ! r !(n − r) ! a. b. c !(5 − 32 ! = 10 9 ! 2 !(9 − 27 ! = 36 9 ! 7 !(9 − 72 ! = 36 Analysis Notice that we obtained the same result for parts (b) and (c). If you look closely at the solution for these two parts, you will see that you end up with the same two factorials in the denominator, but the order is reversed, just as with combinations. n () ⎛ ⎝ ⎞ ⎠ 13.50 Find each binomial coefficient. a. b. ⎞ ⎠ ⎛ 7 ⎝ 3 ⎛ 11 ⎝ 4 ⎞ ⎠ Using the Binomial Theorem n by multiplying, the result is called a binomial expansion, and it includes binomial coefficients. When we expand (x + y) If we wanted to expand (x + y)52, we might multiply (x + y) by itself fifty-two times. This could take hours! If we examine some simple binomial expansions, we can find patterns that will lead us to a shortcut for finding more complicated binomial expansions. (x + y)2 = x2 + 2xy + y2 (x + y)3 = x3 + 3x2 y + 3xy2 + y3 (x + y)4 = x4 + 4x3 y + 6x2 y2 + 4xy3 + y4 First, let’s examine the exponents. With each successive term, the exponent for x decreases and the exponent for y increases. The sum of the two exponents is n for each term. Next, let’s examine the coefficients. Notice that the coefficients increase and then decrease in a symmetrical pattern. The coefficients follow a pattern: n ⎛ ⎝ 0 ⎞ ⎠, n ⎛ ⎝ 1 ⎞ ⎠, n ⎛ ⎝ 2 ⎞ ⎠, ..., n ⎛ n ⎝ ⎞ ⎠. These patterns lead us to the Binomial Theorem, which can be used to expand any binomial. (x + y) n n = ∑ k = 0 = xn + n ⎛ ⎞ ⎠xn − kyk ⎝ k n ⎛ ⎝ 1 ⎞ ⎠xn − 1 y + n ⎛ ⎝ 2 ⎞ ⎠xn − 2 y2 + ... + n ⎛ ⎞ ⎠xyn − 1 + yn ⎝ n − 1 1516 Chapter 13 Sequences, Probability, and Counting Theory Another way to see the coefficients is to examine the expansion of a binomial in general form, x + y, to successive powers 1, 2, 3, and 4. (x + y)1 = x + y (x + y)2 = x2 + 2xy + y2 (x + y)3 = x3 + 3x2 y + 3xy2 + y3 (x + y)4 = x4 + 4x3 y + 6x2 y2 + 4xy3 + y4 Can you guess the next expansion for the binomial (x + y)5 ? Figure 13.20 See Figure 13.20, which illustrates the following: • There are n + 1 terms in the expansion of (x + y) n . • The degree (or sum of the exponents) for each term is n. • The powers on x begin with n and decrease to 0. • The powers on y begin with 0 and increase to n. • The coefficients are symmetric. To determine the expansion on (x + y)5, we see n = 5, degree of 5. In descending order for powers of x, the pattern is as follows: thus, there will be 5+1 = 6 terms. Each term has a combined • • Introduce x5, and then for each successive term reduce the exponent on x by 1 until x0 = 1 is reached. Introduce y0 = 1, and then increase the exponent on y by 1 until y5 is reached. The next expansion would be (x + y)5 = x5 + 5x4 y + 10x3 y2 + 10x2 y3 + 5xy4 + y5. x5, x4 y, x3 y2, x2 y3, xy4, y5 But where do those coefficients come from? The binomial coefficients are symmetric. We can see these coefficients in an array known as Pascal's Triangle, shown in Figure 13.21. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1517 Figure 13.21 To generate Pascal’s Triangle, we start by writing a 1. In the row below, row 2, we write two 1’s. In the 3rd row, flank the ends of the rows with 1’s, and add 1 + 1 to find the middle number, 2. In the nth row, flank the ends of the row with 1’s. Each element in the triangle is the sum of the two elements immediately above it. To see the connection between Pascal’s Triangle and binomial coefficients, let us revisit the expansion of the binomials in general form. The Binomial Theorem The Binomial Theorem is a formula that can be used to expand any binomial. (x + y) n n = ∑ k = 0 = xn + n ⎛ ⎞ ⎠xn − k yk ⎝ k n ⎛ ⎝ 1 ⎞ ⎠xn − 1 y + n ⎛ ⎝ 2 ⎞ ⎠xn − 2 y2 + ... + n ⎛ ⎞ ⎠xyn − 1 + yn ⎝ n − 1 (13.12) Given a binomial, write it in expanded form. 1. Determine the value of n according to the exponent. 2. Evaluate the k = 0 through k = n using the Binomial Theorem formula. 3. Simplify. Example 13.38 Expanding a Binomial Write in expanded form. a. (x + y)5 1518 Chapter 13 Sequences, Probability, and Counting Theory b. ⎛ ⎝3x − y⎞ ⎠ 4 Solution a. Substitute n = 5 into the formula. Evaluate the k = 0 through k = 5 terms. Simplify. (x + y)5 = ⎞ ⎞ ⎠x2 y3 + ⎠x4 y1 + (x + y)5 = x5 + 5x4 y + 10x3 y2 + 10x2 y3 + 5xy4 + y5 ⎞ ⎠x5 y0 + ⎞ ⎠x3 y2 + ⎛ x1 y4 + ⎛ 5 ⎝ 4 ⎞ ⎠x0 y5 ⎛ 5 ⎝ 5 b. Substitute n = 4 into the formula. Evaluate the k = 0 through k = 4 terms. Notice that 3x is in the place that was occupied by x and that – y is in the place that was occupied by y. So we substitute them. Simplify. (3x − y)4 = ⎛ 4 ⎝ 2 (3x − y)4 = 81x4 − 108x3 y + 54x2 y2 − 12xy3 + y4 ⎞ ⎠(3x)3 ( − y)1 + ⎞ ⎠(3x)4 ( − y) ⎞ ⎠(3x)2 ( − y)2 + ⎞ ⎠(3x)1 ( − y)3 + ⎛ 4 ⎝ 3 ⎞ ⎠(3x)0 ( − y)4 ⎛ 4 ⎝ 4 Analysis Notice the alternating signs in part b. This happens because ( − y) raised to odd powers is negative, but ( − y) raised to even powers is positive. This will occur whenever the binomial contains a subtraction sign. 13.51 Write in expanded form. a. b. (x − y)5 (2x + 5y)3 Using the Binomial Theorem to Find a Single Term Expanding a binomial with a high exponent such as (x + 2y)16 can be a lengthy process. Sometimes we are interested only in a certain term of a binomial expansion. We do not need to fully expand a binomial to find a single specific term. Note the pattern of coefficients in the expansion of (x + y)5. (x + y)5 = x5 + ⎞ ⎠x4 x3 y2 + ⎞ ⎠x2 y3 + ⎛ 5 ⎝ 3 ⎛ 5 ⎝ 4 ⎞ ⎠xy4 + y5 The second term is ⎛ 5 ⎝ 1 ⎠x4 y. The third term is ⎛ ⎞ 5 ⎝ 2 ⎞ ⎠x3 y2. We can generalize this result. n ⎛ r ⎝ ⎞ ⎠xn − r yr The (r+1)th Term of a Binomial Expansion The (r + 1)th term of the binomial expansion of (x + y) n is: n ⎛ r ⎝ ⎞ ⎠xn − r yr (13.13) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1519 Given a binomial, write a specific term without fully expanding. 1. Determine the value of n according to the exponent. 2. Determine (r + 1). 3. Determine r. 4. Replace r in the formula for the (r + 1)th term of the binomial expansion. Example 13.39 Writing a Given Term of a Binomial Expansion Find the tenth term of (x + 2y)16 without fully expanding the binomial. Solution Because we are looking for the tenth term, r + 1 = 10, we will use r = 9 in our calculations. n ⎛ r ⎝ ⎞ ⎠xn − r yr ⎛ 16 ⎝ 9 ⎞ ⎠x16 − 9 (2y)9 = 5,857,280x7 y9 13.52 Find the sixth term of (3x − y)9 without fully expanding the binomial. Access these online resources for additional instruction and practice with binomial expansion. • The Binomial Theorem (http://openstaxcollege.org/l/binomialtheorem) • Binomial Theorem Example (http://openstaxcollege.org/l/btexample) 1520 Chapter 13 Sequences, Probability, and Counting Theory 13.6 EXERCISES Verbal 322. What calculated? is a binomial coefficient, and how it is What role do binomial coefficients play in a binomial 323. expansion? Are they restricted to any type of number? 324. What is the Binomial Theorem and what is its use? When is it an advantage to use the Binomial 325. Theorem? Explain. Algebraic the following exercises, evaluate the binomial For coefficient. (3x − 2y)4 340. (4x − 3y)5 341. 5 ⎛ ⎝ x + 3y⎞ 1 ⎠ 342. (x−1 + 2y−1)4 343. ( x − y)5 For the following exercises, use the Binomial Theorem to write the first three terms of each binomial. 326. 327. 328. 329. 330. 331. 332 ⎛ 10 ⎝ 9 ⎞ ⎠ ⎛ 25 ⎝ 11 ⎞ ⎠ ⎛ 17 ⎝ 6 ⎞ ⎠
333. ⎛ 200 ⎝ 199 ⎞ ⎠ For the following exercises, use the Binomial Theorem to expand each binomial. 334. (4a − b)3 335. (5a + 2)3 336. (3a + 2b)3 337. (2x + 3y)4 338. (4x + 2y)5 339. This content is available for free at https://cnx.org/content/col11758/1.5 344. (a + b)17 345. (x − 1)18 346. (a − 2b)15 347. (x − 2y)8 348. (3a + b)20 349. (2a + 4b)7 350. (x3 − y)8 For the following exercises, find the indicated term of each binomial without fully expanding the binomial. 351. 352. 353. 354. 355. The fourth term of (2x − 3y)4 The fourth term of (3x − 2y)5 The third term of (6x − 3y)7 The eighth term of (7 + 5y)14 The seventh term of (a + b)11 356. The fifth term of (x − y)7 The tenth term of (x − 1)12 The ninth term of (a − 3b2)11 357. 358. 359. Chapter 13 Sequences, Probability, and Counting Theory 1521 • • • ( a + 4 a − 5)8 (x3 + 2y2 − z)5 (3x2 − 2y3) 12 The fourth term of ⎛ ⎝x3 − 1 2 10 ⎞ ⎠ 360. The eighth term of ⎛ ⎝ 9 + 2 x ⎞ ⎠ y 2 Graphical For the following exercises, use the Binomial Theorem f (x) = (x + 3)4. Then find and to expand the binomial graph each indicated sum on one set of axes. 361. Find and graph f1(x), such that f1(x) is the first term of the expansion. 362. Find and graph f2(x), such that f2(x) is the sum of the first two terms of the expansion. 363. Find and graph f3(x), such that f3(x) is the sum of the first three terms of the expansion. 364. Find and graph f4(x), such that f4(x) is the sum of the first four terms of the expansion. 365. Find and graph f5(x), such that f5(x) is the sum of the first five terms of the expansion. Extensions n , each term has the 366. In the expansion of (5x + 3y) n form ⎛ ⎝ k ⎞ ⎠an – k bk n value 0, 1, 2, ..., n. If ⎛ ⎝ k , where k successively takes on the ⎞ ⎠ = ⎞ ⎠, what ⎛ 7 ⎝ 2 is the corresponding term? 367. In the expansion of (a + b) , the coefficient of an − k bk is the same as the coefficient of which other term? n 368. Consider the expansion of (x + b)40. What is the exponent of b in the kth term? 369. n ⎛ ⎝ k n ⎞ ⎞ Find ⎛ ⎠ and write ⎠ + ⎝ k − 1 n ⎞ binomial coefficient in the form ⎛ ⎠. Prove it. Hint: Use ⎝ k integer p, such that answer fact any the as a that, the for p ≥ 1, p ! = p(p − 1) !. Which expression cannot be expanded using the 370. Binomial Theorem? Explain. • (x2 − 2x + 1) 1522 Chapter 13 Sequences, Probability, and Counting Theory 13.7 \| Probability Learning Objectives In this section, you will: 13.7.1 Construct probability models. 13.7.2 Compute probabilities of equally likely outcomes. 13.7.3 Compute probabilities of the union of two events. 13.7.4 Use the complement rule to find probabilities. 13.7.5 Compute probability using counting theory. Figure 13.22 An example of a “spaghetti model,” which can be used to predict possible paths of a tropical storm.[1] Residents of the Southeastern United States are all too familiar with charts, known as spaghetti models, such as the one in Figure 13.22. They combine a collection of weather data to predict the most likely path of a hurricane. Each colored line represents one possible path. The group of squiggly lines can begin to resemble strands of spaghetti, hence the name. In this section, we will investigate methods for making these types of predictions. Constructing Probability Models Suppose we roll a six-sided number cube. Rolling a number cube is an example of an experiment, or an activity with an observable result. The numbers on the cube are possible results, or outcomes, of this experiment. The set of all possible outcomes of an experiment is called the sample space of the experiment. The sample space for this experiment is {1, 2, 3, 4, 5, 6}. An event is any subset of a sample space. The likelihood of an event is known as probability. The probability of an event p is a number that always satisfies 0 ≤ p ≤ 1, where 0 indicates an impossible event and 1 indicates a certain event. A probability model is a mathematical 1. The figure is for illustrative purposes only and does not model any particular storm. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1523 description of an experiment listing all possible outcomes and their associated probabilities. For instance, if there is a 1% chance of winning a raffle and a 99% chance of losing the raffle, a probability model would look much like Table 13.3. Outcome Probability Winning the raffle 1% Losing the raffle 99% Table 13.3 The sum of the probabilities listed in a probability model must equal 1, or 100%. Given a probability event where each event is equally likely, construct a probability model. 1. Identify every outcome. 2. Determine the total number of possible outcomes. 3. Compare each outcome to the total number of possible outcomes. Example 13.40 Constructing a Probability Model Construct a probability model for rolling a single, fair die, with the event being the number shown on the die. Solution Begin by making a list of all possible outcomes for the experiment. The possible outcomes are the numbers that can be rolled: 1, 2, 3, 4, 5, and 6. There are six possible outcomes that make up the sample space. Assign probabilities to each outcome in the sample space by determining a ratio of the outcome to the number of possible outcomes. There is one of each of the six numbers on the cube, and there is no reason to think that any particular face is more likely to show up than any other one, so the probability of rolling any number is 1 6 . Outcome Roll of 1 Roll of 2 Roll of 3 Roll of 4 Roll of 5 Roll of 6 Probability Table 13.4 Do probabilities always have to be expressed as fractions? No. Probabilities can be expressed as fractions, decimals, or percents. Probability must always be a number between 0 and 1, inclusive of 0 and 1. 13.53 Construct a probability model for tossing a fair coin. 1524 Chapter 13 Sequences, Probability, and Counting Theory Computing Probabilities of Equally Likely Outcomes Let S be a sample space for an experiment. When investigating probability, an event is any subset of S. When the outcomes of an experiment are all equally likely, we can find the probability of an event by dividing the number of outcomes in the event by the total number of outcomes in S. Suppose a number cube is rolled, and we are interested in finding the probability of the event “rolling a number less than or equal to 4.” There are 4 possible outcomes in the event and 6 possible outcomes in S, so the probability of the event is 4 6 = 2 3 . Computing the Probability of an Event with Equally Likely Outcomes The probability of an event E in an experiment with sample space S with equally likely outcomes is given by P(E) = number of elements in E number of elements in S = n(E) n(S) (13.14) E is a subset of S, so it is always true that 0 ≤ P(E) ≤ 1. Example 13.41 Computing the Probability of an Event with Equally Likely Outcomes A number cube is rolled. Find the probability of rolling an odd number. Solution The event “rolling an odd number” contains three outcomes. There are 6 equally likely outcomes in the sample space. Divide to find the probability of the event. P(E) = 3 6 = 1 2 13.54 A number cube is rolled. Find the probability of rolling a number greater than 2. Computing the Probability of the Union of Two Events We are often interested in finding the probability that one of multiple events occurs. Suppose we are playing a card game, and we will win if the next card drawn is either a heart or a king. We would be interested in finding the probability of the next card being a heart or a king. The union of two events E and F, written E ∪ F, is the event that occurs if either or both events occur. Suppose the spinner in Figure 13.23 is spun. We want to find the probability of spinning orange or spinning a b. P(E ∪ F) = P(E) + P(F) − P(E ∩ F) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1525 Figure 13.23 There are a total of 6 sections, and 3 of them are orange. So the probability of spinning orange is 3 6 of 6 sections, and 2 of them have a b. So the probability of spinning a b is 2 6 = 1 3 = 1 2 . If we added these two probabilities, . There are a total we would be counting the sector that is both orange and a b twice. To find the probability of spinning an orange or a b, we need to subtract the probability that the sector is both orange and has a b The probability of spinning orange or a b is 2 3 . Probability of the Union of Two Events The probability of the union of two events E and F (written E ∪ F ) equals the sum of the probability of E and the probability of F minus the probability of E and F occurring together ( which is called the intersection of E and F and is written as E ∩ F ). P(E ∪ F) = P(E) + P(F) − P(E ∩ F) (13.15) Example 13.42 Computing the Probability of the Union of Two Events A card is drawn from a standard deck. Find the probability of drawing a heart or a 7. Solution A standard deck contains an equal number of hearts, diamonds, clubs, and spades. So the probability of drawing a heart is 1 . There are four 7s in a standard deck, and there are a total of 52 cards. So the probability of drawing 4 a 7 is 1 13 . The only card in the deck that is both a heart and a 7 is the 7 of hearts, so the probability of drawing both a heart and a 7 is 1 52 and P(H ∩ 7) = 1 52 . Substitute P(H) = 1 4 P(7) = 1 13 into the formula. , , 1526 Chapter 13 Sequences, Probability, and Counting Theory P(E ∪ F) = P(E) + P(F) − P(E ∩ F) − 1 52 + 1 13 = 1 4 = 4 13 The probability of drawing a heart or a 7 is 4 13 . 13.55 A card is drawn from a standard deck. Find the probability of drawing a red card or an ace. Computing the Probability of Mutually Exclusive Events Suppose the spinner in Figure 13.23 is spun again, but this time we are interested in the probability of spinning an orange or a d. There are no sectors that are both orange and contain a d, so these two events have no outco
mes in common. Events are said to be mutually exclusive events when they have no outcomes in common. Because there is no overlap, there is nothing to subtract, so the general formula is P(E ∪ F) = P(E) + P(F) Notice that with mutually exclusive events, the intersection of E and F is the empty set. The probability of spinning an orange is 3 6 and the probability of spinning a d is 1 6 . We can find the probability of spinning an orange or a d simply = 1 2 by adding the two probabilities. P(E ∪ F) = P(E) + P(F The probability of spinning an orange or a d is 2 3 . Probability of the Union of Mutually Exclusive Events The probability of the union of two mutually exclusive events E and F is given by P(E ∪ F) = P(E) + P(F) (13.16) Given a set of events, compute the probability of the union of mutually exclusive events. 1. Determine the total number of outcomes for the first event. 2. Find the probability of the first event. 3. Determine the total number of outcomes for the second event. 4. Find the probability of the second event. 5. Add the probabilities. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1527 Example 13.43 Computing the Probability of the Union of Mutually Exclusive Events A card is drawn from a standard deck. Find the probability of drawing a heart or a spade. Solution The events “drawing a heart” and “drawing a spade” are mutually exclusive because they cannot occur at the same time. The probability of drawing a heart is 1 , so the 4 , and the probability of drawing a spade is also 1 4 probability of drawing a heart or a spade is 1 4 + 1 4 = 1 2 13.56 A card is drawn from a standard deck. Find the probability of drawing an ace or a king. Using the Complement Rule to Compute Probabilities We have discussed how to calculate the probability that an event will happen. Sometimes, we are interested in finding the probability that an event will not happen. The complement of an event E, denoted E′, is the set of outcomes in the sample space that are not in E. For example, suppose we are interested in the probability that a horse will lose a race. If event W is the horse winning the race, then the complement of event W is the horse losing the race. To find the probability that the horse loses the race, we need to use the fact that the sum of all probabilities in a probability model must be 1. P(E′) = 1 − P(E) The probability of the horse winning added to the probability of the horse losing must be equal to 1. Therefore, if the probability of the horse winning the race is 1 9 , the probability of the horse losing the race is simply 1 − 1 9 = 8 9 The Complement Rule The probability that the complement of an event will occur is given by P(E′) = 1 − P(E) (13.17) Example 13.44 Using the Complement Rule to Calculate Probabilities Two six-sided number cubes are rolled. a. Find the probability that the sum of the numbers rolled is less than or equal to 3. b. Find the probability that the sum of the numbers rolled is greater than 3. 1528 Chapter 13 Sequences, Probability, and Counting Theory Solution The first step is to identify the sample space, which consists of all the possible outcomes. There are two number cubes, and each number cube has six possible outcomes. Using the Multiplication Principle, we find that there are 6×6, or 36 total possible outcomes. So, for example, 1-1 represents a 1 rolled on each number cube. 1-1 1-2 1-3 1-4 1-5 1-6 2-1 2-2 2-3 2-4 2-5 2-6 3-1 3-2 3-3 3-4 3-5 3-6 4-1 4-2 4-3 4-4 4-5 4-6 5-1 5-2 5-3 5-4 5-5 5-6 6-1 6-2 6-3 6-4 6-5 6-6 Table 13.5 a. We need to count the number of ways to roll a sum of 3 or less. These would include the following outcomes: 1-1, 1-2, and 2-1. So there are only three ways to roll a sum of 3 or less. The probability is 3 36 b. Rather than listing all the possibilities, we can use the Complement Rule. Because we have already found the probability of the complement of this event, we can simply subtract that probability from 1 to find the probability that the sum of the numbers rolled is greater than 3. = 1 12 P(E′) = 1 − P(E) = 1 − 1 12 = 11 12 Two number cubes are rolled. Use the Complement Rule to find the probability that the sum is less than 13.57 10. Computing Probability Using Counting Theory Many interesting probability problems involve counting principles, permutations, and combinations. In these problems, we will use permutations and combinations to find the number of elements in events and sample spaces. These problems can be complicated, but they can be made easier by breaking them down into smaller counting problems. Assume, for example, that a store has 8 cellular phones and that 3 of those are defective. We might want to find the probability that a couple purchasing 2 phones receives 2 phones that are not defective. To solve this problem, we need to calculate all of the ways to select 2 phones that are not defective as well as all of the ways to select 2 phones. There are 5 phones that are not defective, so there are C(5, 2) ways to select 2 phones that are not defective. There are 8 phones, so there are C(8, 2) ways to select 2 phones. The probability of selecting 2 phones that are not defective is: This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1529 ways to select 2 phones that are not defective ways to select 2 phones = C(5, 2) C(8, 2) = 10 28 = 5 14 Example 13.45 Computing Probability Using Counting Theory A child randomly selects 5 toys from a bin containing 3 bunnies, 5 dogs, and 6 bears. a. Find the probability that only bears are chosen. b. Find the probability that 2 bears and 3 dogs are chosen. c. Find the probability that at least 2 dogs are chosen. Solution a. We need to count the number of ways to choose only bears and the total number of possible ways to select 5 toys. There are 6 bears, so there are C(6, 5) ways to choose 5 bears. There are 14 toys, so there are C(14, 5) ways to choose any 5 toys. C(6,5) C(14,5) = 6 2,002 = 3 1,001 b. We need to count the number of ways to choose 2 bears and 3 dogs and the total number of possible ways to select 5 toys. There are 6 bears, so there are C(6, 2) ways to choose 2 bears. There are 5 dogs, so there are C(5, 3) ways to choose 3 dogs. Since we are choosing both bears and dogs at the same time, we will use the Multiplication Principle. There are C(6, 2) ⋅ C(5, 3) ways to choose 2 bears and 3 dogs. We can use this result to find the probability. C(6,2)C(5,3) C(14,5) = 15 ⋅ 10 2,002 = 75 1,001 c. It is often easiest to solve “at least” problems using the Complement Rule. We will begin by finding the probability that fewer than 2 dogs are chosen. If less than 2 dogs are chosen, then either no dogs could be chosen, or 1 dog could be chosen. When no dogs are chosen, all 5 toys come from the 9 toys that are not dogs. There are C(9, 5) ways to choose toys from the 9 toys that are not dogs. Since there are 14 toys, there are C(14, 5) ways to choose the 5 toys from all of the toys. C(9,5) C(14,5) = 63 1,001 If there is 1 dog chosen, then 4 toys must come from the 9 toys that are not dogs, and 1 must come from the 5 dogs. Since we are choosing both dogs and other toys at the same time, we will use the Multiplication Principle. There are C(5, 1) ⋅ C(9, 4) ways to choose 1 dog and 1 other toy. C(5,1)C(9,4) C(14,5) = 5 ⋅ 126 2,002 = 315 1,001 Because these events would not occur together and are therefore mutually exclusive, we add the probabilities to find the probability that fewer than 2 dogs are chosen. We then subtract that probability from 1 to find the probability that at least 2 dogs are chosen. 63 1,001 + 315 1,001 = 378 1,001 1530 Chapter 13 Sequences, Probability, and Counting Theory 1 − 378 1,001 = 623 1,001 A child randomly selects 3 gumballs from a container holding 4 purple gumballs, 8 yellow gumballs, 13.58 and 2 green gumballs. a. Find the probability that all 3 gumballs selected are purple. b. Find the probability that no yellow gumballs are selected. c. Find the probability that at least 1 yellow gumball is selected. Access these online resources for additional instruction and practice with probability. • Introduction to Probability (http://openstaxcollege.org/l/introprob) • Determining Probability (http://openstaxcollege.org/l/determineprob) this website (http://openstaxcollege.org/l/PreCalcLPC11) Visit Learningpod. for additional practice questions from This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1531 13.7 EXERCISES Verbal 385. Find the probability of tossing two heads. What term is used to express the likelihood of an 371. event occurring? Are there restrictions on its values? If so, what are they? If not, explain. 386. Find the probability of tossing exactly one tail. 387. Find the probability of tossing at least one tail. 372. What is a sample space? 373. What is an experiment? For the following exercises, four coins are tossed. 388. What is the sample space? What is the difference between events and outcomes? 374. Give an example of both using the sample space of tossing a coin 50 times. 389. Find the probability of tossing exactly two heads. 390. Find the probability of tossing exactly three heads. The union of two sets is defined as a set of elements 375. that are present in at least one of the sets. How is this similar to the definition used for the union of two events from a probability model? How is it different? Numeric For the following exercises, use the spinner shown in Figure 13.24 to find the probabilities indicated. 391. Find the probability of tossing four heads or four tails. 392. Find the probability of tossing all tails. 393. Find the probability of tossing not all tails. Find the probability of tossing exactly two heads or at 394. least two tails. Find the probability of tossing either two heads or 395. three heads.
For the following exercises, one card is drawn from a standard deck of 52 cards. Find the probability of drawing the following: Figure 13.24 376. Landing on red 377. Landing on a vowel 378. Not landing on blue 379. Landing on purple or a vowel 380. Landing on blue or a vowel 381. Landing on green or blue 382. Landing on yellow or a consonant 383. Not landing on yellow or a consonant For the following exercises, two coins are tossed. 384. What is the sample space? 396. A club 397. A two 398. Six or seven 399. Red six 400. An ace or a diamond 401. A non-ace 402. A heart or a non-jack For the following exercises, two dice are rolled, and the results are summed. Construct a table showing the sample space of 403. outcomes and sums. 404. Find the probability of rolling a sum of 3. Find the probability of rolling at least one four or a 405. sum of 8. Find the probability of rolling an odd sum less than 406. 9. Find the probability of rolling a sum greater than or 407. equal to 15. 1532 Chapter 13 Sequences, Probability, and Counting Theory 408. Find the probability of rolling a sum less than 15. Find the probability of rolling a sum less than 6 or 409. greater than 9. 425. How much less is a player’s chance of selecting 3 winning numbers than the chance of selecting either 4 or 5 winning numbers? Real-World Applications Find the probability of rolling a sum between 6 and 410. 9, inclusive. Use this data for the exercises that follow: In 2013, there were roughly 317 million citizens in the United States, and about 40 million were elderly (aged 65 and over).[2] 426. If you meet a U.S. citizen, what is the percent chance that the person is elderly? (Round to the nearest tenth of a percent.) 427. If you meet five U.S. citizens, what is the percent chance that exactly one is elderly? (Round to the nearest tenth of a percent.) 428. If you meet five U.S. citizens, what is the percent chance that three are elderly? (Round to the nearest tenth of a percent.) If you meet five U.S. citizens, what is the percent four are elderly? (Round to the nearest 429. chance that thousandth of a percent.) It is predicted that by 2030, one in five U.S. citizens 430. will be elderly. How much greater will the chances of meeting an elderly person be at that time? What policy changes do you foresee if these statistics hold true? 411. Find the probability of rolling a sum of 5 or 6. Find the probability of rolling any sum other than 5 412. or 6. For the following exercises, a coin is tossed, and a card is pulled from a standard deck. Find the probability of the following: 413. A head on the coin or a club 414. A tail on the coin or red ace 415. A head on the coin or a face card 416. No aces For the following exercises, use this scenario: a bag of M&Ms contains 12 blue, 6 brown, 10 orange, 8 yellow, 8 red, and 4 green M&Ms. Reaching into the bag, a person grabs 5 M&Ms. 417. What is the probability of getting all blue M&Ms? 418. What is the probability of getting 4 blue M&Ms? 419. What is the probability of getting 3 blue M&Ms? 420. What is the probability of getting no brown M&Ms? Extensions Use the following scenario for the exercises that follow: In the game of Keno, a player starts by selecting 20 numbers from the numbers 1 to 80. After the player makes his selections, 20 winning numbers are randomly selected from numbers 1 to 80. A win occurs if the player has correctly selected 3, 4, or 5 of the 20 winning numbers. (Round all answers to the nearest hundredth of a percent.) What 421. exactly 3 winning numbers? is the percent chance that a player selects What 422. exactly 4 winning numbers? is the percent chance that a player selects What is the percent chance that a player selects all 5 423. winning numbers? 424. What is the percent chance of winning? 2. United States Census Bureau. http://www.census.gov This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1533 CHAPTER 13 REVIEW KEY TERMS Addition Principle if one event can occur in m ways and a second event with no common outcomes can occur in n ways, then the first or second event can occur in m + n ways annuity an investment in which the purchaser makes a sequence of periodic, equal payments arithmetic sequence a sequence in which the difference between any two consecutive terms is a constant arithmetic series the sum of the terms in an arithmetic sequence binomial coefficient the number of ways to choose r objects from n objects where order does not matter; equivalent to n C(n, r), denoted ⎛ r ⎝ ⎞ ⎠ binomial expansion the result of expanding (x + y) n by multiplying Binomial Theorem a formula that can be used to expand any binomial combination a selection of objects in which order does not matter common difference the difference between any two consecutive terms in an arithmetic sequence common ratio the ratio between any two consecutive terms in a geometric sequence complement of an event the set of outcomes in the sample space that are not in the event E diverge a series is said to diverge if the sum is not a real number event any subset of a sample space experiment an activity with an observable result explicit formula a formula that defines each term of a sequence in terms of its position in the sequence finite sequence a function whose domain consists of a finite subset of the positive integers {1, 2, … n} for some positive integer n Fundamental Counting Principle if one event can occur in m ways and a second event can occur in n ways after the first event has occurred, then the two events can occur in m×n ways; also known as the Multiplication Principle geometric sequence a sequence in which the ratio of a term to a previous term is a constant geometric series the sum of the terms in a geometric sequence index of summation in summation notation, the variable used in the explicit formula for the terms of a series and written below the sigma with the lower limit of summation infinite sequence a function whose domain is the set of positive integers infinite series the sum of the terms in an infinite sequence lower limit of summation the number used in the explicit formula to find the first term in a series Multiplication Principle if one event can occur in m ways and a second event can occur in n ways after the first event has occurred, then the two events can occur in m×n ways; also known as the Fundamental Counting Principle mutually exclusive events events that have no outcomes in common n factorial the product of all the positive integers from 1 to n nth partial sum the sum of the first n terms of a sequence nth term of a sequence a formula for the general term of a sequence outcomes the possible results of an experiment 1534 Chapter 13 Sequences, Probability, and Counting Theory permutation a selection of objects in which order matters probability a number from 0 to 1 indicating the likelihood of an event probability model a mathematical description of an experiment listing all possible outcomes and their associated probabilities recursive formula a formula that defines each term of a sequence using previous term(s) sample space the set of all possible outcomes of an experiment sequence a function whose domain is a subset of the positive integers series the sum of the terms in a sequence summation notation a notation for series using the Greek letter sigma; it includes an explicit formula and specifies the first and last terms in the series term a number in a sequence union of two events the event that occurs if either or both events occur upper limit of summation the number used in the explicit formula to find the last term in a series KEY EQUATIONS Formula for a factorial (n − 1)(n − 2) ⋯ (2)(1), for n ≥ 2 recursive formula for nth term of an arithmetic sequence an = an − 1 + dn ≥ 2 explicit formula for nth term of an arithmetic sequence an = a1 + d(n − 1) recursive formula for nth term of a geometric sequence an = ran − 1, n ≥ 2 explicit formula for nth term of a geometric sequence an = a1 r n − 1 sum of the first n terms of an arithmetic series sum of the first n terms of a geometric series sum of an infinite geometric series with – 1 < r < 1 Sn = n(a1 + an) 2 Sn = a1( Sn = a1 1 − r ⋅ r ≠ 1 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1535 number of permutations of n distinct objects taken r at a time number of combinations of n distinct objects taken r at a time number of permutations of n non-distinct objects P(n, r) = n ! (n − r) ! C(n, r) = n ! r !(n − r) ! n ! r1 !r2 ! … rk ! Binomial Theorem (x + yxn − k yk (r + 1)th term of a binomial expansion n ⎛ r ⎝ ⎞ ⎠xn − r yr probability of an event with equally likely outcomes P(E) = n(E) n(S) probability of the union of two events P(E ∪ F) = P(E) + P(F) − P(E ∩ F) probability of the union of mutually exclusive events P(E ∪ F) = P(E) + P(F) probability of the complement of an event P(E ') = 1 − P(E) KEY CONCEPTS 13.1 Sequences and Their Notations • A sequence is a list of numbers, called terms, written in a specific order. • Explicit formulas define each term of a sequence using the position of the term. See Example 13.1, Example 13.2, and Example 13.3. • An explicit formula for the nth term of a sequence can be written by analyzing the pattern of several terms. See Example 13.4. • Recursive formulas define each term of a sequence using previous terms. • Recursive formulas must state the initial term, or terms, of a sequence. • A set of terms can be written by using a recursive formula. See Example 13.5 and Example 13.6. • A factorial is a mathematical operation that can be defined recursively. • The factorial of n is the product of all integers from 1 to n See Example 13.7. 13.2 Arithmetic Sequences • An arithmetic sequence is a sequence where the difference between any two consecutive terms is a constant. • The constant between two consecutive terms is called the com
mon difference. 1536 Chapter 13 Sequences, Probability, and Counting Theory • The common difference is the number added to any one term of an arithmetic sequence that generates the subsequent term. See Example 13.8. • The terms of an arithmetic sequence can be found by beginning with the initial term and adding the common difference repeatedly. See Example 13.9 and Example 13.10. • A recursive formula for an arithmetic sequence with common difference d is given by an = an − 1 + d, n ≥ 2. See Example 13.11. • As with any recursive formula, the initial term of the sequence must be given. • An explicit formula for an arithmetic sequence with common difference d is given by an = a1 + d(n − 1). See Example 13.12. • An explicit formula can be used to find the number of terms in a sequence. See Example 13.13. • In application problems, we sometimes alter the explicit formula slightly to an = a0 + dn. See Example 13.14. 13.3 Geometric Sequences • A geometric sequence is a sequence in which the ratio between any two consecutive terms is a constant. • The constant ratio between two consecutive terms is called the common ratio. • The common ratio can be found by dividing any term in the sequence by the previous term. See Example 13.15. • The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. See Example 13.16 and Example 13.18. • A recursive formula for a geometric sequence with common ratio r is given by an = ran – 1 for n ≥ 2 . • As with any recursive formula, the initial term of the sequence must be given. See Example 13.17. • An explicit formula for a geometric sequence with common ratio r is given by an = a1 r n – 1. See Example 13.19. In application problems, we sometimes alter the explicit formula slightly to an = a0 r n • . See Example 13.20. 13.4 Series and Their Notations • The sum of the terms in a sequence is called a series. • A common notation for series is called summation notation, which uses the Greek letter sigma to represent the sum. See Example 13.21. • The sum of the terms in an arithmetic sequence is called an arithmetic series. • The sum of the first n terms of an arithmetic series can be found using a formula. See Example 13.22 and Example 13.23. • The sum of the terms in a geometric sequence is called a geometric series. • The sum of the first n terms of a geometric series can be found using a formula. See Example 13.24 and Example 13.25. • The sum of an infinite series exists if the series is geometric with –1 < r < 1. • If the sum of an infinite series exists, it can be found using a formula. See Example 13.26, Example 13.27, and Example 13.28. • An annuity is an account into which the investor makes a series of regularly scheduled payments. The value of an annuity can be found using geometric series. See Example 13.29. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1537 13.5 Counting Principles • • If one event can occur in m ways and a second event with no common outcomes can occur in n ways, then the first or second event can occur in m + n ways. See Example 13.30. If one event can occur in m ways and a second event can occur in n ways after the first event has occurred, then the two events can occur in m×n ways. See Example 13.31. • A permutation is an ordering of n objects. • If we have a set of n objects and we want to choose r objects from the set in order, we write P(n, r). • Permutation problems can be solved using the Multiplication Principle or the formula for P(n, r). See Example 13.32 and Example 13.33. • A selection of objects where the order does not matter is a combination. • Given n distinct objects, the number of ways to select r objects from the set is C(n, r) and can be found using a formula. See Example 13.34. • A set containing n distinct objects has 2 n subsets. See Example 13.35. • For counting problems involving non-distinct objects, we need to divide to avoid counting duplicate permutations. See Example 13.36. 13.6 Binomial Theorem • n ⎛ r ⎝ ⎞ ⎠ is called a binomial coefficient and is equal to C(n, r). See Example 13.37. • The Binomial Theorem allows us to expand binomials without multiplying. See Example 13.38. • We can find a given term of a binomial expansion without fully expanding the binomial. See Example 13.39. 13.7 Probability • Probability is always a number between 0 and 1, where 0 means an event is impossible and 1 means an event is certain. • The probabilities in a probability model must sum to 1. See Example 13.40. • When the outcomes of an experiment are all equally likely, we can find the probability of an event by dividing the number of outcomes in the event by the total number of outcomes in the sample space for the experiment. See Example 13.41. • To find the probability of the union of two events, we add the probabilities of the two events and subtract the probability that both events occur simultaneously. See Example 13.42. • To find the probability of the union of two mutually exclusive events, we add the probabilities of each of the events. See Example 13.43. • The probability of the complement of an event is the difference between 1 and the probability that the event occurs. See Example 13.44. • In some probability problems, we need to use permutations and combinations to find the number of elements in events and sample spaces. See Example 13.45. CHAPTER 13 REVIEW EXERCISES Sequences and Their Notation 431. Write the first four terms of the sequence defined by the recursive formula a1 = 2, an = an − 1 + n. 432. Evaluate 6 ! (5 − 3) !3 ! . 433. Write the first four terms of the sequence defined by n the explicit formula an = 10 + 3. 1538 Chapter 13 Sequences, Probability, and Counting Theory 434. Write the first four terms of the sequence defined by n ! the explicit formula an = n(n + 1) . Arithmetic Sequences 435. Is the sequence 4 7 , 47 21 find the common difference. , 82 21 , 39 7 , ... arithmetic? If so, 448. Write a recursive formula for the geometric sequence 1, 1 3 , 1 27 , 1 9 , … 449. Write an explicit formula for the geometric sequence − 1 5 , − 1 135 , − 1 15 , − 1 45 , … 436. Is the sequence 2, 4, 8, 16, ... arithmetic? If so, find the common difference. −5 59,049 ? 450. How many terms are in the finite geometric sequence 437. An arithmetic sequence has the first term a1 = 18 and common difference d = − 8. What are the first five terms? Series and Their Notation 451. Use summation notation to write the sum of terms 1 2 m + 5 from m = 0 to m = 5. 438. An arithmetic sequence has terms a3 = 11.7 and a8 = − 14.6. What is the first term? 452. Use summation notation to write the sum that results from adding the number 13 twenty times. 439. Write a recursive formula for the arithmetic sequence −20, − 10, 0,10,… 453. Use the formula for the sum of the first n terms of an arithmetic series to find the sum of the first eleven terms of the arithmetic series 2.5, 4, 5.5, … . 440. Write a recursive formula for the arithmetic sequence , … , and then find the 31st term. 0, − 1 2 , − 1, − 3 2 441. Write an explicit formula for the arithmetic sequence 7 8 , 29 24 , 37 24 , 15 8 , … 454. A ladder has 15 tapered rungs, the lengths of which increase by a common difference. The first rung is 5 inches long, and the last rung is 20 inches long. What is the sum of the lengths of the rungs? 455. Use the formula for the sum of the first n terms of a geometric series the series to find for S9 442. How many terms are in the finite arithmetic sequence 12, 20, 28, … , 172 ? 12, 6, 3, 3 2 , … Geometric Sequences 443. Find the common ratio for the geometric sequence 2.5, 5, 10, 20, … 444. Is the sequence 4, 16, 28, 40, … geometric? If so find the common ratio. If not, explain why. 445. A geometric sequence has terms a7 = 16,384 and a9 = 262,144 . What are the first five terms? 446. A geometric sequence has the first term a1 = − 3 and common ratio r = 1 2 . What is the 8th term? 447. What are the first five terms of the geometric sequence a1 = 3, an = 4 ⋅ an − 1 ? This content is available for free at https://cnx.org/content/col11758/1.5 456. The fees for the first three years of a hunting club membership are given in Table 13.6. If fees continue to rise at the same rate, how much will the total cost be for the first ten years of membership? Year Membership Fees 1 2 3 $1500 $1950 $2535 Table 13.6 457. ∞ ∑ k = 1 Find the sum of the infinite geometric series 45 ⋅ ( − 1 3 ) k − 1 . Chapter 13 Sequences, Probability, and Counting Theory 1539 458. A ball has a bounce-back ratio of 3 5 the height of the previous bounce. Write a series representing the total distance traveled by the ball, assuming it was initially dropped from a height of 5 feet. What is the total distance? (Hint: the total distance the ball travels on each bounce is the sum of the heights of the rise and the fall.) 459. Alejandro deposits $80 of his monthly earnings into an annuity that earns 6.25% annual interest, compounded monthly. How much money will he have saved after 5 years? 460. The twins Sarah and Scott both opened retirement accounts on their 21st birthday. Sarah deposits $4,800.00 each year, earning 5.5% annual interest, compounded monthly. Scott deposits $3,600.00 each year, earning 8.5% annual interest, compounded monthly. Which twin will earn the most interest by the time they are 55 years old? How much more? Counting Principles 461. How many ways are there to choose a number from the set { − 10, − 6,4,10,12,18,24,32} that is divisible by either 4 or 6 ? In a group of 20 musicians, 12 play piano, 7 play 462. trumpet, and 2 play both piano and trumpet. How many musicians play either piano or trumpet? 470. A day spa charges a basic day rate that includes use of a sauna, pool, and showers. For an extra charge, guests can choose from the following additional services: massage, body scrub, manicure, pedicure, facial, and straight-razor shave.
How many ways are there to order additional services at the day spa? 471. How many distinct ways can the word DEADWOOD be arranged? 472. How many distinct rearrangements of the letters of the word DEADWOOD are there if the arrangement must begin and end with the letter D? Binomial Theorem 473. Evaluate the binomial coefficient ⎛ 23 ⎝ 8 ⎞ ⎠. 474. Use the Binomial Theorem to expand ⎛ ⎝3x + 1 2 6 . y⎞ ⎠ 475. Use the Binomial Theorem to write the first three terms of (2a + b)17. 476. Find the fourth term of ⎛ ⎝3a2 − 2b⎞ ⎠ 11 without fully expanding the binomial. Probability 463. How many ways are there to construct a 4-digit code if numbers can be repeated? For the following exercises, assume two die are rolled. 477. Construct a table showing the sample space. 464. A palette of water color paints has 3 shades of green, 3 shades of blue, 2 shades of red, 2 shades of yellow, and 1 shade of black. How many ways are there to choose one shade of each color? 478. What is the probability that a roll includes a 2 ? 479. What is the probability of rolling a pair? 465. Calculate P(18, 4). 466. In a group of 5 freshman, 10 sophomores, 3 juniors, and 2 seniors, how many ways can a president, vice president, and treasurer be elected? 467. Calculate C(15, 6). 468. A coffee shop has 7 Guatemalan roasts, 4 Cuban roasts, and 10 Costa Rican roasts. How many ways can the shop choose 2 Guatemalan, 2 Cuban, and 3 Costa Rican roasts for a coffee tasting event? How 469. {1, 3, 5, … , 99} have? many subsets does the set 480. What is the probability that a roll includes a 2 or results in a pair? 481. What is the probability that a roll doesn’t include a 2 or result in a pair? 482. What is the probability of rolling a 5 or a 6? 483. What is the probability that a roll includes neither a 5 nor a 6? For the following exercises, use the following data: An elementary school survey found that 350 of the 500 students preferred soda to milk. Suppose 8 children from the (Show calculations and round to the nearest tenth of a percent.) attending a birthday party. school are 1540 Chapter 13 Sequences, Probability, and Counting Theory 484. What is the percent chance that all the children attending the party prefer soda? 485. What is the percent chance that at least one of the children attending the party prefers milk? 486. What is the percent chance that exactly 3 of the children attending the party prefer soda? 487. What is the percent chance that exactly 3 of the children attending the party prefer milk? CHAPTER 13 PRACTICE TEST 488. Write the first four terms of the sequence defined by the recursive formula a = – 14, an = 2 + an – 1 2 . 498. Use summation notation to write the sum of terms 3k 2 − 5 6 k from k = − 3 to k = 15. 489. Write the first four terms of the sequence defined by the explicit formula an = n2 – n – 1 n ! . 499. A community baseball stadium has 10 seats in the first row, 13 seats in the second row, 16 seats in the third row, and so on. There are 56 rows in all. What is the seating capacity of the stadium? 490. Is the sequence 0.3, 1.2, 2.1, 3, … arithmetic? If so find the common difference. 491. An arithmetic sequence has the first term a1 = − 4 . What is the 6th term? and common difference d = – 4 3 492. Write a recursive formula for the arithmetic sequence , … and then find the 22nd −2, − 7 2 , − 5, − 13 2 term. 493. Write an explicit formula for the arithmetic sequence 15.6, 15, 14.4, 13.8, … and then find the 32nd term. 494. Is the sequence − 2, − 1, − 1 2 , − 1 4 , … geometric? If so find the common ratio. If not, explain why. 495. What is the 11th term of the geometric sequence − 1.5, − 3, − 6, − 12, … ? 496. Write a recursive formula for the geometric sequence 1 , … 497. Write an explicit formula for the geometric sequence 4, − 4 3 , − 4 27 , 4 9 , … This content is available for free at https://cnx.org/content/col11758/1.5 500. Use the formula for the sum of the first n terms of a 7 geometric series to find ∑ k = 1 −0.2 ⋅ (−5) k − 1. 501. ∞ ∑ k = 1 1 3 Find the sum of k − 1 ⋅ ⎛ ⎝− 1 5 ⎞ ⎠ . the infinite geometric series 502. Rachael deposits $3,600 into a retirement fund each year. The fund earns 7.5% annual interest, compounded monthly. If she opened her account when she was 20 years old, how much will she have by the time she’s 55? How much of that amount was interest earned? 503. In a competition of 50 professional ballroom dancers, 22 compete in the fox-trot competition, 18 compete in the tango competition, and 6 compete in both the fox-trot and tango competitions. How many dancers compete in the foxtrot or tango competitions? 504. A buyer of a new sedan can custom order the car by choosing from 5 different exterior colors, 3 different interior colors, 2 sound systems, 3 motor designs, and either manual or automatic transmission. How many choices does the buyer have? 505. To allocate annual bonuses, a manager must choose his top four employees and rank them first to fourth. In how many ways can he create the “Top-Four” list out of the 32 employees? Chapter 13 Sequences, Probability, and Counting Theory 1541 506. A rock group needs to choose 3 songs to play at the annual Battle of the Bands. How many ways can they choose their set if have 15 songs to pick from? chance that exactly 3 are butterscotch? (Show calculations and round to the nearest tenth of a percent.) 507. A self-serve frozen yogurt shop has 8 candy toppings and 4 fruit toppings to choose from. How many ways are there to top a frozen yogurt? How many 508. the word EVANESCENCE be arranged if the anagram must end with the letter E? distinct ways can 509. Use the Binomial Theorem to expand ⎛ ⎝ 3 2 x − 1 2 5 . y⎞ ⎠ 510. Find the seventh term of ⎛ ⎝x2 − 1 2 expanding the binomial. 13 ⎞ ⎠ without fully For the following exercises, use the spinner in Figure 13.25. Figure 13.25 511. Construct a probability model showing each possible outcome and its associated probability. (Use the first letter for colors.) 512. What number? is the probability of landing on an odd 513. What is the probability of landing on blue? 514. What is the probability of landing on blue or an odd number? 515. What is the probability of landing on anything other than blue or an odd number? 516. A bowl of candy holds 16 peppermint, 14 butterscotch, and 10 strawberry flavored candies. Suppose a person grabs a handful of 7 candies. What is the percent 1542 Chapter 13 Sequences, Probability, and Counting Theory This content is available for free at https://cnx.org/content/col11758/1.5 Appendix A 1543 A \| APPENDIX A1 Appendix Important Proofs and Derivations Product Rule loga xy = loga x + loga y Proof: Let m = loga x and n = loga y. Write in exponent form. x = am and y = an . Multiply. xy = am an = am + n am + n = xy loga(xy) = m + n = loga x + logb y Change of Base Rule loga b = logc b logc a loga b = 1 logb a where x and y are positive, and a > 0, a ≠ 1. Proof: Let x = loga b. Write in exponent form. a x = b Take the logc of both sides. logc a x = logc b xlogc a = logc b logc b logc a logc b loga b loga b = x = When c = b, loga b = logb b logb a = 1 logb a Heron’s Formula A = s(s − a)(s − b)(s − c) 1544 where s = a + b + c 2 Proof: Let a, b, and c be the sides of a triangle, and h be the height. Appendix A Figure A1 So s = a + b + c 2 . We can further name the parts of the base in each triangle established by the height such that p + q = c. Figure A2 Using the Pythagorean Theorem, h2 + p2 = a2 and h2 + q2 = b2. Since q = c − p, then q2 = (c − p)2. Expanding, we find that q2 = c2 − 2cp + p2. We can then add h2 to each side of the equation to get h2 + q2 = h2 + c2 − 2cp + p2. Substitute this result into the equation h2 + q2 = b2 yields b2 = h2 + c2 − 2cp + p2. Then replacing h2 + p2 with a2 gives b2 = a2 − 2cp + c2. Solve for p to get p = a2 + b2 − c2 2c Since h2 = a2 − p2, we get an expression in terms of a, b, and c. This content is available for free at https://cnx.org/content/col11758/1.5 Appendix A h2 = a2 − p2 1545 = (a + p)(a − p) ⎝a2 + c2 − b2⎞ ⎛ ⎠ 2c = ⎝a2 + c2 − b2⎞ ⎛ ⎠ 2c ⎡ ⎤ ⎤ ⎡ ⎢a + ⎥ ⎥ ⎢a − ⎣ ⎦ ⎦ ⎣ ⎝2ac − a2 − c2 + b2⎞ ⎛ ⎝2ac + a2 + c2 − b2⎞ ⎛ ⎠ ⎠ 4c2 ⎝b2 − (a − c)2⎞ ⎛ ⎝(a + c)2 − b2⎞ ⎛ ⎠ ⎠ 4c2 (a + b + c)(a + c − b)(b + a − c)(b − a + c) 4c2 (a + b + c)( − a + b + c)(a − b + c)(a + b − c) 4c2 2s ⋅ (2s − a) ⋅ (2s − b)(2s − c) 4c2 = = = = = Therefore, h2 = h = 4s(s − a)(s − b)(s − c) c2 2 s(s − a)(s − b)(s − c) c And since A = 1 2 ch, then A = 1 2 c2 s(s − a)(s − b)(s − c) c = s(s − a)(s − b)(s − c) Properties of the Dot Product u · v = v · u Proof: u · v = 〈 u1, u2, ...un 〉 · 〈 v1, v2, ...vn 〉 = u1 v1 + u2 v2 + ... + un vn = v1 u1 + v2 u2 + ... + vn vn = 〈 v1, v2, ...vn 〉 · 〈 u1, u2, ...un 〉 = v · u u · (v + w) = u · v + u · w Proof: u · (v + w) = 〈 u1, u2, ...un 〉 · ⎛ ⎝ 〈 v1, v2, ...vn 〉 + 〈 w1, w2, ...wn 〉 ⎞ ⎠ = 〈 u1, u2, ...un 〉 · 〈 v1 + w1, v2 + w2, ...vn + wn 〉 = 〈 u1(v1 + w1), u2(v2 + w2), ...un(vn + wn) 〉 = 〈 u1 v1 + u1 w1, u2 v2 + u2 w2, ...un vn + un wn 〉 = 〈 u1 v1, u2 v2, ..., un vn 〉 + 〈 u1 w1, u2 w2, ..., un wn 〉 = 〈 u1, u2, ...un 〉 · 〈 v1, v2, ...vn 〉 + 〈 u1, u2, ...un 〉 · 〈 w1, w2, ...wn 〉 = u\|2 Proof: 1546 Appendix A u · u = 〈 u1, u2, ...un 〉 · 〈 u1, u2, ...un 〉 = u1 u1 + u2 u2 + ... + un un 2 + u2 2 + ... + un = u1 = \| 〈 u1, u2, ...un 〉 \|2 = v · u 2 Standard Form of the Ellipse centered at the Origin y2 b2 1 = x2 a2 + Derivation An ellipse consists of all the points for which the sum of distances from two foci is constant: (x − (−c))2 + ⎛ ⎝y − 0⎞ ⎠ 2 + (x − c)2 + ⎛ ⎝y − 0⎞ ⎠ 2 = constant Figure A3 Consider a vertex. Figure A4 Then, (x − (−c))2 + ⎛ ⎝y − 0⎞ ⎠ 2 + (x − c)2 + ⎛ ⎝y − 0⎞ ⎠ 2 = 2a Consider a covertex. This content is available for free at https://cnx.org/content/col11758/1.5 Appendix A 1547 Figure A5 Then b2 + c2 = a2. (x − ( − c))2 + (y − 0)2 + (x − c)2 + (y − 0)2 = 2a (x + c)2 + y2 = 2a − (x − c)2 + y2 ⎝2a − (x − c)2 + y2⎞ ⎛ ⎠ (x + c)2 + y2 = 2 x2 + 2cx + c2 + y2 = 4a2 − 4a (x − c)2 + y2 + (x − c)2 + y2 x2 + 2cx + c2 + y2 = 4a2 − 4a (x − c)2 + y2 + x2 −
2cx + y2 2cx = 4a2 − 4a (x − c)2 + y2 − 2cx − 1 4a ⎠ = (x − c)2 + y2 4cx − 4a2 = 4a (x − c)2 + y2 ⎛ ⎝4cx − 4a2⎞ c ax = (x − c)2 + y2 x2 = (x − c)2 + y2 a2 − 2xc + a2 − 2xc + a − c2 a2 c2 a2 c2 a2 c2 a2 a2 + a2 + x2 = x2 − 2xc + c2 + y2 x2 = x2 + c2 + y2 x2 = x2 + c2 + y2 Let 1 = a2 a2. a2 − c2 = x2 − c2 a2 x2 + y2 a2 − c2 = x2 ⎛ ⎝1 − c2 a2 ⎞ ⎠ + y2 a2 − c2 = x2 ⎛ a2 − c2 ⎝ a2 ⎞ ⎠ + y2 1 = x2 a2 + y2 a2 − c2 (A1) 1548 Because b2 + c2 = a2, then b2 = a2 − c2. Standard Form of the Hyperbola 1 = 1 = x2 a2 + x2 a2 + y2 a2 − c2 y2 b2 Appendix A (A2) y2 b2 1 = x2 a2 − Derivation A hyperbola is the set of all points in a plane such that the absolute value of the difference of the distances between two fixed points is constant. Figure A6 Diagram 1: The difference of the distances from Point P to the foci is constant: (x − ( − c))2 + (y − 0)2 − (x − c)2 + (y − 0)2 = constant Diagram 2: When the point is a vertex, the difference is 2a. (x − (−c))2 + ⎛ ⎝y − 0⎞ ⎠ 2 − (x − c)2 + ⎛ ⎝y − 0⎞ ⎠ 2 = 2a This content is available for free at https://cnx.org/content/col11758/1.5 Appendix A 1549 (x − ( − c))2 + (y − 0)2 − (x − c)2 + (y − 0)2 = 2a (x + c)2 + y2 − (x − c)2 + y2 = 2a (x + c)2 + y2 = 2a + (x − c)2 + y2 ⎝2a + (x − c)2 + y2⎞ ⎛ (x + c)2 + y2 = ⎠ x2 + 2cx + c2 + y2 = 4a2 + 4a (x − c)2 + y2 x2 + 2cx + c2 + y2 = 4a2 + 4a (x − c)2 + y2 + x2 − 2cx + y2 2cx = 4a2 + 4a (x − c)2 + y2 − 2cx 4cx − 4a2 = 4a (x − c)2 + y2 cx − a2 = a (x − c)2 + y2 2 ⎝cx − a2⎞ ⎛ ⎠ = a2 ⎛ ⎝(x − c)2 + y2⎞ ⎠ c2 x2 − 2a2 c2 x2 + a4 = a2 x2 − 2a2 c2 x2 + a2 c2 + a2 y2 c2 x2 + a4 = a2 x2 + a2 c2 + a2 y2 a4 − a2 c2 = a2 x2 − c2 x2 + a2 y2 a2 ⎛ ⎝a2 − c2⎞ ⎝a2 − c2⎞ ⎠x2 + a2 y2 ⎠x2 − a2 y2 ⎛ ⎝a2 − c2⎞ ⎛ ⎝c2 − a2⎞ ⎠ = ⎠ = a2 ⎛ Define b as a positive number such that b2 = c2 − a2. (A3) a2 b2 = b2 x2 − a2 y2 a2 y2 a2 b2 a2 b2 = a2 b2 b2 x2 a2 b2 − y2 x2 a2 − b2 Trigonometric Identities 1 = Pythagorean Identity Even-Odd Identities Table A1 cos2 t + sin2 t = 1 1 + tan2 t = sec2 t 1 + cot2 t = csc2 t cos( − t) = cos t sec( − t) = sec t sin( − t) = − sin t tan( − t) = − tan t csc( − t) = − csc t cot( − t) = − cot t 1550 Appendix A ⎛ ⎝ tan t = cot ⎛ cos t = sin ⎝ π 2 π ⎛ sin t = cos ⎝ 2 π 2 π 2 π ⎛ sec t = csc ⎝ 2 π 2 ⎛ csc t = sec ⎝ ⎛ cot t = tan ⎝ − t⎞ ⎠ − t⎞ ⎠ − t⎞ ⎠ − t⎞ ⎠ − t⎞ ⎠ − t⎞ ⎠ tan t = sin t cos t sec t = 1 cos t csc t = 1 sin t tan t = cos t cot t = 1 sin t cos(α + β) = cos α cos β − sin α sin β cos(α − β) = cos α cos β + sin α sin β sin(α + β) = sin α cos β + cos α sin β sin(α − β) = sin α cos β − cos α sin β tan(α + β) = tan(α − β) = tan α + tan β 1 − tan α tan β tan α − tan β 1 + tan α tan β sin(2θ) = 2sin θ cos θ cos(2θ) = cos2 θ − sin2 θ cos(2θ) = 1 − 2sin2 θ cos(2θ) = 2cos2 θ − 1 tan(2θ) = 2tan θ 1 − tan2 θ sin cos tan tan tan − cos α = ± 1 + cos α = ± 1 − cos α 1 + cos α = sin α 1 + cos α = 1 − cos α sin α Cofunction Identities Fundamental Identities Sum and Difference Identities Double-Angle Formulas Half-Angle Formulas Table A1 This content is available for free at https://cnx.org/content/col11758/1.5 Appendix A 1551 Reduction Formulas sin2 θ = cos2 θ = tan2 θ = 1 − cos(2θ) 2 1 + cos(2θ) 2 1 − cos(2θ) 1 + cos(2θ) Product-to-Sum Formulas Sum-to-Product Formulas ⎡ ⎡ cosαcosβ = 1 2 sinαcosβ = 1 ⎣sin⎛ 2 sinαsinβ = 1 ⎣cos⎛ 2 cosαsinβ = 1 2 ⎣sin⎛ ⎡ ⎡ ⎣cos⎛ ⎝α − β⎞ ⎠ + cos⎛ ⎝α + β⎞ ⎤ ⎦ ⎠ ⎝α + β⎞ ⎠ + sin⎛ ⎝α − β⎞ ⎤ ⎦ ⎠ ⎝α − β⎞ ⎠ − cos⎛ ⎝α + β⎞ ⎤ ⎦ ⎠ ⎝α + β⎞ ⎠ − sin⎛ ⎝α − β⎞ ⎤ ⎦ ⎠ ⎛ sinα + sinβ = 2sin ⎝ ⎛ sinα − sinβ = 2sin ⎝ cos ⎝ ⎠ cos ⎝ ⎠ 2 2 α − β α + β ⎛ ⎞ ⎛ cosα − cosβ = − 2sin ⎠sin ⎝ ⎝ cos ⎝ 2 2 ⎛ cosα + cosβ = 2cos ⎝ ⎞ ⎠ ⎞ ⎠ Law of Sines Law of Cosines Table A1 sin α a = a sin α = sin β b = b sin β = sin γ c c sin γ a2 = b2 + c2 − 2bc cos α b2 = a2 + c2 − 2ac cos β c2 = a2 + b2 − 2ab cos γ 1552 Appendix A ToolKit Functions Figure A7 Figure A8 This content is available for free at https://cnx.org/content/col11758/1.5 Appendix A 1553 Figure A9 Trigonometric Functions Unit Circle Figure A10 1554 Appendix A Angle 0 π 6 , or 30 ° π 4 , or 45 ° π 3 , or 60 ° π 2 , or 90 ° Cosine Sine Tangent Secant 1 0 0 1 Cosecant Undefined Cotangent Undefined Table A2 Undefined Undefined 1 0 This content is available for free at https://cnx.org/content/col11758/1.5 Index 1555 INDEX A absolute maximum, 294, 379 absolute minimum, 294, 379 absolute value, 272, 353 absolute value equation, 197, 216 absolute value functions, 353, 357 absolute value inequality, 210 addition method, 1217, 1224, 1331 Addition Principle, 1502, 1533 addition property, 206 adjacent side, 835, 892 algebraic expression, 23, 99 altitude, 1060, 1201 ambiguous case, 1063, 1201 amplitude, 907, 967 angle, 810, 892 angle of depression, 843, 892 angle of elevation, 843, 892, 1060 angle of rotation, 1409, 1437 angular speed, 828, 892 annual interest, 1497 annual percentage rate (APR), 656, 793 annuity, 1497, 1533 apoapsis, 1422 arc, 815 arc length, 816, 825, 851, 892 arccosine, 951, 967 Archimedes’ spiral, 1129, 1201 arcsine, 951, 967 arctangent, 951, 967 area, 152, 216 area of a circle, 499 area of a sector, 827, 892 argument, 1138, 1201 arithmetic sequence, 1463, 1465, 1467, 1467, 1488, 1533 arithmetic series, 1489, 1533 arrow notation, 580, 634 associative property of addition, 20, 99 associative property of multiplication, 19, 99 asymptotes, 1365 augmented matrix, 1286, 1290, 1292, 1306, 1331 average rate of change, 282, 379 axes of symmetry, 1365 axis of symmetry, 476, 481, 634, 1395, 1397 B base, 16, 99 binomial, 68, 99, 552 binomial coefficient , 1514, 1533 binomial expansion, 1515, 1518, 1533 Binomial Theorem, 1517, 1533 break-even point, 1225, 1331 C cardioid, 1118, 1201 carrying capacity, 762, 793 Cartesian coordinate system, 108, 216 Cartesian equation, 1103 Celsius, 362 center of a hyperbola, 1365, 1437 center of an ellipse, 1344, 1437 central rectangle, 1365 change-of-base formula, 732, 793 circle, 1249, 1250 circular motion, 917 circumference, 815 co-vertex, 1344 co-vertices, 1345 coefficient, 68, 99, 500, 564, 634 coefficient matrix, 1287, 1289, 1308, 1331 cofunction, 1000 cofunction identities, 841, 1000 column, 1272, 1331 column matrix, 1273 combination, 1533 combinations, 1508, 1514 combining functions, 301 common base, 738 common difference, 1463, 1488, 1533 common logarithm, 690, 793 common ratio, 1476, 1491, 1533 commutative, 303 commutative property of addition, 19, 99 commutative property of multiplication, 19, 99 complement of an event , 1527, 1533 completing the square, 179, 181, 216 complex conjugate, 166, 216 Complex Conjugate Theorem, 570 complex number, 161, 216, 1134 complex plane, 161, 216, 1134 composite function, 302, 379 composition of functions, 301 compound inequality, 209, 216 compound interest, 656, 793 compression, 410, 674, 707 conditional equation, 129, 216 conic, 1342, 1364, 1431 conic section, 1437 conic sections, 1159 conjugate axis, 1365, 1437 consistent system, 1213, 1331 constant, 23, 99 constant of variation, 625, 634 continuous, 525 continuous function, 516, 634 convex limaçons, 1120 convex limaҫon , 1201 coordinate plane, 1387 correlation coefficient, 457, 466 cosecant, 874, 892, 932 cosecant function, 932, 933, 979 cosine, 1022, 1024 cosine function, 851, 892, 903, 905, 907, 917 cost function, 300, 1225, 1331 cotangent, 873, 892, 940 cotangent function, 940, 979 coterminal angles, 821, 824, 892 Cramer’s Rule, 1318, 1321, 1326, 1331 cube root, 500 cubic functions, 612 curvilinear path, 1148 D de Moivre, 1141 De Moivre’s Theorem, 1143, 1144, 1201 decompose a composite function, 311 decomposition, 1260 decreasing function, 289, 379, 397 decreasing linear function, 397, 466 1556 Index degenerate conic sections, 1404, 1437 degree, 68, 99, 507, 634, 811, 892 dependent system, 1213, 1224, 1239, 1331 dependent variable, 226, 379 Descartes, 1134 Descartes’ Rule of Signs, 572, 634 determinant, 1317, 1320, 1321, 1331 difference of squares, 73, 99 dimpled limaçons, 1120 dimpled limaҫon , 1201 direct variation, 625, 634 directrix, 1387, 1393, 1395, 1422, 1430, 1431, 1437 discriminant, 183, 216 displacement, 828 distance formula, 118, 216, 991, 1366, 1388 distributive property, 20, 99 diverge, 1533 diverges, 1493 dividend, 551 Division Algorithm, 552, 562, 634 divisor, 551 domain, 227, 237, 256, 258, 379, 950 domain and range, 256 domain of a composite function, 309 dot product, 1193, 1201 double-angle formulas, 1007, 1049 doubling time, 759, 793 Dürer, 1122 E eccentricity, 1423, 1437 electrostatic force, 286 elimination, 1250 ellipse, 1168, 1250, 1343, 1344, 1345, 1348, 1354, 1386, 1424, 1430, 1437 ellipsis, 1446 end behavior, 502, 592, 634 endpoint, 285, 810 entry, 1273, 1331 equation, 26, 99, 235 equation in quadratic form, 199 equation in two variables, 111, 216 equations in quadratic form, 216 Euler, 1134 even function, 334, 379, 978 even-odd identities, 977, 1049 event, 1522, 1533 experiment, 1522, 1533 explicit formula, 1446, 1468, 1480, 1533 exponent, 16, 99 Exponential decay, 645 exponential decay, 653, 667, 753, 756, 760, 774 exponential equation, 737 exponential function, 645 exponential growth, 645, 649, 668, 753, 759, 762, 793 exponential notation, 16, 99 extraneous solution, 743, 793 extraneous solutions, 194, 216 extrapolation, 454, 466 F factor by grouping, 81, 99 Factor Theorem, 563, 634 factorial, 1457 factoring, 172 Fahrenheit, 362 feasible region, 1254, 1331 finite arithmetic sequence, 1469 finite sequence, 1448, 1533 foci, 1343, 1344, 1345, 1366, 1437 focus, 1343, 1387, 1393, 1395, 1422, 1430, 1431 focus (of a parabola), 1437 focus (of an ellipse), 1437 formula, 27, 99, 235 function, 227, 273, 379 function notation, 229 Fundamental Counting Principle, 1504, 1533 Fundamental Theorem of Algebra, 568, 570, 634 G Gauss, 1134, 1233, 1286 Gaussian elimination, 1233, 1289, 1331 general form, 478 general form of a quadratic function, 481, 634 Generalized Pythagorean Theorem, 1079, 1201 geometric sequence, 1476, 1491, 1533 geometric series, 1492, 1533 global maximum, 541, 634 This content is available for free at https://cnx.org/content/col11758/1.5 global
minimum, 541, 634 graph in two variables, 111, 216 gravity, 1172 greatest common factor, 78, 99, 173 H half-angle formulas, 1014, 1049 half-life, 748, 753, 793 Heaviside method, 1262 Heron of Alexandria, 1086 Heron’s formula, 1086 horizontal asymptote, 583, 590, 634 horizontal compression, 340, 379, 1042 horizontal line, 140, 418, 466 horizontal line test, 244, 379 horizontal reflection, 328, 379 horizontal shift, 321, 379, 671, 702, 904 horizontal stretch, 340, 379 hyperbola, 1364, 1369, 1370, 1371, 1375, 1377, 1380, 1387, 1425, 1428, 1437 hypotenuse, 835, 892 I identities, 882, 892 identity equation, 129, 216 identity matrix, 1301, 1306, 1331 identity property of addition, 21, 99 identity property of multiplication, 21, 99 imaginary number, 161, 216 inconsistent equation, 130, 216 inconsistent system, 1213, 1222, 1238, 1331 increasing function, 289, 379, 397 increasing linear function, 397, 466 independent system, 1213, 1331 independent variable, 226, 379 index, 61, 99 index of summation, 1488, 1533 inequality, 1252 infinite geometric series, 1493 infinite sequence, 1448, 1533 infinite series, 1493, 1533 initial point, 1178, 1182, 1201 initial side, 811, 892 inner-loop limaçon , 1201 inner-loop limaçons, 1122 Index 1558 input, 227, 379 integers, 10, 15, 99 intercepts, 116, 216 Intermediate Value Theorem, 537, 634 interpolation, 454, 466 intersection, 1525 interval, 204, 216 interval notation, 204, 216, 257, 290, 379 inverse cosine function, 951, 967 inverse function, 364, 379, 613 inverse matrix, 1306, 1308 inverse of a radical function, 616 inverse of a rational function, 619 inverse property of addition, 21, 99 inverse property of multiplication, 21, 100 inverse sine function, 951, 967 inverse tangent function, 951, 967 inverse trigonometric functions, 950, 951, 955, 960 inverse variation, 627, 634 inverse variations, 627 inversely proportional, 627, 634 invertible function, 634 invertible functions, 610 invertible matrix, 1301, 1317 irrational numbers, 12, 15, 100 J joint variation, 629, 634 K Kronecker, 1134 L latus rectum, 1387, 1395, 1437 Law of Cosines, 1080, 1201 Law of Sines, 1062, 1081, 1201 leading coefficient, 68, 100, 507, 634 leading term, 68, 100, 507, 635 least common denominator, 92, 100, 131 least squares regression, 455, 466 lemniscate, 1124, 1201 linear equation, 129, 216 Linear Factorization Theorem, 570, 635 linear function, 394, 396, 466 linear growth, 645 linear inequality, 216 linear model, 438, 451 linear relationship, 451 linear speed, 829, 892 local extrema, 288, 379 local maximum, 288, 379, 541 local minimum, 288, 379, 541 logarithm, 687, 793 logarithmic equation, 744 logarithmic model, 778 logistic growth model, 762, 793 long division, 550 lower limit of summation, 1488, 1533 M magnitude, 272, 319, 1135, 1178, 1201 main diagonal, 1288, 1331 major and minor axes, 1345 major axis, 1344, 1348, 1437 matrix, 1272, 1273, 1286, 1331 matrix multiplication, 1279, 1303, 1308 matrix operations, 1273 maximum value, 476 measure of an angle, 811, 892 midline, 907, 967 midpoint formula, 122, 216 minimum value, 476 minor axis, 1344, 1437 model breakdown, 453, 466 modulus, 272, 1138, 1201 monomial, 68, 100 Multiplication Principle, 1504, 1533 multiplication property, 206 multiplicative inverse, 1304, 1304 multiplicative inverse of a matrix, 1301, 1331 multiplicity, 530, 635 mutually exclusive events, 1526, 1533 N n factorial, 1457, 1533 natural logarithm, 692, 742, 793 natural numbers, 10, 15, 100, 226 negative angle, 812, 892 Newton’s Law of Cooling, 760, 793 nominal rate, 656, 793 non-right triangles, 1060 nondegenerate conic section, 1406, 1437 nondegenerate conic sections, 1404 nonlinear inequality, 1252, 1331 nth term of the sequence, 1446 nth partial sum, 1533 nth root of a complex number, 1144 nth term of a sequence, 1533 nth partial sum, 1487 O oblique triangle, 1060, 1201 odd function, 334, 379, 977 one-loop limaçon, 1120 one-loop limaҫon , 1201 one-to-one, 668, 686, 723, 732 one-to-one function, 241, 364, 379, 950 opposite side, 835, 892 order of magnitude, 754, 793 order of operations, 16, 100 ordered pair, 109, 216, 226, 258 ordered triple, 1233 origin, 109, 216, 354 outcomes, 1522, 1533 output, 227, 380 P parabola, 476, 486, 1165, 1246, 1386, 1393, 1397, 1422, 1426, 1437 parallel, 141 parallel lines, 421, 466 parallelograms, 1184 parameter, 1148, 1201 parametric equations, 1148, 1164, 1164 parametric form, 1168 parent function, 702 partial fraction decomposition, 1260, 1331 partial fractions, 1260, 1331 Pascal, 1122 Pascal's Triangle, 1516 perfect square trinomial, 72, 100 periapsis, 1422 perimeter, 152, 217 period, 885, 892, 904, 925, 927, 1031 periodic function, 904, 967 permutation, 1504, 1534 perpendicular, 141 perpendicular lines, 421, 466 pH, 722 1559 Index phase shift, 908, 967 piecewise function, 273, 380 piecewise functions, 1451 point-slope form, 401, 466 point-slope formula, 143, 1372 polar axis, 1095, 1201 polar coordinates, 1095, 1097, 1100, 1111, 1202 polar equation, 1104, 1112, 1114, 1202, 1423, 1437 polar form, 1135 polar form of a complex number, 1137, 1202 polar form of a conic, 1432 polar grid, 1095 pole, 1095, 1202 polynomial, 68, 100, 564 polynomial equation, 192, 217 polynomial function, 506, 523, 533, 539, 635 position vector, 1179, 1181 positive angle, 812, 892 power function, 500, 635 power rule for logarithms, 726, 732, 793 principal nth root, 61, 100 principal square root, 54, 100 probability, 1522, 1534 probability model, 1522, 1534 product of two matrices, 1279 product rule for logarithms, 724, 726, 793 product-to-sum formula, 1049 product-to-sum formulas, 1022, 1024 profit function, 1226, 1331 properties of determinants, 1324 Proxima Centauri, 754 Pythagoras, 1134 Pythagorean identities, 976, 1049 Pythagorean Identity, 854, 881, 892 Pythagorean identity, 991 Pythagorean Theorem, 118, 184, 217, 1009, 1043, 1079, 1170 Q quadrant, 108, 217 quadrantal angle, 813, 892 quadratic, 1264, 1266 quadratic equation, 172, 179, 181, 217, 1037 quadratic formula, 181, 183, 217, 1037 quadratic function, 481, 485 quotient, 551 quotient identities, 979, 1049 quotient rule for logarithms, 725, 793 R radian, 815, 817, 817, 892 radian measure, 817, 892 radical, 54, 100 radical equation, 194, 217 radical expression, 54, 100 radical functions, 610 radicand, 54, 100, 194 radiocarbon dating, 757 range, 227, 380, 952 rate of change, 282, 380, 438 rational equation, 132, 217 rational expression, 89, 100, 131, 1260, 1266 rational function, 585, 595, 601, 635 rational number, 131 rational numbers, 10, 15, 100 Rational Zero Theorem, 564, 635 ray, 810, 892 real number line, 13, 100 real numbers, 13, 100 reciprocal, 141, 363, 500 reciprocal function, 579 reciprocal identities, 979, 1049 reciprocal identity, 931, 940 rectangular coordinates, 1095, 1097, 1100 rectangular equation, 1104, 1158 rectangular form, 1137, 1168 recursive formula, 1454, 1466, 1478, 1534 reduction formulas, 1012, 1049 reference angle, 822, 863, 877, 892 reflection, 676, 711 regression analysis, 774, 778, 781 regression line, 456 relation, 226, 380 remainder, 551 Remainder Theorem, 562, 635 removable discontinuity, 588, 635 Restricting the domain, 373 resultant, 1183, 1202 revenue function, 1225, 1331 Richter Scale, 685 This content is available for free at https://cnx.org/content/col11758/1.5 right triangle, 835, 950 roots, 477, 635 rose curve, 1126, 1202 row, 1272, 1331 row matrix, 1273 row operations, 1288, 1294, 1304, 1305, 1306, 1332 row-echelon form, 1288, 1292, 1332 row-equivalent, 1288, 1332 S sample space, 1522, 1534 SAS (side-angle-side) triangle, 1079 scalar, 1185, 1202, 1276 scalar multiple, 1186, 1276, 1332 Scalar multiplication, 1185 scalar multiplication, 1202, 1276 scatter plot, 451 scientific notation, 46, 48, 100 secant, 873, 892, 931 secant function, 931 sector of a circle, 827 sequence, 1446, 1463, 1534 series, 1487, 1534 set-builder notation, 204, 261, 380 sigma, 1487 sine, 977, 1023, 1025 sine function, 851, 892, 902, 907, 915, 919 sinusoidal function, 905, 967 slope, 136, 217, 395, 466 slope-intercept form, 395, 396, 401, 466 smooth curve, 516, 635 solution set, 129, 217, 1234, 1332 solving systems of linear equations, 1217 special angles, 989 square matrix, 1273, 1317 square root property, 178, 217 SSS (side-side-side) triangle, 1079 standard form, 139 standard form of a quadratic function, 481, 635 standard position, 811, 892, 1179, 1202 stretch, 674 stretching/compressing factor, 928, 929 substitution method, 1216, 1332 Index 1560 vertex, 476, 635, 810, 893, 1344, 1344, 1387, 1395 vertex form of a quadratic function, 479, 635 vertical asymptote, 582, 586, 592, 635, 951 vertical compression, 336, 380 vertical line, 140, 418, 466 vertical line test, 242, 380 vertical reflection, 328, 380 vertical shift, 318, 380, 411, 670, 704, 760, 908 vertical stretch, 336, 380, 410, 707 vertices, 1344, 1345 volume, 152, 217 volume of a sphere, 499 W whole numbers, 10, 15, 100 X x-axis, 108, 217 x-coordinate, 109, 217 x-intercept, 116, 217, 416 Y y-axis, 108, 217 y-coordinate, 109, 217 y-intercept, 116, 217, 395 Z zero-product property, 173, 217 zeros, 477, 525, 530, 566, 635, 1117 sum and difference formulas for cosine, 990 sum and difference formulas for sine, 993 sum and difference formulas for tangent, 996 sum-to-product formula, 1049 sum-to-product formulas, 1025 summation notation, 1488, 1534 surface area, 609 symmetry test, 1112 synthetic division, 555, 566, 635 system of equations, 1287, 1288, 1290, 1292, 1309 system of linear equations, 444, 1212, 1215, 1216, 1332 system of nonlinear equations, 1246, 1332 system of nonlinear inequalities, 1254, 1332 system of three equations in three variables, 1321 T tangent, 873, 893, 925, 926 tangent function, 926, 927, 928, 944, 978 term, 1446, 1463, 1534 term of a polynomial, 68, 100 term of a polynomial function, 506, 635 terminal point, 1178, 1182, 1202 terminal side, 811, 893 transformation, 317, 410 translation, 1348 transverse axis, 1365, 1437 trigonom
etric equations, 1158 trigonometric functions, 877 trigonometric identities, 1080 trinomial, 68, 100 turning point, 513, 534, 635 U union of two events, 1524, 1534 unit circle, 817, 835, 851, 865, 893, 1032 unit vector, 1188, 1202 upper limit of summation, 1488, 1534 upper triangular form, 1233 V variable, 23, 100 varies directly, 625, 635 varies inversely, 627, 635 vector, 1178, 1202 vector addition, 1183, 1202
er cannot be expressed in the form where a and b are integers and b 0. b 18 Number Systems When writing an irrational number, we use three dots (. . .) after a series of digits to indicate that the number does not terminate. The dots do not indicate a pattern, and no raised bar can be placed over any digits. In an irrational number, we are never certain what the next digit will be when these dots (. . .) are used. In this section, we will see more examples of irrational numbers, both positive and negative. First, however, we need to review a few terms you learned in earlier mathematics courses. Squares and Square Roots To square a number means to multiply the number by itself. For example: The square of 3 is 9. The square of 4 is 16. 32 3 · 3 9 42 4 · 4 16 Calculators have a special key, x2 , that will square a number. ENTER: 5 x2 ENTER DISPLAY: 5 2 2 5 To find a square root of a number means to find a number that, when mul- tiplied by itself, gives the value under the radical sign, . For example: 9 16 3 4 " " A square root of 9 equals 3 because 3 · 3 9. A square root of 16 equals 4 because 4 · 4 16. " Calculators also have a key, This key is often the second function of the ¯ x2 key. For example: , that will display the square root of a number. ENTER: 2nd ¯ 25 ENTER DISPLAY: √ ( 2 5 5 When the square root key is pressed, the calculator displays a square root sign followed by a left parenthesis. It is not necessary to close the parentheses if the entire expression that follows is under the radical sign. However, when other numbers and operations follow that are not part of the expression under the radical sign, the right parenthesis must be entered to indicate the end of the radical expression. More Irrational Numbers The Irrational Numbers 19 2 When a square measures 1 unit on every side, its diagonal measures units. You can use a ruler to " measure the diagonal and then on a show the placement of number line. " 2 What is the value of 2 ? Can we find a decimal number that, when multiplied by itself, equals 2? 2 We expect to be somewhere between 1 and 2. " " 1 2 1 1 2 Use a calculator to find the 0 1 2 value. ENTER: 2nd ¯ 2 ENTER DISPLAY Check this answer by multiplying: 1.414213562 1.414213562 1.999999999, too small. 1.414213563 1.414213563 2.000000002, too large. Note that if, instead of rewriting the digits displayed on the screen, we square the answer using 2nd ANS , the graphing calculator will display 2 because in that case it uses the value of calculator, which has more decimal places than are displayed on the screen. that is stored in the memory of the " 2 No matter how many digits can be displayed on a calculator, no terminating decimal, nor any repeating decimal, can be found for 2 because " 2 is an irrational number. " In the same way, an infinite number of square roots are irrational numbers, for example: 3 " 5 " 3.2 " 0.1 " 2 2 " 3 2 " 20 Number Systems The values displayed on a calculator for irrational square roots are called rational approximations. A rational approximation for an irrational number is a rational number that is close to, but not equal to, the value of the irrational number. The symbol ≈ means approximately equal to. Therefore, it is not correct to write 3 5 1.732 , but it is correct to write 3 < 1.732 . " Another interesting number that you have encountered in earlier courses is p, read as “pi.” Recall that p equals the circumference of a circle divided by its " diameter, or p 5 C d . C d p is an irrational number. There are many rational approximations for p, including: p ≈ 3.14 p < 22 7 If p is doubled, or divided by two, or if a rational number is added to or subtracted from p, the result is again an irrational number. There are infinitely many such irrational numbers, for example: p ≈ 3.1416 2p p 2 p 7 p – 3 Approximation Scientific calculators have a key that, when pressed, will place in the display a rational approximation for p that is more accurate than the ones given above. p On a graphing calculator, when the key is accessed, the screen shows the symbol p but a rational approximation is used in the calculation. On a graphing calculator: ENTER: 2nd p ENTER DISPLAY The Irrational Numbers 21 With a calculator, however, you must be careful how you interpret and use the information given in the display. At times, the value shown is exact, but, more often, displays that fill the screen are rational approximations. To write a rational approximation to a given number of decimal places, round the number. Procedure To round to a given decimal place: 1. Look at the digit in the place at the immediate right of the decimal place to which you are rounding the number. 2. If the digit being examined is less than 5, drop that digit and all digits to the right. (Example: 3.1415927 . . . rounded to two decimal places is 3.14 because the digit in the third decimal place, 1, is less than 5.) 3. If the digit being examined is greater than or equal to 5, add 1 to the digit in the place to which you are rounding and then drop all digits to the right. (Example: 3.1415927 . . . rounded to four decimal places is 3.1416 because the digit in the fifth decimal place, 9, is greater than 5.) EXAMPLE 1 True or False: 5 1 " Solution Use a calculator. 5 5 " " 10 ? Explain why. ENTER: 2nd ¯ 5 ) 2nd ¯ 5 ENTER DISPLAY ENTER: 2nd ¯ 10 ENTER DISPLAY Use these rational approximations to conclude that the values are not equal. Answer False. 5 1 5 2 " " " 10 because 5 1 " " 5 . 4 while 10 , 4 . " 22 Number Systems EXAMPLE 2 Find a rational approximation for each irrational number, to the nearest hundredth. a. 0.1 b. 3 " " Solution Use a calculator. a. ENTER: 2nd ¯ 3 ENTER b. ENTER: 2nd ¯ .1 ENTER DISPLAY: √ ( 3 DISPLAY Use the rules for rounding. The digit in the thousandths place, 2, is less than 5. Drop this digit and all digits to the right of it. The digit in the thousandths place, 6, is greater than or equal to 5. Add 1 to the digit in the hundredths place and drop all digits to the right of it. Answer: 3 ≈ 1.73 " Answer: " 0.1 < 0.32 EXAMPLE 3 The circumference C of a circle with a diameter d is found by using the formula C pd. a. Find the exact circumference of a circle whose diameter is 8. b. Find, to the nearest thousandth, a rational approximation of the circumfer- ence of this circle. Solution a. C pd C p · 8 or 8p b. Use a calculator. ENTER: 2nd p 8 ENTER DISPLAY Round the number in the display to three decimal places: 25.133. Answers a. 8p is the exact circumference, an irrational number. b. 25.133 is the rational approximation of the circumference, to the nearest thousandth. The Irrational Numbers 23 EXAMPLE 4 Which of the following four numbers is an irrational number? In each case, the . . . that follows the last digit indicates that the established pattern of digits repeats. (1) 0.12 (2) 0.12121212 . . . (3) 0.12111111 . . . (4) 0.12112111211112 . . . Solution Each of the first three numbers is a repeating decimal. Choice (1) is a terminating decimal that can be written with a repeating zero. Choice (2) repeats the pair of digits 12 from the first decimal place and choice (3) repeats the digit 1 from the third decimal place. In choice (4), the pattern increases the number of times the digit 1 occurs after each 2. Therefore, (4) is not a repeating decimal and is irrational. Answer (4) 0.12112111211112 . . . is irrational. EXERCISES Writing About Mathematics 1. Erika knows that the sum of two rational numbers is always a rational number. Therefore, she concludes that the sum of two irrational numbers is always an irrational number. Give some examples that will convince Erika that she is wrong. 2. Carlos said that 3.14 is a better approximation for p than 22 7 . Do you agree with Carlos? Explain your answer. Developing Skills In 3–22, tell whether each number is rational or irrational. 3. 0.36 4. 0.36363636 . . . 5. 0.36 6. 0.363363336 . . . 7. 8 " 11. 0.989989998 . . . 15. 5.28 8. 10p 12. 0.725 16. 0.14141414 . . . 19. 48 20. 49 " " 9. 0.12131415 . . . 13. 17. 21. 121 " 2 5 " 0.24682 16 10. " 14. p 30 18. –p 22. p – 2 23. Determine which of the following irrational numbers are between 1 and 4. (1) p 2 (2) 5 " 2 (3) " 4 (4) 11 " (5) 2 3 " 24 Number Systems In 24–43 write the rational approximation of each given number: a. as shown on a calculator display, b. rounded to the nearest thousandth (three decimal places) c. rounded to the nearest hundredth (two decimal places). 24. 29. 34. 39. 5 90 " " 12 " 2 " 82 25. 30. 35. 40. 7 " 2 14 " 16 6.5 " " 44. A rational approximation for 22 19 " 2 " 17 " 3 26. 31. 36. 41. 2 55 " is 1.732. 3 " 27. 32. 37. 42. 75 0.2 " " p 3 1,732 " 28. 33. 38. 43. 63 0.3 " " 0.17 " 241 " a. Multiply 1.732 by 1.732. 45. a. Find (3.162)2. b. Which is larger, b. Find (3.163)2. 3 or 1.732? " c. Is 3.162 or 3.163 a better approximation for 10 ? Explain why. " In 46–50, use the formula C pd to find, in each case, the circumference C of a circle when the diameter d is given. a. Write the exact value of C by using an irrational number. b. Find a rational approximation of C to the nearest hundredth. 46. d 7 47. d = 15 48. d 72 49. d 1 2 50. d 31 3 51. True or False: 52. True or False: 4 1 " 18 1 4 5 " 18 5 " " " 8 ? Explain why or why not. 36 ? Explain why or why not. " Hands-On Activity Cut two squares, each of which measures 1 foot on each side. Cut each square along a diagonal (the line joining opposite corners of the square). Arrange the four pieces of the squares into a larger square. a. What is the area of each of the two squares that you cut out? b. What is the area of the larger square formed by using the pieces of the smaller squares? c. What should be the length of each side of the larger square? Is this length rational or irrational? d. Measure the length of each side of the larger square? Is this measurement rational or irrational? e. Should the answers to parts c and d be the same? Explain your answer. 1-4 THE REAL NUMBERS The Real Numbers 25 Recall that rational numbers
can be written as repeating decimals, and that irrational numbers are decimals that do not repeat. Taken together, rational and irrational numbers make up the set of all numbers that can be written as decimals. The set of real numbers is the set that consists of all rational numbers and all irrational numbers. The accompanying diagram shows that the rational numbers are a subset of the real numbers, and the irrational numbers are also a subset of the real numbers. Notice, however, that the rationals and the irrationals take up different spaces in the diagram because they have no numbers in common. Together, these two sets of numbers form the real numbers. The cross-hatched shaded portion in the diagram contains no real numbers. The cross-hatched shading indicates that no other numbers except the rationals and irrationals are real numbers. Real Numbers Irrational Numbers Rational Numbers We have seen that there are an infinite number of rational numbers and an infinite number of irrationals. For every rational number, there is a corresponding point on the number line, and, for every irrational number, there is a corresponding point on the number line. All of these points, taken together, make up the real number line. Since there are no more holes in this line, we say that the real number line is now complete. The completeness property of real numbers may be stated as follows: Every point on the real number line corresponds to a real number, and every real number corresponds to a point on the real number line. Ordering Real Numbers There are two ways in which we can order real numbers: 1. Use a number line. On the standard horizontal real number line, the graph of the greater number is always to the right of the graph of the smaller number. 26 Number Systems 2. Use decimals. Given any two real numbers that are not equal, we can express them in decimal form (even using rational approximations) to see which is greater. EXAMPLE 1 The number line that was first seen in Section 1-1 is repeated below. –2 13 – 6 –1 2– 0 — –0.43 .8 Of the numbers shown here, tell which are: a. counting numbers b. whole numbers c. integers d. rational numbers e. irrational numbers f. real numbers. Solution a. Counting numbers: b. Whole numbers: c. Integers: d. Rational Numbers: e. Irrational numbers f. Real numbers: All: 1, 2, 3, 4 0, 1, 2, 3, 4 2, 1, 0, 1, 2, 3, 4 213 6 , 22, 21, 20.43, 0, 1 2, 1, 2, 23 4, 3, 3.8, 4 3, p 2, 2 " " 213 6 , 22, 2 " 2, 21, 20.43, 0, 1 2, 1, 3, 2, 23 4, 3, p, 3.8, 4 " EXAMPLE 2 Order these real numbers from least to greatest, using the symbol . 0.3 0.3 " 0.3 Solution STEP 1. Write each real number in decimal form: 0.3 0.3000000 . . . 0.3 ≈ 0.547722575 (a rational approximation, displayed on a calculator) " 0.3 0.3333333 . . . STEP 2. Compare these decimals: 0.3000000 . . . 0.3333333 . . . 0.547722575 STEP 3. Replace each decimal with the number in its original form: 0.3 0.3 0.3 " Answer 0.3 0.3 0.3 " The Real Numbers 27 EXERCISES Writing About Mathematics 1. There are fewer than 6 persons in my family. The board is less than 6 feet long. Each of the given statements can be designated by the inequality x 6. How are the numbers that make the first statement true different from those that make the second statement true? How are they the same? 2. Dell said that it is impossible to decide whether p is larger or smaller than 10 because the calculator gives only rational approximations for these numbers. Do you agree with Dell? Explain. " 3. The decimal form of a real number consists of two digits that repeat for the first one- hundred decimal places. The digits in the places that follow the one-hundredth decimal place are random, form no pattern, and do not terminate. Is the number rational or irrational? Explain. Developing Skills 4. Twelve numbers have been placed on a number line as shown here. –2 3– – –2.7 –1 0 –0.63 1 0.5 1 3 2 π 2 6 Of these numbers, tell which are: a. counting numbers b. whole numbers c. integers d. rational numbers e. irrational numbers f. real numbers 5. Given the following series of numbers: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 " " " " " " " " " " Of these ten numbers, tell which is (are): a. rational b. irrational c. real 6. Given the following series of numbers: p, 2p, 3p, 4p, 5p Of these five numbers, tell which is (are): a. rational b. irrational c. real In 7–18, determine, for each pair, which is the greater number. 7. 2 or 2.5 11. 0.7 or 0.7 15. 3.14 or p 8 8. 8 or " 12. 5.6 or 5.9 16. 0.5 or 0.5 " 0.2 or 0.22 9. 13. 0.43 or 0.431 2 or 1.414 17. " 10. 0.2 or 0.23 14. 0.21 or 0.2 18. p or .22 7 28 Number Systems In 19–24, order the numbers in each group from least to greatest by using the symbol . 19. 0.202, 0.2 , 0.2022 20. 0.4 , 0.45, 0.4499 21. 0.67 , 0.6 , 0.667 22. 2 " 2, 2 " 3, 21.5 23. 0.5, 0.5 , 0.3 " 24. p, , 3.15 10 " In 25–34, tell whether each statement is true or false. 25. Every real number is a rational number. 26. Every rational number is a real number. 27. Every irrational number is a real number. 28. Every real number is an irrational number. 29. Every rational number corresponds to a point on the real number line. 30. Every point on the real number line corresponds to a rational number. 31. Every irrational number corresponds to a point on the real number line. 32. Every point on the real number line corresponds to an irrational number. 33. Some numbers are both rational and irrational. 34. Every repeating decimal corresponds to a point on the real number line. Hands-On Activity a. Using a cloth or paper tape measure, find, as accurately as you can, the distance across and the distance around the top of a can or other object that has a circular top. If you do not have a tape measure, fit a narrow strip of paper around the circular edge and measure the length of the strip with a yardstick. b. Divide the measure of the circumference, the distance around the circular top, by the measure of the diameter, the distance across the circular top at its center. c. Repeat steps a and b for other circular objects and compare the quotients obtained in step b. Compare your results from step b with those of other members of your class. What conclusions can you draw? 1-5 NUMBERS AS MEASUREMENTS In previous sections, we defined the subsets of the real numbers. When we use a counting number to identify the number of students in a class or the number of cars in the parking lot, these numbers are exact. However, to find the length of a block of wood, we must use a ruler, tape measure, or some other measuring instrument. The length that we find is dependent upon the instrument we use to measure and the care with which we make the measurement. Numbers as Measurements 29 For example, in the diagram, a block of wood is placed along the edge of a ruler that is marked in tenths of an inch. We might say that the block of wood is 2.7 inches in length but is this measure exact? Inches 1 2 3 All measurements are approximate. When we say that the length of the block of wood is 2.7 inches, we mean that it is closer to 2.7 inches than it is to 2.6 inches or to 2.8 inches. Therefore, the true measure of the block of wood whose length is given as 2.7 inches is between 2.65 and 2.75 inches. In other words, the true measure is less than 0.05 inches from 2.7 and can be written as 2.7 0.05 inches. The value 0.05 is called the greatest possible error (GPE) of measurement and is half of the place value of the last digit. Significant Digits The accuracy of measurement is often indicated in terms of the number of significant digits. Significant digits are those digits used to determine the measure and excludes those zeros that are used as place holders at the beginning of a decimal fraction and at the end of an integer. Rules for Determining Significant Digits RULE 1 All nonzero digits are significant. 135.6 has four significant digits. All digits are significant. RULE 2 All zeros between significant digits are significant. 130.6 has four significant digits. The zero is significant because it is between significant digits. RULE 3 All zeros at the end of a decimal fraction are significant. 135.000 has six significant digits. The three zeros at the end of the decimal fraction are significant. Zeros that precede the first nonzero digit in a decimal fraction are RULE 4 not significant. 0.00424 has three significant digits. The zeros that precede the nonzero digits in the decimal fraction are placeholders and are not significant. 30 Number Systems Zeros at the end of an integer may or may nor be significant. RULE 5 Sometimes a dot is placed over a zero if it is significant. 4,500 has two significant digits. Neither zero is significant. 4,50˙ 0 has three significant digits. The zero in the tens place is significant but the zero in the ones place is not. 4,500˙ has four significant digits. The zero in the ones place is significant. Therefore, the zero in the tens place is also significant because it is between significant digits. In any problem that uses measurement, the rules of greatest possible error and significant digits are used to determine how the answer should be stated. We can apply these rules to problems of perimeter and area. Recall the formulas for perimeter and area that you learned in previous courses. Let P represent the perimeter of a polygon, C the circumference of a circle, and A the area of any geometric figure. Triangle Rectangle Square Circle P a b c P 2l 2w P 4s C pd or C 2pr 2bh A 5 1 A lw A s2 A pr2 Precision The precision of a measurement is the place value of the last significant digit in the number. The greatest possible error of a measurement is one-half the place value of the last significant digit. In the measurement 4,500 feet, the last significant digit is in the hundreds place. Therefore, the greatest possible error is 100 50. We can write the measurement as 4,500 50 feet. One number is said to be more precise than another if the place value of its last significant digit is smaller. F
or example, 3.40 is more precise than 3.4 because 3.40 is correct to the nearest hundredth and 3.4 is correct to the nearest tenth. 1 2 When measures are added, the sum can be no more precise than the least precise number of the given values. For example, how should the perimeter of a triangle be stated if the measures of the sides are 34.2 inches, 27.52 inches, and 29 inches? P a b c P 34.2 27.52 29 90.72 Since the least precise measure is 29 which is precise to the nearest integer, the perimeter of the triangle should be given to the nearest integer as 91 inches. Numbers as Measurements 31 Accuracy The accuracy of a measure is the number of significant digits in the measure. One number is said to be more accurate than another if it has a larger number of significant digits. For example, 0.235 is more accurate than 0.035 because 0.235 has three significant digits and 0.035 has two, but 235 and 0.235 have the same degree of accuracy because they both have three significant digits. When measures are multiplied, the product can be no more accurate than the least accurate of the given values. For example, how should the area of a triangle be stated if the base measures 0.52 meters and the height measures 0.426 meters? A A (0.52)(0.426) 0.5(0.52)(0.426) 0.11076 1 2bh 1 2 Since the less accurate measure is 0.52, which has two significant digits, the area should be written with two significant digits as 0.11 square meters. Note that the 1 or 0.5 is not a measurement but an exact value that has been determined by 2 counting or by reasoning and therefore is not used to determine the accuracy of the answer. One last important note: when doing multi-step calculations, make sure to keep at least one more significant digit in intermediate results than is needed in the final answer. For example, if a computation requires three significant digits, then use at least four significant digits in your calculations. Otherwise, you may encounter what is known as round-off error, which is the phenomena that occurs when you discard information contained in the extra digit, skewing your calculations. In this text, you will often be asked to find the answer to an exercise in which the given numbers are thought of as exact values and the answers are given as exact values. However, in certain problems that model practical applications, when the given data are approximate measurements, you may be asked to use the precision or accuracy of the data to determine how the answer should be stated. EXAMPLE 1 State the precision and accuracy of each of the following measures. a. 5.042 cm b. 12.0 ft c. 93,000,000 mi 32 Number Systems Solution a. 5.042 cm b. 12.0 ft c. 93,000,000 mi EXAMPLE 2 Precision thousandths tenths millions Accuracy 4 significant digits 3 significant digits 2 significant digits Of the measurements 125 feet and 6.4 feet, a. which is the more precise? b. which is the more accurate? Solution The measurement 125 feet is correct to the nearest foot, has an error of 0.5 feet, and has three significant digits. The measurement 6.4 feet is correct to the nearest tenth of a foot, has an error of 0.05 feet, and has two significant digits. Answers a. The measure 6.4 feet is more precise because it has the smaller error. b. The measure 125 feet is more accurate because it has the larger number of significant digits. EXAMPLE 3 The length of a rectangle is 24.3 centimeters and its width is 18.76 centimeters. Using the correct number of significant digits in the answer, express a. the perimeter b. the area. Solution a. Use the formula for the perimeter of a rectangle. P 2l 2w P 2(24.3) 2(18.76) P 86.12 Perimeter is a sum since 2l means l l and 2w means w w. The answer should be no more precise than the least precise measurement. The least precise measurement is 24.3, given to the nearest tenth. The perimeter should be written to the nearest tenth as 86.1 centimeters. b. To find the area of a rectangle, multiply the length by the width. A lw A (24.3)(18.76) A 455.868 Area is a product and the answer should be no more accurate than the least accurate of the given dimensions. Since there are three significant digits in 24.3 and four significant digits in 18.76, there should be three significant digits in the answer. Therefore, the area should be written as 456 square centimeters. Answers a. 86.1 cm b. 456 sq cm Numbers as Measurements 33 EXERCISES Writing about Mathematics 1. If 12.5 12.50, explain why a measure of 12.50 inches is more accurate and more precise than a measurement of 12.5 inches. 2. A circular track has a radius of 63 meters. Mario rides his bicycle around the track 10 times. Mario multiplied the radius of the track by 2p to find the circumference of the track. He said that he rode his bicycle 4.0 kilometers. Olga said that it would be more correct to say that he rode his bicycle 4 kilometers. Who is correct? Explain your answer. Developing Skills In 3–10, for each of the given measurements, find a. the accuracy b. the precision c. the error. 3. 24 in. 4. 5.05 cm 5. 2,400 ft 6. 454 lb 7. 0.0012 kg 8. 1.04 yd 9. 1.005 m 10. 900 mi In 11–14, for each of the following pairs, select the measure that is a. the more precise b. the more accurate. 11. 57 in. and 4,250 in. 12. 2.50 ft and 2.5 ft 13. 0.0003 g and 32 g 14. 500 cm and 0.055 m Applying Skills In 15–18, express each answer to the correct number of significant digits. 15. Alicia made a square pen for her dog using 72.4 feet of fencing. a. What is the length of each side of the pen? b. What is the area of the pen? 16. Corinthia needed 328 feet of fencing to enclose her rectangular garden. The length of the garden is 105 feet. a. Find the width of the garden. b. Find the area of the garden. 17. Brittany is making a circular tablecloth. The diameter of the tablecloth is 10.5 inches. How much lace will she need to put along the edge of the tablecloth? 18. The label on a can of tomatoes is a rectangle whose length is the circumference of the can and whose width is the height of the can. If a can has a diameter of 7.5 centimeters and a height of 10.5 centimeters, what is the area of the label? 34 Number Systems CHAPTER SUMMARY A set is a collection of distinct objects or elements. The counting numbers or natural numbers are {1, 2, 3, 4, . . .}. The whole numbers are {0, 1, 2, 3, 4, . . .}. The integers are {. . . , 4, 3, 2, 1, 0, 1, 2, 3, 4, . . .}. These sets of numbers form the basis for a number line, on which the length of a segment from 0 to 1 is called the unit measure of the line. a The rational numbers are all numbers that can be expressed in the form b where a and b are integers and b 0. Every rational number can be expressed as a repeating decimal or as a terminating decimal (which is actually a decimal in which 0 is repeated). The irrational numbers are decimal numbers that do not terminate and do not repeat. On calculators and in the solution of many problems, rational approximations are used to show values that are close to, but not equal to, irrational numbers. The real numbers consist of all rational numbers and all irrational numbers taken together. On a real number line, every point represents a real number and every real number is represented by a point. The precision of a measurement is determined by the place value of the last significant digit. The accuracy of a measurement is determined by the number of significant digits in the measurement. VOCABULARY 1-1 Mathematics • Real number • Number • Numeral • Counting numbers • Natural numbers • Successor • Whole numbers • Set • Finite set • Digit • Infinite set • Empty set • Null set • Numerical expression • Simplify • Negative numbers • Opposites • Integers • Subset • Number line • Graph • Standard number line • Unit measure • Absolute value • Inequality 1-2 Rational numbers • Everywhere dense • Common fraction • Decimal fraction • Terminating decimal • Repeating decimal • Periodic decimal 1-3 Irrational numbers • Square • Square root • Radical sign • Rational approximation • Pi (p) • Round 1-4 Real numbers • Real number line • Completeness property of real numbers 1-5 Greatest possible error (GPE) • Significant digits • Precision • Accuracy REVIEW EXERCISES In 1–5, use a calculator to evaluate each expression and round the result to the nearest hundredth. Review Exercises 35 1. 29.73 14.6 " 6. Order the numbers 5, 3, and 1 using the symbol . 2. 38 9 3. 12.232 4. 216 5. p 12 In 7–10, state whether each sentence is true of false. 7. 7 8 8. –7 2 9. 4 8 10. 9 9 In 11–16, write each rational number in the form , where a and b are integers and b 0. a b 11. 0.9 12. 0.45 13. 81 2 14. 14 15. 0.3 16. 63 17. Find a rational number between 19.9 and 20. In 18–22, tell whether each number is rational or irrational. 18. 0.64 19. 22. 0.040040004 . . . 6 " 20. 64 " 21. p In 23–27, write a rational approximation of each given number: a. as shown on a calculator display b. rounded to the nearest hundredth. 23. 11 " 24. 0.7 " 25. 905 " 1,599 26. " 27. p In 28–32, determine which is the greater number in each pair. 20 28. 5 or " 31. 0.41 or 0.4 32. 0.12 or 0.121 29. 12 8 or 12 8 30. 3.2 or p In 33–37, tell whether each statement is true or false. 33. Every integer is a real number. 34. Every rational number is an integer. 35. Every whole number is a counting number. 36. Every irrational number is a real number. 37. Between 0 and 1, there is an infinite number of rational numbers. 38. Draw a number line, showing the graphs of these numbers: 0, 1, 4, 3, 1.5, and p. 36 Number Systems In 39 and 40, use the given number line where the letters are equally spaced 39. Find the real number that corresponds to each point indicated by a letter shown on the number line when C 0 and E 1. 40. Between what two consecutive points on this number line is the graph of: a. 1.8 b. 0.6 c. 2 d. p e. 6 " " 41. The distance across a circular fountain (the diameter of the fountain) is 445 centimeters. The distance in centimeters around the fountain (the circu
mference of the fountain) can be found by multiplying 445 by p. a. Find the circumference of the fountain in centimeters. Round your answer to the nearest ten centimeters. b. When the circumference is rounded to the nearest ten centimeters, are the zeros significant? Exploration Using only the digits 5 and 6, and without using a radical sign: a. Write an irrational number. b. Write three irrational numbers that are between 5 and 6 in increasing order. c. Write three irrational numbers that are between 0.55 and 0.56 in increasing order. d. Write three irrational numbers that are between 0.556 and 0.556 in increas- ing order. OPERATIONS AND PROPERTIES Jesse is fascinated by number relationships and often tries to find special mathematical properties of the five-digit number displayed on the odometer of his car. Today Jesse noticed that the number on the odometer was a palindrome and an even number divisible by 11, with 2 as three of the digits.What was the five-digit reading? (Note: A palindrome is a number, word, or phrase that is the same read left to right as read right to left, such as 57375 or Hannah.) In this chapter you will review basic operations of arithmetic and their properties. You will also study operations on sets. CHAPTER 2 CHAPTER TABLE OF CONTENTS 2-1 Order of Operations 2-2 Properties of Operations 2-3 Addition of Signed Numbers 2-4 Subtraction of Signed Numbers 2-5 Multiplication of Signed Numbers 2-6 Division of Signed Numbers 2-7 Operations With Sets 2-8 Graphing Number Pairs Chapter Summary Vocabulary Review Exercises Cumulative Review 37 38 Operations and Properties 2-1 ORDER OF OPERATIONS The Four Basic Operations in Arithmetic Bicycles have two wheels. Bipeds walk on two feet. Biceps are muscles that have two points of origin. Bilingual people can speak two languages. What do these bi-words have in common with the following examples? 6.3 0.9 7.2 113 7 5 92 7 7 2 21 21.4 3 64.2 9 4 2 5 41 2 The prefix bi- means “two.” In each example above, an operation or rule was followed to replace two rational numbers with a single rational number. These familiar operations of addition, subtraction, multiplication, and division are called binary operations. Each of these operations can be performed with any pair of rational numbers, except that division by zero is meaningless and is not allowed. In every binary operation, two elements from a set are replaced by exactly one element from the same set. There are some important concepts to remember when working with binary operations: 1. A set must be identified, such as the set of whole numbers or the set of rational numbers. When no set is identified, use the set of all real numbers. 2. The rule for the binary operation must be clear, such as the rules you know for addition, subtraction, multiplication, and division. 3. The order of the elements is important. Later in this chapter, we will use the notation (a, b) to indicate an ordered pair in which a is the first element and b is the second element. For now, be aware that answers may be different depending on which element is first and which is second. Consider subtraction. If 8 is the first element and 5 is the second element, then: 8 5 3. But if 5 is the first element and 8 is the second element, then: 5 8 3 4. Every problem using a binary operation must have an answer, and there must be only one answer. We say that each answer is unique, meaning there is one and only one answer. DEFINITION A binary operation in a set assigns to every ordered pair of elements from the set a unique answer from the set. Note that, even when we find the sum of three or more numbers, we still add only two numbers at a time, indicating the binary operation: 4 9 7 (4 9) 7 13 7 20 Order of Operations 39 Factors When two or more numbers are multiplied to give a certain product, each number is called a factor of the product. For example: • Since 1 16 16, then 1 and 16 are factors of 16. • Since 2 8 16, then 2 and 8 are factors of 16. • Since 4 4 16, then 4 is a factor of 16. • The numbers 1, 2, 4, 8, and 16 are all factors of 16. Prime Numbers A prime number is a whole number greater than 1 that has no whole number factors other than itself and 1. The first seven prime numbers are 2, 3, 5, 7, 11, 13, 17. Whole numbers greater than 1 that are not prime are called composite numbers. Composite numbers have three or more whole number factors. Some examples of composite numbers are 4, 6, 8, 9, 10. Bases, Exponents, Powers When the same number appears as a factor many times, we can rewrite the expression using exponents. For example, the exponent 2 indicates that the factor appears twice. In the following examples, the repeated factor is called a base. 4 4 16 can be written as 42 16. 42 is read as “4 squared,” or “4 raised to the second power,” or “the second power of 4.” Exponent Base 42 = 16 Power The exponent 3 indicates that a factor is used three times. 4 4 4 64 can be written as 43 64. 43 is read as “4 cubed,” or “4 raised to the third power,” or “the third power of 4.” Exponent Base 43 = 64 Power 40 Operations and Properties The examples shown above lead to the following definitions: DEFINITION A base is a number that is used as a factor in the product. An exponent is a number that tells how many times the base is to be used as a factor. The exponent is written, in a smaller size, to the upper right of the base. A power is a number that is a product in which all of its factors are equal. A number raised to the first power is equal to the number itself, as in 61 6. Also, when no exponent is shown, the exponent is 1, as in 9 91. EXAMPLE 1 Compute the value of 45. Solution 4 4 4 4 4 1,024 Calculator Solution Use the exponent key, ^ , on a calculator. ENTER: 4 ^ 5 ENTER DISPLAY: 4 ^ 5 1 0 2 4 Answer 1,024 EXAMPLE 2 Find 2 3 A Solution a. The exact value is B 3 a. as an exact value b. as a rational approximation. a fraction 27 3 2 3 A B b. Use a calculator. ENTER: 2 ( 3 ) ^ 3 ENTER DISPLAY Note: The exact value is a rational number that can also be written as the repeating decimal 0.296 . Answers a. 8 27 b. 0.2962962963 Order of Operations 41 Computations With More Than One Operation When a numerical expression involves two or more different operations, we need to agree on the order in which they are performed. Consider this example: 11 3 2 Suppose that one person multiplied first. Suppose another person subtracted first. 11 3 2 11 6 5 11 3 2 8 2 16 In order that there will be one and only one correct answer to problems like this, mathematicians have agreed to follow this order of operations: Who is correct? 1. Simplify powers (terms with exponents). 2. Multiply and divide, from left to right. 3. Add and subtract, from left to right. Therefore, we multiply before we subtract, and 11 3 2 11 6 5 is correct. A different problem involving powers is solved in this way: 1. Simplify powers: 5 23 + 3 5 8 3 2. Multiply and divide: 3. Add and subtract: 40 3 43 Expressions with Grouping Symbols In mathematics, parentheses ( ) act as grouping symbols, giving different meanings to expressions. For example, (4 6) 7 means “add 7 to the product of 4 and 6,” while 4 (6 7) means “multiply the sum of 6 and 7 by 4.” When simplifying any numerical expression, always perform the operations within parentheses first. (4 6) 7 24 7 31 4 (6 7) 4 13 52 Besides parentheses, other symbols are used to indicate grouping, such as brackets [ ]. The expressions 2(5 9) and 2[5 9] have the same meaning: 2 is multiplied by the sum of 5 and 9. A bar, or fraction line, also acts as a symbol of grouping, telling us to perform the operations in the numerator and/or denominator first. 20 2 8 3 5 12 11 2 42 Operations and Properties However, when entering expressions such as these into a calculator, the line of the fraction is usually entered as a division and a numerator or denominator that involves an operation must be enclosed in parentheses. ENTER: ( 20 8 ) 3 ENTER DISPLAY: ( 2 0 8 ) / 3 4 ENTER: 6 ( 3 1 ) ENTER DISPLAY When there are two or more grouping symbols in an expression, we perform the operations on the numbers in the innermost symbol first. For example: 5 2[6 (3 1)3] = 5 2[6 23] = 5 2[6 8] = 5 2[14] = 5 28 = 33 Procedure To simplify a numerical expression, follow the correct order of operations: 1. Simplify any numerical expressions within parentheses or within other grouping symbols, starting with the innermost. 2. Simplify any powers. 3. Do all multiplications and divisions in order from left to right. 4. Do all additions and subtractions in order from left to right. EXAMPLE 3 Simplify the numerical expression 80 4(7 5). Solution Remember that, in the given expression, 4(7 5) means 4 times the value in Order of Operations 43 the parentheses. How to Proceed (1) Write the expression: (2) Simplify the value within the parentheses: (3) Multiply: (4) Subtract: 80 4(7 5) 80 4(2) 80 8 72 Calculator Solution ENTER: 80 4 ( 7 5 ) ENTER DISPLAY Answer 72 EXERCISES Writing About Mathematics 1. Explain why 2 is the only even prime. 2. Delia knows that every number except 2 that ends in a multiple of 2 is composite. Therefore, she concludes that every number except 3 that ends in a multiple of 3 is composite. Is Delia correct? Explain how you know. Developing Skills In 3–10, state the meaning of each expression in part a and in part b, and simplify the expression in each part. 3. a. 20 (6 1) 5. a. 12 (3 0.5) 7. a. (12 8 ) 4 9. a. 7 52 b. 20 6 1 b. 12 3 0.5 b. 12 8 4 b. (7 5)2 4. a. 18 (4 3) 6. a. 15 (2 1) 8. a. 48 (8 4 ) 10. a. 4 32 11. Noella said that since the line of a fraction indicates division, 10 15 5 3. Do you agree with Noella? Explain why or why not. 10 3 15 5 3 3 b. 18 4 3 b. 15 2 1 b. 48 8 4 b. (4 3)2 is the same as 44 Operations and Properties In 12–15: a. Find, in each case, the value of the three given powers. b. Name, in each case, the expression that has the greatest value. 12. 52, 53, 54 13. (0.5)2, (0.5)3, (0.5)4 14. (0.5)2, (0.6)2, (0.7)2 15. (1.1
)2, (1.2)2, (1.3)2 In 16–23: a. List all of the whole numbers that are factors of each of the given numbers. b. Is the number prime, composite, or neither? 16. 82 20. 1 17. 101 21. 808 18. 71 22. 67 19. 15 23. 397 Applying Skills In 24–28, write a numerical expression for each of the following and find its value to answer the question. 24. What is the cost of two chocolate chip and three peanut butter cookies if each cookie costs 28 cents? 25. What is the cost of two chocolate chip cookies that cost 30 cents each and three peanut but- ter cookies that cost 25 cents each? 26. How many miles did Ms. McCarthy travel if she drove 30 miles per hour for hour and 55 3 4 miles per hour for 11 2 hours? 27. What is the cost of two pens at $0.38 each and three notebooks at $0.69 each? 28. What is the cost of five pens at $0.29 each and three notebooks at $0.75 each if ordered from a mail order company that adds $1.75 in postage and handling charges? In 29–30, use a calculator to find each answer. 29. The value of $1 invested at 6% for 20 years is equal to (1.06)20. Find, to the nearest cent, the value of this investment after 20 years. 30. The value of $1 invested at 8% for n years is equal to (1.08)n. How many years will be required for $1 invested at 8% to double in value? (Hint: Guess at values of n to find the value for which (1.08)n is closest to 2.00.) 31. In each box insert an operational symbol +, –, , , and then insert parentheses if needed to make each of the following statements true. a. 3 □ 2 □ 1 4 d. 1 □ 3 □ 1 4 e-2 PROPERTIES OF OPERATIONS Properties of Operations 45 When numbers behave in a certain way for an operation, we describe this behavior as a property. You are familiar with these operations from your study of arithmetic. As we examine the properties of operations, no proofs are given, but the examples will help you to see that these properties make sense and to identify the sets of numbers for which they are true. The Property of Closure A set is said to be closed under a binary operation when every pair of elements from the set, under the given operation, yields an element from that set. 1. Add any two numbers. The sum of two whole numbers is a whole number. 23 11 34 7.8 4.8 12.6 The sum of two rational numbers is a rational number. The sum of p and its opposite, p, two irrational nump (p) 0 bers, is 0, a rational number. Even though the sum of two irrational numbers is usually an irrational number, the set of irrational numbers is not closed under addition. There are some pairs of irrational numbers whose sum is not an irrational number. However, p, p, 0, and each of the other numbers used in these examples are real numbers and the sum of two real numbers is a real number. The sets of whole numbers, rational numbers, and real numbers are each closed under addition. 2. Multiply any two numbers. (2)(4 The product of two whole numbers is a whole number. The product of two rational numbers is a rational number. 2 , two irrational numThe product of " bers, is 2, a rational number. and " 2 Though the product of two irrational numbers is usually an irrational number, there are some pairs of irrational numbers whose product is not an irrational number. The set of irrational numbers is not closed under multiplication. 2 , 2, and each of the other numbers used in these examples are real However, numbers and the product of two real numbers is a real number. " The sets of whole numbers, rational numbers, and real numbers are each closed under multiplication. 46 Operations and Properties 3. Subtract any two numbers. 7 12 5 12.7 8.2 4.5 3 2 " 3 5 0 " The difference of two whole numbers is not a whole number, but these whole numbers are also integers and the difference between two integers is an integer. The difference of two rational numbers is a rational number. 3 The difference of , two irrational num" bers, is 0, a rational number. 3 " and Even though the difference of two irrational numbers is usually an irrational number, there are some pairs of irrational numbers whose difference is not an irrational number. The set of irrational numbers is not closed under 3 , 0, and each of the other numbers used in these examsubtraction. However, ples are real numbers and the difference of two real numbers is a real number. " The sets of integers, rational numbers, and real numbers are each closed under subtraction. 4. Divide any two numbers by a nonzero number. (Remember that division by 0 is not allowed.) 9 2 4. The quotient of two whole numbers or two integers is not always a whole number or an integer. The quotient of two rational numbers is a rational number. 5 4 " 5 5 1 " 5 The quotient of " bers, is 1, a rational number. 5 " and , two irrational num- Though the quotient of two irrational numbers is usually an irrational number, there are some pairs of irrational numbers whose quotient is not an irrational number. The set of irrational numbers is not closed under division. 5 , 1, and each of the other numbers used in these examples are real However, numbers and the quotient of two nonzero real numbers is a nonzero real number. " The sets of nonzero rational numbers, and nonzero real numbers are each closed under division. Later in this book, we will study operations with signed numbers and operations with irrational numbers in greater detail. For now, we will simply make these observations: The set of whole numbers is closed under the operations of addition and multiplication. Properties of Operations 47 The set of integers is closed under the operations of addition, subtraction, and multiplication. The set of rational numbers is closed under the operations of addition, subtraction, and multiplication, and the set of nonzero rational numbers is closed under division. The set of real numbers is closed under the operations of addition, subtraction, and multiplication, and the set of nonzero real numbers is closed under division. Commutative Property of Addition When we add rational numbers, we assume that we can change the order in which two numbers are added without changing the sum. For example, 4 5 5 4 and the commutative property of addition . These examples illustrate In general, we assume that for every number a and every number b: a b b a Commutative Property of Multiplication In the same way, when we multiply rational numbers, we assume that we can change the order of the factors without changing the product. 2 3 1 1 For example, 5 4 4 5, and the commutative property of multiplication. 4 5 1 4 3 1 2 . These examples illustrate In general, we assume that for every number a and every number b: a b b a Subtraction and division are not commutative, as shown by the following counterexample. 12 7 7 12 5 5 12 3 3 12 4 3 12 Associative Property of Addition Addition is a binary operation; that is, we add two numbers at a time. If we wish to add three numbers, we find the sum of two and add that sum to the third. For example: 48 Operations and Properties 2 5 8 (2 5) 8 or 2 5 8 2 (5 8) 7 8 15 2 13 15 The way in which we group the numbers to be added does not change the sum. Therefore, we see that (2 5) 8 2 (5 8). This example illustrates the associative property of addition. In general, we assume that for every number a, every number b, and every number c: (a b) c a (b c) Associative Property of Multiplication In a similar way, to find a product that involves three factors, we first multiply any two factors and then multiply this result by the third factor. We assume that we do not change the product when we change the grouping. For example: 5 4 2 (5 4) 2 or 5 4 2 5 (4 2) 20 2 40 5 8 40 Therefore, (5 4) 2 5 (4 2). This example illustrates the associative property of multiplication. In general, we assume that for every number a, every number b, and every number c: a (b c) (a b) c Subtraction and division are not associative, as shown in the following coun- terexamples. (15 4) 3 15 (4 3) 11 3 15 1 8 14 (8 4) 2 8 (4 2) 2 2 8 2 1 4 The Distributive Property We know 4(3 2) 4(5) 20, and also 4(3) 4(2) 12 8 20. Therefore, we see that 4(3 2) 4(3) 4(2). This result can be illustrated geometrically. Recall that the area of a rectan- gle is equal to the product of its length and its width. Properties of Operations 49 3 2 4 4(3+2) = 4 4(3) + 4 4(2) (3+2) 3 2 This example illustrates the distributive property of multiplication over addition, also called the distributive property. This means that the product of one number times the sum of a second and a third number equals the product of the first and second numbers plus the product of the first and third numbers. In general, we assume that for every number a, every number b, and every number c: a(b c) ab ac and (a b)c ac bc The distributive property is also true for multiplication over subtraction: a(b c) ab ac and (a b)c ac bc The distributive property can be useful for mental computations. Observe how we can use the distributive property to find each of the following products as a sum: 1. 6 23 6(20 3) 6 20 6 3 120 18 138 9 3 31 3 2. 3. 6.5 8 (6 0.5)8 6 8 0.5 8 48 4 52 1 9 3 9 27 3 30 3 311 3 9 B A Working backward, we can also use the distributive property to change the form of an expression from a sum or a difference to a product: 1. 5(12) 5(8) 5(12 8) 5(20) 100 2. 7(14) 7(4) 7(14 4) 7(10) 70 Addition Property of Zero and the Additive Identity Element The equalities 5 + 0 5 and 0 2.8 2.8 are true. They illustrate that the sum of a rational number and zero is the number itself. These examples lead us to observe that: 50 Operations and Properties 1. The addition property of zero states that for every number a: a 0 a and 0 a a 2. The identity element of addition, or the additive identity, is 0. Thus, for any number a: If a x a, or if x a a, it follows that x 0. Additive Inverses (Opposites) When we first studied integers, we learned about opposites. For example, the opposite of 4 is 4, and the opposite of 10 is 10. Every rational number a has an opposite, –a, such that their sum
is 0, the identity element in addition. The opposite of a number is called the additive inverse of the number. In general, for every rational number a and its opposite a: a (a) 0 On a calculator, the (-) key, is used to enter the opposite of a number. The following example shows that the opposite of 4.5 is 4.5. ENTER: (-) (-) 4.5 ENTER DISPLAY: - - 4 . 5 4 . 5 Multiplication Property of One and the Multiplicative Identity Element The sentences 5 1 5 and 1 4.6 4.6 are true. They illustrate that the product of a number and one is the number itself. These examples lead us to observe that: 1. The multiplication property of one states that for every number a: a 1 a and 1 a a 2. The identity element of multiplication, or the multiplicative identity, is 1. Properties of Operations 51 Multiplicative Inverses (Reciprocals) When the product of two numbers is 1 (the identity element for multiplication), then each of these numbers is called the multiplicative inverse or reciprocal of the other. Consider these examples: 4 ? 1 4 5 1 1 The reciprocal of 4 is . 4 The reciprocal of 1 4 is 4. 21 The multiplicative inverse of B A The multiplicative inverse of 21 2 or 5 2 2 is . 5 2 5 5 is or 2 21 2 . Since there is no number that, when multiplied by 0, gives 1, the number 0 has no reciprocal, or no multiplicative inverse. In general, for every nonzero number a, there is a unique number 1 a such that: a ? 1 a 5 1 On the calculator, a special key, displays the reciprocal. For example, if each of the numbers shown above is entered and the reciprocal key is pressed, the reciprocal appears in decimal form. x1 ENTER: 4 x1 ENTER ENTER: ( 5 2 ) x1 ENTER DISPLAY: 4 – 1 DISPLAY Note: Parentheses must be used when calculating the reciprocal of a fraction. For many other numbers, however, the decimal form of the reciprocal is not shown in its entirety in the display. For example, we know that the reciprocal of 1 6 is , but what appears is a rational approximation of . 6 1 6 ENTER: 6 x1 ENTER DISPLAY The display shows the rational approximation of rounded to the last decimal place displayed by the calculator. A calculator stores more decimal places in its operating system than it has in its display. The decimal displayed times the original number will equal 1. 1 6 52 Operations and Properties To display the reciprocal of 6 from the example on the previous page as a common fraction, we use 6 x1 MATH ENTER ENTER . Multiplication Property of Zero The sentences 7 0 0 and 0 0 are true. They illustrate that the product of a rational number and zero is zero. This property is called the multiplication property of zero: 3 4 In general, for every number a: a 0 0 and 0 a 0 EXAMPLE 1 Write, in simplest form, the opposite (additive inverse) and the reciprocal (multiplicative inverse) of each of the following: a. 7 b. d. 0.2 e. p c. 23 8 11 5 Solution Number Opposite Reciprocal a. 7 b. 23 8 –7 3 8 1 7 28 3 5 222 3 c. 5 5 6 11 5 211 5 5 2 6 5 d. 0.2 e. p 0.2 p 5 6 5 1 p EXAMPLE 2 Express 6t + t as a product and give the reason for each step. Solution Step Reason (1) 6t t 6t 1t (6 1)t 7t (2) (3) Multiplication property of 1. Distributive property. Addition. Answer 7t Properties of Operations 53 EXERCISES Writing About Mathematics 1. If x and y represent real numbers and xy x: a. What is the value of y if the equation is true for all x? Explain your answer. b. What is the value of x if the equation is true for all y? Explain your answer. 2. Cookies and brownies cost $0.75 each. In order to find the cost of 2 cookies and 3 brownies Lindsey added 2 3 and multiplied the sum by $0.75. Zachary multiplied $0.75 by 2 and then $0.75 by 3 and added the products. Explain why Lindsey and Zachary both arrived at the correct cost of the cookies and brownies. Developing Skills 3. Give the value of each expression. a. 9 0 1 (p) g. p b. 9 0 h. 1 4.5 A B c. 9 1 0 1 i. 7 " 2 3 3 0 1.63 3 0 d. j. e. 0 1 3 k. 2 3 5 " f. 1 0 3 l. 2 3 225 " In 4–13: a. Replace each question mark with the number that makes the sentence true. b. Name the property illustrated in each sentence that is formed when the replacement is made. 4. 8 6 6 ? 6. (3 9) 15 3 (9 ?) 8. (0.5 0.2) 0.7 0.5 (? 0.7) 10. (3 7) 5 (? 3) 5 12. 7(4 ?) 7(4) 5. 17 5 ? 17 7. 6(5 8) 6(5) ?(8) 9. 4 0 ? 11. (?)(8 2) (8 2)(9) 13. ?x x In 14–25: a. Name the additive inverse (opposite) of each number. b. Name the multiplicative inverse (reciprocal) of each number. 14. 17 21 3 20. 15. 1 21. 2p 16. –10 3 7 22. 17. 2.5 23. 1.780 18. –1.8 2 1 11 24. 19. 25. 1 9 3 5 71 In 26–31, state whether each sentence is a correct application of the distributive property. If you believe that it is not, state your reason. 26. 6(5 8) 6(5) 6(8) 28. 5 (8 6) (5 8) (5 6) 30. 14a 4a (14 4)a 10 5 10 3 1 2 1 1 2 1 1 1 27. 5 5 B 29. 3(x 5) 3x 3 5 31. 18(2.5) 18(2) 18(0.5) A 54 Operations and Properties In 32–35: a. Tell whether each sentence is true or false. b. Tell whether the commutative property holds for the given operation. 32. 357 19 19 357 34. 25 7 7 25 33. 2 1 1 2 35. 18(3.6) 3.6(18) In 36–39: a. Tell whether each sentence is true of false. b. Tell whether the associative property holds for the given operation 36. (73 68) 92 73 (68 92) 38. (19 8) 5 19 (8 5) 37. (24 6) 2 24 (6 2) 39. 9 (0.3 0.7) (9 0.3) 0.7 40. Insert parentheses to make each statement true. a. 3 2 1 3 3 d. 3 3 3 3 3 1 b. 4 3 2 2 3 e. 3 3 3 3 3 0 c. 8 8 8 8 8 8 f. 0 12 3 16 8 0 Applying Skills 41. Steve Heinz wants to give a 15% tip to the taxi driver. The fare was $12. He knows that 10% of $12 is $1.20 and that 5% would be half of $1.20. Explain how this information can help Steve calculate the tip. What mathematical property is he using to determine the tip? 42. Juana rides the bus to and from work each day. Each time she rides the bus the fare is $1.75. She works five days a week. To find what she will spend on bus fare each week, Juana wants to find the product 2(1.75)(5). Juana rewrote the product as 2(5)(1.75). a. What property of multiplication did Juana use when she changed 2(1.75)5 to 2(5)(1.75)? b. What is her weekly bus fare? 2-3 ADDITION OF SIGNED NUMBERS Adding Numbers That Have the Same Signs The number line can be used to find the sum of two numbers. Start at 0. To add a positive number, move to the right. To add a negative number, move to the left. EXAMPLE 1 Add 3 and 2. Addition of Signed Numbers 55 +3 +2 –1 0 +1 +2 +3 +4 +5 +6 Solution Start at 0 and move 3 units to the right to 3; then move 2 more units to the right, arriving at +5. Calculator Solution ENTER: 3 2 ENTER DISPLAY: 3 + 2 5 Answer 5 The sum of two positive integers is the same as the sum of two whole numbers. The sum +5 is a number whose absolute value is the sum of the absolute values of 3 and 2 and whose sign is the same as the sign of 3 and 2. EXAMPLE 2 Add 3 and 2. Solution Start at 0 and move 3 units to the left to 3: then move 2 more units to the left, arriving at 5. Calculator Solution ENTER: (-) 3 (-) 2 ENTER DISPLAY: - 3 + - 2 - 5 Answer 5 –2 –3 –6 –5 –4 –3 –2 –1 0 +1 The sum 5 is a number whose absolute value is the sum of the absolute values of 3 and 2 and whose sign is the same as the sign of 3 and 2. Examples 1 and 2 illustrate that the sum of two numbers with the same sign is a number whose absolute value is the sum of the absolute values of the numbers and whose sign is the sign of the numbers. Procedure To add two numbers that have the same sign: 1. Find the sum of the absolute values. 2. Give the sum the common sign. 56 Operations and Properties Adding Numbers That Have Opposite Signs EXAMPLE 3 Add: 3 and 2. Solution Start at 0 and move 3 units to the right to 3; then move 2 units to the left, arriving at 1. –2 +3 Calculator Solution ENTER: 3 (-) 2 ENTER –1 0 +1 +2 +3 +4 DISPLAY: 3 + - 2 1 Answer 1 This sum can also be found by using properties. In the first step, substitution is used, replacing (3) with the sum (1) (2). (3) ( 2) [(1) (2)] (2) Substitution (1) [(2) ( 2)] Associative property 1 0 1 Addition property of zero Addition property of opposites The sum 1 is a number whose absolute value is the difference of the absolute values of 3 and 2 and whose sign is the same as the sign of 3, the number with the greater absolute value. EXAMPLE 4 Add: 3 and 2. Solution Start at 0 and move 3 units to the left to 3; then move 2 units to the right, arriving at 1. +2 –3 Calculator Solution ENTER: (-) 3 2 ENTER –4 –3 –2 –1 0 +1 DISPLAY: - 3 + 2 - 1 Answer 1 Addition of Signed Numbers 57 This sum can also be found by using properties. In the first step, substitution is used, replacing (3) with the sum (1) (2). (3) (2) [(1) (2)] (2) Substitution (1) [(2) (2)] Associative property 1 0 1 Addition property of zero Addition property of opposites The sum 1 is a number whose absolute value is the difference of the absolute values of 3 and 2 and whose sign is the same as the sign of 3, the number with the greater absolute value. Examples 3 and 4 illustrate that the sum of a positive number and a negative number is a number whose absolute value is the difference of the absolute values of the numbers and whose sign is the sign of the number having the larger absolute value. Procedure To add two numbers that have different signs: 1. Find the difference of the absolute values of the numbers. 2. Give this difference the sign of the number that has the greater absolute value. 3. The sum is 0 if both numbers have the same absolute value. EXAMPLE 5 Find the sum of 233 4 and 11 4 . Solution How to Proceed (1) Since the numbers have different signs, find the difference of their absolute values: (2) Give the difference the sign of the number with the greater absolute value: 33 4 2 11 4 5 22 4 222 4 5 221 2 Answer 58 Operations and Properties Calculator Solution 33 The number 4 33 233 , is the sum of 3 and site of 4 4 , 23 4 . is the sum of the whole number 3 and the fraction . The oppo- 3 4 Enclose the absolute value of the sum of 3 and 3 4 in parentheses. ENTER: (-) ( 3 3 4 ) 1 1 4 ENTER DISPLAY Answer 2.5 or 221
2 The method used in Example 5 to enter the negative mixed number illustrate the following property: 233 4 Property of the Opposite of a Sum For all real numbers a and b: (a b) (a) (b) When adding more than two signed numbers, the commutative and associative properties allow us to arrange the numbers in any order and to group them in any way. It may be helpful to add positive numbers first, add negative numbers next, and then add the two results. EXERCISES Writing About Mathematics 1. The sum of two numbers is positive. One of the numbers is a positive number that is larger than the sum. Is the other number positive or negative? Explain your answer. 2. The sum of two numbers is positive. One of the numbers is negative. Is the other number positive or negative? Explain your answer. 3. The sum of two numbers is negative. One of the numbers is a negative number that is smaller than the sum. Is the other number positive or negative? Explain your answer. 4. The sum of two numbers is negative. One of the numbers is positive. Is the other number positive or negative? Explain your answer. 5. Is it possible for the sum of two numbers to be smaller than either of the numbers? If so, give an example. Developing Skills In 6–10, find each sum or difference. 7. 10 5 6. 6 4 8. 4.5 4.5 9. 6 4 10. 6 4 Subtraction of Signed Numbers 59 In 11–27, add the numbers. Use a calculator to check your answer. 13. 87 (87) 12. 23 (35) 11. 17 (28) 15. 2331 3 192 3 19. (47) (35) (47) 16. 253 4 22. 2143 4 2173 4 B 25. 6.25 (0.75) A 81 2 17. +7 20. 343 8 (73) 262 7 A B 23. 13 (13) 26. (12.4) 13.0 14. 2.06 1.37 18. (3.72) (5.28) 21. 2731 2 86 24. 42 43 B A 27. 12.4 13.0 Applying Skills 28. In 1 hour, the temperature rose 4° Celsius and in the next hour it dropped 6° Celsius. What was the net change in temperature during the two-hour period? 29. An elevator started on the first floor and rose 30 floors. Then it came down 12 floors. At which floor was it at that time? 30. A football team gained 7 yards on the first play, lost 2 yards on the second, and lost 8 yards on the third. What was the net result of the three plays? 31. Fay has $250 in a bank. During the month, she made a deposit of $60 and a withdrawal of $80. How much money did Fay have in the bank at the end of the month? 32. During a four-day period, the dollar value of a share of stock rose $1.50 on the first day, dropped $0.85 on the second day, rose $0.12 on the third day, and dropped $1.75 on the fourth day. What was the net change in the stock during this period? 2-4 SUBTRACTION OF SIGNED NUMBERS In arithmetic, to subtract 3 from 7, we find the number that, when added to 3, gives 7. We know that 7 3 4 because 3 4 7. Subtraction is the inverse operation of addition. DEFINITION In general, for every number c and every number b, the expression c b is the number a such that b a c. We use this definition in order to subtract signed numbers. To subtract 2 from 3, written as (3) (2), we must find a number that, when added to 2, will give 3. We write: (2) (?) 3 60 Operations and Properties We can use a number line to find the answer to this open sentence. From a point 2 units to the left of 0, move to the point that represents 3, that is, 3 units to the right of 0. We move 5 units to the right, a motion that represents 5. +5 –2 –3 –2 –1 0 +1 +2 +3 +4 Therefore, (3) (2) 5 because (2) (5) 3. Notice that (3) (2) can also be represented as the directed distance from 2 to 3 on the number line. Subtraction can be written vertically as follows: (3) or (2) 5 Subtract: (3) (2) 5 minuend subtrahend difference When you first learned to subtract numbers, you learned to check your answer. The answer (the difference) plus the number being subtracted (the subtrahend) must be equal to the number from which you are subtracting (the minuend). Check each of the following examples using: subtrahend difference minuend: Subtract Now, consider another way in which addition and subtraction are related. In each of the following examples, compare the result obtained when a signed number is subtracted with the result obtained when the opposite of that signed number is added. Subtract Add 9 9 6 6 3 3 Subtract Add 7 7 2 2 5 5 Subtract Add 5 5 2 2 7 7 Subtract Add 3 3 1 1 4 4 Observe that, in each example, adding the opposite (the additive inverse) of a signed number gives the same result as subtracting that signed number. It therefore seems reasonable to define subtraction as follows: DEFINITION If a is any signed number and b is any signed number, then: a b a (b) Procedure To subtract one signed number from another, add the opposite (additive inverse) of the subtrahend to the minuend. Subtraction of Signed Numbers 61 Uses of the Symbol We have used the symbol in three ways: 1. To indicate that a number is negative: 2 2. To indicate the opposite of a number: (4) Negative 2 Opposite of negative 4 3. To indicate subtraction: Opposite of a a 4 (3) Difference between 4 and 3 Note that an arithmetic expression such as 3 7 can mean either the dif- ference between 3 and 7 or the sum of 3 and 7. 3 7 3 (7) 3 (7) When writing an arithmetic expression, we use the same sign for both a neg, is ative number and subtraction. On a calculator, the key for subtraction, not the same key as the key for a negative number, will result in an error message. (-) . Using the wrong key EXAMPLE 1 Perform the indicated subtractions. a. (30) (12) b. (19) (7) c. (4) (0) d. 0 8 Answers a. 18 b. 12 c. 4 d. 8 EXAMPLE 2 From the sum of 2 and 8, subtract 5. Solution (2 8) (5) 6 (5) 11 Calculator Solution ENTER: (-) 2 8 (-) 5 ENTER DISPLAY Answer 11 62 Operations and Properties EXAMPLE 3 Subtract the sum of 7 and 2 from 4. Solution 4 (7 2) 4 (5) 4 (5) 1 Calculator Solution ENTER: (-) 4 ( (-) 7 2 ) ENTER DISPLAY Answer 1 EXAMPLE 4 How much greater than 3 is 9? Solution 9 (3) 9 3 12 Answer 9 is 12 greater than 3. +12 –3 0 +9 Note that parentheses need not be entered in the calculator in Example 2 since the additions and subtractions are to be done in the order in which they occur in the expression. Parentheses are needed, however, in Example 3 since the sum of 7 and 2 is to be found first, before subtracting this sum from 4. EXERCISES Writing About Mathematics 1. Does 8 12 mean the difference between 8 and 12 or the sum of 8 and 12? Explain your answer. 2. How is addition used to check subtraction? Developing Skills In 3–10, perform each indicated subtraction. Check your answers using a calculator. 3. 23 (35) 4. 87 (87) 5. 5.4 (8.6) 6. 8.8 (3.7) 7. 2.06 (1.37) 8. 2331 3 1192 3 A B 9. 253 4 281 2 A B 10. 7 62 7 A B In 11–18, Find the value of each given expression. Subtraction of Signed Numbers 63 13. (12) (57 12) 16. 2143 4 2 2173 4 A B A B 18. 15. 8 2 2343 12. (3.72) (5.28) 2731 86 2 32 (32) 11. (18) (14) 14. (47) (35 47) 273 2 723 17. 8 19. How much is 18 decreased by 7? 20. How much greater than 15 is 12? 21. How much greater than 4 is 1? 22. What number is 6 less than 6? 23. Subtract 8 from the sum of 6 and 12. 24. Subtract 7 from the sum of 18 and 10. 25. State whether each of the following sentences is true or false: a. (5) (3) (3) (5) b. (7) (4) (4) (7) 26. If x and y represent real numbers: a. Does x y y x for all replacements of x and y? Justify your answer. b. Does x y y x for any replacements of x and y? For which values of x and y? c. What is the relation between x y and y x for all replacements of x and y? d. Is the operation of subtraction commutative? In other words, for all signed numbers x and y, does x y y x? 27. State whether each of the following sentences is true or false: a. (15 9) 6 15 (9 6) b. [(10) (4)] (8) (10) [(4) (8)] 28. Is the operation of subtraction associative? In other words, for all signed numbers x, y, and z, does (x y) z x (y z)? Justify your answer. Applying Skills 29. Express as a signed number the increase or decrease when the Celsius temperature changes from: a. 5° to 8° b. 10° to 18° c. 6° to 18° d. 12° to 4° 30. Find the change in altitude when you go from a place that is 15 meters below sea level to a place that is 95 meters above sea level. 31. In a game, Sid was 35 points “in the hole.” How many points must he make in order to have a score of 150 points? 32. The record high Fahrenheit temperature in New City is 105°; the record low is 9°. Find the difference between these temperatures. 33. At one point, the Pacific Ocean is 0.50 kilometers in depth; at another point it is 0.25 kilo- meters in depth. Find the difference between these depths. 64 Operations and Properties 2-5 MULTIPLICATION OF SIGNED NUMBERS Four Possible Cases in the Multiplication of Signed Numbers We will use a common experience to illustrate the various cases that can arise in the multiplication of signed numbers. 1. Represent a gain in weight by a positive number and a loss of weight by a negative number. 2. Represent a number of months in the future by a positive number and a number of months in the past by a negative number. Multiplying a Positive Number by a Positive Number CASE 1 If a boy gains 2 pounds each month, 4 months from now he will be 8 pounds heavier than he is now. Using signed numbers, we may write: (2)(4) 8 The product of the two positive numbers is a positive number. Multiplying a Negative Number by a Positive Number CASE 2 If a boy loses 2 pounds each month, 4 months from now he will be 8 pounds lighter than he is now. Using signed numbers, we may write: (2)(4) 8 The product of the negative number and the positive number is a negative number. Multiplying a Positive Number by a Negative Number CASE 3 If a girl gained 2 pounds each month, 4 months ago she was 8 pounds lighter than she is now. Using signed numbers, we may write: (2)(4) 8 The product of the positive number and the negative number is a negative number. Multiplying a Negative Number by a Negative Number CASE 4 If a girl lost 2 pounds each month, 4 months ago she was 8 pounds heavier than she is now. Using signed numbers, we may write: (2)(4) 8 The product of t
he two negative numbers is a positive number. In all four cases, the absolute value of the product, 8, is equal to the product of the absolute values of the factors, 4 and 2. Multiplication of Signed Numbers 65 Using the Properties of Real Numbers to Multiply Signed Numbers You know that the sum of any real number and its inverse is 0, a (a) 0, and that for all real numbers a, b, and c, a(b c) ab ac. These two facts will enable us to demonstrate the rules for multiplying signed numbers. 4[7 (7)] 4(7) 4(7) 4(0) 28 4(7) 0 28 4(7) In order for this to be a true statement, 4(7) must equal 28, the opposite of 28. Since addition is a commutative operation, 4(7) 7(4). Then 7[4 (4)] 7(4) (7)(4) 7(0) 28 (7)(4) 0 28 (7)(4) In order for this to be a true statement, 7(4) must be 28, the additive inverse of 28. Rules for Multiplying Signed Numbers RULE 1 The product of two positive numbers or of two negative numbers is a positive number whose absolute value is the product of the absolute values of the numbers. The product of a positive number and a negative number is a negative RULE 2 number whose absolute value is the product of the absolute values of the numbers. In general, if a and b are both positive or are both negative, then: ab a · b If one of the numbers, a or b, is positive and the other is negative, then: ab (a · b)) Procedure To multiply two signed numbers: 1. Find the product of the absolute values. 2. Write a plus sign before this product when the two numbers have the same sign. 3. Write a minus sign before this product when the two numbers have different signs. 66 Operations and Properties EXAMPLE 1 Find the product of each of the given pairs of numbers. a. (12)(4) c. (18)(3) Answers 48 54 e. (3.4)(3) 10.2 b. (13)(5) d. (15)(6) 271 8 f. A B (23) Answers 5 65 5 90 5 1213 8 Use the distributive property of multiplication over addition to find the product 8(72). 8(72) 8[(70) (2)] 8(70) 8(2) (560) (16) 576 Answer EXAMPLE 2 Solution EXAMPLE 3 Find the value of (2)3. Solution Answer 8 (2)3 (2)(2)(2) 4(2) 8 Note: The product of an odd number (3) of negative factors is negative. EXAMPLE 4 Find the value of (3)4. Solution (3)4 (3)(3)(3)(3) [(3)(3)][(3)(3)] (9)(9) 81 Answer 81 Note: The product of an even number (4) of negative factors is positive. In this example, the value of (3)4 was found to be 81. This is not equal to 34, which is the opposite of 34 or 1(34). To find the value of 34, first find the value of 34, which is 81, and then write the opposite of this power, 81. Thus, (3)4 81 and 34 81. Multiplication of Signed Numbers 67 On a calculator, evaluate (3)4. On a calculator, evaluate 34. ENTER: ( (-) 3 ) ^ 4 ENTER ENTER: (-) 3 ^ 4 ENTER DISPLAY: ( - 3 ) ^ 4 DISPLAY EXERCISES Writing About Mathematics 1. Javier said that (5)(4)(2) 40 because the product of numbers with the same sign is positive. Explain to Javier why he is wrong. 2. If a(b) means the opposite of ab, explain how knowing that 3(4) 12 can be used to show that 3(4) 12. Developing Skills In 3–12, find the product of each pair of numbers. Check your answers using a calculator. 3. 17(6) 4. +27(6) 5. 9( 27) 6. 23(15) 7. +4(4) 8. +5.4(0.6) 9. 2.6( 0.05) 10. 231 3 112 3 A B 11. 253 4 21 2 A B 12. 262 3(7) In 13–18, find the value of each expression. 13. (18)(4)(5) 16. (12)(7 3) 14. (3.72)(0.5)(0.2) 273 1 723 8 28 17. A B 15. (4)(35 7) 18. 4 23 4 1 1 4 In 19–28, find the value of each power. 19. (3)2 24. 24 20. 32 25. (0.5)2 21. (5)3 26. (0.5)2 22. (2)4 27. 0.52 23. (2)4 28. (3)3 In 29–34, name the property that is illustrated in each statement. 29. 2(1 3) 2(1) 2(3) 31. (1 2) 3 1 (2 3) 33. (5)(7) (7)(5) 30. (2) (1) (1) (2) 32. 2 0 2 34. 3 (2 3) (3 2) (3) 68 Operations and Properties 2-6 DIVISION OF SIGNED NUMBERS Using the Inverse Operation in Dividing Signed Numbers Division may be defined as the inverse operation of multiplication, just as subtraction is defined as the inverse operation of addition. To divide 6 by 2 means to find a number that, when multiplied by 2, gives 6. That number is 3 because 3(2) 6. Thus, 3, or 6 2 3. The number 6 is the dividend, 2 is the divisor, and 3 is the quotient. 6 2 It is impossible to divide a signed number by 0; that is, division by 0 is undefined. For example, to solve (9) 0, we would have to find a number that, when multiplied by 0, would give 9. There is no such number since the product of any signed number and 0 is 0. In general, for all signed numbers a and b (b 0), a b or means to find a b the unique number c such that cb a In dividing nonzero signed numbers, there are four possible cases. Consider the following examples: CASE 1 CASE 2 CASE 3 CASE 4 16 13 26 23 26 13 16 23 ? implies (?)(3) 6. Since (2)(3) 6, 2 16 13 ? implies (?)(3) 6. Since (2)(3) 6, 2 26 23 ? implies (?)(3) 6. Since (2)(3) 6, 2 26 13 ? implies (?)(3) 6. Since (2)(3) 6, 2 16 23 In the preceding examples, observe that: 1. When the dividend and divisor are both positive or both negative, the quo- tient is positive. 2. When the dividend and divisor have opposite signs, the quotient is nega- tive. 3. In all cases, the absolute value of the quotient is the absolute value of the dividend divided by the absolute value of the divisor. Rules for Dividing Signed Numbers The quotient of two positive numbers or of two negative numbers is a RULE 1 positive number whose absolute value is the absolute value of the dividend divided by the absolute value of the divisor. Division of Signed Numbers 69 The quotient of a positive number and a negative number is a nega- RULE 2 tive number whose absolute value is the absolute value of the dividend divided by the absolute value of the divisor. In general, if a and b are both positive or are both negative, then: a b a b or a b \|a\| \|b\| If one of the numbers, a or b, is positive and the other is negative, then: a b (a b) or a b \|a\| \|b\| A B Procedure To divide two signed numbers: 1. Find the quotient of the absolute values. 2. Write a plus sign before this quotient when the two numbers have the same sign. 3. Write a minus sign before this quotient when the two numbers have different signs. Rule for Dividing Zero by a Nonzero Number If the expression 0 (5) or ?, then (?)(5) 0. Since 0 is the only number that can replace ? and result in a true statement, 0 (5) or 0. This illustrates that 0 divided by any nonzero number is 0. In general, if a is a nonzero number (a 0), then 0 25 0 25 EXAMPLE 1 0 a 0. 0 a Perform each indicated division, if possible. 160 a. 115 a. 4 b. b. 110 290 21 9 227 c. 23 c. 9 d. (45) 9 d. 5 e. 0 (9) f. 3 0 e. 0 f. Undefined Answers Using the Reciprocal in Dividing Signed Numbers In Section 2-2, we learned that for every nonzero number a, there is a unique number , called the reciprocal or multiplicative inverse, such that a ? 1 a 5 1 1 a . 70 Operations and Properties Using the reciprocal of a number, we can define division in terms of multi- plication as follows: For all numbers a and b (b 0): a b a · (b 0) 1 b a b Procedure To divide a signed number by a nonzero signed number, multiply the dividend by the reciprocal of the divisor. Notice that we exclude division by 0. The set of nonzero real numbers is closed with respect to division because every nonzero real number has a unique reciprocal, and multiplication by this reciprocal is always possible. EXAMPLE 2 Perform each indicated division by using the reciprocal of the divisor. a. 130 12 230 190 b. c. (54) 6 21 3 B 23 5 e. A f. 0 (9) (13) 4 227 4 d. A B Answers 11 2 B 1 1 90 B 1 9 6 5 115 5 21 3 5 181 B 23 1 B 5 25 A B 5 0 (130) (254) (230) A A A 227 3 25 3 13 0 A A 21 9 B EXERCISES Writing About Mathematics 1. If x and y represent nonzero numbers, what is the relationship between x y and y x? 2. If x y, are there any values of x and y for which x y y x? Developing Skills In 3–10, name the reciprocal (the multiplicative inverse) of each given number. 3. 6 7. 1 2 4. 5 2 1 10 8. 5. 1 9. 23 4 6. 1 10. x if x 0 In 11–26, find the indicated quotients or write “undefined” if no quotient exists. Operations with Sets 71 14. 0 213 10.01 20.001 13. 210 110 20.25 22.5 18. 17. 20. (75) (0) 22. (1.5) (0.03) 12. 248 116 13.6 20.12 11. 263 29 215 245 16. 15. 19. (100) (2.5) 21. (0.5) (0.25) 23. 25. (112) 4 17 8 4 A B A 21 3 A 221 32 B B 24. 26. A 23 4 B 211 4 A 4 (16) 221 2 4 B A B In 27–28, state whether each sentence is true or false: 27. a. [(16) (4)] (2) (16) [(4) (2)] b. [(36) (6)] (2) (36) [(6) (2)] c. Division is associative. 28. a. (12 6) 2 12 2 6 2 b. [(25) (10)] (5) (25) (5) (10) (5) c. 2 (3 5) 2 3 2 5 d. Division is distributive over addition and subtraction. 2-7 OPERATIONS WITH SETS Recall that a set is simply a collection of distinct objects or elements, such as a set of numbers in arithmetic or a set of points in geometry. And, just as there are operations in arithmetic and in geometry, there are operations with sets. Before we look at these operations, we need to understand one more type of set. The universal set, or the universe, is the set of all elements under consider- ation in a given situation, usually denoted by the letter U. For example: 1. Some universal sets, such as all the numbers we have studied, are infinite. Here, U = {real numbers}. 2. In other situations, such as the scores on a classroom test, the universal set can be finite. Here, using whole-number grades, U = {0, 1, 2, 3, . . ., 100}. Three operations with sets are called intersection, union, and complement. Intersection of Sets The intersection of two sets, A and B, denoted by ments that belong to both sets, A and B. For example: A d B , is the set of all ele- 72 Operations and Properties 1. When A = {1, 2, 3, 4, 5}, and B = {2, 4, 6, 8, 10}, then A d B is {2, 4}. 2. In the diagram, two lines called 4 AB 4 CD and inter- sect. Each line is an infinite set of points although only three points are marked on each line. The intersection is a set that has one element, point E, the point that is on line AB ( ) and on line CD 4 AB ( ). We write the intersection of the lines in the 4
d CD example shown as 4 AB 4 CD A D E C B 3. Intersection is a binary operation, are subsets of the universal set and example: 5 E . F d G 5 H , where sets F, G, and H d is the operation symbol. For U set of natural numbers {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, . . .} F multiples of 2 {2, 4, 6, 8, 10, 12, . . .} G multiples of 3 {3, 6, 9, 12, 15, 18, . . .} multiples of 6 {6, 12, 18, 24, 30, . . .} F d G 4. Two sets are disjoint sets if their intersection is the empty set ( or {}); that is, if they do not have a common element. For example, when K = {1, 3, 5, 7, 9, 11, 13} and L = {2, 4, 6, 8}, . Therefore, K and L are disjoint sets. K d L 5 Union of Sets The union of two sets, A and B, denoted by belong to set A or to set B, or to both set A and set B. For example: A < B , is the set of all elements that 1. If A = {1, 2, 3, 4} and B = {2, 4, 6}, then {1, 2, 3, 4, 6}. Note that an element is not repeated in the union of two sets even if it is an element of each set. A < B 2. In the diagram, both region R (gray shading) and region S (light color shading) represent sets of points. The shaded parts of both R < S regions represents , and the dark color shading where the regions overlap represents R d S . R S 3. If A {1, 2} and B = {1, 2, 3, 4, 5}, then the union of A and B is {1, 2, 3, 4, 5}. A < B B. Once again we have We can write an example of a binary operation, where the elements are taken from a universal set and where the operation here is union. {1, 2, 3, 4, 5}, or A < B Operations with Sets 73 4. The union of the set of all rational numbers and the set of all irrational numbers is the set of real numbers. {real numbers} {rational numbers} {irrational numbers} Complement of a Set The complement of a set A, denoted by , is the set of all elements that belong to the universe U but do not belong to set A. Therefore, before we can determine the complement of A, we must know U. For example: A 1. If A {3, 4, 5} and U = {1, 2, 3, 4, 5}, then {1, 2} because 1 and 2 A belong to the universal set U but do not belong to set A. 2. If the universe is {whole numbers} and A = {even whole numbers} then {odd whole numbers} because the odd whole numbers belong to the A universal set but do not belong to set A. A Although it seems at first that only one set is being considered in writing the , actually there are two sets. This fact suggests a binary complement of A as operation, in which the universe U and the set A are the pair of elements, complement is the operation, and the unique result is . The complement of any universe is the empty set. Note that the complement can also be written as U \ A to emphasize that it is a binary operation. A EXAMPLE 1 If U {1, 2, 3, 4, 5, 6, 7}, A {6, 7}, and B {3, 5, 7}, determine A d B . Solution U = {1, 2, 3, 4, 5, 6, 7} Since A {6, 7}, then {1, 2, 3, 4, 5}. Since B = {3, 5, 7}, then {1, 2, 4, 6}. A B Since 1, 2, and 4 are elements in both , we can write: B A and {1, 2, 4} Answer EXAMPLE 2 A d B Using sets U, A, and B given for Example 1, find the complement of the set A < B , that is, determine A < B . Solution Since A < B contains all of the elements that are common to A and B, A < B 5 3, 5, 6, 7 6 5 . Therefore, A < B 5 1, 2, 4 6 5 Answer 74 Operations and Properties EXERCISES Writing About Mathematics 1. A line is a set of points. Can the intersection of two lines be the empty set? Explain. 2. Is the union of the set of prime numbers and the set of composite numbers equal to the set of counting numbers? Explain. Developing Skills In 3–10, A {1, 2, 3}, B {3, 4, 5, 6}, and C = {1, 3, 4, 6}. In each case, perform the given operation and list the element(s) of the resulting set. 3. 7. A d B B d C 4. A < B 8. B < C 5. 9. A d C B < 6. 10. A < C B d 11. Using the sets A, B, and C given for Exercises 3–10, list the element(s) of the smallest possi- ble universal set of which A, B, and C are all subsets. In 12–19, the universe U = {1, 2, 3, 4, 5}, A {1, 5}, B = {2, 5}, and C {2}. In each case, perform the given operation and list the element(s) of the resulting set. 12. A A d B 13. B 14. C A d B 15. A < B A < B 17. A < B 16. 20. If U {2, 4, 6, 8} and {6}, what are the elements of A? 21. If U {2, 4, 6, 8}, A {2}, and {2, 4}, what are the elements of B 22. If U {2, 4, 6, 8}, A {2}, and {2, 4}, what is the set A d B ? 18. 19. A B A < B ? 23. Suppose that the set A has two elements and the set B has three elements. a. What is the greatest number of elements that A < B can have? b. What is the least number of elements that A < B can have? c. What is the greatest number of elements that A d B can have? d. What is the least number of elements that A d B can have? 24. Let the universe U {2, 4, 6, 8, 10, 12}, A {2, 8, 12} and B {4, 10}. a. A b. A (the complement of A ) c. B d. B B (the complement of ) 25. Let the universe U {1, 2, 3, 4, 5, 6, 7, 8}. a. Find the elements of A d B , A d B , and A < B when A and B are equal to: (1) A = {1, 2, 3, 4}; B {5, 6, 7, 8} (3) A {1, 3, 5, 7}; B {2, 4, 6, 8} (2) A {2, 4}; B {6, 8} (4) A {2}; B {4} A d B . b. When A and B are disjoint sets, describe, in words, the set c. If A and B are disjoint sets, what is the set A < B ? Explain. 2-8 GRAPHING NUMBER PAIRS Even though we know that the surface of the earth is approximately the surface of a sphere, we often model the earth by using maps that are plane surfaces. To locate a place on a map, we choose two reference lines, the equator and the prime meridian. The location of a city is given in term of east or west longitude (distance from the prime meridian) and north or south latitude (distance from the equator). For example, the city of Lagos in Nigeria is located at 3° east longitude and 6° north latitude, and the city of Dakar in Senegal is located 17° west longitude and 15° north latitude. Points on a Plane Graphing Number Pairs 75 20° W 10° W 0° 10° E 20° E 30° E 40° E Dakar Equator Lagos 50° E 30°N 20°N 10°N 0° 10°S 20°S 30°s The method used to locate cities on a map can be used to locate any point on a plane. The reference lines are a horizontal number line called the x-axis and a vertical number line called the y-axis. These two number lines, which have the same scale and are drawn perpendicular to each other, are called the coordinate axes. The plane determined by the axes is called the coordinate plane. In a coordinate plane, the intersection of the two axes is called the origin and is indicated as point O. This point of intersection is assigned the value 0 on both the x- and y-axes. Moving to the right and moving up are regarded as movements in the positive direction. In the coordinate plane, points to the right of O on the x-axis and on lines parallel to the x-axis and points above O on the y-axis and on lines parallel to the y-axis are assigned positive values. Moving to the left and moving down are regarded as movements in the negative direction. In the coordinate plane, points to the left of O on the x-axis and on lines parallel to the xaxis and points below O on the y-axis and on lines parallel to the y-axis are assigned negative values. y 4 3 2 1 O –4 –3 –2 –1 1 2 3 4 x –1 –2 –3 –4 76 Operations and Properties Quadrant II y 3 2 1 –3 –2 –1 –1 –2 –3 Quadrant III The x-axis and the y-axis separate the plane into four regions called quadrants. These quadrants are numbered I, II, III, and IV in a counterclockwise order, beginning at the upper right, as shown in the accompanying diagram. The points on the axes are not in any quadrant. Quadrant I O 1 2 3 x Quadrant IV Coordinates of a Point Every point on the plane can be described by two numbers, called the coordinates of the point, usually written as an ordered pair. The first number in the pair is called the x-coordinate or the abscissa. The second number is the y-coordinate or the ordinate. In general, the coordinates of a point are represented as (x, y). In the graph at the right, point A, which is the graph of the ordered pair (2, 3), lies in quadrant I. Here, A lies a distance of 2 units to the right of the origin (in a positive direction along the x-axis) and then up a distance of 3 units (in a positive direction parallel to the y-axis). Point B, the graph of (4, 1) in quadrant II, lies a distance of 4 units to the left of the origin (in a negative direction along the x-axis) and then up 1 unit (in a positive direction parallel to the y-axis). B(–4,+1) s i x a y O A(+2,+3) x-axis C(–2,–3) D(+3,–5) In quadrant III, every ordered pair (x, y) consists of two negative numbers. For example, C, the graph of (2, 3), lies 2 units to the left of the origin (in the negative direction along the x-axis) and then down 3 units (in the negative direction parallel to the y-axis). Point D, the graph of (3, 5) in quadrant IV, lies 3 units to the right of the origin (in a positive direction along the x-axis) and then down 5 units (in a negative direction parallel to the y-axis). Point O, the origin, has the coordinates (0, 0). Locating a Point on the Coordinate Plane An ordered pair of signed numbers uniquely determines the location of a point. Graphing Number Pairs 77 Procedure To find the location of a point on the coordinate plane: 1. Starting from the origin O, move along the x-axis the number of units given by the x-coordinate. Move to the right if the number is positive or to the left if the number is negative. If the x-coordinate is 0, there is no movement along the x-axis. 2. Then, from the point on the x-axis, move parallel to the y-axis the number of units given by the y-coordinate. Move up if the number is positive or down if the number is negative. If the y-coordinate is 0, there is no movement in the y direction. To locate the point A(3, 4), from O, move 3 units to the left along the x-axis, then 4 units down, parallel to the y-axis. To locate the point B(4, 0), from O, move 4 units to the right along the xaxis. There is no movement parallel to the y-axis. To locate the point C(0, 5), there is no movement along the x-axis. From O, move 5 units down along
the y-axis. y 2 1 O –1 –2 –3 –4 –5 –4 –3 –2 –1 A(–3, –4) B(4, 0) 1 2 3 4 x5 C(0, –5) Finding the Coordinates of a Point on a Plane The location of a point on the coordinate plane uniquely determines the coordinates of the point. Procedure To find the coordinates of a point: 1. From the point, move along a vertical line to the x-axis.The number assigned to that point on the x-axis is the x-coordinate of the point. 2. From the point, move along a horizontal line to the y-axis.The number assigned to that point on the y-axis is the y-coordinate of the point. 78 Operations and Properties To find the coordinates of point R, from point R, move in the vertical direction to 5 on the x-axis and in the horizontal direction to 6 on the yaxis. The coordinates of R are (5, 6) To find the coordinates of point S, from point S, move in the vertical direction to 2 on the x-axis and in the horizontal direction to 4 on the y-axis. The coordinates of S are (2, 4) Point T, lies at 5 on the x-axis. Therefore, the x-coordinate is 5 and the y-coordinate is 0. The coordinates of T are (5, 0). Note that if a point lies on the yaxis, the x-coordinate is 0. If a point lies on the x-axis, the y-coordinate is 0. Graphing Polygons A polygon is a closed figure whose sides are line segments. A quadrilateral is a polygon with four sides. The endpoints of the sides are called vertices. A quadrilateral can be represented in the coordinate plane by locating its vertices and then drawing the sides, connecting the vertices in order. The graph at the right shows the rectangle ABCD. The vertices are A(3, 2), B(3, 2), C(3, 2) and D(3, 2). From the graph, note the following5 –4 –3 –2 –1 O –1 –2 –3 –4 –5 –6 1 2 3 4 x5 R y 1 O 1 –1 –1 B(–3, 2) C(–3, –2) A(3, 2) x D(3, –2) 1. Points A and B have the same y-coordinate and are on a line parallel to the x-axis. 2. Points C and D have the same y-coordinate and are on a line parallel to the x-axis. 3. Lines parallel to the x-axis are parallel to each other. 4. Lines parallel to the x-axis are perpendicular to the y-axis. 5. Points B and C have the same x-coordinate and are on a line parallel to the y-axis. Graphing Number Pairs 79 6. Points A and D have the same x-coordinate and are on a line parallel to the y-axis. 7. Lines parallel to the y-axis are parallel to each other. 8. Lines parallel to the y-axis are perpendicular to the x-axis. Now, we know that ABCD is a rectangle, because it is a parallelogram with right angles. From the graph, we can find the dimensions of this rectangle. To find the length of the rectangle, we can count the number of units from A to B or from C to D. AB CD 6. Because points on the same horizontal line have the same y-coordinate, we can also find AB and CD by subtracting their xcoordinates. AB CD 3 (3) 3 3 6 To find the width of the rectangle, we can count the number of units from B to C or from D to A. BC DA 4. Because points on the same vertical line have the same x-coordinate, we can find BC and DA by subtracting their ycoordinates. BC DA 2 (2) 2 2 4 EXAMPLE 1 Graph the following points: A(4, 1), B(1, 5), C(2, 1). Then draw ABC and find its area. Solution The graph at the right shows ABC. To find the area of the triangle, we need to know the lengths of the base and of the altitude drawn to that base. The base of ABC is AC . AC 4 (2) 4 2 6 y B(1, 5) C(–2,–1) D(1, 1) O 1 –1 –1 A(4, 1) x The line segment drawn from B perpendicular to is the altitude AC BD . BD 5 1 4 1 2(AC)(BD) 1 2(6)(4) Area 12 Answer The area of ABC is 12 square units. 80 Operations and Properties EXERCISES Writing About Mathematics 1. Mark is drawing triangle ABC on the coordinate plane. He locates points A(2, 4) and C(5, 4). He wants to make AC BC, C a right angle, and point B lie in the first quadrant. What must be the coordinates of point B? Explain how you found your answer. 2. Phyllis graphed the points D(3, 0), E(0, 5), F(2, 0), and G(0, 4) on the coordinate plane and joined the points in order. Explain how Phyllis can find the area of this polygon, then find the area. Developing Skills 3. Write as ordered number pairs the coordinates of points A, B, C, D, E, F, G, H, and O in the graph. B y F O 2 1 G –2 C –1 –1 –2 H A E x 21 D In 4–15, draw a pair of coordinate axes on graph paper and locate the point associated with each ordered number pair. Label each point with its coordinates. 4. (5, 7) 8. (1, 6) 12. (3, 0) 5. (3, 2) 9. (8, 5) 13. (0, 4) 6. (2, 6) 10. (4, 4) 14. (0, 6) 7. (4, 5) 11. (5, 0) 15. (0, 0) In 16–20, name the quadrant in which the graph of each point described appears. 16. (5, 7) 17. (3, 2) 18. (7, 4) 19. (1, 3) 20. (2, 3) 21. Graph several points on the x-axis. What is the value of the y-coordinate for every point in the set of points on the x-axis? 22. Graph several points on the y-axis. What is the value of the x-coordinate for every point in the set of points on the y-axis? 23. What are the coordinates of the origin in the coordinate plane? Applying Skills In 24–33: a. Graph the points and connect them with straight lines in order, forming a polygon. b. Identify the polygon. c. Find the area of the polygon. 24. A(1, 1), B(8, 1), C(1, 5) 26. C(8, 1), A(9, 3), L(4, 3), F(3, 1) 25. P(0, 0), Q(5, 0), R(5, 4), S(0, 4) 27. H(4, 0), O(0, 0), M(0, 4), E(4, 4) Graphing Number Pairs 81 28. H(5, 3), E(5, 3), N(2, 0) 30. B(3, 2), A(2, 2), R(2, 2), N(3, 2) 32. R(4, 2), A(0, 2), M(0, 7) 29. F(5, 1), A(5, 5), R(0, 5), M(2, 1) 31. P(3, 0), O(0, 0), N(2, 2), D(1, 2) 33. M(1, 1), I(3, 1), L(3, 3), K(1, 3) 34. Graph points A(1, 1), B(5, 1), and C(5, 4). What must be the coordinates of point D if ABCD is a rectangle? 35. Graph points P(1, 4) and Q(2, 4). What are the coordinates of R and S if PQRS is a square? (Two answers are possible.) 36. a. Graph points S(3, 0), T(0, 4), A(3, 0), and R(0, 4), and draw the quadrilateral STAR. b. Find the area of STAR by adding the areas of the triangles into which the axes divide it. 37. a. Graph points P(2, 0), L(1, 1), A(1, 1), N(2, 0), E(1, 1), and T(1, 1). Draw PLANET, a six-sided polygon called a hexagon. b. Find the area of PLANET. (Hint: Use the x-axis to separate the hexagon into two parts.). CHAPTER SUMMARY A binary operation in a set assigns to every ordered pair of elements from the set a unique answer from that set. The general form of a binary operation is a b c, where a, b, and c are elements of the set and is the operation symbol. Binary operations exist in arithmetic, in geometry, and in sets. Operations in arithmetic include addition, subtraction, multiplication, divi- sion, and raising to a power. Powers are the result of repeated multiplication of the same factor, as in 53 125. Here, the base 5 with an exponent of 3 equals 5 5 5 or 125, the power. Numerical expressions are simplified by following a clear order of opera- tions: (1) simplify within parentheses or other grouping symbols; (2) simplify powers; (3) multiply and divide from left to right; (4) add and subtract from left to right. Many properties are used in operations with real numbers, including: • closure under addition, subtraction, and multiplication; • commutative properties for addition and multiplication, a b b a; • associative properties for addition and multiplication, (a b) c a (b c); 82 Operations and Properties • distributive property of multiplication over addition or subtraction, a (b c) a b a c and (a b) c a c b c; • additive identity (0); • additive inverses (opposites), called a for every element a; • multiplicative identity (1); • multiplicative inverses (reciprocals), 1 a for every nonzero element a; the nonzero real numbers are closed under division. Operations with sets include the intersection of sets, the union of sets, and the complement of a set. Basic operations with signed numbers: • To add two numbers that have the same sign, find the sum of the absolute values and give this sum the common sign. • To add two numbers that have different signs, find the difference of the absolute values of the numbers. Give this difference the sign of the number that has the greater absolute value. The sum is 0 if both numbers have the same absolute value. • To subtract one signed number from another, add the opposite (additive inverse) of the subtrahend to the minuend. • To multiply two signed numbers, find the product of the absolute values. Write a plus sign before this product when the two numbers have the same sign. Write a minus sign before this product when the two numbers have different signs. • To divide two signed numbers, find the quotient of the absolute values. Write a plus sign before this quotient when the two numbers have the same sign. Write a minus sign before this quotient when the two numbers have different signs. When 0 is divided by any number, the quotient is 0. Division by 0 is not defined. The location of a point on a plane is given by an ordered pair of numbers that indicate the distance of the point from two reference lines, a horizontal line and a vertical line, called coordinate axes. The horizontal line is called the xaxis, the vertical line is the y-axis, and their intersection is the origin. The pair of numbers that are used to locate a point on the plane are called the coordinates of the point. The first number in the pair is called the x-coordinate or abscissa, and the second number is the y-coordinate or ordinate. The coordinates of a point are represented as (x, y). VOCABULARY 2-1 Binary operation • Factor • Prime • Composite • Base • Exponent • Power • Order of operations • Parentheses • Brackets Review Exercises 83 2-2 Property • Closure • Commutative property • Associative property • Distributive property of multiplication over addition and subtraction • Addition property of zero • Additive identity • Additive inverse (opposite) • Multiplication property of one • Multiplicative inverse (reciprocal) • Multiplication property of zero 2-3 Property of the opposite of a sum 2-7 Universal set • Intersection • Disjoint sets • Empty set • Union • Comp
lement 2-8 x-axis • y-axis • Coordinate axes • Coordinate plane • Origin • Quadrants • Coordinates • Ordered pair • x-coordinate • Abscissa • y-coordinate • Ordinate • Graph of the ordered pair 2-9 Polygon • Quadrilateral • Vertices REVIEW EXERCISES In 1–8, simplify each numerical expression. 1. 20 3 4 4. (8 16) (4 2) 7. (6 8)2 2. (20 3) ( 4) 5. (0.16)2 8. 7(9 7)3 3. –8 16 4 2 6. 62 82 In 9–14: a. Replace each question mark with a number that makes the sentence true. b. Name the property illustrated in each sentence that is formed when the replacement is made. 9. 8 (2 9) 8 (9 ?) 11. 3(?) 3 13. 5(7 4) 5(7) ?(4) 10. 8 (2 9) (8 ?) 9 12. 3(?) 0 14. 5(7 4) (7 4)? In 15–24, the universe U {1, 2, 3, 4, 5, 6}, set A {1, 2, 4, 5}, and set B {2, 4, 6}. In each case, perform the given operation and list the element(s) of the resulting set. 15. A d B 20. A < A 16. 21. A < B A d B 17. A 22. A < B 18. 23. A d B A d B 19. A d A 24. A < B In 25–32, find each sum or difference. 25. 6 6 29. 23 0 26. 6 6 30. 54 52 27. 3.2 4.5 31. 100 25 28. 4 5 32. 0 7 84 Operations and Properties In 33–40, to each property named in Column I, match the correct application of the property found in Column II. Column I Column II 33. Associative property of multiplication 34. Associative property of addition 35. Commutative property of addition 36. Commutative property of multiplication 37. Identity element of multiplication 38. Identity element of addition 39. Distributive property 40. Multiplication property of zero a. 3 4 4 3 b. 3 1 3 c. 0 4 0 d. 3 0 3 e. 3 4 4 3 f. 3(4 5) 3(4) 3(5) g. 3(4 5) (3 4)5 h. (3 4) 5 3 (4 5) 41. a. Find the number on the odometer of Jesse’s car described in the chapter opener on page 37. b. What would the number be if it were an odd number? 42. a. Rectangle ABCD is drawn with two sides parallel to the x-axis. The coordinates of vertex A are (2, 4) and the coordinates of C are (3, 5). Find the coordinates of vertices B and D. b. What is the area of rectangle ABCD? 43. Maurice answered all of the 60 questions on a multiple-choice test. The where S is the score on test was scored by using the formula S the test, R is the number right, and W is the number wrong. R 2 W 4 a. What is the lowest possible score? b. How many answers did Maurice get right if his score was 5? c. Is it possible for a person who answers all of the questions to get a score of 4? Explain why not or find the numbers of right and wrong answers needed to get this score. d. Is it possible for a person who does not answer all of the questions to get a score of 4? Explain why not or find the numbers of right and wrong answers needed to get this score. 44. Solve the following problem using signed numbers. Doug, the team’s replacement quarterback, started out on his team’s 30-yard line. On the first play, one of his linemen was offsides for a loss of 5 yards. On the next play, Doug gave the ball to the runningback who made a gain of 8 yards. He then made a 17-yard pass. Then Doug was tackled, for a loss of 3 yards. Where was Doug on the field after he was tackled? Cumulative Review 85 Exploration In this activity, you will derive a rule to determine if a number is divisible by 3. We will start our exploration with the number 23,568. For steps 1–4, fill in the blanks with a digit that will make the equality true. STEP 1. Write the number as a sum of powers of ten. 23,568 20,000 3,000 500 60 8 □ 10,000 □ 1,000 □ 100 □ 10 □ STEP 2. Rewrite the powers of ten as a multiple of 9, plus 1. □ 1,000 □ □ (9,999 1) □ (999 1) □ (99 1) □ (9 1) □ □ 10,000 □ 100 □ 10 STEP 3. Use the distributive property to expand the product terms. Do not multiply out products involving the multiples of 9. □ (9,999 1) □ (999 1) □ (99 1) □ (9 1) □ (□ 9,999 □) (□ 999 □) (□ 99 □) (□ 9 □) □ STEP 4. Group the products involving the multiples of 9 first and then the remaining digit terms. (□ 9,999 □) (□ 999 □) (□ 99 □) (□ 9 □) □ (□ 9,999 □ 999 □ 99 □ 9) (□ □ □ □ □) STEP 5. Compare the expression involving the digit terms with the original number. What do they have in common? STEP 6. The expression involving the multiples of 9 is divisible by 3. Why? STEP 7. If the expression involving the digit terms is divisible by 3, will the entire expression be divisible by 3? Explain. STEP 8. Based on steps 1–7 write the rule to determine if a number is divisible by 3. CUMULATIVE REVIEW CHAPTERS 1–2 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. Which number is not an integer? (1) 7 (2) 2 (3) 0.2 (4) 9 " 86 Operations and Properties 2. Which inequality is false? (1) 4 3 (2) 3 3 3. What is the opposite of 4? (1) 21 4 (2) 1 4 (3) 4 3 (4) 3 3 4. Which of the following numbers has the greatest value? (1) 5 7 (2) 0.7 (3) p 5 5. Under which operation is the set of integers not closed? (3) 4 (4) 4 (4) 25 36 (1) Addition (2) Subtraction (3) Multiplication (4) Division 6. Which of the following numbers has exactly three factors? (1) 1 (2) 2 (3) 9 (4) 15 7. The graph of the ordered pair (3, 5) is in which quadrant? (1) I (2) II (3) III (4) IV 8. Put the following numbers in order, starting with the smallest: 21 , 3 21 3 21 3 (1) (2) 21 5 22 , 7 22 7 21 5 21 5 22 7 (3) (4) 21 5 21 5 21 3 22 7 22 7 21 3 9. When rounded to the nearest hundredth, 3 is approximately equal to (1) 1.732 (2) 1.730 10. For which value of t is t t2? 1 t (2) 2 (1) 1 Part II " (3) 1.73 (3) 0 (4) 1.74 (4) 1 2 Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. Mrs. Ling spends more than $4.90 and less than $5.00 for meat for tonight’s dinner. Write the set of all possible amounts that she could have paid for the meat. Is this a finite or an infinite set? Explain. 12. In a basketball league, 100 students play on 8 teams. Each team has at least 12 players. What is the largest possible number of players on any one team? Cumulative Review 87 Part III Answer all questions in this part. Each correct answer will receive 3 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. A teacher wrote the sequence 2, 4, 6, . . . and asked the class what the next number could be. Three students each gave a different answer. The teacher said that each of the answers was correct. a. Josie said 8. Explain the rule that she used. b. Emil said 10. Explain the rule that he used. c. Ross said 12. Explain the rule that he used. 14. Evaluate the following expression without using a calculator. Show each step in your computation. 4(7 3) 8 (2 6)2 Part IV Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. The vertices of triangle ABC are A(2, 3), B(5, 3), and C(0, 4). Draw triangle ABC on the coordinate plane and find its area. 16. A survey to which 250 persons responded found that 140 persons said that they watch the news on TV at 6 o’clock, 120 persons said that they watch the news on TV at 11 o’clock and 40 persons said that they do not watch the news on TV at any time. a. How many persons from this group watch the news both at 6 and at 11? b. How many persons from this group watch the news at 6 but not at 11? c. How many persons from this group watch the news at 11 but not at 6? CHAPTER 3 CHAPTER TABLE OF CONTENTS 3-1 Using Letters to Represent Numbers 3-2 Translating Verbal Phrases Into Symbols 3-3 Algebraic Terms and Vocabulary 3-4 Writing Algebraic Expressions in Words 3-5 Evaluating Algebraic Expressions 3-6 Open Sentences and Solution Sets 3-7 Writing Formulas Chapter Summary Vocabulary Review Exercises Cumulative Review 88 ALGEBRAIC EXPRESSIONS AND OPEN SENTENCES An express delivery company will deliver a letter or package locally, within two hours.The company has the following schedule of rates. In addition to the basic charge of $25, the cost is $3 per mile or part of a mile for the first 10 miles or less and $4.50 per mile or part of a mile for each additional mile over 10. Costs such as those described, that vary according to a schedule, are often shown by a formula or set of formulas. Formulas can be used to solve many different problems. In this chapter, you will learn to write algebraic expressions and formulas, to use algebraic expressions and formulas to solve problems, and to determine the solution set of an open sentence. 3-1 USING LETTERS TO REPRESENT NUMBERS Using Letters to Represent Numbers 89 Eggs are usually sold by the dozen, that is, 12 in a carton. Therefore, we know that: In 1 carton, there are 12 1 or 12 eggs. In 2 cartons, there are 12 2 or 24 eggs. In 3 cartons, there are 12 3 or 36 eggs. In n cartons, there are 12 n or 12n eggs. Here, n is called a variable or a placeholder that can represent different numbers from the set of whole numbers, {1, 2, 3, . . .}. The set of numbers that can replace a variable is called the domain or the replacement set of that variable. Recall that a numerical expression contains only numbers. An algebraic expression, such as 12n, however, is an expression or phrase that contains one or more variables. In this section, you will see how verbal phrases are translated into algebraic expressions, using letters as variables and using symbols to represent operations. Verbal Phrases Involving Addition The algebraic expression a b may be used to represent several different verbal phrases, such as: a plus b the sum of a and b a added to b b is added to a a is increased by b b mor
e than a The word exceeds means “is more than.” Thus, the number that exceeds 5 by 2 can be written as “2 more than 5” or 5 2. Compare the numerical and algebraic expressions shown below. A numerical expression: An algebraic expression: The number that exceeds 5 by 2 is 5 2, or 7. The number that exceeds a by b is a b. Verbal Phrases Involving Subtraction The algebraic expression a b may be used to represent several different verbal phrases, such as: a minus b b subtracted from a a decreased by b a diminished by b b less than a a reduced by b the difference between a and b Verbal Phrases Involving Multiplication The algebraic expressions a b, a b, (a)(b) and ab may be used to represent several different verbal phrases, such as: a times b the product of a and b b multiplied by a 90 Algebraic Expressions and Open Sentences The preferred form to indicate multiplication in algebra is ab. Here, a product is indicated by using no symbol between the variables being multiplied. The multiplication symbol is avoided in algebra because it can be confused with the letter or variable x. The raised dot, which is sometimes mistaken for a decimal point, is also avoided. Parentheses are used to write numerical expressions: (3)(5)(2) or 3(5)(2). Note that all but the first number must be in parentheses. In algebraic expressions, parentheses may be used but they are not needed: 3(b)(h) 3bh. Verbal Phrases Involving Division The algebraic expressions a b and may be used to represent several different verbal phrases, such as: a b a divided by b the quotient of a and b a 4 The symbols a 4 and mean one-fourth of a as well as a divided by 4. Phrases and Commas In some verbal phrases, using a comma can prevent misreading. For example, in “the product of x and y, decreased by 2,” the comma after y makes it clear that the x and y are to be multiplied before subtracting 2 and can be written as (xy) 2 or xy 2. Without the comma, the phrase, “the product of x and y decreased by 2,” would be written x(y 2). EXAMPLE 1 Use mathematical symbols to translate the following verbal phrases into algebraic language: a. w more than 3 b. w less than 3 c. r decreased by 2 Answers 3 w 3 w r 2 d. the product of 5r and s e. twice x, decreased by 10 f. 25, diminished by 4 times n g. the sum of t and u, divided by 6 h. 100 decreased by twice (x 5) 5rs 2x 10 25 4n t1u 6 100 2(x 5) Translating Verbal Phrases into Symbols 91 EXERCISES Writing About Mathematics 1. Explain why the sum of a and 4 can be written as a 4 or as 4 a. 2. Explain why 3 less than a can be written as a 3 but not as 3 a. Developing Skills In 3–20, use mathematical symbols to translate the verbal phrases into algebraic language. 3. y plus 8 6. x times 7 4. 4 minus r 5. 7 times x 7. x divided by 10 8. 10 divided by x 9. c decreased by 6 10. one-tenth of w 11. the product of x and y 12. 5 less than d 13. 8 divided by y 14. y multiplied by 10 15. t more than w 16. one-third of z 17. twice the difference of p and q 18. a number that exceeds m by 4 19. 5 times x, increased by 2 20. 10 decreased by twice a In 21–30, using the letter n to represent “number,” write each verbal phrase in algebraic language. 21. a number increased by 2 22. 20 more than a number 23. 8 increased by a number 24. a number decreased by 6 25. 2 less than a number 26. 3 times a number 27. three-fourths of a number 28. 4 times a number, increased by 3 29. 3 less than twice a number 30. 10 times a number, decreased by 2 In 31–34, use the given variable(s) to write an algebraic expression for each verbal phrase. 31. the number of baseball cards, if b cards are added to a collection of 100 cards 32. Hector’s height, if he was h inches tall before he grew 2 inches 33. the total cost of n envelopes that cost $0.39 each 34. the cost of one pen, if 12 pens cost d dollars 3-2 TRANSLATING VERBAL PHRASES INTO SYMBOLS A knowledge of arithmetic is important in algebra. Since the variables represent numbers that are familiar to you, it will be helpful to solve each problem by first using a simpler related problem; that is, relate similar arithmetic problems to the given algebraic one. 92 Algebraic Expressions and Open Sentences Procedure To write an algebraic expression involving variables: 1. Think of a similar problem in arithmetic. 2. Write an expression for the arithmetic problem, using numbers. 3. Write a similar expression for the problem, using letters or variables. EXAMPLE 1 Represent each phrase by an algebraic expression. a. a distance that is 20 meters shorter than x meters b. a bill for n baseball caps, each costing d dollars c. a weight that is 40 pounds heavier than p pounds d. an amount of money that is twice d dollars Solution a. How to Proceed (1) Think of a similar problem in arithmetic: (2) Write an expression for this arithmetic problem: Think of a distance that is 20 meters shorter than 50 meters. 50 20 (3) Write a similar expression using x 20 Answer the letter x in place of 50. b. How to Proceed (1) Think of a similar problem in arithmetic: Think of a bill for 6 caps, each costing 5 dollars. (2) Write an expression for this arithmetic problem. Multiply the number of caps by the cost of one cap: 6(5) (3) Write a similar expression nd Answer using n and d: Note: After some practice, you will be able to do steps (1) and (2) mentally. c. (p 40) pounds or (40 p) pounds d. 2d dollars Answers a. (x 20) meters c. (p 40) or (40 p) pounds b. nd dollars d. 2d dollars Translating Verbal Phrases into Symbols 93 EXAMPLE 2 Brianna paid 17 dollars for batteries and film for her camera. If the batteries cost x dollars, express the cost of the film in terms of x. Solution If Brianna had spent 5 dollars for the batteries, the amount that was left is found by subtracting the 5 dollars from the 17 dollars, (17 5) dollars. This would have been the cost of the film. If Brianna spent x dollars for the batteries, then the difference, (17 x) dollars would have been the cost of the film. Answer (17 x) dollars Note: In general, if we know the sum of two quantities, then we can let x represent one of these quantities and (the sum x) represent the other. EXERCISES Writing About Mathematics 1. a. Represent the number of pounds of grapes you can buy with d dollars if each pound costs b dollars. b. Does the algebraic expression in part a always represent a whole number? Explain your answer by showing examples using numbers. 2. a. If x apples cost c cents, represent the cost of one apple. b. If x apples cost c cents, represent the cost of n apples. c. Do the algebraic expressions in parts a and b always represent whole numbers? Explain your answer. Developing Skills In 3–18, represent each answer in algebraic language, using the variable mentioned in the problem. 3. The number of kilometers traveled by a bus is represented by x. If a train traveled 200 kilo- meters farther than the bus, represent the number of kilometers traveled by the train. 4. Mr. Gold invested $1,000 in stocks. If he lost d dollars when he sold the stocks, represent the amount he received for them. 5. The cost of a mountain bike is 5 times the cost of a skateboard. If the skateboard costs x dollars, represent the cost of the mountain bike. 6. The length of a rectangle is represented by l. If the width of the rectangle is one-half of its length, represent its width. 7. After 12 centimeters had been cut from a piece of lumber, c centimeters were left. Represent the length of the original piece of lumber. 94 Algebraic Expressions and Open Sentences 8. Paul and Martha saved 100 dollars. If the amount saved by Paul is represented by x, repre- sent the amount saved by Martha. 9. A ballpoint pen sells for 39 cents. Represent the cost of x pens. 10. Represent the cost of t feet of lumber that sells for g cents a foot. 11. If Hilda weighed 45 kilograms, represent her weight after she had lost x kilograms. 12. Ronald, who weighs c pounds, is d pounds overweight. Represent the number of pounds Ronald should weigh. 13. A woman spent $250 for jeans and a ski jacket. If she spent y dollars for the ski jacket, rep- resent the amount she spent for the jeans. 14. A man bought an article for c dollars and sold it at a profit of $25. Represent the amount for which he sold it. 15. The width of a rectangle is represented by w meters. Represent the length of the rectangle if it exceeds the width by 8 meters. 16. The width of a rectangle is x centimeters. Represent the length of the rectangle if it exceeds twice the width by 3 centimeters. 17. If a plane travels 550 kilometers per hour, represent the distance it will travel in h hours. 18. If a car traveled for 5 hours at an average rate of r kilometers per hour, represent the dis- tance it traveled. 19. a. Represent the total number of days in w weeks and 5 days. b. Represent the total number of days in w weeks and d days. Applying Skills 20. An auditorium with m rows can seat a total of c people. If each row in the auditorium has the same number of seats, represent the number of seats in one row. 21. Represent the total number of calories in x peanuts and y potato chips if each peanut con- tains 6 calories and each potato chip contains 14 calories. 22. The charges for a telephone call are $0.45 for the first 3 minutes and $0.09 for each additional minute or part of a minute. Represent the cost of a telephone call that lasts m minutes when m is greater than 3. 23. A printing shop charges a 75-cent minimum for the first 8 photocopies of a flyer. Additional copies cost 6 cents each. Represent the cost of c copies if c is greater than 8. 24. A utility company measures gas consumption by the hundred cubic feet, CCF. The company has a three-step rate schedule for gas customers. First, there is a minimum charge of $5.00 per month for up to 3 CCF of gas used. Then, for the next 6 CCF, the charge is $0.75 per CCF. Finally, after 9 CCF, the charge is $0.55 per CCF. Represent the cost of g CCF of gas if g is greater than 9. 3-3 ALGEBRAIC TERMS AND VOCABULARY
Terms Algebraic Terms and Vocabulary 95 A term is a number, a variable, or any product or quotient of numbers and vari5 c ables. For example: 5, x, 4y, 8ab, , and are terms. k2 5 An algebraic expression that is written as a sum or a difference has more than one term. For example, 4a 2b 5c has three terms. These terms, 4a, 2b, and 5c, are separated by and signs. Factors of a Term If a term contains two or more numbers or variables, then each number, each variable, and each product of numbers and variables is called a factor of the term, or factor of the product. For example, the factors of 3xy are 1, 3, x, y, 3x, 3y, xy, and 3xy. When we factor whole numbers, we write only factors that are integers. Any factor of an algebraic term is called the coefficient of the remaining factor, or product of factors, of that term. For example, consider the algebraic term 3xy: 3 is the coefficient of xy 3y is the coefficient of x 3x is the coefficient of y xy is the coefficient of 3 When an algebraic term consists of a number and one or more variables, the number is called the numerical coefficient of the term. For example: In 8y, the numerical coefficient is 8. In 4abc, the numerical coefficient is 4. When the word coefficient is used alone, it usually means a numerical coefficient. Also, since x names the same term as 1x, the coefficient of x is understood to be 1. This is true of all terms that contain only variables. For example: 7 is the coefficient of b in the 1 is the coefficient of b in the term b. term 7b. 2.25 is the coefficient of gt in 1 is the coefficient of gt in the term gt. the term 2.25gt. Bases, Exponents, and Powers You learned in Chapter 2 that a power is the product of equal factors. A power has a base and an exponent. The base is one of the equal factors of the power. 96 Algebraic Expressions and Open Sentences The exponent is the number of times the base is used as a factor. (If a term is written without an exponent, the exponent is understood to be 1.) 42 4(4): x3 = x(x)(x): 35m: 5d2 = 5(d)(d): base 4 base x base m base d exponent 2 exponent 3 exponent 1 exponent 2 power 42 16 power x3 power m1 or m power d2 An exponent refers only to the number or variable that is directly to its left, as seen in the last example, where 2 refers only to the base d. To show the product 5d as a base (or to show any sum, difference, product, or quotient as a base), we must enclose the base in parentheses. (5d)2 (5d)(5d): base 5d exponent 2 power (5d)2 exponent 2 power (a 4)2 (a + 4)2 (a + 4)( a + 4): base (a + 4) Note that –d4 is not the same as (d)4. –d 4 –1(d)(d)(d)(d) is always a negative number. (d)4 1(d)(d)(d)(d) is always a positive number since the exponent is even. EXAMPLE 1 For each term, name the coefficient, base, and exponent. a. 4x5 b. w8 c. 2pr coefficient 4 coefficient –1 coefficient 2p Answers base x base w base r exponent 5 exponent 8 exponent 1 Note: Remember that coefficient means numerical coefficient, and that 2p is a real number. EXERCISES Writing About Mathematics 1. Does squaring distribute over multiplication, that is, does (ab)2 = (a2)(b2)? Write (ab)2 as (ab)(ab) and use the associative and commutative properties of multiplication to justify your answer. 2. Does squaring distribute over addition, that is, does (a b)2 a2 b2? Substitute values for a and b to justify your answer. Algebraic Terms and Vocabulary 97 Developing Skills In 3–6, name the factors (other than 1) of each product. 3. xy 4. 3a 5. 7mn 6. 1st In 7–14, name, in each case, the numerical coefficient of x. 7. 8x 11. 1.4x 8. (5 2)x 12. 2 7x 9. 1 2x 13. 3.4x In 15–22, name, in each case, the base and exponent of the power. 15. m2 19. 106 16. s3 20. (5y)4 17. t 21. (x y)5 10. x 14. x 18. (a)4 22. 12c3 In 23–29, write each expression, using exponents. 23. b b b b b 26. 7 r r r s s 24. p r r 27. (6a)(6a)(6a) 29. the fourth power of (m + 2n) 25. a a a a b b 28. (a b)(a b)(a b) In 30–33, write each term as a product without using exponents. 30. r6 31. 5x4 32. 4a4b2 33. (3y)5 In 34–41, name, for each given term, the coefficient, base, and exponent. 34. 3k 38. 2y " Applying Skills 35. k3 39. 0.0004t12 36. pr2 3 2a4 40. 37. (ax)5 41. (b)3 42. If x represents the cost of a can of soda, what could 5x represent? 43. If r represents the speed of a car in miles per hour, what could 3r represent? 44. If n represents the number of CDs that Alice has, what could n 5 represent? 45. If d represents the number of days until the end of the year, what could 46. If s represents the length of a side of a square, what could 4s represent? d 7 represent? 47. If r represents the measure of the radius of a circle, what could 2r represent? 48. If w represents the number of weeks in a school year, what could 49. If d represents the cost of one dozen bottles of water, what could w 4 d 12 represent? represent? 50. If q represents the point value of one field goal, what could 7q represent? 98 Algebraic Expressions and Open Sentences 3-4 WRITING ALGEBRAIC EXPRESSIONS IN WORDS In Section 1 of this chapter, we listed the words that can be represented by each of the four basic operations. We can use these same lists to write algebraic expressions in words and to write problems that can be represented by a given algebraic expression. For an algebraic expression such as 2n 3, n could be any real number. That is, associated with any real number n, there is exactly one real number that is the value of 2n 3. However, if n and 2n 3 represent the number of cans of tuna that two customers buy, then n must be a whole number greater than or equal to 2 in order for both n and 2n 3 to be whole numbers. For this situation, the domain or replacement set would be the set of whole numbers. EXAMPLE 1 If n represents the number of points that Hradish scored in a basketball game and 2n 3 represents the number of points that his friend Brad scored, describe in words the number of points that Brad scored. What is a possible domain for the variable n? Solution The number of points scored is always a whole number. In order that 2n 3 be a whole number, n must be at least 2. Answer The number of points that Brad scored is 3 less than twice the number that Hradish scored. A possible domain for n is the set of whole numbers greater than or equal to 2. EXAMPLE 2 Molly earned d dollars in July and the number of dollars that Molly earned in August. 1 2d 1 10 dollars in August. Describe in words Answer In August, Molly earned 10 more than half the number of dollars that she earned in July. EXAMPLE 3 Describe a situation in which x and 12 x can be used to represent variable quantities. List the domain or replacement set for the answer. Solution If x eggs are used from a full dozen of eggs, there will be 12 x eggs left. Answer The domain or replacement set is the set of whole numbers less than or equal to 12. Writing Algebraic Expressions in Words 99 Another Solution The distance from my home to school is 12 miles. On my way to school, after I have traveled x miles, I have 12 x miles left to travel. Answer The domain or replacement set is the set of non-negative real numbers that are less than or equal to 12. Many other answers are possible. EXERCISES Writing About Mathematics 1. a. If 4 n represents the number of books Ken read in September and 4 n represents the number of books he read in October, how many books did he read in these two months? b. What is the domain of the variable n? 2. Pedro said that the replacement set for the amount that we pay for any item is the set of rational numbers of the form 0.01x where x is a whole number. Do you agree with Pedro? Explain why or why not. Developing Skills In 3–14: a. Write in words each of the given algebraic expressions. b. Describe a possible domain for each variable. 3. By one route, the distance that Ian walks to school is d miles. By a different route, the dis- tance is d 0.2 miles. 4. Juan pays n cents for a can of soda at the grocery store. When he buys soda from a machine, he pays n 15 cents. 5. Yesterday Alexander spent a minutes on leisurely reading and 3a 10 minutes doing homework. 6. The width of a rectangle is w meters and the length is 2w 8 meters. 7. During a school day, Abby spends h hours in class, hours at lunch and hours on sports. d 8. Jen spends d hours at work and 12 9. Alicia’s score for 18 holes of golf was g and her son’s score was 10 g. 10. Tom paid d cents for a notebook and 5d 30 cents for a pen. h 6 hours driving to and from work. h 3 11. Seema’s essay for English class had w words and Dominic’s had 12. Virginia read r books last month and Anna read 3r 5 books. 3 4 w 1 80 words. 13. Mario and Pete are playing a card game where it is possible to have a negative score. Pete’s score is s and Mario’s score is s 220. 100 Algebraic Expressions and Open Sentences 14. In the past month, Agatha has increased the time that she walks each day from m minutes to 3m 10 minutes. 3-5 EVALUATING ALGEBRAIC EXPRESSIONS Benjamin has 1 more tape than 3 times the number of tapes that Julia has. If Julia has n tapes, then Benjamin has 3n 1 tapes. The algebraic expression 3n 1 represents an unspecified number. Only when the variable n is replaced by a specific number does 3n 1 become a specific number. For example: If n 10, then 3n 1 3(10) 1 30 1 31. If n 15, then 3n 1 3(15) 1 45 1 46. Since in this example, n represents the number of tapes that Julia has, only whole numbers are reasonable replacements for n. Therefore, the replacement set is the set of whole numbers or some subset of the set of whole numbers. When we substitute specific values for the variables in an algebraic expression and then determine the value of the resulting expression, we are evaluating the algebraic expression. When we determine the number that an algebraic expression represents for specific values of its variables, we are evaluating the algebraic expression. Procedure To evaluate an algebraic expression, replace the variables by the given values, and then follow the r
ules for the order of operations. 1. Replace the variables by the given values. 2. Evaluate the expression within the grouping symbols such as parentheses, always simplifying the expressions in the innermost groupings first. 3. Simplify all powers and roots. 4. Multiply and divide, from left to right. 5. Add and subtract, from left to right. EXAMPLE 1 Evaluate 50 3x when x 7. Solution How to Proceed (1) Write the expression: 50 3x Evaluating Algebraic Expressions 101 (2) Replace the variable by its given value: (3) Multiply: (4) Subtract: 50 3(7) 50 21 29 Answer 29 EXAMPLE 2 Evaluate 2x2 5x 4 when: a. x = 7 b. x = 1.2 Solution How to Proceed a. (1) Write the expression: (2) Replace the variable by the value 7: (3) Evaluate the power: (4) Multiply: (5) Add: b. (1) Write the expression: (2) Replace the variable by the value 1.2: (3) Evaluate the power: (4) Multiply: (5) Add and subtract: 2x2 5x 4 2(7)2 5(7) 4 2(49) 5(7) 4 98 35 4 137 2x2 5x 4 2(1.2)2 5(1.2) 4 2(1.44) 5(1.2) 4 2.88 6 4 0.88 Answers a. 137 b. 0.88 EXAMPLE 3 Evaluate 2a 5 1 (n 2 1)d when a 4, n 10, and d 3. Solution How to Proceed (1) Write the expression: (2) Replace the variables with their given values: (3) Simplify the expressions grouped by parentheses or fraction bar: (4) Multiply and divide: (5) Add: 2a 5 1 (n 2 1)d 2(24) 5 1 (10 – 1)(3) 28 5 1(9)(3) 213 213 252 5 5 1 27 5 1 265 5 Answer 102 Algebraic Expressions and Open Sentences Calculator Solution The values given for the variables can be stored in the calculator. ENTER: (-) 4 STO ALPHA A ENTER 10 STO ALPHA N ENTER 3 STO ALPHA D ENTER DISPLAY Now enter the algebraic expression to be evaluated. ENTER: 2 ALPHA A 5 ( ALPHA N 1 ) ALPHA D ENTER DISPLAY Answer 252 5 5 25.4 EXAMPLE 4 Evaluate (2x)3 2x3 when x 0.40. Solution How to Proceed (1) Write the expression: (2) Replace the variable by its given value: (3) Simplify the expression within brackets: (4) Evaluate the powers: (5) Multiply: (6) Subtract: Answer 0.384 (2x)3 2x3 [2(0.40)]3 2(0.40)3 [0.80]3 2(0.40)3 0.512 2(0.064) 0.512 0.128 0.384 Evaluating Algebraic Expressions 103 EXERCISES Writing About Mathematics 1. Explain why, in an algebraic expression such as 12ab, 12 is called a constant and a and b are called variables? 2. Explain why, in step 2 of Example 1, parentheses were needed when x was replaced by its value. Developing Skills To understand this topic, you should first evaluate the expressions in Exercises 3 to 27 without a calculator. Then, store the values of the variables in the calculator and enter the given algebraic expressions to check your work. In 3–27, find the numerical value of each expression. Use a 8, b 6, d 3, x 4, and y 0.5. 15. 3. 5a 7. b 2 11. 7xy3 3 4x3 19. (ay)3 23. 3y (x d) 26. (2a 5d)2 Applying Skills 4. 1 2x 8. ax2 12. ab dx 16. (3y)2 20. x(y 2) 9. 13. 5. 0.3y 3bd 9 5a 1 1 2 5b 1 4x2y 17. 21. 4(2x 3y) 6. a 3 10. 5x 2y 14. 0.2d 0.3b 18. a2 3d2 22. 1 2x(y 1 0.1)2 24. 2(x y) 5 27. (2a)2 (5d)2 25. (x d)5 28. At one car rental agency, the cost of a car for one day can be determined by using the alge- braic expression $32.00 $0.10m where m represents the number of miles driven. Determine the cost of rental for each of the following: a. Mike Baier drove the car he rented for 35 miles. b. Dana Morse drove the car he rented for 435 miles. c. Jim Szalach drove the car he rented for 102 miles. 29. The local pottery co-op charges $40.00 a year for membership and $0.75 per pound for firing pottery pieces made by the members. The algebraic expression 40 0.75p represents the yearly cost to a member who brings p pounds of pottery to be fired. Determine the yearly cost for each of the following: a. Tiffany is an amateur potter who fired 35 pounds of work this year. b. Nia sells her pottery in a local craft shop and fired 485 pounds of work this year. 30. If a stone is thrown down into a deep gully with an initial velocity of 30 feet per second, the distance it has fallen, in feet, after t seconds can be found by using the algebraic expression 16t2 30t. Find the distance the stone has fallen: a. after 1 second. b. after 2 seconds. c. after 3 seconds. 104 Algebraic Expressions and Open Sentences 31. The Parkside Bread Company sells cookies and scones as well as bread. Bread (b) costs $4.50 a loaf, cookies (c) cost $1.10 each, and scones (s) cost $1.50 each. The cost of a bakery order can be represented by 4.50b 1.10c 1.50s. Determine the cost of each of the following orders: a. six cookies and two scones b. three loaves of bread and one cookie c. one loaf of bread, a dozen cookies, and a half-dozen scones 32. A Green Thumb volunteer can plant shrubbery at a rate of 6 shrubs per hour and a Friendly Garden volunteer can plant shrubbery at a rate of 8 shrubs per hour. The total number of shrubs that g Green Thumb volunteers and f Friendly Garden volunteers can plant in h hours is given by the algebraic expression 6gh 8fh. Determine the number of shrubs planted: a. in 3 hours by 2 Green Thumb and 1 Friendly Garden volunteers. b. in 2 hours by 4 Green Thumb and 4 Friendly Garden volunteers. 3-6 OPEN SENTENCES AND SOLUTION SETS In this chapter, you learned how to translate words into algebraic expressions. The value of an algebraic expression depends on the value of the variables. When the values of the variables change, the value of the algebraic expression changes. For example, x 6 is an algebraic expression. The value of x 6 depends on the value of x. If one value is assigned to an algebraic expression, an algebraic sentence is formed. These sentences may be formulas, equations, or inequalities. For example, when the value 9 is assigned to the algebraic expression x 6, we can write the sentence “Six more than x is 9.” This sentence can be written in symbols as x 6 9. Every sentence that contains a variable is called an open sentence. x 6 9 An open sentence is neither true nor false. x 5 8 3y 12 2n 0 The sentence will be true or false only when the variables are replaced by numbers from a domain or a replacement set, such as {0, 1, 2, 3}. The numbers from the domain that make the sentence true are the elements of the solution set of the open sentence. A solution set, as seen below, can contain one or more numbers or, at times, no numbers at all, from the replacement set. EXAMPLE 1 Using the domain {0, 1, 2, 3}, find the solution set of each open sentence: a. x 6 9 b. 2n 0 Solution a. Procedure: Replace x in the open sentence with numbers from the domain {0, 1, 2, 3}. x + 6 9 Let x 0. Then 0 6 9 is false. Let x 1. Then l 6 9 is false. Let x 2. Then 2 6 9 is false. Let x 3. Then 3 6 9 is true. Here, only when x 3 does the open sentence become a true sentence. Open Sentences and Solution Sets 105 b. Procedure: Replace n in the open sentence with numbers from the domain {0, 1, 2, 3}. 2n 0 Let n 0. Then 2(0) 0 or 0 0 is false. Let n 1. Then 2(l) 0 or 2 0 is true. Let n 2. Then 2(2) 0 or 4 0 is true. Let n 3. Then 2(3) 0 or 6 0 is true. Here, three elements of the domain make the open sentence true. Answer: a. Solution set {3}. Answer: b. Solution set {1, 2, 3} EXAMPLE 2 Find the solution set for the open sentence 3y 12 using: a. the domain {3, 5, 7} b. the domain {whole numbers} Solution a. Procedure: Replace y with 3, 5, and 7. If y 3, then 3(3) 12 is false. If y 5, then 3(5) 12 is false. If y 7, then 3(7) 12 is false. When y is replaced by each of the numbers from the domain, no true statement is found. The solution set for this domain is the empty set or the null set, written in symbols as { } or as . Answer: a. The solution set is { } or . b. Procedure: Of course, you cannot replace y with every whole number, but you can use multiplication facts learned previously. You know that 3(4) 12. Let y 4. Then 3(4) 12 is true. No other whole number would make the open sentence 3y 12 a true sentence. Answer: b. The solution set is {4}. 106 Algebraic Expressions and Open Sentences EXERCISES Writing About Mathematics 1. For the open sentence x 7 12, write a domain for which the solution set is the empty set. 2. For the open sentence x 7 12, write a domain for which the solution set has only one element. 3. For the open sentence x 7 12, write a domain for which the solution set is an infinite set. Developing Skills In 4–11, tell whether each is an open sentence, a true sentence, or an algebraic expression. 4. 2 3 5 0 8. n 7 5. x 10 14 9. 3 2 10 6. y 4 10. 3r 2 7. 3 7 2(5) 11. 2x 7 15 In 12–15, name the variable in each open sentence. 12. x 5 9 13. 4y 20 14. r 6 2 15. 7 3 a In 16–23, using the domain {5, 4, 3, 2, 1, 0, 1, 2, 3, 4, 5}, find the solution set for each open sentence. 16. n 3 7 20. 2n 1 8 Applying Skills 17. x x 0 2n 1 1 3 5 4 21. 18. 5 n 2 22. 3x 2 , x 19. n 3 9 23. 2x 4 24. Pencils sell for $0.19 each. Torry wants to buy at least one but not more than 10 pencils and has $1.50 in his pocket. a. Use the number of pencils that Torry wants to buy to write a domain for this problem. b. The number of pencils that Torry might buy, x, can be found using the open sentence 0.19x 1.50. Find the solution set of this open sentence using the domain from part a. c. How many pencils can Torry buy? 25. The local grocery store has frozen orange juice on sale for $0.99 a can but limits the number of cans that a customer may buy at the sale price to no more than 5. a. The domain for this problem is the number of cans of juice that a customer may buy at the sale price. Write the domain. b. If Mrs. Dajhon does not want to spend more than $10, the number of cans that she might buy at the sale price, y, is given by the equation 0.99y 10. Find the solution set of this equation using the domain from part a. c. How many cans can Mrs. Dajhon buy if she does not want to spend more than $10? Writing Formulas 107 26. Admission to a recreation park is $17.50. This includes all rides except for a ride called The Bronco that costs $1.50 for each ride. Ian has $25 to spend. a. Find the domain for this problem, the number of times a person might ride The Bronco.
b. The number of times Ian might ride The Bronco, z, can be found using the open sen- tence 17.50 1.50z 25. Find the solution set of this open sentence using the domain from part a. c. How many times can Ian ride The Bronco? 3-7 WRITING FORMULAS A formula uses mathematical language to express the relationship between two or more variables. Some formulas are found by the strategy of looking for patterns. For example, how many square units are shown in the rectangle on the left? This rectangle, measuring 4 units in length and 3 units in width, contains a total of 12 square units of area. Many such examples led to the conclusion that the area of a rectangle is equal to the product of its length and width. This relationship is expressed by the formula A lw where A, l, and w are variables that represent, respectively, the area, the length, and the width of a rectangle. A formula is an open sentence that states that two algebraic expressions are equal. In formulas, the word is is translated into the symbol . EXAMPLE 1 Write a formula for each of the following relationships. a. The perimeter P of a square is equal to 4 times the length of one side. b. The total cost C of an article is equal to its price p plus an 8% tax on the price. c. The sum S of the measures of the interior angles of an n-sided polygon is 180 times 2 less than the number of sides. Solution a. Let s represent the length of each side of a square. b. 8% (or 8 percent) means 8 hundredths, written as 0.08 or 8 100 . P 4s Answer 100p c. “2 less than the number of sides” means (n 2). C p 0.08p or C 5 p 1 8 Answer S 180(n 2) Answer 108 Algebraic Expressions and Open Sentences EXAMPLE 2 Write a formula that expresses the number of months m that are in y years. Solution Look for a pattern. In 1 year, there are 12 months. In 2 years, there are 12(2) or 24 months. In 3 years, there are 12(3) or 36 months. In y years, there are 12(y) or 12y months. This equals m, the number of months. Answer m 12y EXAMPLE 3 The Short Stop Diner pays employees $6.00 an hour for working 40 hours a week or less. For working overtime, an employee is paid $9.00 for each hour over 40 hours. Write a formula for the wages, W, of an employee who works h hours in a week. Solution Two formulas are needed, one for h 40 and the other for h 40. If h 40, the wage is 6.00 times the number of hours, h. W 6.00h If h 40, the employee has worked 40 hours at $6.00 an hour and the remaining hours, h 40, at $9.00 an hour. W 6.00(40) 9.00(h 40). Answer W 6h if h 40 and W 6(40) 9(h 40) if h 40. (Note that the formula for h 40 may also be given as W 240 9(h 40).) EXERCISES Writing About Mathematics 1. Fran said that a recipe is a type of formula. Do you agree or disagree with Fran? Explain your answer. 2. a. Is an algebraic expression a formula? Explain why or why not. b. Is a formula an open sentence? Explain why or why not. Writing Formulas 109 Developing Skills In 3–17, write a formula that expresses each relationship. 3. The total length l of 10 pieces of lumber, each m meters in length, is 10 times the length of each piece of lumber. 4. An article’s selling price S equals its cost c plus the margin of profit m. 5. The perimeter P of a rectangle is equal to the sum of twice its length l and twice its width w. 6. The average m of three numbers, a, b, and c is their sum divided by 3. 7. The area A of a triangle is equal to one-half the length of the base b multiplied by the length of the altitude h. 8. The area A of a square is equal to the square of the length of a side s. 9. The volume V of a cube is equal to the cube of the length of an edge e. 10. The surface area S of a cube is equal to 6 times the square of the length of an edge e. 11. The surface area S of a sphere is equal to the product of 4p and the square of the radius r. 12. The average rate of speed r is equal to the distance that is traveled d divided by the time spent on the trip t. 13. The Fahrenheit temperature F is 32° more than nine-fifths of the Celsius tempera- ture C. 14. The Celsius temperature C is equal to five-ninths of the difference between the Fahrenheit temperature F and 32°. 15. The dividend D equals the product of the divisor d and the quotient q plus the remainder r. 16. A sales tax T that must be paid when an article is purchased is equal to 8% of the price of the article v. 17. A salesman’s weekly earnings F is equal to his weekly salary s increased by 2% of his total volume of sales v. Applying Skills 18. A ferry takes cars, drivers, and passengers across a body of water. The total ferry charge C in dollars is $20.00 for the car and driver, plus d dollars for each passenger. a. Write a formula for C in terms of d and the number of passengers, n. b. Find the cost of the ferry for a car if d $15 and there are 5 persons in the car. c. Find the cost of the ferry for a car with only the driver. 110 Algebraic Expressions and Open Sentences 19. The cost C in cents of an internet telephone call lasting m minutes is x cents for the first 3 minutes and y cents for each additional minute. a. Write two formulas for C, one for the cost of calls lasting 3 minutes or less (m 3), and another for the cost of calls lasting more than 3 minutes (m 3). b. Find the cost of a 2.5 minute telephone call if x $0.25 and y $0.05. c. Find the cost of a 10 minute telephone call if x $0.25 and y $0.05. 20. The cost D in dollars of sending a fax of p pages is a dollars for sending the first page and b dollars for each additional page. a. Write two formulas for D, one for the cost of faxing 1 page (p 1), and another for the cost of faxing more than 1 page (p 1). b. Find the cost of faxing 1 page if a $1.00 and b $0.60. c. Find the cost of faxing 5 pages if a $1.00 and b $0.60. 21. A gasoline dealer is allowed a profit of 12 cents a gallon for each gallon sold. If more than 25,000 gallons are sold in a month, an additional profit of 3 cents for every gallon over that number is given. a. Write two formulas for the gasoline dealer’s profit, P, one for when the number of gallons sold, n, is not more than 25,000 (n 25,000), and another for when more than 25,000 gallons are sold (n 25,000). b. Find P when 21,000 gallons of gasoline are sold in one month. c. Find P when 30,000 gallons of gasoline are sold in one month. 22. Gabriel earns a bonus of $25 for each sale that he makes if the number of sales, s, in a month is 20 or less. He earns an extra $40 for each additional sale if he makes more than 20 sales in a month. a. Write a formula for Gabriel’s bonus, B, when s 20. b. Write a formula for B when s 20. c. In August, Gabriel made 18 sales. Find his bonus for August. d. In September, Gabriel made 25 sales. Find his bonus for September. 23. Mrs. Lucy is selling cookies at a local bake sale. If she sells exactly 3 dozen cookies, the cost of ingredients will equal her earnings. If she sells more than 3 dozen cookies, Mrs. Lucy will make a profit of 25 cents for each cookie sold. a. Write a formula for Mrs. Lucy’s earnings, E, when the number of cookies sold, c, is equal to 36. b. Write a formula for E when c 36. c. Find the number of cookies Mrs. Lucy sold if she makes a profit of $2.00. CHAPTER SUMMARY Chapter Summary 111 An algebraic expression, such as x 6, is an expression or a phrase that contains one or more variables, such as x. The variable is a placeholder for numbers. To evaluate an expression, replace each variable with a number and follow the order of operations. A term is a number, a variable, or any product or quotient of numbers and variables. In the term 6by, 6 is the numerical coefficient. In the term n3, the base is n, the exponent is 3, and the power is n3. The power n3 means that base n is used as a factor 3 times. An open sentence, which can be an equation or an inequality, contains a variable. When the variable is replaced by numbers from a domain, the numbers that make the open sentence true are the elements of the solution set of the sentence. A formula is a sentence that shows the relationship between two or more variables. VOCABULARY 3-1 Variable • Placeholder • Domain • Replacement set • Algebraic expression 3-3 Term • Factor • Coefficient • Numerical coefficient 3-5 Evaluating an algebraic expression 3-6 Open sentence • Solution set 3-7 Formula REVIEW EXERCISES 1. Explain the difference between an algebraic expression and an open sentence. 2. Explain the difference between 2a2 and (2a)2. In 3–6, use mathematical symbols to translate the verbal phrases into algebraic language. 3. x divided by b 4. 4 less than r 5. q decreased by d 6. 3 more than twice g In 7–14, find the value of each expression. 7. 6ac d when a 10, c 8, and d 5 8. 4b2 when b 2.5 112 Algebraic Expressions and Open Sentences 9. 3b c when b 7 and c 14 10. km 9 when k 15 and m 0.6 11. when a 5, b 3, and c 12 bc a 12. 2a2 2a when a 5 1 4 a 5 1 13. (2a)2 2a when 4 14. a(b c) when a 2.5, b 1.1, and c 8.9 15. Write an algebraic expression for the total number of cents in n nickels and q quarters. 16. In the term 2xy3 what is the coefficient? 17. In the term 2xy3 what is the exponent of y? 18. In the term 2xy3, what is the base that is used 3 times as a factor? 19. If distance is the product of rate and time, write a formula for distance, d, in terms of rate, r, and time, t. 20. What is the smallest member of the solution set of 19.4 y 29 if the domain is {46.25, 47.9, 48, 48.5, 49, 50, 51.3 }? 21. What is the smallest member of the solution set of 19.4 y 29 if the domain is the set of whole numbers? 22. What is the smallest member of the solution set of 19.4 y 29 if the domain is the set of real numbers? 23. In a baseball game, the winning team scored n runs and the losing team scored 2n 5 runs. a. Describe in words the number of runs that the losing team scored. b. What could have been the score of the game? Is there more than one answer? c. What are the possible values for n? 24. A mail order book club offers books for $8.98 each plus $3.50 for shipping and handli
ng on each order. The cost of Bethany’s order, which totaled less than $20, can be expressed as 8.98b 3.50 20 where b represents the number of books Bethany ordered. a. What could be the domain for this problem? b. What is the solution set for this open sentence? 25. Write two algebraic expressions to represent the cost of sending an express delivery based on the rates given in the chapter opener on page 88, the first if the delivery distance is 10 miles or less, and the second if the delivery distance is more than 10 miles. 26. A list of numbers that follows a pattern begins with the numbers 2, 5, 8, 11, . . . . Chapter Summary 113 a. Find the next number in the list. b. Write a rule or explain how the next number is determined. c. What is the 25th number in the list? 27. Each of the numbers given below is different from the others, that is, it belongs to a set of numbers to which the others do not belong. Explain why each is different. 3 6 9 35 28. Two oranges cost as much as five bananas. One orange costs the same as a banana and an apple. How many apples cost the same as three bananas? Exploration STEP 1. Write a three-digit multiple of 11 by multiplying any whole number from 10 to 90 by 11. Add the digits in the hundreds and the ones places. If the sum is greater than or equal to 11, subtract 11. Compare this result to the digit in the tens place. Repeat the procedure for other three-digit multiples of 11. STEP 2. Write a three-digit number that is not a multiple of 11 by adding any counting number less than 11 to a multiple of 11 used in step 1. Add the digits in the hundreds and the ones places. If the sum is greater than or equal to 11, subtract 11. Compare this result to the digit in the tens place. Repeat the procedure with another number. STEP 3. Based on steps 1 and 2, can you suggest a way of determining whether or not a three-digit number is divisible by 11? STEP 4. Write a four-digit multiple of 11 by multiplying any whole number from 91 to 909 by 11. Add the digits in the hundreds and ones places. Add the digits in the thousands and tens places. If one sum is greater than or equal to 11, subtract 11. Compare these results. Repeat the procedure for another four-digit multiple of 11. STEP 5. Write a four-digit number that is not a multiple of 11 by adding any counting number less than 11 to a multiple of 11 used in step 4. Add the digits in the hundreds place and ones place. Add the digits in the thousands place and tens place. If one sum is greater than or equal to 11, subtract 11. Compare these results. Repeat the procedure starting with another number. STEP 6. Based on steps 4 and 5, can you suggest a way of determining whether or not a four-digit number is divisible by 11? STEP 7. Write a rule for determining whether or not any whole number is divis- ible by 11. 114 Algebraic Expressions and Open Sentences CUMULATIVE REVIEW CHAPTERS 1–3 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. Which of the following is the set of negative integers greater than –3? (1) {4, 5, 6, 7, . . .} (2) {3, 4, 5, 6, . . .} (3) {3, 2, 1} (4) {2, 1} 5 3 (3) 1.666666667 2. The exact value of the rational number can be written as (1) 1.6 (2) 1.6 (4) 1.666666666 3. Rounded to the nearest hundredth, (1) 2.23 (2) 2.236 5 is approximately equal to (3) 2.24 (4) 2.240 " 4. Which of the following numbers is rational? (1) p (2) (3) 1.42 2 " (4) 0.4 " 5. Which of the following inequalities is a true statement? (1) 0.026 0.25 0.2 (3) 0.2 0.25 0.026 (2) 0.2 0.026 0.25 (4) 0.026 0.2 0.25 6. The length of a rectangle is given as 30.02 yards. This measure has how many significant digits? (2) 2 (1) 1 7. Which of the following is not a prime? (1) 7 (2) 23 (3) 3 (3) 37 (4) 4 (4) 51 8. The arithmetic expression 8 5(0.2)2 10 is equal to (1) 0.9 (2) 0.78 (3) 7.98 (4) 8.02 9. Which of the following is a correct application of the distributive prop- erty? (1) 4(8 0.2) 4(8) 4(0.2) (2) 6(3 1) 6(1 3) 10. The additive inverse of 7 is (1) –7 (2) 0 Part II (3) 8(5 4) 8(5) 4 (4) 8(5)(4) 8(5) 8(4) (3) 1 7 (4) 7 Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. Cumulative Review 115 11. Of the 80 students questioned about what they had read in the past month, 35 had read nonfiction, 55 had read fiction, and 22 had read neither fiction nor nonfiction. How many students had read both fiction and nonfiction? 12. What is the largest number that is the product of three different two-digit primes? Part III Answer all questions in this part. Each correct answer will receive 3 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. sides of equal measure) is 13. The formula for the area of an equilateral triangle (a triangle with three 3 4 s2 A 5 " triangle if the measure of one side is 12.6 centimeters. Express your answer to the number of significant digits determined by the given data. . Find the area of an equilateral 14. A teacher wrote the sequence 1, 2, 4, . . . and asked what the next number could be. Three students each gave a different answer and the teacher said that all three answers were correct. a. Adam said 7. Explain what rule Adam used. b. Bette said 8. Explain what rule Bette used. c. Carlos said 5. Explain what rule Carlos used. Part IV Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. c 5 1 15. If a 7, b –5, and 3 , evaluate the expression a2b c 1 3(b 2 2) . Do not use a calculator. Show each step in your calculation. 16. Michelle bought material to make a vest and skirt. She used half of the material to make the skirt and two-thirds of what remained to make the vest. She had yards of material left. 11 4 a. How many yards of material did she buy? b. How many yards of material did she use for the vest? c. How many yards of material did she use for the skirt? CHAPTER 4 CHAPTER TABLE OF CONTENTS 4-1 Solving Equations Using More Than One Operation 4-2 Simplifying Each Side of an Equation 4-3 Solving Equations That Have the Variable in Both Sides 4-4 Using Formulas to Solve Problems 4-5 Solving for a Variable in Terms of Another Variable 4-6 Transforming Formulas 4-7 Properties of Inequalities 4-8 Finding and Graphing the Solution Set of an Inequality 4-9 Using Inequalities to Solve Problems Chapter Summary Vocabulary Review Exercises Cumulative Review 116 FIRST DEGREE EQUATIONS AND INEQUALITIES IN ONE VARIABLE An equation is an important problem-solving tool. A successful business person must make many decisions about business practices. Some of these decisions involve known facts, but others require the use of information obtained from equations based on expected trends. For example, an equation can be used to represent the following situation. Helga sews hand-made quilts for sale at a local craft shop. She knows that the materials for the last quilt that she made cost $76 and that it required 44 hours of work to complete the quilt. If Helga received $450 for the quilt, how much did she earn for each hour of work, taking into account the cost of the materials? Most of the problem-solving equations for business are complex. Before you can cope with complex equations, you must learn the basic principles involved in solving any equation. 4-1 SOLVING EQUATIONS USING MORE THAN ONE OPERATION Solving Equations Using More Than One Operation 117 Some Terms and Definitions An equation is a sentence that states that two algebraic expressions are equal. For example, x 3 9 is an equation in which x 3 is called the left side, or left member, and 9 is the right side, or right member. An equation may be a true sentence such as 5 2 7, a false sentence such as 6 3 4, or an open sentence such as x 3 9. The number that can replace the variable in an open sentence to make the sentence true is called a root, or a solution, of the equation. For example, 6 is a root of x + 3 9. As discussed in Chapter 3, the replacement set or domain is the set of possible values that can be used in place of the variable in an open sentence. If no replacement set is given, the replacement set is the set of real numbers. The set consisting of all elements of the replacement set that are solutions of the open sentence is called the solution set of the open sentence. For example, if the replacement set is the set of real numbers, the solution set of x 3 9 is {6}. If no element of the replacement set makes the open sentence true, the solution set is the empty or null set, or {}. If every element of the domain satisfies an equation, the equation is called an identity. Thus, 5 x x (5) is an identity when the domain is the set of real numbers because every element of the domain makes the sentence true. Two equations that have the same solution set are equivalent equations. To solve an equation is to find its solution set. This is usually done by writing simpler equivalent equations. If not every element of the domain makes the sentence true, the equation is called a conditional equation, or simply an equation. Therefore, x 3 9 is a conditional equation. Properties of Equality When two numerical or algebraic expressions are equal, it is reasonable to assume that if we change each in the same way, the resulting expressions will be equal. For example: 5 7 12 (5 7) 3 12 3 (5 7) 8 12 8 2(5 7) 2(12) 5 1 7 3 5 12 3 These examples suggest the following properti
es of equality: 118 First Degree Equations and Inequalities in One Variable Properties of Equality 1. The addition property of equality. If equals are added to equals, the sums are equal. 2. The subtraction property of equality. If equals are subtracted from equals, the differences are equal. 3. The multiplication property of equality. If equals are multiplied by equals, the products are equal. 4. The division property of equality. If equals are divided by nonzero equals, the quotients are equal. 5. The substitution principle. In a statement of equality, a quantity may be substituted for its equal. To solve an equation, you need to work backward or “undo” what has been done by using inverse operations. To undo the addition of a number, add its opposite. For example, to solve the equation x 7 19, use the addition property of equality. Add the opposite of 7 to both sides. x 1 7 5 19 27 27 x 5 12 The variable x is now alone on one side and it is easy to read the solution, x 12. To solve an equation in which the variable has been multiplied by a number, either divide by that number or multiply by its reciprocal. (Remember multiplying by the reciprocal is the same as dividing by the number.) To solve 1 6x 24, divide both sides by 6 or multiply both sides by . 6 6x 24 6x 6 5 24 6 x 4 or 6x 24 6(6x) 5 1 1 x 4 6(24) To solve x 3 5 5 , multiply each side by the reciprocal of which is 3. 1 3 (3)x x 3 5 5 3 5 (3)5 x 15 In the equation 2x 3 15, there are two operations in the left side: multiplication and addition. In forming the left side of the equation, x was first multiplied by 2, and then 3 was added to the product. To solve this equation, we must undo these operations by using the inverse elements in the reverse order. Since the last operation was to add 3, the first step in solving the equation is to add its opposite, 3, to both sides of the equation or subtract 3 from both sides Solving Equations Using More Than One Operation 119 of the equation. Here we are using either the addition or the subtraction property of equality. 2x 1 3 5 15 2x 1 3 1 (23) 5 15 1 (23) 2x 5 12 or 2x 1 3 23 2x 5 5 15 23 12 Now we have a simpler equation that has the same solution set as the original and includes only multiplication by 2. To solve this simpler equation, we multiply both sides of the equation by , the reciprocal of 2, or divide both sides of the equation by 2. Here we can use either the multiplication or the division property of equality. 1 2 2x 5 12 2(2x) 5 1 1 x 5 6 2(12) or 2x 5 12 2x 2 5 12 2 x 5 6 After an equation has been solved, we check the equation, that is, we verify that the solution does in fact make the given equation true by replacing the variable with the solution and performing any computations. Check: 2x 3 15 2(6) 3 15 12 3 15 15 15 ✔ To find the solution of the equation 2x 3 15, we used several properties of the four basic operations and of equality. The solution below shows the mathematical principle that we used in each step. 2x 3 15 Given (2x 3) (3) 15 (3) Addition property of equality 2x [3 (3)] 15 (3) Associative property of addition 2x 0 12 2x 12 1 2(12) 1 2(12) 1 2(2x) 1 x 2(2) D 1x 6 x 6 C Additive inverse property Additive identity property Multiplication property of equality Associative property of multiplication Multiplicative inverse property Multiplicative identity property These steps and properties are necessary to justify the solution of an equation of this form. However, when solving an equation, we do not need to write each of the steps, as shown in the examples that follow. 120 First Degree Equations and Inequalities in One Variable EXAMPLE 1 Solve and check: 7x 15 71 Solution How to Proceed (1) Write the equation: (2) Add 15, the opposite of 15 to each side: (3) Since multiplication and division are inverse operations, divide each side by 7: (4) Check the solution. Write the solution in place of x and perform the computations: 7x 7x 15 71 15 15 56 7x 56 7 7 x 8 7(8) 1 15 5? 56 1 15 5? 7x 15 71 71 71 71 71 ✔ Answer x 8 Note: The check is based on the substitution principle. EXAMPLE 2 Solution Find the solution set and check: 3 5x 2 6 18 3 5x 2 6 5 218 16 16 3 12 5x 5 5 3(212) 3 3 5x A B x 20 Addition property of equality Multiplication property of equality Check 6 18 3 5x 3 5(220) 2 6 5? 212 2 6 5? 218 218 18 18 ✔ Answer The solution set is {20}. EXAMPLE 3 Solve and check: 7 x 9 Solution METHOD 1. Think of 7 x as 7 (1x). Solving Equations Using More Than One Operation 121 27 71(2x) 5 9 27 2x 5 2 21x 21 5 2 21 x 2 Addition property of equality Division property of equality Check 7 x 9 7 2 (2) 5? 9 7 1 2 5? 9 9 9 ✔ METHOD 2. Add x to both sides of the equation so that the variable has a positive coefficient. How to Proceed (1) Write the equation: (2) Add x to each side of the equation: (3) Add 9 to each side of the equation: The check is the same as for Method 1. Answer {2} or EXERCISES Writing About Mathematics 1. Is it possible for the equation 2x 5 0 to have a solution in the set of positive real num- bers? Explain your answer. 2. Max wants to solve the equation 7x 15 71. He begins by multiplying both sides of the equation by , the reciprocal of the coefficient of x. 1 7 a. Is it possible for Max to solve the equation if he begins in this way? If so, what 1 would be the result of multiplying by and what would be his next step? 7 b. In this section you learned to solve the equation 7x 15 71 by first adding the opposite of 15, 15, to both sides of the equation. Which method do you think is better? Explain your answer. Developing Skills In 3 and 4, write a complete solution for each equation, listing the property used in each step. 3. 3x 5 35 4. 1 2x21 15 122 First Degree Equations and Inequalities in One Variable 5. 55 6a 7 9. 15 a 3 In 5–32, solve and check each equation. 6. 17 8c 7 10. 11 6d 1 3 4y 5 14. 12 2 13. 8 3x 17. 7.2 21. 4a 0.2 5 18. 22. 4 3t 0.2 a 4 1 9 4m 5 25. 29. 47 4 5t 1 7 1 7 5 14 2 x 26. 0.04c 1.6 0 30. 0.8r 19 20 15. 7. 9 1x 7 11. 8 y = 1 4 5 45 5t 2 y 19. 2 5 1 3 1 5 4x 1 11 27. 15x 14 19 23. 16. 8. 11 15t 16 3a 12. 12 8 23 5m 9d 2 1 30 2 5 171 2 5 2 2 3y 28. 8 18c 1 20. 24. 13 31. 1 3w 1 6 5 22 32. 842 162m 616 Applying Skills 33. The formula F 9 5C132 gives the relationship between the Fahrenheit temperature F and the Celsius temperature C. Solve the equation 59 degrees Celsius when the Fahrenheit temperature is 59°. 9 5C132 to find the temperature in 34. When Kurt orders from a catalog, he pays $3.50 for shipping and handling in addition to the cost of the goods that he purchases. Kurt paid $33.20 when he ordered six pairs of socks. Solve the equation 6x 3.50 33.20 to find x, the price of one pair of socks. 35. When Mattie rents a car for one day, the cost is $29.00 plus $0.20 a mile. On her last trip, Mattie paid $66.40 for the car for one day. Find the number of miles, m, that Mattie drove by solving the equation 29 0.20x 66.40. 36. On his last trip to the post office, Hal paid $4.30 to mail a package and bought some 39-cent stamps. He paid a total of $13.66. Find s, the number of stamps that he bought, by solving the equation 0.39s 4.30 13.66. 4-2 SIMPLIFYING EACH SIDE OF AN EQUATION An equation is often written in such a way that one or both sides are not in simplest form. Before starting to solve the equation by using additive and multiplicative inverses, you should simplify each side by removing parentheses if necessary and adding like terms. Recall that an algebraic expression that is a number, a variable, or a product or quotient of numbers and variables is called a term. First-degree equations in one variable contain two kinds of terms, terms that are constants and terms that contain the variable to the first power only. Simplifying Each Side of an Equation 123 Like and Unlike Terms Two or more terms that contain the same variable or variables, with corresponding variables having the same exponents, are called like terms or similar terms. For example, the following pairs are like terms. 6k and k 5x2 and 7x2 9ab and 0.4ab 9 2x2y3 and 211 3 x2y3 Two terms are unlike terms when they contain different variables, or the same variable or variables with different exponents. For example, the following pairs are unlike terms. 3x and 4y 5x2 and 5x3 9ab and 0.4a 8 3x3y2 and 4 7x2y3 To add like terms, we use the distributive property of multiplication over addition. 9x 2x (9 2)x 11x 16d 3d (–16 3)d 13d Note that in the above examples, when like terms are added: 1. The sum has the same variable factor as the original terms. 2. The numerical coefficient of the sum is the sum of the numerical coeffi- cients of the terms that were added. The sum of like terms can be expressed as a single term. The sum of unlike terms cannot be expressed as a single term. For example, the sum of 2x and 3 cannot be written as a single term but is written 2x 3. EXAMPLE 1 Solve and check: 2x 3x 4 6 Solution How to Proceed Check (1) Write the equation: (2) Simplify the left side by 2x 3x 4 6 5x 4 6 2x 3x 4 6 26 2(22) 1 3(22) 1 4 5? combining like terms: (3) Add 4, the additive inverse of 4, to each side: (4) Multiply by , the 1 5 multiplicative inverse of 5: (5) Simplify each side. Answer 2 24 2 6 1 4 5? 26 6 6 ✔ 4 4 10 5x 5(5x) 5 1 1 5(210) x 2 124 First Degree Equations and Inequalities in One Variable Note: When solving equations, remember to check the answer in the original equation and not in the simplified one. The algebraic expression that is on one side of an equation may contain parentheses. Use the distributive property to remove the parentheses solving the equation. The following examples illustrate how the distributive and associative properties are used to do this. EXAMPLE 2 Solve and check: 27x 3(x 6) 6 Solution Since 3(x 6) means that (x 6) is to be multiplied by 3, we will use the distributive property to remove parentheses and then combine like terms. Note that for this solution, in the first three steps the left side is being simplified. These steps apply only to the left si
de and only change the form but not the numerical value. The next two steps undo the operations of addition and multiplication that make up the expression 24x 18. Since adding 18 and dividing by 24 will change the value of the left side, the right side must be changed in the same way to retain the equality. How to Proceed (1) Write the equation: (2) Use the distributive property: (3) Combine like terms: (4) Use the addition property of equality. Add 18, the additive inverse of 18, to each side: (5) Use the division property of equality. Divide each side by 24: (6) Simplify each side: Check (1) Write the equation: (2) Replace x by 21 2 (3) Perform the indicated computation: Answer x 21 2 27x 3(x 6) 6 27x 3x 18 6 24x 18 6 18 18 12 24x 24x 24 5 212 24 21 x 2 27 21 2 A 27 A B 21 2 3 B 21 2 27x 3(x 6) 6 5? 6 2 2 6 B A 261 5? 6 2 3 2 B A 2 1 183 227 2 5? 6 2 1 39 2 5? 6 227 12 2 5? 6 6 6 ✔ Simplifying Each Side of an Equation 125 Representing Two Numbers with the Same Variable Problems often involve finding two or more different numbers. It is useful to express these numbers in terms of the same variable. For example, if you know the sum of two numbers, you can express the second in terms of the sum and the first number. • If the sum of two numbers is 12 and one of the numbers is 5, then the other number is 12 5 or 7. • If the sum of two numbers is 12 and one of the numbers is 9, then the other number is 12 9 or 3. • If the sum of two numbers is 12 and one of the numbers is x, then the other number is 12 x. A problem can often be solved algebraically in more than one way by writing and solving different equations, as shown in the example that follows. The methods used to obtain the solution are different, but both use the facts stated in the problem and arrive at the same solution. EXAMPLE 3 The sum of two numbers is 43. The larger number minus the smaller number is 5. Find the numbers. Solution This problem states two facts: FACT 1 The sum of the numbers is 43. FACT 2 larger number is 5 more than the smaller. The larger number minus the smaller number is 5. In other words, the (1) Represent each number in terms of the same variable using Fact 1: the sum of the numbers is 43. Let x the larger number. Then, 43 x the smaller number. (2) Write an equation using Fact 2: The larger number minus the smaller number is 5. \|_________________\| ↓ ↓ ↓ 5 x \|__________________\| ↓ (43 x) ↓ 126 First Degree Equations and Inequalities in One Variable (3) Solve the equation. (a) Write the equation: (b) To subtract (43 x), add its opposite: (c) Combine like terms: (d) Add the opposite of 43 to each side: (e) Divide each side by 2: (4) Find the numbers. The larger number x 24. The smaller number 43 x 43 24 19. x (43 x) 5 x (43 x) 5 2x 43 5 43 43 48 2x 2 5 48 2x 2 x 24 Check A word problem is checked by comparing the proposed solution with the facts stated in the original wording of the problem. Substituting numbers in the equation is not sufficient since the equation formed may not be correct. The sum of the numbers is 43: 24 19 43. The larger number minus the smaller number is 5: 24 19 5. Alternate Solution Reverse the way in which the facts are used. (1) Represent each number in terms of the same variable using Fact 2: the larger number is 5 more than the smaller. Let x the smaller number. Then, x 5 the larger number. (2) Write an equation using the first fact. The sum of the numbers is 43. \|______________________\| ↓ ↓ ↓ 43 x (x 5) (3) Solve the equation. (a) Write the equation: (b) Combine like terms: (c) Add the opposite of 5 to each side: (d) Divide each side by 2: (4) Find the numbers. The smaller number x 19. The larger number x 5 19 5 24. (5) Check. (See the first solution.) x (x 5) 43 2x 5 43 5 5 38 2 5 38 2x 2 x 19 2x Answer The numbers are 24 and 19. Simplifying Each Side of an Equation 127 EXERCISES Writing About Mathematics 1. Two students are each solving a problem that states that the difference between two num- bers is 12. Irene represents one number by x and the other number by x 12. Henry represents one number by x and the other number by x 12. Explain why both students are correct. 2. A problem states that the sum of two numbers is 27. The numbers can be represented by x and 27 x. Is it possible to determine which is the larger number and which is the smaller number? Explain your answer. Developing Skills In 3–28, solve and check each equation. 3. x (x 6) 20 5. (15x 7) 12 4 7. x (4x 32) 12 9. 5(x 2) 20 11. 8(2c 1) 56 13. 30 2(10 y) 15. 25 2(t 5) 19 17. 55 4 3(m 2) 19. 3(2b 1) 7 50 21. 7r (6r 5) 7 23. 5m 2(m 5) 17 25. 3(a 5) 2(2a 1) 0 4. x (12 x) 38 6. (14 3c) 7c 94 8. 7x (4x 39) 0 10. 3(y 9) 30 12. 6(3c 1) 42 14. 4(c 1) 32 16. 18 6x 4(2x 3) 18. 5(x 3) 30 10 20. 5(3c 2) 8 43 22. 8b 4(b 2) 24 24. 28y 6(3y 5) 40 26. 0.04(2r 1) 0.03(2r 5) 0.29 27. 0.3a (0.2a 0.5) 0.2(a 2) 1.3 28. 3 4(8 1 4x) 2 1 3(6x 1 3) 5 9 Applying Skills In 29–33, write and solve an equation for each problem. Follow these steps: a. List two facts in the problem. b. Choose a variable to represent one of the numbers to be determined. c. Use one of the facts to write any other unknown numbers in terms of the chosen variable. d. Use the second fact to write an equation. e. Solve the equation. 128 First Degree Equations and Inequalities in One Variable f. Answer the question. g. Check your answer using the words of the problem. 29. Sandi bought 6 yards of material. She wants to cut it into two pieces so that the difference between the lengths of the two pieces will be 1.5 yards. What should be the length of each piece? 30. The Tigers won eight games more than they lost, and there were no ties. If the Tigers played 78 games, how many games did they lose? 31. This month Erica saved $20 more than last month. For the two months, she saved a total of $70. How much did she save each month? 32. On a bus tour, there are 100 passengers on three buses. Two of the buses each carry four fewer passengers than the third bus. How many passengers are on each bus? 33. For a football game, of the seats in the stadium were filled. There were 31,000 empty seats 4 5 at the game. What is the stadium’s seating capacity? 4-3 SOLVING EQUATIONS THAT HAVE THE VARIABLE IN BOTH SIDES A variable represents a number. As you know, any number may be added to both sides of an equation without changing the solution set. Therefore, the same variable (or the same multiple of the same variable) may be added to or subtracted from both sides of an equation without changing the solution set. For instance, to solve 8x 30 5x, write an equivalent equation that has only a constant in the right side. To do this, eliminate 5x from the right side by adding its opposite, 5x, to each side of the equation. METHOD 1 8x 30 5x 5x 5x 3x 30 x 10 METHOD 2 8x 30 5x 8x (5x) 30 5x (5x) 3x 30 x 10 Check 8x 30 5x 5? 30 1 5(10) 5? 30 1 50 8(10) 80 80 80 ✔ Answer: x 10 To solve an equation that has the variable in both sides, transform it into an equivalent equation in which the variable appears in only one side. Then, solve the equation. Solving Equations That Have the Variable in Both Sides 129 EXAMPLE 1 Solve and check: 7x 63 2x Solution How to Proceed (1) Write the equation: (2) Add 2x to each side of the equation: 7x 63 2x 2x 2x 9x 63 Check 7x 63 2x 5? 7(7) 5? 49 49 49 ✔ 63 2 2(7) 63 2 14 (3) Divide each side of the equation by 9: (4) Simplify each side: 9x 9 5 63 9 x 7 Answer x 7 EXAMPLE 2 Solution To solve an equation that has both a variable and a constant in both sides, first write an equivalent equation with only a variable term on one side. Then solve the simplified equation. The following example shows how this can be done. Solve and check: 3y 7 5y 3 METHOD 1 3y 7 5y 3 5y 5y 2y 7 7 2y 3 7 10 3y METHOD 2 3y 7 5y 3 3y 7 2y 3 3 3 10 2y Check 3y 7 5y 3 3(5) 7 15 7 5? 5(5) 2 3 5? 25 2 3 22 22 ✔ 22y 22 210 22 y 5 2y 10 2 5 2 y 5 Answer y 5 A graphing calculator can be used to check an equation. The calculator can determine whether a given statement of equality or inequality is true or false. If the statement is true, the calculator will display 1; if the statement is false, the calculator will display 0. The symbols for equality and inequality are found in the menu. TEST 130 First Degree Equations and Inequalities in One Variable To check that y 5 is the solution to the equation 3y 7 5y 3, first store 5 as the value of y. then enter the equation to be checked. ENTER: 5 STO ALPHA Y ENTER 3 ALPHA Y 7 2nd TEST ENTER 5 ALPHA Y 3 ENTER DISPLAY The calculator displays 1 which indicates that the statement of equality is true for the value that has been stored for y. 5 1 EXAMPLE 3 The larger of two numbers is 4 times the smaller. If the larger number exceeds the smaller number by 15, find the numbers. Note: When s represents the smaller number and 4s represents the larger number, “the larger number exceeds the smaller by 15” has the following meanings. Use any one of them. 1. The larger equals 15 more than the smaller, written as 4s = 15 s. 2. The larger decreased by 15 equals the smaller, written as 4s 15 s. 3. The larger decreased by the smaller is 15, written as 4s s 15. Solution Let s = the smaller number. Then 4s = the larger number. The larger is 15 more than the smaller. \|_________\| \|__________\| \|__________\| ↓ 4s ↓ ↓ s ↓ ↓ 15 4s 15 s 4s 15 s s s 3s 15 s 5 4s 4(5) 20 Check The larger number, 20, is 4 times the smaller number, 5. The larger number, 20, exceeds the smaller number, 5, by 15. Answer The larger number is 20; the smaller number is 5. Solving Equations That Have the Variable in Both Sides 131 EXAMPLE 4 In his will, Uncle Clarence left $5,000 to his two nieces. Emma’s share is to be $500 more than Clara’s. How much should each niece receive? Solution (1) Use the fact that the sum of the two shares is $5,000 to express each share in terms of a variable. Let x Clara’s share. Then 5,000 x Emma’s share. (2) Use the fact that Emma’s share is $500 more than Clara’s share to write an equation. Emma’s share is $
500 more than Clara’s share. \|____________\| \|_____________\| ↓ ↓ x ↓ ↓ 5,000 x 500 \|__________\| ↓ (3) Solve the equation to find Clara’s share. x 5,000 x 500 x x 500 2x 500 2x 5,000 500 4,500 2,250 x Clara’s share is x $2,250. (4) Find Emma’s share: 5,000 x 5,000 2,250 $2,750. Alternate Solution (1) Use the fact that Emma’s share is $500 more than Clara’s share to express each share in terms of a variable. Let x Clara’s share. Then x 500 Emma’s share. (2) Use the fact that the sum of the two shares is $5,000 to write an equation. Clara’s share plus Emma’s share is $5,000. \|____________\| ↓ x (x 500) 5,000 \|______________\| ↓ ↓ ↓ ↓ (3) Solve the equation to find Clara’s share x (x 500) 5,000 2x 500 5,000 500 500 4,500 2,250 2x x Clara’s share is x $2,250. 132 First Degree Equations and Inequalities in One Variable (4) Find Emma’s share: x 500 2250 500 $2,750. Check $2,750 is $500 more than $2,250, and $2,750 $2,250 $5,000. Answer Clara’s share is $2,250, and Emma’s share is $2,750. EXERCISES Writing About Mathematics 1. Milus said that he finds it easier to work with integers than with fractions. Therefore, in 2a 1 3 , he began by multiplying both sides of the 3 4a 2 7 5 1 order to solve the equation equation by 4. 4 A 3 1 2a 1 3 5 4 4a 2 7 3a 28 2a 12 A B B Do you agree with Milus that this is a correct way of obtaining the solution? If so, what mathematical principle is Milus using? 2. Katie said that Example 3 could be solved by letting equal the smaller number and x equal the larger number. Is Katie correct? If so, what equation would she write to solve the problem? x 4 Developing Skills In 3–36, solve and check each equation. 3. 7x 10 2x 6. y 4y 30 9. 0.8m 0.2m 24 23 4x 1 24 5 3x 12. 15. x 9x 72 18. 7r 10 3r 50 21. x 4 9x 4 24. c 20 55 4c 27. 3m (m 1) 6m 1 3t 2 11 5 4(16 2 t) 2 1 2 3t 30. 32. 8c 1 7c 2(7 c) 34. 4(3x 5) 5x 2( x 15) 36. 5 3(a 6) a 1 8a 4. 9x 44 2x 7. 2d 36 5d 10. 8y 90 2y 13. 5a 40 3a 16. 0.5m 30 1.1m 19. 4y 20 5y 9 22. 9x 3 2x 46 25. 2d 36 3d 54 28. x 3(1 x) 47 x 4y 2 8 5. 5c 28 c 4y 5 11 21 8. 11. 2.3x 36 0.3x 14. 5c 2c 81 4c 5 93 41 4c 1 44 17. 20. 7x 8 6x 1 23. y 30 12y 14 26. 7y 5 9y 29 29. 3b 8 10 (4 8b) 31. 18 4n 8 2(1 8n) 33. 8a 3(5 2a) 85 3a 35. 3m 5m 12 7m 88 5 Solving Equations That Have the Variable in Both Sides 133 In 37–42, a. write an equation to represent each problem, and b. solve the equation to find each number. 37. Eight times a number equals 35 more than the number. Find the number. 38. Six times a number equals 3 times the number, increased by 24. Find the number. 39. If 3 times a number is increased by 22, the result is 14 less than 7 times the number. Find the number. 40. The greater of two numbers is 1 more than twice the smaller. Three times the greater exceeds 5 times the smaller by 10. Find the numbers. 41. The second of three numbers is 6 more than the first. The third number is twice the first. The sum of the three numbers is 26. Find the three numbers. 42. The second of three numbers is 1 less than the first. The third number is 5 less than the sec- ond. If the first number is twice as large as the third, find the three numbers. Applying Skills In 43–50, use an algebraic solution to solve each problem. 43. It took the Gibbons family 2 days to travel 925 miles to their vacation home. They traveled 75 miles more on the first day than on the second. How many miles did they travel each day? 44. During the first 6 month of last year, the interest on an investment was $130 less than dur- ing the second 6 months. The total interest for the year was $1,450. What was the interest for each 6-month period? 45. Gemma has 7 more five-dollar bills than ten-dollar bills. The value of the five-dollar bills equals the value of the ten-dollar bills. How many five-dollar bills and ten-dollar bills does she have? 46. Leonard wants to save $100 in the next 2 months. He knows that in the second month he will be able to save $20 more than during the first month. How much should he save each month? 47. The ABC Company charges $75 a day plus $0.05 a mile to rent a car. How many miles did Mrs. Kiley drive if she paid $92.40 to rent a car for one day? 48. Kesha drove from Buffalo to Syracuse at an average rate of 48 miles per hour. On the return trip along the same road she was able to travel at an average rate of 60 miles per hour. The trip from Buffalo to Syracuse took one-half hour longer than the return trip. How long did the return trip take? 49. Carrie and Crystal live at equal distances from school. Carries walks to school at an average rate of 3 miles per hour and Crystal rides her bicycle at an average rate of 9 miles per hour. It takes Carrie 20 minutes longer than Crystal to get to school. How far from school do Crystal and Carrie live? 50. Emmanuel and Anthony contributed equal amounts to the purchase of a gift for a friend. Emmanuel contributed his share in five-dollar bills and Anthony gave his share in onedollar bills. Anthony needed 12 more bills than Emmanuel. How much did each contribute toward the gift? 134 First Degree Equations and Inequalities in One Variable 4-4 USING FORMULAS TO SOLVE PROBLEMS To solve for the subject of a formula, substitute the known values in the formula and perform the required computation. For example, to find the area of a triangle when b 4.70 centimeters and h 3.20 centimeters, substitute the given values in the formula for the area of a triangle: 1 2bh 1 2(4.70 cm) A 7.52 cm2 (3.20 cm) A is the subject of the formula. Now that you can solve equations, you will be able to find the value of any variable in a formula when the values of the other variables are known. To do this: 1. Write the formula. 2. Substitute the given values in the formula. 3. Solve the resulting equation. The values assigned to the variables in a formula often have a unit of measure. It is convenient to solve the equation without writing the unit of measure, but the answer should always be given in terms of the correct unit of measure. EXAMPLE 1 The perimeter of a rectangle is 48 centimeters. If the length of the rectangle is 16 centimeters, find the width to the nearest centimeter. Solution You know that the perimeter of a geometric figure is the sum of the lengths of all of its sides. When solving a perimeter problem, it is helpful to draw and label a figure to model the region. Use the formula P 2l 2w. P 2l 2w 48 2(16) 2w 48 32 2w 32 32 16 16 2w 8 w 2w Answer 8 centimeters Check P 2l 2w 2(16) 1 2(8) 48 5? 5? 48 48 48 ✔ 32 1 16 16 cm w w 16 cm Using Formulas to Solve Problems 135 EXAMPLE 2 A garden is in the shape of an isosceles triangle, a triangle that has two sides of equal measure. The length of the third side of the triangle is 2 feet greater than the length of each of the equal sides. If the perimeter of the garden is 86 feet, find the length of each side of the garden. Solution Let x the length of each of the two equal sides. Then, x 2 the length of the third side. The perimeter is the sum of the lengths of the sides. \|__________________________________\| \|_____________\| ↓ ↓ x x (x 2) 86 ↓ x x x + 2 86 3x 2 84 3x 28 x The length of each of the equal sides x 28. The length of the third side x 2 28 2 30. Check Perimeter 28 28 30 86 ✔ Answer The length of each of the equal sides is 28 feet. The length of the third side (the base) is 30 feet. EXAMPLE 3 The perimeter of a rectangle is 52 feet. The length is 2 feet more than 5 times the width. Find the dimensions of the rectangle. Solution Use the formula for the perimeter of a rectangle, P 2l 2w, to solve this problem. Let w the width, in feet, of the rectangle. Then 5w 2 the length, in feet, of the rectangle. P 2l 2w 52 2(5w 2) 2w 52 10w 4 2w 52 12w 4 48 12w 4 w 5? 5? 44 1 8 52 52 52 ✔ 2(22) 1 2(4) 52 Check The length, 22, is 2 more than 5 times the width, 4. ✔ P 2l 2w Answer The width is 4 feet; the length is 5(w) 2 5(4) 2 22 feet. 136 First Degree Equations and Inequalities in One Variable EXAMPLE 4 Sabrina drove from her home to her mother’s home which is 150 miles away. For the first half hour, she drove on local roads. For the next two hours she drove on an interstate highway and increased her average speed by 15 miles per hour. Find Sabrina’s average speed on the local roads and on the interstate highway. Solution List the facts stated by this problem: FACT 1 Sabrina drove on local roads for hour or 0.5 hour. 1 2 FACT 2 Sabrina drove on the interstate highway for 2 hours. FACT 3 Sabrina’s rate or speed on the interstate highway was 15 mph more than her rate on local roads. This problem involves rate, time, and distance. Use the distance formula, d rt, where r is the rate, or speed, in miles per hour, t is time in hours, and d is distance in miles. (1) Represent Sabrina’s speed for each part of the trip in terms of r. Let r Sabrina’s speed on the local roads. Then r 15 Sabrina’s speed on the interstate highway. (2) Organize the facts in a table, using the distance formula. Rate Time Distance Local Roads Interstate highway r r 15 0.5 2 0.5r 2(r 15) (3) Write an equation. The distance on the local roads plus the distance on the highway is 150 miles. \|________\| \|_____________________________\| ↓ ↓ 150 \|____________________________\| ↓ 2(r 15) ↓ 0.5r ↓ (4) Solve the equation. (a) Write the equation: (b) Use the distributive property: 0.5r 2(r 15) 150 0.5r 2r 30 150 Using Formulas to Solve Problems 137 (c) Combine like terms: (d) Add 30, the opposite of 30 to each side of the equation: (e) Divide each side by 2.5: 2.5r 2.5r 30 150 30 30 120 2.5 5 120 2.5r 2.5 r 48 (5) Find the average speed for each part of the trip. Sabrina’s speed on local roads r 48 mph. Sabrina’s speed on the highway r 15 48 15 63 mph. Check On local roads: On the interstate highway: The total distance traveled: 0.5(48) 24 miles 2(63) 126 miles 150 miles ✔ Answer Sabrina traveled at an average speed of 48 miles per hour on local roads and 63 miles per hour on the interstate highway. EXERCISES Writing About Mathematics 1. In Example 4, step 4 uses
the equivalent equation 2.5r 30 150. Explain what each term in the left side of this equation represents in the problem. 2. Antonio solved the problem in Example 4 by letting r represent Sabrina’s rate of speed on the interstate highway. To complete the problem correctly, how should Antonio represent her rate of speed on the local roads? Explain your answer. Developing Skills In 3–19, state the meaning of each formula, tell what each variable represents, and find the required value. In 3–12, express each answer to the correct number of significant digits. 3. If P a b c, find c when P 85 in., a 25 in., and b 12 in. 4. If P 4s, find s when P 32 m. 5. If P 4s, find s when P 6.8 ft. 6. If P 2l 2w, find w when P 26 yd and l 8 yd. 7. If P 2a b, find b when P 80 cm and a 30 cm. 8. If P 2a b, find a when P 18.6 m and b 5.8 m. 9. If A bh, find b when A 240 cm2 and h 15 cm. 138 First Degree Equations and Inequalities in One Variable 10. If A bh, find h when A 3.6 m2 and b 0.90 m. 1 2bh 11. If A , find h when A 24 sq ft and b 8.0 ft. 12. If V lwh, find w when V 72 yd3, l 0.75 yd, and h 12 yd. 13. If d rt, find r when d 120 mi and t 3 hr. 14. If I prt, find the principal, p, when the interest, I, is $135, the yearly rate of interest, r, is 2.5%, and the time, t, is 3 years. 15. If I prt, find the rate of interest, r, when I $225, p $2,500, and t 2 years. 16. If T nc, find the number of items purchased, n, if the total cost, T, is $19.80 and the cost of one item, c, is $4.95. 17. If T nc, find the cost of one item purchased, c, if T $5.88 and n 12. 18. If S nw, find the hourly wage, w, if the salary earned, S, is $243.20 and the number of hours worked, n, is 38. 19. If S nw, find the number of hours worked, n, if S $315.00 and w $8.40. In 20–32, a. write a formula that can be used to solve each problem, b. use the formula to solve each problem and check the solution. All numbers may be considered to be exact values. 20. Find the length of a rectangle whose perimeter is 34.6 centimeters and whose width is 5.7 centimeters. 21. The length of the second side of a triangle is 2 inches less than the length of the first side. The length of the third side is 12 inches more than the length of the first side. The perimeter of the triangle is 73 inches. Find the length of each side of the triangle. 22. Two sides of a triangle are equal in length. The length of the third side exceeds the length of one of the other sides by 3 centimeters. The perimeter of the triangle is 93 centimeters. Find the length of each side of the triangle. 23. The length of a rectangle is 5 meters more than its width. The perimeter is 66 meters. Find the dimensions of the rectangle. 24. The width of a rectangle is 3 yards less than its length. The perimeter is 130 yards. Find the length and the width of the rectangle. 25. The length of each side of an equilateral triangle is 5 centimeters more than the length of each side of a square. The perimeters of the two figures are equal. Find the lengths of the sides of the square and of the triangle. 26. The length of each side of a square is 1 centimeter more than the width of a rectangle. The length of the rectangle is 1 centimeter less than twice its width. The perimeters of the two figures are equal. Find the dimensions of the rectangle. 27. The area of a triangle is 36 square centimeters. Find the measure of the altitude drawn to the base when the base is 8 centimeters. Using Formulas to Solve Problems 139 28. The altitude of a triangle is 4.8 meters. Find the length of the base of the triangle if the area is 8.4 square meters. 29. The length of a rectangle is twice the width. If the length is increased by 4 inches and the width is decreased by 1 inch, a new rectangle is formed whose perimeter is 198 inches. Find the dimensions of the original rectangle. 30. The length of a rectangle exceeds its width by 4 feet. If the width is doubled and the length is decreased by 2 feet, a new rectangle is formed whose perimeter is 8 feet more than the perimeter of the original rectangle. Find the dimensions of the original rectangle. 31. A side of a square is 10 meters longer than the side of an equilateral triangle. The perimeter of the square is 3 times the perimeter of the triangle. Find the length of each side of the triangle. 32. The length of each side of a hexagon is 4 inches less than the length of a side of a square. The perimeter of the hexagon is equal to the perimeter of the square. Find the length of a side of the hexagon and the length of a side of the square. Applying Skills 33. The perimeter of a rectangular parking lot is 146 meters. Find the dimensions of the lot, using the correct number of significant digits, if the length is 7.0 meters less than 4 times the width. 34. The perimeter of a rectangular tennis court is 228 feet. If the length of the court exceeds twice its width by 6.0 feet, find the dimensions of the court using the correct number of significant digits. In 35–48, make a table to organize the information according to the formula to be used. All numbers may be considered to be exact values. 35. Rahul has 25 coins, all quarters and dimes. Copy the table given below and organize the facts in the table using the answers to a through c. Number of coins in one denomination b a Value of one coin a b Total value of the coins of that denomination Number of Coins Value of One Coin Total Value Dimes Quarters 140 First Degree Equations and Inequalities in One Variable a. If x is the number of dimes Rahul has, express, in terms of x, the number of quarters he has. b. Express the value of the dimes in terms of x. c. Express the value of the quarters in terms of x. d. If the total value of the dimes and quarters is $4.90, write and solve an equation to find how many dimes and how many quarters Rahul has. e. Check your answer in the words of the problem. 36. If the problem had said that the total value of Rahul’s 25 dimes and quarters was $5.00, what conclusion could you draw? 37. When Ruth emptied her bank, she found that she had 84 coins, all nickels and dimes. The value of the coins was $7.15. How many dimes did she have? (Make a table similar to that given in exercise 35.) 38. Adele went to the post office to buy stamps and postcards. She bought a total of 25 stamps, some 39-cent stamps and the rest 23-cent postcards. If she paid $8.47 altogether, how many 39-cent stamps did she buy? 39. Carlos works Monday through Friday and sometimes on Saturday. Last week Carlos worked 38 hours. Copy the table given below and organize the facts in the table using the answers to a through c. Hours Worked Wage Per Hour Earnings Monday–Friday Saturday a. If x is the total number of hours Carlos worked Monday through Friday, express, in terms of x, the number of hours he worked on Saturday. b. Carlos earns $8.50 an hour when he works Monday through Friday. Express, in terms of x, his earnings Monday through Friday. c. Carlos earns $12.75 an hour when he works on Saturday. Express, in terms of x, his earnings on Saturday. d. Last week Carlos earned $340. How many hours did he work on Saturday? 40. Janice earns $6.00 an hour when she works Monday through Friday and $9.00 an hour when she works on Saturday. Last week, her salary was $273 for 42 hours of work. How many hours did she work on Saturday? (Make a table similar to that given in exercise 39.) 41. Candice earns $8.25 an hour and is paid every two weeks. Last week she worked 4 hours longer than the week before. Her pay for these two weeks, before deductions, was $594. How many hours did she work each week? Using Formulas to Solve Problems 141 42. Akram drove from Rochester to Albany, a distance of 219 miles. After the first 1.5 hours of travel, it began to snow and he reduced his speed by 26 miles per hour. It took him another 3 hours to complete the trip. Copy the table given below and fill in the entries using the answers to a through c. Rate Time Distance First part of the trip Last part of the trip a. If r is the average speed at which Akram traveled for the first part of the trip, express, in terms of r, his average speed for the second part of the trip. b. Express, in terms of r, the distance that Akram traveled in the first part of the trip. c. Express, in terms of r, the distance that Akram traveled in the second part of the trip. d. Find the speed at which Akram traveled during each part of the trip. 43. Vera walked from her home to a friend’s home at a rate of 3 miles per hour. She rode to 5 6 of an hour work with her friend at an average rate of 30 miles per hour. It took Vera a total of 50 minutes tance of 16 miles. How long did she walk and how long did she ride with her friend to get to work? (Make a table similar to that given in exercise 42.) to walk to her friend’s home and to get to work, traveling a total dis- A B 44. Peter drove a distance of 189 miles. Part of the time he averaged 65 miles per hour and for the remaining time, 55 miles per hour. The entire trip took 3 hours. How long did he travel at each rate? 45. Shelly and Jack left from the same place at the same time and drove in opposite directions along a straight road. Jack traveled 15 miles per hour faster than Shelly. After 3 hours, they were 315 miles apart. Find the rate at which each traveled. 46. Carla and Candice left from the same place at the same time and rode their bicycles in the same direction along a straight road. Candice bicycled at an average speed that was threequarters of Carla’s average speed. After 2 hours they were 28 miles apart. What was the average speed of Carla and Candice? 47. Nolan walked to the store from his home at the rate of 5 miles per hour. After spending one-half hour in the store, his friend gave him a ride home at the rate of 30 miles per hour. 1 1 12 hours He arrived home 1 hour and 5 minutes after he left. How far is the store from Nolan’s home? A B 48. Mrs. Dang drove her daughter to school at an average rate of 45 miles per hour. She returned home
by the same route at an average rate of 30 miles per hour. If the trip took one-half hour, How long did it take to get to school? How far is the school from their home? 142 First Degree Equations and Inequalities in One Variable 4-5 SOLVING FOR A VARIABLE IN TERMS OF ANOTHER VARIABLE An equation may contain more than one variable. For example, the equation ax b 3b contains the variables a, b, and x. To solve this equation for x means to express x in terms of the other variables. To plan the steps in the solution, it is helpful to use the strategy of using a simpler related problem, that is, to compare the solution of this equation with the solution of a simpler equation that has only one variable. In Example 1, the solution of ax b 3b is compared with the solution of 2x 5 15. The same operations are used in the solution of both equations. EXAMPLE 1 Solve for x in ax b = 3b. Solution Compare with 2x 5 15. Check 2x 5 15 5 5 10 2x 2x 2 5 10 2 x 5 ax b 3b b 2b b ax a 5 2b ax a x 5 2b a Answer x 5 2b a EXAMPLE 2 Solve for x in x a b. Solution Compare with x 5 9. x 5 9 5 5 14 x Answer EXAMPLE 3 Solve for x in 2ax 10a2 3ax (a 0). ax b 3b 2b ? 3b 1 b 5 a B ? 2b 1 b 5 3b a A 3b 3b ✔ Check x a b b 1 a 2 a 5? b b b ✔ Solution Compare with 2x 10 3x. 2x 10 3x 3x 3ax 3x 2ax 10a2 3ax 3ax 5x 10 5x 5 5 10 5 x 2 5ax 10a2 5a 5 10a2 5ax 5a x 2a Transforming Formulas 143 Check 2ax 10a2 3ax 2a(2a) 5? 10a2 2 3a(2a) 5? 10a2 2 6a2 4a2 4a2 4a2 ✔ Answer x 2a EXERCISES Writing About Mathematics 1. Write a simpler related equation in one variable that can be used to suggest the steps needed to solve the equation a(x b) 4ab for x. 2. Write a simpler related equation in one variable that can be used to suggest the steps needed to solve the equation 5cy d 2cy for y. Developing Skills In 3–24, solve each equation for x or y and check. 3. 5x b 7. x 5r 7r 11. d y 9 15. ax b 3b 19. bx 9b2 22. m2x 3m2 12m2 4. sx 8 8. x a 4a 12. 3x q 5q 16. dx 5c 3c 5. ry s 9. y c 9c 13. 3x 8r r 17. r sy t 6. hy m 10. 4 x k 14. cy d 4d 18. m 2(x n) 20. cx c2 5c2 7cx 23. 9x 24a 6a 4x 21. rsx rs2 0 24. 8ax 7a2 19a2 5ax 4-6 TRANSFORMING FORMULAS A formula is an equation that contains more than one variable. Sometimes you want to solve for a variable in the formula that is different from the subject of the formula. For example, the formula for distance, d, in terms of rate, r, and time, t, is d = rt. Distance is the subject of the formula, but you might want to rewrite the formula so that it expresses time in terms of distance and rate. You do this by solving the equation d = rt for t in terms of d and r. 144 First Degree Equations and Inequalities in One Variable EXAMPLE 1 a. Solve the formula d = rt for t. b. Use the answer obtained in part a to find the value of t when d 200 miles and r 40 miles per hour. Solution a. d rt d r 5 rt r d r 5 t b. t 5 d r 5 200 40 5 Answers a. t d r b. t 5 hours Note that the rate is 40 miles per hour, that is, 40 miles 1 hour . Therefore, 200 miles 40 miles 1 hour 200 miles 3 1 hour 40 miles 5 hours We can think of canceling miles in the numerator and the denominator of the fractions being multiplied. EXAMPLE 2 a. The formula for the volume of a cone is V . Solve this formula for h. b. Find the height of a cone that has a volume of 92.0 cubic centimeters and a circular base with a radius of 2.80 centimeters. Express the answer using the correct number of significant digits. 1 3Bh Solution a. V 3V 1 3Bh 1 3 3Bh B A 3V Bh 3V B 5 Bh B 3V h B b. Find B, the area of the base of the cone. Since the base is a circle, its area is p times the square of the radius, r. B pr2 p(2.80)2 ENTER: 2nd p 2.80 x2 ENTER Transforming Formulas 145 DISPLAY Now use the answer to part a to find h: h 5 3V B < 3(92.0) 24.63 ENTER: 3 92.0 24.63 ENTER DISPLAY Since each measure is given to three significant digits, round the answer to three significant digits. Answers a. h 5 3V B b. The height of the cone is 11.2 centimeters. EXERCISES Developing Skills In 1–14, transform each given formula by solving for the indicated variable. 3. d rt for r 6. I prt for t 2. A bh for h 5. P br for r 1. P 4s for s 4. V lwh for l 1 7. A 2bh 10. P 2a b for b for h 13. F 9 5C 1 32 for C 1 8. V 3Bh 11. P 2a b for a for B 14. 2S n(a l) for a 1 9. s 2gt 12. P 2l + 2w for w for g Applying Skills 15. The concession stand at a movie theater wants to sell popcorn in containers that are in the shape of a cylinder. The volume of the cylinder is given by the formula V = pr2h, where V is the volume, r is the radius of the base, and h is the height of the container. a. Solve the formula for h. b. If the container is to hold 1,400 cubic centimeters of popcorn, find, to the nearest tenth, the height of the container if the radius of the base is: (1) 4.0 centimeters (2) 5.0 centimeters (3) 8.0 centimeters c. The concession stand wants to put an ad with a height of 20 centimeters on the side of the container. Which height from part b do you think would be the best for the container? Why? 146 First Degree Equations and Inequalities in One Variable 16. A bus travels from Buffalo to Albany, stopping at Rochester and Syracuse. At each city there is a 30-minute stopover to unload and load passengers and baggage. The driving distance from Buffalo to Rochester is 75 miles, from Rochester to Syracuse is 85 miles, and from Syracuse to Albany is 145 miles. The bus travels at an average speed of 50 miles per hour. a. Solve the formula d rt for t to find the time needed for each part of the trip. b. Make a schedule for the times of arrival and departure for each city if the bus leaves Buffalo at 9:00 A.M. 4-7 PROPERTIES OF INEQUALITIES The Order Property of Real Numbers If two real numbers are graphed on the number line, only one of the following three situations can be true: x is to the left of y x and y are at the same point x is to the right of These three cases illustrate the order property of real numbers: If x and y are two real numbers, then one and only one of the following can be true: x y or x y or x y Let y be a fixed point, for example, y 3. Then y separates the real numbers into three sets. For any real number, one of the following must be true: x 3, x 3, x 3. The real numbers, x, that make the inequality x 3 true are to the left of 3 on the number line. The circle at 3 indicates that 3 is the boundary value of the set. The circle is not filled in, indicating that 3 does not belong to this set. 0 3 x < 3 Properties of Inequalities 147 The real number that makes the corresponding equality, x 3, true is a single point on the number line. This point, x 3, is also the boundary between the values of x that make x 3 true and the values of x that make x 3 true. The circle is filled in, indicating that 3 belongs to this set. Here, 3 is the only element of the set. The real numbers, x, that make x 3 true are to the right of 3 on the number line. Again, the circle at 3 indicates that 3 is the boundary value of the set. The circle is not filled in, indicating that 3 does not belong to this set The Transitive Property of Inequality From the graph at the right, you can see that, if x lies to the left of y, and y lies to the left of z, then x lies to the left of z. The graph illustrates the transitive property of inequality: x y z For the real numbers x, y, and z: If x y and y z, then x z; and if z y and y x, then z x. The Addition Property of Inequality The following table shows the result of adding a number to both sides of an inequality. True Sentence 9 2 Order is “greater than.” 9 2 Order is “greater than.” 2 9 Order is “less than.” 2 9 Order is “less than.” Number to Add to Both Sides 9 3 ? 2 3 Add a positive number. 9 (–3) ? 2 (–3) Add a negative number. 2 3 ? 9 3 Add a positive number. 2 (–3) ? 9 (–3) Add a negative number. Result 12 5 Order is unchanged. 6 1 Order is unchanged. 5 12 Order is unchanged 1 6 Order is unchanged 148 First Degree Equations and Inequalities in One Variable The table illustrates the addition property of inequality: For the real numbers x, y, and z: If x y, then x z y z; and if x y, then x z y z. Since subtracting the same number from both sides of an inequality is equivalent to adding the additive inverse to both sides of the inequality, the following is true: When the same number is added to or subtracted from both sides of an inequality, the order of the new inequality is the same as the order of the original one. EXAMPLE 1 Use the inequality 5 9 to write a new inequality: a. by adding 6 to both sides b. by adding 9 to both sides Solution a. 5 6 9 6 11 15 b. 5 (–9) 9 (–9) –4 0 Answers a. 11 15 b. 4 0 The Multiplication Property of Inequality The following table shows the result of multiplying both sides of an inequality by the same number. True Sentence 9 2 Order is “greater than.” 5 9 Order is “less than.” 9 2 Order is “greater than.” 5 9 Order is “less than.” Number to Multiply Both Sides 9(3) ? 2(3) Multiply by a positive number 5(3) ? 9(3) Multiply by a positive number. 9(–3) ? 2(–3) Multiply by a negative number 5(–3) ? 9(–3) Multiply by a negative number. Result 27 6 Order is unchanged. 15 27 Order is unchanged. 27 6 Order is changed. 15 27 Order is changed. Properties of Inequalities 149 The table illustrates that the order does not change when both sides are multiplied by the same positive number, but does change when both sides are multiplied by the same negative number. In general terms, the multiplication property of inequality states: For the real numbers x, y, and z: If z is positive (z 0) and x y, then xz yz. If z is positive (z 0) and x y, then xz yz. If z is negative (z 0) and x y, then xz yz. If z is negative (z 0) and x y, then xz yz. Dividing both sides of an inequality by a number is equivalent to multiplying both sides by the multiplicative inverse of the number. A number and its multiplicative inverse always have the same sign. Therefore, the following is true: When both sides of an inequality are multiplied or divided by the same positive number, the o
rder of the new inequality is the same as the order of the original one. When both sides of an inequality are multiplied or divided by the same negative number, the order of the new inequality is the opposite of the order of the original one. EXAMPLE 2 Use the inequality 6 9 to write a new inequality: a. by multiplying both sides by 2. 21 3 b. by multiplying both sides by . Solution a. 6 9 b. 6 9 6(2) ? 9(2) 12 18 6 A Answers a. 12 18 b. 2 3 21 3 21 ? 9 3 A 2 3 B B 150 First Degree Equations and Inequalities in One Variable EXERCISES Writing About Mathematics 1. Sadie said that if 5 4, then it must be true that 5x 4x. Do you agree with Sadie? Explain why or why not. 2. Lucius said that if x y and a b then x a y b. Do you agree with Lucius? Explain why or why not. 3. Jason said that if x y and a b then x a y b. Do you agree with Jason? Explain why or why not. Developing Skills In 4–31, replace each question mark with the symbol or so that the resulting sentence will be true. 4. Since 8 2, 8 1 ? 2 1. 8. Since 7 3, 6. Since 9 5, 9 2 ? 5 2. 2 2 3(3) 3(7) ? 21 3 10. Since 9 6, 9 4 . ? 6 12. If y 6, then y 2 ? 6 2. A B 5. Since 6 2, 6 (–4) ? 2 (–4). 1 4 B 9. Since 8 4, (8) (4) ? (4) (4). 7. Since 2 8, ? 28 2 22 2 1 4 A A B . 21 3 . B A 11. If 5 x, then 5 7 ? x 7. 13. If 20 r, then 4(20) ? 4(r). 14. If t 64, then t 8 ? 64 8. 15. If x 8, then 2x ? 2(8). 16. If y 8, then y (–4) ? 8 (–4). 17. If x 2 7, the x 2 (–2) ? 7 (–2) or x ? 5. 18. If y 3 12, then y 3 3 ? 12 3 or y ? 15. 19. If a 5 14, then a 5 5 ? 14 5 or a ? 9. 8 2 20. If 2x 8, then or x ? 4. 1 3y 21. If 4, then A 22. If 3x 36, then 2x 2 ? 1 3 3y B 23x 36 23 23 2(22x) ? 21 21 24. If x 5 and 5 y, then x ? y. 23. If 2x 6, then ? ? 3(4) or y ? 12. or x ? 12. 2(6) 26. If 3 7, then 7 ? 3. 28. 1f 9 x, then x ? 9. or x ? 3. 25. If m 7 and 7 a, then m ? a. 27. If 4 12, then 12 ? 4. 29. If 7 a, then a ? 7. 30. If x 10 and 10 z, then x ? z. 31. If a b and c b, then a ? c. 4-8 FINDING AND GRAPHING THE SOLUTION OF AN INEQUALITY Finding and Graphing the Solution of an Inequality 151 When an inequality contains a variable, the domain or replacement set of the inequality is the set of all possible numbers that can be used to replace the variable. When an element from the domain is used in place of the variable, the inequality may be true or it may be false. The solution set of an inequality is the set of numbers from the domain that make the inequality true. Inequalities that have the same solution set are equivalent inequalities. To find the solution set of an inequality, solve the inequality by methods similar to those used in solving an equation. Use the properties of inequalities to transform the given inequality into a simpler equivalent inequality whose solution set is evident. In Examples 1–5, the domain is the set of real numbers. EXAMPLE 1 Find and graph the solution of the inequality x 4 1. Solution How to Proceed (1) Write the inequality: (2) Use the addition property of inequality. Add 4 to each side: x 4 1 4 4 5 x –2 – The graph above shows the solution set. The circle at 5 indicates that 5 is the boundary between the numbers to the right, which belong to the solution set, and the numbers to the left, which do not belong. Since 5 is not included in the solution set, the circle is not filled in. Check (1) Check one value from the solution set, for example, 7. This value will make the inequality true. 7 4 1 is true. (2) Check the boundary value, 5. This value, which separates the values that make the inequality true from the values that make it false, will make the corresponding equality, x 4 1, true. 5 4 1 is true. Answer x 5 An alternative method of expressing the solution set is interval notation. When this notation is used, the solution set is written as (5, ). The first number, 5 names the lower boundary. The symbol , often called infinity, indicates that there is no upper boundary, that is, that the set of real numbers continues without end. The parentheses indicate that the boundary values are not elements of the set. 152 First Degree Equations and Inequalities in One Variable EXAMPLE 2 Find and graph the solution of 5x 4 11 2x. Solution The solution set of 5x 4 11 2x includes all values of the domain for which either 5x 4 11 2x is true or 5x 4 11 2x is true. How to Proceed (1) Write the inequality: (2) Add 2x to each side: (3) Add 4 to each side: (4) Divide each side by 7: 5x 4 11 2x 2x 2x 7x 4 11 4 4 7 7x 7 # 7 7x 7 x 1 The solution set includes 1 and all of the real numbers less than 1. This is shown on the graph below by filling in the circle at 1 and drawing a heavy line to the left of 1. Answer x 1 –4 –3 –2 –1 0 1 2 3 The solution set can also be written in interval notation as (, 1]. The symbol , often called negative infinity, indicates that there is no lower boundary, that is, all negative real numbers less than the upper boundary are included. The number, 1, names the upper boundary. The right bracket indicates that the upper boundary value is an element of the set. EXAMPLE 3 Find and graph the solution set: 2(2x 8) 8x 0. Solution How to Proceed (1) Write the inequality: (2) Use the distributive property: (3) Combine like terms in the left side: (4) Add 16 to each side: 2(2x 8) 8x 0 4x 16 8x 0 4x 16 0 16 16 16 4x Finding and Graphing the Solution of an Inequality 153 (5) Divide both sides by 4. Dividing by a negative number reverses the inequality: (6) The graph of the solution set includes 4 and all of the real numbers to the right of 4 on the number line: Answer x 4 or [4, ) 24x 24 $ 16 24 x 4 –5 –4 –3 –2 –1 0 1 2 3 Graphing the Intersection of Two Sets The inequality 3 x 6 is equivalent to (3 x) and (x 6). This statement is true when both simple statements are true and false when one or both statements are false. The solution set of this inequality consists of all of the numbers that are in the solution set of both simple inequalities. The graph of 3 x 6 can be drawn as shown below. How to Proceed (1) Draw the graph of the solution set of the first inequality, 3 x, a few spaces above the number line: (2) Draw the graph of the solution set of the second inequality, x 6, above the number line, but below the graph of the first inequality: Solution 3 < x –2 –2 –3) Draw the graph of the intersection of these two sets by shading, on the number line, the points that belong to the solution set of both simple inequalities: Since 3 is in the solution set of x 6 but not in the solution set of 3 x, 3 is not in the intersection of the two sets. Also, since 6 is in the solution set of 3 x but not in the solution set of x 6, 6 is not in the intersection of the two sets. Therefore, the circles at 3 and 6 are not filled in, indicating that these boundary values are not elements of the solution set of 3 x 6. –2 – This set can also be written as (3, 6), a pair of numbers that list the left and right boundaries of the set. The parentheses indicate that the boundary values do not belong to the set. Similarly, the set of numbers 3 x 6 can be written as [3, 6]. The brackets indicate that the boundary values do belong to the set. 154 First Degree Equations and Inequalities in One Variable Although this notation is similar to that used for an ordered pair that names a point in the coordinate plane, the context in which the interval or ordered pair is used will determine the meaning. EXAMPLE 4 Solve the inequality and graph the solution set: 7 x 5 0. Solution How to Proceed (1) First solve the inequalities for x2) Draw the graphs of 2 x and x 5 above the number line: –2 < x x < 5 –4 –3 –2 –3) Draw the graph of all points that are common to the graphs of 2 x and x 5: –2 < x < 5 –4 –3 –2 – Answer 2 x 5 or (–2, 5) Graphing the Union of Two Sets The inequality (x 3) or (x 6) is true when one or both of the simple statements are true. It is false when both simple statements are false. The solution set of the inequality consists of the union of the solution sets of the two simple statements. The graph of the solution set can be drawn as shown below. How to Proceed (1) Draw the graph of the solution set of the first inequality a few spaces above the number line: (2) Draw the graph of the solution set of the second inequality above the number line, but below the graph of the first inequality: Solution x > 3 12 2 – EXAMPLE 5 Solution Finding and Graphing the Solution of an Inequality 155 (3) Draw the graph of the union by shading, on the number line, the points that belong to the solution of one or both of the simple inequalities: –2 – Since 3 is not in the solution set of either inequality, 3 is not in the union of the two sets. Therefore, the circle at 3 is not filled in. Answer: x 3 or (3, ) Solve the inequality and graph the solution set: (x 2 0) or (x 3 0). How to Proceed (1) Solve each inequality for x: (2) Draw the graphs of x 2 and x 3 above the number line: (3) Draw the graph of all points of the graphs of x 2 or x 3: x 2 0 2 2 2 x or 2 x > 3 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 Answer (x 2) or (x 3) or (, 2) or (3, ) Note: Since the solution is the union of two sets, the answer can also be (2`,22) < (3,`) expressed using set notation: (x , 22) < (x . 3) or . EXERCISES Writing About Mathematics 1. Give an example of a situation that can be modeled by the inequality x 5 in which a. the solution set has a smallest value, and b. the solution set does not have a smallest value. 2. Abram said that the solution set of (x 4) or (x 4) is the set of all real numbers. Do you agree with Abram? Explain why or why not. 156 First Degree Equations and Inequalities in One Variable Developing Skills In 3–37, find and graph the solution set of each inequality. The domain is the set of real numbers. 3. x 2 4 7. x 3 6 11. y 4 4 15. 15 3y 19. –10x 20 4. z 6 4 8. 19 y 17 12. 25 d 22 16. –10 4h 20. 12 1.2r 24. –10 2.5z 28. 5x 4 4 3x 5 . 31 9. 4 13. 3t 6 17. –6y 24 1 3x . 2 21. 25. 2x 1 5 29. 8y 1 3y 29 6. x 1.5 3.5 10. –3.5 c 0.5 14. 2x 12 18. 27 9y 22 3z $ 6 22. 26.
3y 6 12 30. 6x 2 8x 14 x 2 . 1 23. 27. 5y 3 13 31. 8m 2(2m 3) 34. 0 2x 4 6 37. (x 5 2x) or ( x 8 3x) 38. Which of the following is equivalent to y 4 9? (3) y 13 32. 0 x 3 6 35. (x 1 3) or (x 1 9) (2) y 5 39. Which of the following is equivalent to 4x 5x 6? (2) x 6 40. The smallest member of the solution set of 3x 7 8 is (1) x 6 (1) y 5 (3) x 6 (1) 3 (2) 4 (3) 5 41. The largest member of the solution set of 4x 3x 2 is (1) 1 (2) 2 (3) 3 33. –5 x 2 7 36. (2x 2) or (x 5 10) (4) y 13 (4) x 6 (4) 6 (4) 4 In 42–47, write an inequality for each graph using interval notation. 42. 43. 44. 45. 46. –4 –3 –2 –1 –4 –3 –2 –1 –4 –3 –2 –1 –4 –3 –2 –1 –4 –3 –2 – 47. –4 –3 4 –1 48. a. Graph the inequality (x 2) or (x 2). –2 3 0 1 2 b. Write an inequality equivalent to (x 2) or (x 2). Using Inequalities to Solve Problems 157 4-9 USING INEQUALITIES TO SOLVE PROBLEMS Many problems can be solved by writing an inequality that describes how the numbers in the problem are related and then solving the inequality. An inequality can be expressed in words in different ways. For example: x 12 A number is more than 12. A number exceeds 12. A number is greater than 12. A number is over 12. x 12 A number is less than 12. A number is under 12. x 12 A number is at least 12. A number has a minimum value of 12. A number is not less than 12. A number is not under 12. x 12 A number is at most 12. A number has a maximum value of 12. A number is not greater than 12. A number does not exceed 12. A number is not more than 12. Procedure To solve a problem that involves an inequality: 1. Choose a variable to represent one of the unknown quantities in the problem. 2. Express other unknown quantities in terms of the same variable. 3. Choose an appropriate domain for the problem. 4. Write an inequality using a relationship given in the problem, a previously known relationship, or a formula. 5. Solve the inequality. 6. Check the solution using the words of the problem. 7. Use the solution of the inequality to answer the question in the problem. EXAMPLE 1 Serafina has $53.50 in her pocket and wants to purchase shirts at a sale price of $14.95 each. How many shirts can she buy? Solution (1) Choose a variable to represent the number of shirts Serafina can buy and the cost of the shirts. Let x the number of shirts that she can buy. Then, 14.95x the cost of the x shirts. The domain is the set of whole numbers, since she can only buy a whole number of shirts. 158 First Degree Equations and Inequalities in One Variable (2) Write an inequality using a relationship given in the problem. The cost of the shirts is less than or equal to $53.50. \|___________________\| \|_____________________\| ↓ 14.95x ↓ ↓ $53.50 (3) Solve the inequality. 14.95x 53.50 14.95x 14.95 # 53.50 14.95 Use a calculator to complete the computation. ENTER: 53.50 14.95 ENTER DISPLAY Therefore, x 3.578595318. Since the domain is the set of whole numbers, the solution set is {x : x is a counting number less than or equal to 3} or {0, 1, 2, 3}. (4) Check the solution in the words of the problem. 0 shirt costs $14.95(0) $0 1 shirt costs $14.95(1) $14.95 2 shirts cost $14.95(2) $29.90 3 shirts cost $14.95(3) $44.85 4 or more shirts cost more than $14.95 (4) or more than $59.80 Answer Serafina can buy 0, 1, 2, or 3 shirts. EXAMPLE 2 The length of a rectangle is 5 centimeters more than its width. The perimeter of the rectangle is at least 66 centimeters. Find the minimum measures of the length and width. Solution If the perimeter is at least 66 centimeters, then the sum of the measures of the four sides is either equal to 66 centimeters or is greater than 66 centimeters. Let x the width of the rectangle. Then, x + 5 the length of the rectangle. The domain is the set of positive real numbers. Using Inequalities to Solve Problems 159 The perimeter of the rectangle is at least 66 centimeters. \|____________________________\| \|_________\| \|_____________\| ↓ x (x 5) x (x 5) ↓ ↓ 66 4x 10 66 4x 10 10 66 10 4x 56 x 14 The width can be any real number that is greater than or equal to 14 and the length is any real number that is 5 more than the width. Since we are looking for the minimum measures, the smallest possible width is 14 and the smallest possible length is 14 5 or 19. Answer The minimum width is 14 centimeters, and the minimum length is 19 centimeters. EXERCISES Writing About Mathematics 1. If there is a number x such that x 3, is it true that x 3? Explain why or why not. 2. Is the solution set of x 3 the same as the solution set of x 3? Explain why or why not. Developing Skills In 3–12, represent each sentence as an algebraic inequality. 3. x is less than or equal to 15. 4. y is greater than or equal to 4. 5. x is at most 50. 6. x is more than 50. 7. The greatest possible value of 3y is 30. 9. The maximum value of 4x 6 is 54. 11. The product of 3x and x 1 is less than 35. 8. The sum of 5x and 2x is at least 70. 10. The minimum value of 2x 1 is 13. 12. When x is divided by 3 the quotient is greater than 7. In 13–19, in each case write and solve the inequality that represents the given conditions. Use n as the variable. 13. Six less than a number is less than 4. 14. Six less than a number is greater than 4. 15. Six times a number is less than 72. 16. A number increased by 10 is greater than 50. 160 First Degree Equations and Inequalities in One Variable 17. A number decreased by 15 is less 18. Twice a number, increased by 6, is less than 35. than 48. 19. Five times a number, decreased by 24, is greater than 3 times the number. Applying Skills In 20–29, in each case write an inequality and solve the problem algebraically. 20. Mr. Burke had a sum of money in a bank. After he deposited an additional sum of $100, he had at least $550 in the bank. At least how much money did Mr. Burke have in the bank originally? 21. The members of a club agree to buy at least 250 tickets for a theater party. If they expect to buy 80 fewer orchestra tickets than balcony tickets, what is the least number of balcony tickets they will buy? 22. Mrs. Scott decided that she would spend no more than $120 to buy a jacket and a skirt. If the price of the jacket was $20 more than 3 times the price of the skirt, find the highest possible price of the skirt. 23. Three times a number increased by 8 is at most 40 more than the number. Find the greatest value of the number. 24. The length of a rectangle is 8 meters less than 5 times its width. If the perimeter of the rec- tangle is at most 104 meters, find the greatest possible width of the rectangle. 25. The length of a rectangle is 12 centimeters less than 3 times its width. If the perimeter of the rectangle is at most 176 centimeters, find the greatest possible length of the rectangle. 26. Mrs. Diaz wishes to save at least $1,500 in 12 months. If she saves $300 during the first 4 months, what is the least possible average amount that she must save in each of the remaining 8 months? 27. Two consecutive even numbers are such that their sum is greater than 98 decreased by twice the larger. Find the smallest possible values for the integers. 28. Minou wants $29 to buy music online. Her father agrees to pay her $6 an hour for gardening in addition to her $5 weekly allowance for helping around the house. What is the minimum number of hours Minou must work at gardening to receive at least $29 this week? 29. Allison has more than 2 but less than 3 hours to spend on her homework. She has work in math, English, and social studies. She plans to spend equal amounts of time studying English and studying social studies, and to spend twice as much time studying math as in studying English. a. What is the minimum number of minutes she can spend on English homework? b. What is the maximum number of minutes she can spend on social studies? c. What is the maximum number of minutes she can devote to math? CHAPTER SUMMARY Chapter Summary 161 The properties of equality allow us to write equivalent equations to solve an equation. 1. The addition property of equality: If equals are added to equals the sums are equal. 2. The subtraction property of equality: If equals are subtracted from equals the differences are equal. 3. The multiplication property of equality: If equals are multiplied by equals the products are equal. 4. The division property of equality: If equals are divided by nonzero equals the quotients are equal. 5. The substitution property: In any statement of equality, a quantity may be substituted for its equal. Before solving an equation, simplify each side if necessary. To solve an equation that has the variable in both sides, transform it into an equivalent equation in which the variable appears in only one side. Do this by adding the opposite of the variable term on one side to both sides of the equation. Use the properties of equality. Any equation or formula containing two or more variables can be transformed so that one variable is expressed in terms of all other variables. To do this, think of solving a simpler but related equation that contains only one variable. Order property of numbers: For real numbers x and y, one and only one of the following can be true: x y, x y, or x y. Transitive property of inequality: For real numbers x, y, and z, if x y and y z, then x z, and if x y and y z, then x z. Addition property of inequality: When the same number is added to or subtracted from both sides of an inequality, the order of the new inequality is the same as the order of the original one. Multiplication property of inequality: When both sides of an inequality are multiplied or divided by the same positive number, the order of the new inequality is the same as the order of the original one. When both sides of an inequality are multiplied or divided by the same negative number, the order of the new inequality is the opposite of the order of the original one. The domain or replacement set of an inequality is the set of all possible numbers that can be used to replace the variable. The solution set of an
inequality is the set of numbers from the domain that make the inequality true. Inequalities that have the same solution set are equivalent inequalities. 162 Algebraic Expressions and Open Sentences VOCABULARY 4-1 Equation • Left side • Left member • Right side • Right member • Root • Solution • Solution set • Identity • Equivalent equations • Solve an equation • Conditional equation • Check 4-2 First degree equation in one variable • Like terms • Similar terms • Unlike terms 4-4 Perimeter • Distance formula 4-7 Order property of the real numbers • Transitive property of inequality • Addition property of inequality • Multiplication property of inequality 4-8 Domain of an inequality • Replacement set of an inequality • Solution set of an inequality • Equivalent inequalities • Interval notation REVIEW EXERCISES 1. Compare the properties of equality with the properties of inequality. Explain how they are alike and how they are different. In 2–9, solve for the variable and check. 2. 8w 60 4w 4. 4h 3 23 h 6. 8a (6a 5) 1 8. 3(4x 1) 2 17x 10 3. 8w 4w 60 5. 5y 3 2y 7. 2(b 4) 4(2b 1) 9. (x 2) (3x 2) x 3 In 10–15, solve each equation for x in terms of a, b, and c. 10. a x bc 11. cx a b c 14. ax b 12. bx a 5a c 15. ax 2b c 13. a1c 2 x b 16. a. Solve A 1 2bh for h in terms of A and b. b. Find h when A 5.4 and b 0.9. 17. If P 2l + 2w, find w when P 17 and l 5. 18. If F 9 5C 32, find C when F 68. In 19–26, find and graph the solution set of each inequality. 20. 2x 3 5 19. 6 x 3 23. 3 x 1 2 22. x 4 1 1 3x 21. 24. (x 2 5) and (2x 14) Review Exercises 163 25. (x 2) or (x 0) 26. (x 4 1) and (2x 18) In 27–30, tell whether each statement is sometimes, always, or never true. Justify your answer by stating a property of inequality or by giving a counterexample. 27. If x y, then a x a y. 29. If x y and y z, then x z. 28. If x y, then ax ay. 30. If x y, then x y. In 31–33, select the answer choice that correctly completes the statement or answers the question. 31. An inequality that is equivalent to 4x 3 5 is 1 2 (1) x 2 (2) x 2 (3) x (4) x 1 2 –1 0 1 2 3 4 The solution set of which inequality is shown in the graph above? (1) x 2 0 (3) x 2 0 (2) x 2 0 (4) x 2 0 32. 33. –5 –4 –3 –2 –1 0 1 2 The above graph shows the solution set of which inequality? (1) 4 x 1 (2) 4 x 1 (3) 4 x l (4) 4 x 1 34. The figure on the right consists of two squares and two isosceles right triangles. Express the area of the figure in terms of s, the length of one side of a square. 35. Express in terms of w the number of days in w weeks and 4 days. 36. The length of a rectangular room is 5 feet more than 3 times the width. The perimeter of the room is 62 feet. Find the dimensions of the room. 37. A truck must cross a bridge that can support a maximum weight of 24,000 pounds. The weight of the empty truck is 1,500 pounds, and the driver weighs 190 pounds. What is the weight of a load that the truck can carry? 38. In an apartment building there is one elevator, and the maximum load that it can carry is 2,000 pounds. The maintenance supervisor wants to move a replacement part for the air-conditioning unit to the roof. The part weighs 1,600 pounds, and the mechanized cart on which it is being moved weighs 250 pounds. When the maintenance supervisor drives the cart onto the elevator, the alarm sounds to signify that the elevator is overloaded. 164 Algebraic Expressions and Open Sentences a. How much does the maintenance supervisor weigh? b. How can the replacement part be delivered to the roof if the part can- not be disassembled? 39. A mail-order digital photo developer charges 8 cents for each print plus a $2.98 shipping fee. A local developer charges 15 cents for each print. How many digital prints must be ordered in order that: a. the local developer offers the lower price? b. the mail-order developer offers the lower price? 40. a. What is an appropriate replacement set for the problem in the chapter opener on page 116 of this chapter? b. Write and solve the equations suggested by this problem. c. Write the solution set for this problem. Exploration The figure at the right shows a circle inscribed in a square. Explain how pr2 , (2r)2 . this figure shows that CUMULATIVE REVIEW CHAPTERS 1–4 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. The rational numbers are a subset of (1) the integers (2) the counting numbers (3) the whole numbers (4) the real numbers 2. If x 12.6 8.4 0.7x, then x equals (1) 0.07 (2) 0.7 (3) 7 (4) 70 3. The solution set of 2x 4 5x 14 is (1) 210 3 U V (2) 10 3 U V (3) {6} (4) {–6} 4. Which of the following inequalities is false? 2 (3) 0.6 3 2 (1) 0.6 3 2 3 # 0.6 (2) 2 (4) 0.6 3 Cumulative Review 165 5. Which of the following identities is an illustration of the commutative property of addition? (1) (x 3) 2 x (3 2) (2) x 3 3 x (3) 5(x 3) 5x 15 (4) x 0 x 6. Which of the following sets is closed under division? (1) nonzero whole numbers (2) nonzero integers (3) nonzero even integers (4) nonzero rational numbers 7. The measure of one side of a rectangle is 20.50 feet. This measure is given to how many significant digits? (1) 1 (2) 2 (3) 3 (4) 4 8. In the coordinate plane, the vertices of quadrilateral ABCD have the coor- dinates A(–2, 0), B(7, 0), C(7, 5), and D(0, 5). The quadrilateral is (1) a rhombus (2) a rectangle (3) a parallelogram (4) a trapezoid 9. One element of the solution set of (x 3) or (x 5) is (1) 4 (4) None of the above. The solution set is the empty set. (2) 2 (3) 5 10. When x 3, x2 is (1) 6 (2) 6 (3) 9 (4) 9 Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. A quadrilateral has four sides. Quadrilateral ABCD has three sides that have equal measures. The measure of the fourth side is 8.0 cm longer than each of the other sides. If the perimeter of the quadrilateral is 28.0 m, find the measure of each side using the correct number of significant digits. 12. To change degrees Fahrenheit, F, to degrees Celsius, C, subtract 32 from the Fahrenheit temperature and multiply the difference by five-ninths. a. Write an equation for C in terms of F. b. Normal body temperature is 98.6° Fahrenheit. What is normal body temperature in degrees Celsius? 166 Algebraic Expressions and Open Sentences Part III Answer all questions in this part. Each correct answer will receive 3 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. Is it possible for the remainder to be 2 when a prime number that is greater than 2 is divided by 4? Explain why or why not. 14. A plum and a pineapple cost the same as three peaches. Two plums cost the same as a peach. How many plums cost the same as a pineapple? Part IV Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. h 2(b11b2) 15. A trapezoid is a quadrilateral with only one pair of parallel sides called the bases of the trapezoid. The formula for the area of a trapezoid is A where h represents the measure of the altitude to the bases, and b1 and b2 represent the measures of the bases. Find the area of a trapezoid if h 5.25 cm, b1 9.50 cm. Express your answer 12.75 cm, and b2 to the number of significant digits determined by the given data. 16. Fred bought three shirts, each at the same price, and received less than $12.00 in change from a $50.00 bill. a. What is the minimum cost of one shirt? b. What is the maximum cost of one shirt? OPERATIONS WITH ALGEBRAIC EXPRESSIONS Marvin is planning two rectangular gardens that will have the same width. He wants one to be 5 feet longer than it is wide and the other to be 8 feet longer than it is wide. How can he express the area of each of the gardens and the total area of the two gardens in terms of w, the width of each? Problems like this often occur in many areas of business, science and technology as well as every day life. When we use variables and the rules for adding and multiplying expressions involving variables, we can often write general expressions that help us investigate many possibilities in the solution of a problem. In this chapter, you will learn to add, subtract, multiply, and divide algebraic expressions. CHAPTER 5 CHAPTER TABLE OF CONTENTS 5-1 Adding and Subtracting Algebraic Expressions 5-2 Multiplying Powers That Have the Same Base 5-3 Multiplying by a Monomial 5-4 Multiplying Polynomials 5-5 Dividing Powers That Have the Same Base 5-6 Powers With Zero and Negative Exponents 5-7 Scientific Notation 5-8 Dividing by a Monomial 5-9 Dividing by a Binomial Chapter Summary Vocabulary Review Exercises Cumulative Review 167 168 Operations with Algebraic Expressions 5-1 ADDING AND SUBTRACTING ALGEBRAIC EXPRESSIONS Recall that an algebraic expression that is a number, a variable, or a product or quotient of numbers and variables is called a term. Examples of terms are: 7 a 2b 24 7y2 0.7ab5 2 5 w Two or more terms that contain the same variable or variables with corresponding variables having the same exponents, are called like terms or similar terms. For example, the following pairs are like terms. 6k and k 5x2 and 7x2 9ab and 0.4ab 9 2x2y3 and 211 3 x2y3 Two terms are unlike terms when they contain different variables, or the same variable or variables with different exponents. For example, the following pairs are unlike terms. 3x and 4y 5x2 and 5x3 9ab and 0.4a 8 3x3y2 and 4 7x2y3 To add or subtract like terms,
we use the distributive property of multipli- cation over addition or subtraction. 9x 2x (9 2)x 11x 16cd 3cd (16 3)cd 13cd 18y2 5y2 (18 5)y2 13y2 7ab ab 7ab 1ab (7 1)ab 6ab Since the distributive property is true for any number of terms, we can express the sum or difference of any number of like terms as a single term. 3ab2 4ab2 2ab2 (3 4 2)ab2 l ab2 ab2 x3 11x3 8x3 4x3 (1 11 8 4)x3 0x3 0 Recall that when like terms are added: 1. The sum or difference has the same variable or variables as the original terms. 2. The numerical coefficient of the sum or difference is the sum or difference of the numerical coefficients of the terms that were added. The sum of unlike terms cannot be expressed as a single term. For example, the sum of 2x and 3y cannot be written as a single term but is written 2x 3y. EXAMPLE 1 Add: a. 3a (8a) b. 12b2 (5b2) c. 15abc 6abc d. 8x2 y x2y e. 9y 9y f. 2(a b) 6(a b) EXAMPLE 2 Adding and Subtracting Algebraic Expressions 169 Answers [3 (8)]a 5a [12 (5)]b2 [12 5]b2 7b2 (15 6)abc 9abc (8 1)x2y 7x2y (9 9)y 0y 0 (2 6)(a b) 8(a b) An isosceles triangle has two sides that are equal in length. The length of each of the two equal sides of an isosceles triangle is twice the length of the third side of the triangle. If the length of the third side is represented by n, represent in simplest form the perimeter of the triangle. Solution n represents the length of the base. 2n represents the length of one of the equal sides. 2n represents the length of the other equal side. Perimeter n 2n 2n (1 2 2)n 5n. Note that the length of a side of a geometric figure is a positive number. Therefore, the variable n must represent a positive real number, that is, the replacement set for n must be the set of positive real numbers. Answer 5n Monomials and Polynomials A term that has no variable in the denominator is called a monomial. For example, 5, 5w, and 5 w A monomial or the sum of monomials is called a polynomial. A polynomial may have one or more terms. Some polynomials are given special names to indicate the number of terms. are monomials, but is not a monomial. 3w2 5 • A monomial such as 4x2 may be considered to be a polynomial of one term. (Mono- means “one”; poly- means “many.”) • A polynomial of two unlike terms, such as 10a 12b, is called a binomial. (Bi- means “two.”) 170 Operations with Algebraic Expressions • A polynomial of three unlike terms, such as x2 3x 2, is called a trinomial. (Tri- means “three.”) • A polynomial such as 5x2 (2x) (4) is usually written as 5x2 2x 4. A polynomial has been simplified or is in simplest form when it contains no like terms. For example, 5x3 8x2 5x3 7, when expressed in simplest form, becomes 8x2 7. A polynomial is said to be in descending order when the exponents of a particular variable decrease as we move from left to right. The polynomial x3 5x2 4x 9 is in a descending order of powers of x. A polynomial is said to be in ascending order when the exponents of a particular variable increase as we move from left to right. The polynomial 4 5y y2 is in an ascending order of powers of y. To add two polynomials, we use the commutative, associative, and distribu- tive properties to combine like terms. EXAMPLE 3 Solution Simplify: 3ab + 5b ab 4ab 2b How to Proceed (1) Write the expression: (2) Group like terms together by using the commutative and associative properties: (3) Use the distributive property: (4) Simplify the numerical expressions that are in parentheses: 3ab 5b ab 4ab 2b 3ab ab 4ab 5b 2b (3ab ab 4ab) (5b 2b) (3 1 4)ab (5 2)b 6ab 3b Answer 6ab 3b EXAMPLE 4 Solution Find the sum: (3x2 5) (6x2 8) How to Proceed (1) Write the expression: (2) Use the associative property: (3) Use the commutative property: (4) Use the associative property: (5) Add like terms: Answer 9x2 13 (3x2 5) (6x2 8) 3x2 (5 6x2) 8 3x2 (6x2 5) 8 (3x2 6x2) (5 8) 9x2 13 Adding and Subtracting Algebraic Expressions 171 The sum of polynomials can also be arranged vertically, placing like terms under one another. The sum of 3x2 5 and 6x2 8 can be arranged as shown at the right. Addition can be checked by substituting any convenient value for the variable and evaluating each polynomial and the sum. 3x2 5 6x2 8 9x2 13 Check Let x 4 3x2 5 3(4)2 5 53 6x2 8 6(4)2 8 104 9x2 13 9(4)2 13 157 ✔ EXAMPLE 5 Simplify: 6a [5a (6 3a)] Solution When one grouping symbol appears within another, first simplify the expres- sion within the innermost grouping symbol. How to Proceed (1) Write the expression (2) Use the commutative property: (3) Use the associative property: (4) Combine like terms: (5) Use the associative property: (6) Combine like terms: 6a [5a (6 3a)] 6a [5a (3a 6)] 6a [(5a 3a) 6] 6a [2a 6] (6a 2a) 6 8a 6 Answer 8a 6 EXAMPLE 6 Solution Express the difference (4x2 2x 3) (2x2 5x 3) in simplest form. How to Proceed (1) Write the subtraction problem: (2) To subtract, add the opposite (4x2 2x 3) (2x2 5x 3) (4x2 2x 3) (2x2 5x 3) of the polynomial to be subtracted: (3) Use the commutative and associative properties to group like terms: (4) Add like terms: Answer 2x2 7x (4x2 2x2) (2x 5x) (3 3) 2x2 7x 0 2x2 7x 172 Operations with Algebraic Expressions EXERCISES Writing About Mathematics 1. Christopher said that 3x x 3. a. Use the distributive property to show Christopher that his answer is incorrect. b. Substitute a numerical value of x to show Christopher that his answer is incorrect. 2. Explain how the procedure for adding like terms is similar to the procedure for adding frac- tions. Developing Skills In 327, write each algebraic expression in simplest form. 3. (8c) (7c) 5. (20r) (5r) 7. (5ab) (9ab) 9. 5y 6y 9y 14y 11. (8x2) (x2) (12x2) (2x2) 13. 7b (4b 6) 15. (6x 4) 5x 17. 8d2 (6d2 4d) 19. 9y [7 (6y 7)] 21. 5a [3b (2a 4b)] 23. 3y2 [6y2 (3y 4)] 25. d2 [9d (2 4d2)] 27. (x3 9x 5) (4x2 12x 5) 4. (4a) (6a) 6. (7w) (7w) 8. (6x) (4x) (5x) (10x) 10. 4m 9m 12m m 12. 4a (9a 3) 14. 8c (7 9c) 16. r (s 2r) 18. (5x 3) (6x 5) 20. (5 6y) (9y 2) 22. (5x2 4) (3x2 9) 24. (x3 3x2) (2x2 9) 26. (x2 5x 24) (x2 4x 9) In 28–31, state whether each expression is a monomial, a binomial, a trinomial, or none of these. 30. 2a2 3a 6 31. x3 2x2 x 7 28. 8x 3 29. 7y 32. a. Give an example of the sum of two binomials that is a binomial. b. Give an example of the sum of two binomials that is a monomial. c. Give an example of the sum of two binomials that is a trinomial. d. Give an example of the sum of two binomials that has four terms. e. Can the sum of two binomials have more than four terms? Multiplying Powers That Have the Same Base 173 Applying Skills In 33–41, write each answer as a polynomial in simplest form. 33. A cheeseburger costs 3 times as much as a soft drink, and an order of fries costs twice as much as a soft drink. If a soft drink costs s cents, express the total cost of a cheeseburger, an order of fries, and a soft drink in terms of s. 34. Jack deposited some money in his savings account in September. In October he deposited twice as much as in September, and in November he deposited one-half as much as in September. If x represents the amount of money deposited in September, represent, in terms of x, the total amount Jack deposited in the 3 months. 35. On Tuesday, Melita read 3 times as many pages as she read on Monday. On Wednesday she read 1.5 times as many pages as on Monday, and on Thursday she read half as many pages as on Monday. If Melita read p pages on Monday, represent in terms of p, the total number of pages she read in the 4 days. 36. The cost of 12 gallons of gas is represented by 12x, and the cost of a quart of oil is repre- sented by 2x 30. Represent the cost of 12 gallons of gas and a quart of oil. 37. In the last basketball game of the season, Tom scored 2x points, Tony scored x 5 points, Walt scored 3x 1 points, Dick scored 4x 7 points, and Dan scored 2x 2 points. Represent the total points scored by these five players. 38. Last week, Greg spent twice as much on bus fare as he did on lunch, and 3 dollars less on entertainment than he did on bus fare. If x represents the amount, in dollars, spent on lunch, express in terms of x the total amount Greg spent on lunch, bus fare, and entertainment. 39. The cost of a chocolate shake is 40 cents less than the cost of a hamburger. If h represents the cost, in cents, of a hamburger, represent in terms of h the cost of a hamburger and a chocolate shake in dollars. 40. Rosie spent 12 dollars more for fabric for a new dress than she did for buttons, and 1 dollar less for thread than she did for buttons. If b represents the cost, in dollars, of the buttons, represent in terms of b the total cost of the materials needed for the dress. 41. The length of a rectangle is 7z2 3 inches and the width is 9z2 2 inches. Represent the perimeter of the rectangle. 5-2 MULTIPLYING POWERS THAT HAVE THE SAME BASE Finding the Product of Powers We know that y2 means y y and y3 means y y y. Therefore, 2 3 5 y2 y3 (y y)(y y y) (y y y y y) y5 174 Operations with Algebraic Expressions Similarly, and 2 4 6 c2 c4 (c c)(c c c c) (c c c c c c) c6 1 4 3 x x3 (x)(x x x) (x x x x) x4 The exponent in each product is the sum of the exponents in the factors, as shown in these examples. In general, when x is a real number and a and b are positive integers: xa xb xa b EXAMPLE 1 Simplify each of the following products: a. x5 x2 b. a7 a c. 32 34 Answers a. x5 x2 x52 x7 b. a7 a a71 a8 c. 32 34 324 36 Note: When we multiply powers with like bases, we do not actually perform the operation of multiplication but rather count up the number of times that the base is to be used as a factor to find the product. In Example 1c above, the answer does not give the value of the product but indicates only the number of times that 3 must be used as a factor to obtain the product. We can use to evaluate the products 32 34 and 36 to show that they the power key are equal. ^ ENTER: 3 ^ 2 3 ^ 4 ENTER ENTER: 3 ^ 6 ENTER DISPLAY: 3 ^ 2 * 3 ^ 4 DISPLAY Finding a Power of a Power Since (x3)4 x3 x3 x3 x3, then (x3)4 x12. The exponent 1
2 can be obtained by addition: 3 3 3 3 12 or by multiplication: 4 3 12. In general, when x is a real number and a and c are positive integers: (xa)c xac Multiplying Powers That Have the Same Base 175 An expression such as (x5y2)3 can be simplified by using the commutative and associative properties: (x5y2)3 (x5y2)(x5y2)(x5y2) (x5 x5 x5)(y2 y2 y2) x15y6 When the base is the product of two or more factors, we apply the rule for the power of a power to each factor. (x5y2)3 (x5)3 (y2)3 x5(3)y2(3) x15y6 Thus, (xayb)c (xa)c(yb)c xacybc The expression (5 4)3 can be evaluated in two ways. (5 4)3 53 43 125 64 8,000 (5 4)3 203 8,000 EXAMPLE 2 Simplify each expression in two ways. a. (a2)3 b. (ab2)4 c. (32 42)3 Solution a. (a2)3 a2 a2 a2 a222 a6 or (a2)3 a2(3) a6 Answer a6 b. (ab2)4 ab2 ab2 ab2 ab2 (a a a a)(b2 b2 b2 b2) a4b8 or (ab2)4 a1(4)b2(4) a4b8 Answer a4b8 c. (32 42)3 (32)3 (42)3 36 46 (3 4)6 126 or (32 42)3 ((3 4)2)3 (122)3 126 Answer 126 176 Operations with Algebraic Expressions To evaluate the expression in Example 2c, use a calculator. Evaluate (32 42)3: Evaluate (12)6: ENTER: ( ^ 3 3 x2 4 x2 ) ENTER: 12 ^ 6 ENTER ENTER DISPLAY DISPLAY: 1 2 ^ 6 EXERCISES Writing About Mathematics 1. Does 53 53 253? Use the commutative and associative properties of multiplication to explain why or why not. 2. Does 24 4 26? Use the commutative and associative properties of multiplication to explain why or why not. Developing Skills In 3–26, multiply in each case. 3. a2 a3 7. z3 z3 z5 11. e4 e5 e 15. 43 4 19. (z3)2 (z4)2 23. (22 32)3 4. b3 b4 8. t8 t 4 t2 12. 23 22 16. 24 25 2 20. (x2y3)2 24. (5 23)4 5. r2 r4 r5 9. x x 13. 34 33 17. (x3)2 21. (ab2)4 25. (1002 103)5 6. r3 r3 10. a2 a 14. 52 54 18. (a4)2 22. (rs)3 26. (a2)5 a In 27–31, multiply in each case. (All exponents are positive integers.) 27. xa x2a 28. yc y2 29. cr c2 30. xm x 31. (3y)a (3y)b In 32–39, state whether each sentence is true or false. 32. 104 103 107 36. 33 22 66 33. 24 22 28 37. 54 5 55 34. 33 22 65 38. (22)3 25 35. 1480 1410 1490 39. (63)4 (64)3 Multiplying by a Monomial 177 Applying Skills 40. Two students attended the first meeting of the Chess Club. At that meeting, they decided that each person would bring one additional person to the next meeting, doubling the membership. At the second meeting, they again decided that each person would bring one additional person to the next meeting, again doubling the membership. If this plan was carried out for n meetings, the membership would equal 2n persons. a. How many persons attended the fifth meeting? b. At which meeting would the membership be twice as large as at the fifth meeting? 41. In the metric system, 1 meter 102 centimeters and 1 kilometer 103 meters. How many centimeters equal one kilometer? 5-3 MULTIPLYING BY A MONOMIAL Multiplying a Monomial by a Monomial We know that the commutative property of multiplication makes it possible to arrange the factors of a product in any order and that the associative property of multiplication makes it possible to group the factors in any combination. For example: (5x)(6y) (5)(6)(x)(y) (5 6)(x y) 30xy (3x)(7x) (3)(7)(x)(x) (3 7) (x x) 21x2 (2x2)(+5x4) (2)(x2)(+5)(x4) [(2)(+5)] [(x2)(x4)] 10x6 (3a2b3)(4a4b) (3)(a2) (b3)(4)(a4) (b) [(3)(4)][(a2)(a4)][(b3)(b)] 12a6b4 In the preceding examples, the factors may be rearranged and grouped mentally. Procedure To multiply a monomial by a monomial: 1. Use the commutative and associative properties to rearrange and group the fac- tors.This may be done mentally. 2. Multiply the numerical coefficients. 3. Multiply powers with the same base by adding exponents. 4. Multiply the products obtained in Steps 2 and 3 and any other variable factors by writing them with no sign between them. 178 Operations with Algebraic Expressions EXAMPLE 1 Multiply: Answers 24xyz a. (8xy)(3z) c. (6y3)(y) 6y4 e. (5x2y3)(2xy2) 10x3y5 g. (3x2)3 b. (4a3)(5a5) Answers 20a8 12a5b7 d. (3a2b3)(4a3b4) f. (6c2d4)(0.5d) 3c2d5 (3x2)(3x2)(3x2) 27x6 or (3)3(x2)3 27x6 EXAMPLE 2 Solution Represent the area of a rectangle whose length is 3x and whose width is 2x. How to Proceed (1) Write the area formula: (2) Substitute the values of l and w: (3) Perform the multiplication: A lw (3x)(2x) (3 2)(x x) 6x2 Answer 6x2 Multiplying a Polynomial by a Monomial The distributive property of multiplication over addition is used to multiply a polynomial by a monomial. Therefore, a(b c) ab ac x(4x 3) x(4x) x(3) 4x2 3x This result can be illustrated geometrically. Let us separate a rectangle, whose length is 4x 3 and whose width is x, into two smaller rectangles such that the length of one rectangle is 4x and the length of the other is 3. x 3 4x x(4x + 3) 4x + 3 = x (x)(4x) + x (x)(3) 4x 3 Multiplying by a Monomial 179 Since the area of the largest rectangle is equal to the sum of the areas of the two smaller rectangles: x(4x 3) x(4x) x(3) 4x2 3x Procedure To multiply a polynomial by monomial, use the distributive property: Multiply each term of the polynomial by the monomial and write the result as the sum of these products. Multiplication and Grouping Symbols When an algebraic expression involves grouping symbols such as parentheses, we follow the general order of operations and perform operations with algebraic terms. In the example at the right, first simplify the expression within parentheses: Next, multiply: Finally, combine like terms by addition 8y 2(7y 4y) 5 8y 2(3y) 5 8y 6y 5 2y 5 or subtraction: In many expressions, however, the terms within parentheses cannot be combined because they are unlike terms. When this happens, we use the distributive property to clear parentheses and then follow the order of multiplying before adding. Here, clear the parentheses by using the distributive property: Next, multiply: Finally, combine like terms by addition: 3 7(2x 3) 3 7(2x) 7(3) 3 14x 21 24 14x The multiplicative identity property states that a l a. By using this property, we can say that 5 (2x 3) 5 1(2x 3) and then follow the procedures shown above. 5 (2x 3) 5 1(2x 3) 5 1(2x) 1(3) 5 2x 3 2 2x Also, since a l a, we can use this property to simplify expressions in which a parentheses is preceded by a negative sign: 6y (9 7y) 6y 1(9 7y) 6y 1(9) 1(7y) 6y 9 7y 13y 9 180 Operations with Algebraic Expressions EXAMPLE 3 Multiply: a. 5(r 7) b. 8(3x 2y 4z) c. 5x(x2 2x 4) d. 3a2b2(4ab2 3b2) Answers 5r 35 24x 16y 32z 5x3 10x2 20x 12a3b4 9a2b4 EXAMPLE 4 Simplify: a. 5x(x2 2) 7x b. 3a (5 7a) a. How to Proceed (1) Write the expression: (2) Use the distributive property: 5x(x2) 5x(2) 7x (3) Multiply: (4) Add like terms: 5x3 10x 7x 5x3 3x 5x(x2 2) 7x Answer 5x3 3x b. How to Proceed (1) Write the expression: (2) Use the distributive property: (3) Add like terms: 3a (5 7a) 3a 1(5 7a) 3a 5 7a 10a 5 Answer 10a 5 EXERCISES Writing About Mathematics 1. In an algebraic term, how do you show the product of a constant times a variable or the product of different variables? 2. In the expression 2 3(7y), which operation is performed first? Explain your answer. 3. In the expression (2 3)(7y), which operation is performed first? Explain your answer. 4. In the expression 5y(y 3), which operation is performed first? Explain your answer. 5. Can the sum x2 x3 be written in simpler form? Explain your answer. 6. Can the product x2(x3) be written in simpler form? Explain your answer. Multiplying by a Monomial 181 Developing Skills In 7–29, find each product. A B A 18b 7. (4b)(6b) 10. (8r)(2r) 23 4a 13. 16. (7r)(5st) 19. (3s)(4s)(5t) 22. (20y3))(7y2) 25. (8y5)(5y) 28. (7a3b)(5a2b2) B 8. (5)(2y)(3y) 11. (7x)(2y)(3z) A B A 1 2y 26x 21 3z 14. B A 17. (2)(6cd)(e) 20. (5a2)(4a2) 23. (18r5)(5r2) 26. (9z)(8z4)(z3) 29. (4ab2)(2a2b3) B 9. (4a)(5b) 12. (6x)(0.5y) 15. (5ab)(3c) 18. (9xy)(2x) 21. (6x4)(3x3) 24. (3z2)(4z) 27. (6x2y3)(4x4y2) 30. 3(6c 3d) 33. 10(2x 0.2y) 4c 2 5 3 8d In 30–47, write each product as a polynomial. 31. 5(4m 6n) 2 3m 2 4n 34. A 37. 4x(5x 6) 40. mn(m n) 43. r3s3(2r4s 3s4) 46. 3xy(x2 xy y2) 36. B A 39. 5c2(15c 4c) 42. 3ab(5a2 7b2) 45. 8(2x2 3x 5) 216 12 B 32. 2(8a 6b) B 28 4r 2 1 4s 35. A 38. 5d(d2 3d) 41. ab(a b) 44. 10d(2a 3c 4b) 47. 5r2s2(2r2 3rs 4s2) In 48–50, represent the area of each rectangle whose length l and width w are given. 48. l 5y, w 3y 51. The dimensions of the outer rectangle pictured at the right are 4x by 3x 6. The dimen- 50. l 3c, w 8c 2 49. l 3x, w 5y sions of the inner rectangle are 2x by x 2. a. Express the area of the outer rectangle in terms of x. 3x – 6 b. Express the area of the inner rectangle in terms of x. c. Express as a polynomial in simplest form the area of the 4x x + 2 2x shaded region. In 52–73, simplify each expression. 52. 5(d 3) 10 55. 2(x 1) 6 53. 3(2 3c) 5c 56. 4(3 6a 8s) 54. 7 2(7x 5) 57. 5 4(3e 5) 182 Operations with Algebraic Expressions 58. 8 (4e 2) 61. 9 (5t 6) 59. a (b a) 62. 4 (2 8s) 60. (6b 4) 2b 63. (6x 7) 14 64. 5x(2x 3) 9x 65. 12y 3y(2y 4) 66. 7x 3(2x 1) 8 67. 7c 4d 2(4c 3d) 68. 3a 2a(5a a) a2 69. (a 3b) (a 3b) 70. 4(2x 5) 3(2 7x) 71. 3(x y) 2(x 3y) 72. 5x(2 3x) x(3x 1) 73. y(y 4) y(y 3) 9y Applying Skills In 74–86, write each answer as a polynomial in simplest form. 74. If 1 pound of grass seed costs 25x cents, represent in terms of x the cost of 7 pounds of seed. 75. If a bus travels at the rate of 10z miles per hour for 4 hours, represent in terms of z the dis- tance traveled. 76. If Lois has 2n nickels, represent in terms of n the number of cents she has. 77. If the cost of a notebook is 2x 3, express the cost of five notebooks. 78. If the length of a rectangle is 5y 7 and the width is 3y, represent the area of the rectangle. 79. If the measure of the base of a triangle is 3b 2 and the height is 4b, represent the area of the triangle. 80. Represent the distance traveled in 3 hours by a car traveling at 3x 7 miles per hour. 81. Represent in terms of x and y the amount saved in 3y weeks if x 2 dollars are saved each week. 82. The length of a rectangular skating rink is 2 less than 3 times the width. If w represents the width of the rink, represent th
e area in terms of w. 83. An internet bookshop lists used books for 3x 5 dollars each. The cost for shipping and handling is 2 dollars for five books or fewer. Represent the total cost of an order for four used books. 84. A store advertises skirts for x 5 dollars and allows an additional 10-dollar reduction on the total purchase if three or more skirts are bought. Represent the cost of five skirts. 85. A store advertises skirts for x 5 dollars and allows an additional two-dollar reduction on each skirt if three or more skirts are purchased. Represent the cost of five skirts. 86. A store advertises skirts for x 5 dollars and tops for 2x 3 dollars. Represent the cost of two skirts and three tops. Multiplying Polynomials 183 5-4 MULTIPLYING POLYNOMIALS As discussed in Section 5-3, to find the product (x 4)(a), we use the distributive property of multiplication over addition: (x 4)(a) x(a) 4(a) Now, let us use this property to find the product of two binomials, for example, (x 4)(x 3). (a) x (a) 4 (a) (x 4) (x 4)(x 3) x(x 3) 4(x 3) x2 3x 4x 12 x2 7x 12 This result can also be illustrated geometricallyx + 4)(x + 3) = 3 + x x(x + 3) 4(x + 3) = 3 + x x 3x x2 4 12 4x 3 x (x + 4)(x + 3) x(x + 3) + 4(x + 3) x2 + 3x + 4x + 12 In general, for all a, b, c, and d: (a b)(c d) a(c d) b(c d) ac ad bc bd Notice that each term of the first polynomial multiplies each term of the second. At the right is a convenient vertical arrangement of the preceding multiplication, similar to the arrangement used in arithmetic multiplication. Note that multiplication is done from left to right. x 3 x 4 x(x 3) → x2 3x 4(x 3) → 4x 12 x2 7x 12 The word FOIL serves as a convenient way to remember the steps neces- sary to multiply two binomials. Last First Outside (2x 5)(x 4) 2x(x) 2x(4) 5(x) 5(4) Inside ➤ ➤ ➤ ➤ 2x2 8x 5x 20 2x2 3x 20 184 Operations with Algebraic Expressions Procedure To multiply a polynomial by a polynomial, first arrange each polynomial in descending or ascending powers of the same variable.Then use the distributive property: multiply each term of the first polynomial by each term of the other. EXAMPLE 1 Solution Simplify: (3x 4)(4x 5) METHOD 1 ➤ ➤ (3x 4)(4x 5) 3x(4x 5) 4(x 5) 12x2 15x 16x 20 12x2 x 20 ➤ ➤ Answer 12x2 x 20 EXAMPLE 2 Solution Simplify: (x2 3xy 9y2)(x 3y) METHOD 2 3x 4 4x 5 12x2 16x 15x 20 12x2 x 20 ➤ ➤ ➤ ➤ (x2 3xy 9y2)(x 3y) x2(x 3y) 3xy(x 3y) 9y2(x 3y) x3 3x2y 3x2y 9xy2 9xy2 27y3 x3 0x2y 0xy2 27y3 ➤ ➤ Answer x3 27y3 EXAMPLE 3 Solution Simplify: (2x 5)2 (x 3) (2x 5)2 (x 3) (2x 5)(2x 5) (x 3) 2x(2x) 2x(5) 5(2x) 5(5) (x) (+3) 4x2 10x 10x 25 x 3 4x2 21x 28 Answer 4x2 21x 28 Multiplying Polynomials 185 EXERCISES Writing About Mathematics 1. The product of two binomials in simplest form can have four terms, three terms, or two terms. a. When does the product of two binomials have four terms? b. When is the product of two binomials a trinomial? c. When is the product of two binomials a binomial? 2. Burt wrote (a 3)2 as a2 9. Prove to Burt that he is incorrect. Developing Skills In 3–35, write each product as a polynomial. 3. (a 2)(a 3) 6. (x 7)(x 2) 9. (b 8)(b 10) 12. (12 r)(6 r) 15. (5a 9)(5a 9) 18. (2x 3)(2x 3) 21. (a b)(a b) 24. (x 4y)(x 4y) 27. (r2 5)(r2 2) 30. (2c 1)(2c2 3c 1) 33. (x 4)(x 4)(x 4) 4. (x 5)(x 3) 7. (m 3)(m 7) 10. (6 y)(5 y) 13. (x 5)(x 5) 16. (2x 1)(x 6) 19. (3d 8)(3d 8) 22. (a + b)(a b) 25. (x 4y)2 28. (x2 y2)(x2 y2) 31. (3 2a a2)(5 2a) 34. (a 5)3 5. (d 9)(d 3) 8. (t 15)(t 6) 11. (8 e)(6 e) 14. (2y 7)(2y 7) 17. (5y 2)(3y 1) 20. (x y)(x y) 23. (a b)2 26. (9x 5y)(2x 3y) 29. (x 2)(x2 3x 5) 32. (2x 1)(3x 4)(x 3) 35. (x y)3 In 36–43, simplify each expression. 36. (x 7)(x 2) x2 38. r(r 2) (r 5) 40. (x 4)(x 3) (x 2)(x 5) 42. (y 4)2 (y 3)2 37. 2(3x 1)(2x 3) 14x 39. 8x2 (4x 3)(2x 1) 41. (3y 5)(2y 3) (y 7)(5y 1) 43. a[(a 2)(a 2) 4] Applying Skills In 44–46, use grouping symbols to write an algebraic expression that represents the answer. Then, express each answer as a polynomial in simplest form. 44. The length of a rectangle is 2x 5 and its width is x 7. Express the area of the rectangle as a trinomial. 186 Operations with Algebraic Expressions 45. The dimensions of a rectangle are represented by 11x 8 and 3x 5. Represent the area of the rectangle as a trinomial. 46. A train travels at a rate of (15x 100) kilometers per hours. a. Represent the distance it can travel in (x 3) hours as a trinomial. b. If x 2, how fast does the train travel? c. If x 2, how far does it travel in (x 3) hours? 5-5 DIVIDING POWERS THAT HAVE THE SAME BASE We know that any nonzero number divided by itself is 1. Therefore, x x 1 and y3 y3 1. In general, when x 0 and a is a positive integer: Therefore, xa xa 1 x5 x3 5 x2 ? x3 x3 y5 ? y4 y9 y4 5 y4 c5 c 5 c4 ? c c x2 1 x2 y5 1 y5 c4 1 c4 These same results can be obtained by using the relationship between divi- sion and multiplication: If a b c, then c b a. • Since x2 x3 x5, then x5 x3 x2. • Since y5 y4 y9, then y9 y4 y5. • Since c4 c c5, then c5 c1 c4. Observe that the exponent in each quotient is the difference between the exponent of the dividend and the exponent of the divisor. In general, when x 0 and a and b are positive integers with a b: xa xb xab Procedure To divide powers of the same base, find the exponent of the quotient by subtracting the exponent of the divisor from the exponent of the dividend.The base of the quotient is the same as the base of the dividend and of the divisor. Dividing Powers That Have the Same Base 187 EXAMPLE 1 Simplify by performing each indicated division. b. y5 y a. x9 x5 c. c5 c5 d. 105 103 Answers a. x95 x4 b. y51 y4 c. 1 d. 1053 102 EXAMPLE 2 Write 57 ? 54 58 in simplest form: a. by using the rules for multiplying and dividing powers with like bases. b. by using a calculator. Solution a. First simplify the numerator. Then, apply the rule for division of powers with the same base. 57 b. On a calculator: ? 54 58 5 5714 58 5 511 58 5 511–8 5 53 5 125 ENTER ENTER DISPLAY Answer 125 EXERCISES Writing About Mathematics 1. Coretta said that 54 5 14. Do you agree with Coretta? Explain why or why not. 2. To evaluate the expression , 38 35 ? 32 a. in what order should the operations be performed? Explain your answer. b. does 38 35 ? 32 38 35 32? Explain why or why not. Developing Skills In 3–18, divide in each case. 3. x8 x2 7. e9 e3 4. a10 a5 5. c5 c4 8. m12 m4 9. n10 n9 6. x7 x7 10. r6 r6 188 Operations with Algebraic Expressions 11. x8 x 15. 106 104 12. z10 z 16. 34 32 13. t5 t 17. 53 5 14. 25 22 18. 104 10 In 19–24, divide in each case. (All exponents are positive integers.) 19. x5a x2a 22. sx s2 (x 2) 20. y10b y2b 23. ab ab 21. rc rd (c d) 24. 2a 2b (a b) In 25–32: a. Simplify each expression by using the rules for multiplying and dividing powers with like bases. b. Evaluate the expression using a calculator. Compare your answers to parts a and b. 25. 29. 23 ? 24 22 106 102 ? 104 26. 30. 58 54 ? 5 108 ? 102 (105)2 27. 31. 102 ? 103 104 64 ? 69 62 ? 63 28. 32. 33 ? 32 32 45 ? 45 (42)4 In 33–35, tell whether each sentence is true or false. 33. 10099 1098 1002 34. a6 a2 a4 (a 0) 35. 45045 45040 15 5-6 POWERS WITH ZERO AND NEGATIVE EXPONENTS Integers that are negative or zero, such as 0, 1, and 2 can also be used as exponents. We will define powers having zero and negative integral exponents in such a way that the properties that were valid for positive integral exponents will also be valid for zero and negative integral exponents. In other words, the following properties will be true when the exponents a and b are positive integers, negative integers, or 0: xa xb xa b xa xb xa b (xa)b xab The Zero Exponent x3 We know that, for x 0, 1. If x3 ment, we must let x0 1 since x0 and 1 are each equal to lowing definition: x3 5 x3 – 3 5 x0 x3 x3 x3 is to be a meaningful state- . This leads to the fol- DEFINITION x0 1 if x is a number such that x 0. It can be shown that all the laws of exponents remain valid when x0 is defined as 1. For example: • Using the definition 100 1, we have 103 100 103 1 103 Powers with Zero and Negative Exponents 189 • Using the law of exponents, we have 103 100 1030 103. The two procedures result in the same product. • Using the definition 100 1, we have 103 100 103 1 103 • Using the law of exponents, we have 103 100 1030 103. The two procedures result in the same quotient. The definition x0 1 (x 0) permits us to say that the zero power of any number except 0 equals 1. 40 1 (4)0 1 (4x)0 1 (4x)0 1 A calculator will return this value. For example, to evaluate 40: ENTER: 4 ^ 0 ENTER DISPLAY: 4 ^ 0 1 Note that 4x0 41 x0 4 1 4 but (4x)0 40 x0 1 1 1. The Negative Integral Exponent We know that, for x 0, x3 x5 5 1 ? x3 ? x3 5 1 x2 x2 ? is to be a meaningful statement, we must let x2 ? 1 5 1 x2 x3 5 1 x3 . x22 5 1 x2 since x2 x3 If x5 5 x3–5 5 x22 1 x2 x3 x5 and are each equal to . This leads to the following definition: DEFINITION xn if x 0. 1 xn A graphing calculator will return equal values for x2 and . For example, 1 x2 let x 5. Evaluate 52. ENTER: 5 ^ (-) 2 ENTER DISPLAY: 5 ^ - 2 Evaluate ENTER: 1 . 1 52 5 ^ 2 ENTER DISPLAY 190 Operations with Algebraic Expressions as It can be shown that all the laws of exponents remain valid if xn is defined 1 . For example: xn 24 5 22 • Using the definition 24 , we have 22 24 • Using the law of exponents, we have 22 24 22(4) 22. 22 5 222 . 24 5 1 1 24 ? 22 1 The two procedures give the same result. Now we can say that, for all integral values of a and b, xa xb 5 xa2b (x 2 0) EXAMPLE 1 Transform each given expression into an equivalent one with a positive exponent. Answers a. 43 b. 101 c. d. 1 225 5 3 A B 22 10 1 43 1 101 5 1 1 4 225 5 1 4 1 32 5 1 52 4 1 322 5 1 522 25 5 1 3 25 52 3 32 1 5 32 1 5 25 52 5 2 3 5 B A EXAMPLE 2 Compute the value of each expression. a. 30 b. 102 c. (5)0 24 d. 6(33) EXAMPLE 3 Answers 1 1 100 102 5 1 11 1 24 5 1 1 1 1 5 6 33 1 27 6 A B A B 16 5 1 1 16 5 6 27 5 2 9 Use the laws of exponents to perform the indicated operations. Answers Answers a. 27 23 27(3) 24 b. 36 32 36(2) 3
62 34 c. (x4)3 x4(3) x12 d. (y2)4 y2(4) y8 Scientific Notation 191 EXERCISES Writing About Mathematics 1. Sasha said that for all x 0, x2 is a positive number less than 1. Do you agree with Sasha? Explain why or why not. 2. Brandon said that, when n is a whole number, the number 10n when written in ordinary dec- imal notation uses n 1 digits. Do you agree with Brandon? Explain why or why not. Developing Skills In 3–7, transform each given expression into an equivalent expression involving a positive exponent. 3. 104 4. 21 5. 22 6. m6, m 0 7. r3, r 0 2 3 A B In 8–19, compute each value using the definitions of zero and negative exponents. Compare your answers with the results obtained using a calculator. 8. 100 12. 101 9. (4)0 13. 102 10. 40 14. 103 16. 1.5(10)3 17. 70 62 18. 0 1323 1 2 A B 11. 32 15. 4(10)2 19. 2 41 In 20–27, use the laws of exponents to perform each indicated operation. 21. 34 32 25. (41)2 20. 102 105 24. 34 30 28. Find the value of 7x0 (6x)0, (x 0). 29. Find the value of 5x0 2x1 when x 4. 22. 103 105 26. (33)2 23. (42)2 44 27. 20 25 5-7 SCIENTIFIC NOTATION Scientists and mathematicians often work with numbers that are very large or very small. In order to write and compute with such numbers more easily, these workers use scientific notation. A number is expressed in scientific notation when it is written as the product of two quantities: the first is a number greater than or equal to 1 but less than 10, and the second is a power of 10. In other words, a number is in scientific notation when it is written as where 1 a 10 and n is an integer. a 10n 192 Operations with Algebraic Expressions Writing Numbers in Scientific Notation To write a number in scientific notation, first write it as the product of a number between 1 and 10 times a power of 10. Then express the power of 10 in exponential form. The table at the right shows some integral powers of 10. When the exponent is a positive integer, the power can be written as 1 followed by the number of 0’s equal to the exponent of 10. When the exponent is a negative integer, the power can be written as a decimal value with the number of decimal places equal to the absolute value of the exponent of 10. 3,000,000 3 1,000,000 3 106 780 7.8 100 7.8 102 3 3 1 3 100 0.025 2.5 0.01 2.5 102 0.0003 3 0.0001 3 104 Powers of 10 105 100,000 104 10,000 103 1,000 102 100 101 10 100 1 101 102 103 104 101 5 1 102 5 1 103 5 1 104 5 1 10 100 1 1 1 1 0.1 0.01 0.001 0.0001 1,000 10,000 When writing a number in scientific notation, keep in mind the following: • A number equal to or greater than 10 has a positive exponent of 10. • A number equal to or greater than 1 but less than 10 has a zero exponent of 10. • A number between 0 and 1 has a negative exponent of 10. EXAMPLE 1 The distance from the earth to the sun is approximately 93,000,000 miles. Write this number in scientific notation. Solution How to Proceed (1) Write the number, placing a decimal point after 93,000,000. the last digit. (2) Place a caret (^) after the first nonzero digit so that replacing the caret with a decimal point will give a number between 1 and 10. (3) Count the number of digits between the caret and the decimal point. This is the exponent of 10 in scientific notation. The exponent is positive because the given number is greater than 10. 9 3,000,000. ^ 9 3,000,000. ^ 7 (4) Write the number in the form a 10n, where where a is found by replacing the caret with a decimal point and n is the exponent found in Step 3. Answer 9.3 107 Scientific Notation 193 9.3 107 EXAMPLE 2 Express 0.0000029 in scientific notation. Solution Since the number is between 0 and 1, the exponent will be negative. Place a caret after the first nonzero digit to indicate the position of the dec- imal point in scientific notation. Answer 0.0000029 0.000002 9 2.9 106 ^ 6 Graphing calculators can be placed in scientific notation mode and will return the results shown in Examples 1 and 2 when the given numbers are entered. ENTER: MODE ENTER CLEAR .0000029 ENTER DISPLAY This display is read as 2.9 106, where the integer following “E” is the exponent to the base 10 used to write the number in scientific notation. Changing to Ordinary Decimal Notation We can change a number that is written in scientific notation to ordinary decimal notation by expanding the power of 10 and then multiplying the result by the number between 1 and 10. 194 Operations with Algebraic Expressions EXAMPLE 3 The approximate population of the United States is 2.81 108. Find the approximate number of people in the United States. Solution How to Proceed (1) Evaluate the second factor, which 2.81 108 2.81 100,000,000 is a power of 10: (2) Multiply the factors: 2.81 108 281,000,000 Answer 281,000,000 people Note: We could have multiplied 2.81 by 108 quickly by moving the decimal point in 2.81 eight places to the right. EXAMPLE 4 The diameter of a red blood corpuscle is expressed in scientific notation as 7.5 104 centimeters. Write the number of centimeters in the diameter as a decimal fraction. Solution How to Proceed (1) Evaluate the second factor, which 7.5 104 7.5 0.0001 is a power of 10: (2) Multiply the factors: 7.5 104 0.00075 Answer 0.00075 cm Note: We could have multiplied 7.5 by 104 quickly by moving the decimal point in 7.5 four places to the left. EXAMPLE 5 Calculator Solution Use a calculator to find the product: 45,000 570,000. ENTER: 45000 570000 ENTER DISPLAY calculator will shift to scientific notation when the number is too large or too small for the display. The number in this display can be changed to decimal notation by using the procedure shown in Examples 3 and 4. Answer 2.565 1010 2.565 10,000,000,000 25,650,000,000 Scientific Notation 195 EXAMPLE 6 Use a calculator to find the mass of 2.70 1015 hydrogen atoms if the mass of one hydrogen atom is 1.67 1024 grams. Round the answer to three significant digits. Solution Multiply the mass of one hydrogen atom by the number of hydrogen atoms. (1.67 1024) (2.70 1015) (1.67 2.70) (1024 1015) (1.67 2.70) (1024 15) 4.509 109 Round 4.509 to 4.51, which has three significant digits. Answer 4.51 109 4.51 0.000000001 0.00000000451 grams Calculator Solution Use a calculator to multiply the mass of one hydrogen atom by the number of hydrogen atoms. Enter the numbers in scientific notation. ENTER: 1.67 2nd EE (-) 24 2.7 2nd EE 15 ENTER DISPLAY Round 4.509 to three significant digits. Answer 4.51 109 4.51 0.000000001 0.00000000451 grams EXERCISES Writing About Mathematics 1. Jared said that when a number is in scientific notation, a 10n, the number of digits in a is the number of significant digits. Do you agree with Jared? Explain why or why not. 2. When Corey wanted to enter 2.54 105 into his calculator, he used this sequence of keys: 2.54 or why not. 2nd EE 5 ENTER . Is this a correct way to enter the number? Explain why 196 Operations with Algebraic Expressions Developing Skills In 3–8, write each number as a power of 10. 3. 100 6. 0.0001 4. 10,000 7. 1,000,000,000 5. 0.01 8. 0.0000001 In 9–20, find the number that is expressed by each numeral. 11. 103 15. 6 101 19. 1.27 103 10. 1010 14. 4 108 18. 8.3 1010 9. 107 13. 3 105 17. 1.3 104 12. 105 16. 9 107 20. 6.14 102 In 21–32, find the value of n that will make each resulting statement true. 21. 120 1.2 10n 24. 0.00161 1.61 10n 27. 0.00000000375 3.75 10n 30. 2.54 2.54 10n 22. 9,300 9.3 10n 25. 0.0000760 7.60 10n 28. 872,000,000 8.72 10n 31. 0.00456 4.56 10n 23. 5,280 5.28 10n 26. 52,000 5.2 10n 29. 0.800 8.00 10n 32. 7,123,000 7.123 10n In 33–44, express each number in scientific notation. 33. 8,400 37. 0.00061 41. 453,000 34. 27,000 38. 0.0000039 42. 0.00381 35. 54,000,000 36. 320,000,000 39. 0.0000000140 40. 0.156 43. 375,000,000 44. 0.0000763 In 45–48, compute the result of each operation. Using the correct number of significant digits: a. write the result in scientific notation, b. write the result in ordinary decimal notation. 45. (2.9 103)(3.0 103) 47. (7.50 104) (2.5 103) 46. (2.55 102)(3.00 103) 48. (6.80 105) (3.40 108) Applying Skills In 49–52, express each number in scientific notation. 49. A light-year, which is the distance light travels in 1 year, is approximately 9,500,000,000,000 kilometers. 50. A star that is about 12,000,000,000,000,000,000,000 miles away can be seen by the Palomar telescope. 51. The radius of an electron is about 0.0000000000005 centimeters. 52. The diameter of some white blood corpuscles is approximately 0.0008 inches. Dividing by a Monomial 197 In 53–57, express each number in ordinary decimal notation. 53. The diameter of the universe is 2 109 light-years. 54. The distance from the earth to the moon is 2.4 105 miles. 55. In a motion-picture film, the image of each picture remains on the screen approximately 6 102 seconds. 56. Light takes about 2 108 seconds to cross a room. 57. The mass of the earth is approximately 5.9 1024 kilograms. 5-8 DIVIDING BY A MONOMIAL Dividing a Monomial by a Monomial We know that a b ? c We can rewrite this equality interchanging the left and right members. bd 5 a ac d 5 ac bd b ? c d Using this relationship, we can write: 2x4 5 230 2 a5 5 221 a4 ? ? 23 y2 3y0z1 3 1 z 3z y2 ? 7a1b3 7ab3 221a5b4 23a4b 15x2 12y2z2 4y2z x6 x4 b4 b 230x6 ? ? 5 12 z2 4 z Procedure To divide a monomial by a monomial: 1. Divide the numerical coefficients. 2. When variable factors are powers of the same base, divide by subtracting exponents. 3. Multiply the quotients from steps 1 and 2. If the area of a rectangle is 42 and its length is 6, we can find its width by dividing the area, 42, by the length, 6. Thus, 42 6 7, which is the width. 198 Operations with Algebraic Expressions Similarly, if the area of a rectangle is represented by 42x2 and its length by 6x, we can find its width by dividing the area, 42x2, by the length, 6x: Therefore, the width can be represented by 7x. 42x2 6x 7x EXAMPLE 1 Divide: Answers a. b. c. 24a5 23a2 218x3y2 26x2y 20a3c4d2 25a3c3 EXAMPLE 2 24 23 218 26 20 25 ? ? ? a5 a2 5 28a3 x3 y
2 x2 ? y c4 a3 a3 ? c3 ? d2 3xy 4(1)cd2 4cd2 The area of a rectangle is 24x4y3. Express, in terms of x and y, the length of the rectangle if the width is 3xy2. Solution The length of a rectangle can be found by dividing the area by the width. 24x4y3 3xy2 5 8x3y Answer Dividing a Polynomial by a Monomial We know that to divide by a number is the same as to multiply by its reciprocal. Therefore, a 1 c b 5 1 b(a 1 c) 5 a b 1 c b Similarly, and 2x 1 2y 2 5 1 2(2x 1 2y) 5 2x 2 1 2y 2 5 x 1 y 21a2b 2 3ab 3ab 5 1 3ab(21a2b 2 3ab) 5 21a2b 3ab 2 3ab 3ab 5 7a 2 1 Usually, the two middle steps are done mentally. Procedure To divide a polynomial by a monomial, divide each term of the polynomial by the monomial. Dividing by a Monomial 199 EXAMPLE 3 Divide: a. (8a5 6a4) 2a2 b. 24x3y4 2 18x2y2 2 6xy 26xy Answers 4a3 3a2 4x2y3 3xy 1 EXERCISES Writing About Mathematics 1. Mikhail divided (12ab2 6ab) by (6ab) and got 2b for his answer. Explain to Mikhail why his answer is incorrect. 2. Angelique divided (15cd 11c) by 5c and got (3d 2.2) as her answer. Do you agree with Angelique? Explain why or why not. Developing Skills In 3–26, divide in each case. 3. 14x2y2 7 4. 36y10 6y2 7. 249c4b3 7c2b2 11. (14x 7) 7 15. p 1 prt p 19. 9y9 2 6y6 23y3 23. 22a2 2 3a 1 1 21 8. 12. 224x2y 23xy cm 1 cn c 16. y2 2 5y 2y 20. 24. 8a3 2 4a2 24a2 2.4y5 1 1.2y4 2 0.6y3 20.6y3 Applying Skills 5. 9. 13. 18x6 2x2 256abc 8abc tr 2 r r 17. 18d3 1 12d2 6d 21. 25. 3ab2 2 4a2b ab a3 2 2a2 0.5a2 6. 10. 14. 18. 22. 26. 5x2y3 25y3 227xyz 9xz 8c2 2 12d2 24 18r5 1 12r3 6r2 4c2d 2 12cd2 4cd 1.6cd 2 4.0c2d 0.8cd 27. If five oranges cost 15y cents, represent the average cost of one orange. 28. If the area of a triangle is 32ab and the base is 8a, represent the height of the triangle. 29. If a train traveled 54r miles in 9 hours, represent the average distance traveled in 1 hour. 30. If 40ab chairs are arranged in 5a rows with equal numbers of chairs in each row, represent the number of chairs in one row. 200 Operations with Algebraic Expressions 5-9 DIVIDING BY A BINOMIAL When we divide 736 by 32, we use repeated subtraction of multiples of 32 to determine how many times 32 is contained in 736. To divide a polynomial by a binomial, we will use a similar procedure to divide x2 6x 8 by x 2. How to Proceed (1) Write the usual division form: (2) Divide the first term of the dividend by the first term of the divisor to obtain the first term of the quotient: (3) Multiply the whole divisor by the first term of the quotient. Write each term of the product under the like term of the dividend: (4) Subtract and bring down the next term of the dividend to obtain a new dividend: (5) Divide the first term of the new dividend by the first term of the divisor to obtain the next term of the quotient: (6) Repeat steps (3) and (4), multiplying the whole divisor by the new term of the quotient. Subtract this product from the new dividend. Here the remainder is zero and the division is complete: x 1 2qx2 1 6x 1 8 x x 1 2qx2 1 6x 1 8 x x 1 2qx2 1 6x 1 8 x2 2x x x 1 2qx2 1 6x 1 8 x2 2x 4x 8 x 4 x 1 2qx2 1 6x 1 8 x2 2x 4x 8 x 4 x 1 2qx2 1 6x 1 8 x2 2x 4x 8 4x 8 0 The division can be checked by multiplying the quotient by the divisor to obtain the dividend: (x 4)(x 2) x(x 2) 4(x 2) x2 2x 4x 8 x2 6x 8 EXAMPLE 1 Divide 5s 6s2 6 by 2s 3 and check. Solution First arrange the terms of the dividend in descending order: 6s2 5s 6 Dividing by a Binomial 201 3s 2 2s 1 3q6s2 1 5s 2 6 6s2 9s (-) 4s 6 4s 6 0 Check (3s 2)(2s 3) 3s(2s 3) 2(2s 3) 6s2 9s 4s 6 6s2 5s 6 ✔ Note that we subtracted 9s from 5s by adding 9s to 5s. Answer 3s 2 EXERCISES Writing about Mathematics 1. Nate said that 2. Mason wrote x3 1 as x3 0x2 0x 1 before dividing by x 1. 1 5 x2 2 1 x 1 1 5 x3 x3 2 1 x 1 21 . Is Nate correct? Explain why or why not. a. Does x3 1 x3 0x2 0x 1? b. Divide x3 1 by x 1 by writing x3 0x2 0x 1 as the dividend. Check your answer to show that your computation is correct. Developing Skills In 3–14, divide and check. 3. (b2 5b 6) (b 3) 4. (y2 3y 2) (y 2) 5. (m2 8m 7) (m 1) 6. w2 1 2w 2 15 w 1 5 7. y2 1 21y 1 68 y 1 17 8. 9. (3a2 8a 4) (3a 2) 10. (15t2 19t 56) (5t 7) 11. x2 1 7x 1 10 x 1 5 10y2 2 y 2 24 2y 1 3 8 – 22c 1 12c2 4c 2 2 13. (17x 66 x2) (x 6) 12. 15. One factor of x2 4x 21 is x 7. Find the other factor. 14. x2 2 64 x 2 8 Applying Skills 16. The area of a rectangle is represented by x2 8x 9. If its length is represented by x 1, how can the width be represented? 17. The area of a rectangle is represented by 3y2 8y 4. If its length is represented by 3y 2, how can the width be represented? 202 Operations with Algebraic Expressions CHAPTER SUMMARY Two or more terms that contain the same variable, with corresponding variables having the same exponents, are called like terms. The sum of like terms is the sum of the coefficients of the terms times the common variable factor of the terms. A term that has no variable in the denominator is called a monomial. A polynomial is the sum of monomials. To subtract one polynomial from another, add the opposite of the polynomial to be subtracted (the subtrahend) to the polynomial from which it is to be subtracted (the minuend). When x is a nonzero real number and a and b are integers: xa xb xa b (xa)b xab xa xb xa b x0 1 x2a 5 1 xa A number is in scientific notation when it is written as a 10n, where 1 a 10 and n is an integer. If x a 10n. Then: • When x 10, n is positive. • When 1 x 10, n is zero. • When 0 x 1, n is negative. To multiply a polynomial by a polynomial, multiply each term of one polynomial by each term of the other polynomial and write the product as the sum of these results in simplest form. To divide a polynomial by a monomial, divide each term of the polynomial by the monomial and write the quotient as the sum of these results. To divide a polynomial by a binomial, subtract multiples of the divisor from the dividend until the remainder is 0 or of degree less than the degree of the divisor. VOCABULARY 5-1 Term • Like terms (similar terms) • Unlike terms • Monomial • Polynomial • Binomial • Trinomial • Simplest form • Descending order • Ascending order 5-4 FOIL 1 5-5 Zero exponent (x0 1) • Negative integral exponent (xn ) xn 5-7 Scientific notation Review Exercises 203 REVIEW EXERCISES 1. Explain why scientific notation is useful. 2. Is it possible to write a general rule for simplifying an expression such as an + bn? In 3–17, simplify each expression. 3. 5bc bc 6. 8mg(3g) 9. (6ab3)2 12. (2a 5)2 6y4 22y3 5y 1 15. 4. 3y2 2y y2 8y 2 7. 3x2(4x2 2x 1) 10. (6a b)2 13. 2x x(2x 5) 16. 6w3 2 8w2 1 2w 2w 5. 5t (4 8t) 8. (4x 3)(2x 1) 11. (2a 5)(2a 5) 14. 17. 40b3c6 28b2c x2 1 x 2 30 x 2 5 In 18–21, use the laws of exponents to perform the operations, and simplify. 18. 35 34 21. 120 122 12 20. [2(102)]3 19. (73)2 In 22–25, express each number in scientific notation. 22. 5,800 23. 14,200,000 24. 0.00006 25. 0.00000277 In 26–29, find the decimal number that is expressed by each given numeral. 26. 4 104 27. 3.06 103 29. 1.03 104 28. 9.7 108 30. Express the area of each of the gardens and the total area of the two gar- dens described in the chapter opener on page 167. 31. If the length of one side of a square is 2h 3, express in terms of h: a. the perimeter of the square. b. the area of the square. 32. The perimeter of a triangle is 41px. If the lengths of two sides are 18px and 7px, represent the length of the third side. 33. If the length of a rectangle can be represented by x 5, and the area of the rectangle by x2 7x 10, find the polynomial that represents: a. the width of the rectangle. b. the perimeter of the rectangle. 204 Operations with Algebraic Expressions 34. The cost of a pizza is 20 cents less than 9 times the cost of a soft drink. If x represents the cost, in cents, of a soft drink, express in simplest form the cost of two pizzas and six soft drinks. Exploration Study the squares of two-digit numbers that end in 5. From what you observe, can you devise a method for finding the square of such a number mentally? Can this method be applied to the square of a three-digit number that ends in 5? Study the squares of the integers from 1 to 12. From what you observe, can you devise a method that uses the square of an integer to find the square of the next larger integer? CUMULATIVE REVIEW CHAPTERS 1–5 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. Which of the numbers listed below has the largest value? (1) 12 3 (2) 1.67 (3) 1.67 (4) 12 7 1 2. For which of the following values of x is x2 x ? x (1) 1 (2) 0 (3) 3 (4) 3. Which of the numbers given below is not a rational number? (2) 11 2 (3) 1.3 (4) 2 3 7 3 (1) 2 " 4. Which of the following inequalities is false? (1) 1.5 11 2 (2) 1.5 11 2 (3) 1.5 1.5 (4) 1.5 1 5. Which of the following identities is an illustration of the associative prop- erty? (1) x 7 7 x (2) 3(x 7) 3x 3(7) (3) (x 7) 3 3 (7 x) (4) (x 7) 3 x ( 7 3) 6. The formula C 5 9(F 2 32) can be used to find the Celsius temperature, C, for a given Fahrenheit temperature, F. What Celsius temperature is equal to a Fahrenheit temperature of 68°? (1) 3° (2) 20° (3) 35° (4) 180° 7. If the universe is the set of whole numbers, the solution set of x 3 is (4) {1, 2, 3} (2) {0, 1, 2, 3} (1) {0, 1, 2} (3) {1, 2} 8. The perimeter of a square whose area is 81 square centimeters is Cumulative Review 205 (1) 9 cm (2) 18 cm 9. In simplest form, (2x 4)2 3(x 1) is equal to (3) 20.25 cm (4) 36 cm (1) 4x2 13x 13 (2) 4x2 13x 19 (3) 4x2 3x 19 (4) 4x2 3x 13 10. To the nearest tenth of a meter, the circumference of a circle whose radius is 12.0 meters is (1) 37.6 m (2) 37.7 m (3) 75.3 m (4) 75.4 m Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. The f
ormula for the volume V of a cone is V 5 1 3Bh where B is the area of the base and h is the height. Solve the formula for h in terms of V and B. 12. Each of the numbers given below is different from the others. Explain in what way each is different. 2 7 77 84 Part III Answer all questions in this part. Each correct answer will receive 3 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. Solve the given equation for x. Show each step of the solution and name the property that is used in each step. 3(x 4) 5x 8 14. Simplify the following expression. Show each step of the simplification and name the property that you used in each step. 4a 7 (7 3a) 206 Operations with Algebraic Expressions Part IV Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. A small park is in the shape of a rectangle that measures 525 feet by 468 feet. a. Find the number of feet of fencing that would be needed to enclose the park. Express your answer to the nearest foot. b. If the entire park is to be planted with grass seed, find the number of square feet to be seeded. Express your answer to the correct number of significant digits based on the given dimensions. c. The grass seed to be purchased is packaged in sacks, each of which holds enough seed to cover 25,000 square feet of ground. How many sacks of seed are needed to seed the park? 16. An ice cream stand sells single-dip cones for $1.75 and double-dip cones for $2.25. Yesterday, 500 cones were sold for $930. How many single-dip and how many double-dip cones were sold? RATIO AND PROPORTION Everyone likes to save money by purchasing something at a reduced price. Because merchants realize that a reduced price may entice a prospective buyer to buy on impulse or to buy at one store rather than another, they offer discounts and other price reductions.These discounts are often expressed as a percent off of the regular price. When the Acme Grocery offers a 25% discount on frozen vegetables and the Shop Rite Grocery advertises “Buy four, get one free,” the price-conscious shopper must decide which is the better offer if she intends to buy five packages of frozen vegetables. In this chapter, you will learn how ratios, and percents which are a special type of ratio, are used in many everyday problems. CHAPTER 6 CHAPTER TABLE OF CONTENTS 6-1 Ratio 6-2 Using a Ratio to Express a Rate 6-3 Verbal Problems Involving Ratio 6-4 Proportion 6-5 Direct Variation 6-6 Percent and Percentage Problems 6-7 Changing Units of Measure Chapter Summary Vocabulary Review Exercises Cumulative Review 207 208 Ratio and Proportion 6-1 RATIO A ratio, which is a comparison of two numbers by division, is the quotient obtained when the first number is divided by the second, nonzero number. Since a ratio is the quotient of two numbers divided in a definite order, care must be taken to write each ratio in its intended order. For example, the ratio of 3 to 1 is written 3 1 (as a fraction) while the ratio of 1 to 3 is written 1 3 (as a fraction) or or 3 : 1 (using a colon) 1 : 3 (using a colon) In general, the ratio of a to b can be expressed as a b or a b or a : b To find the ratio of two quantities, both quantities must be expressed in the same unit of measure before their quotient is determined. For example, to compare the value of a nickel and a penny, we first convert the nickel to 5 pennies and then find the ratio, which is or 5 : 1. Therefore, a nickel is worth 5 times as much as a penny. The ratio has no unit of measure. 5 1 Equivalent Ratios is a fraction, we can use the multiplication property of 1 to find 5 Since the ratio 1 many equivalent ratios. For example: 5 2 5 10 15 3 1 5 5 5 x 5 5x 1x 1 3 x (x 0) From the last example, we see that 5x and lx represent two numbers whose ratio is 5 : 1. In general, if a, b, and x are numbers (b 0, x 0), ax and bx represent two numbers whose ratio is a : b because b 3 x 5 ax x bx a b 5 a 24 16 b 3 1 5 a is a fraction, we can divide the numerator and the Also, since a ratio such as denominator of the fraction by the same nonzero number to find equivalent ratios. For example: 24 16 5 24 4 2 16 4 4 5 6 4 A ratio is expressed in simplest form when both terms of the ratio are whole numbers and when there is no whole number other than 1 that is a factor of 16 5 24 4 8 24 16 4 2 5 12 8 24 16 5 24 4 4 16 4 8 5 3 2 both of these terms. Therefore, to express the ratio both terms by 8, the largest integer that will divide both 24 and 16. Therefore, 3 in simplest form is . 2 in simplest form, we divide 24 16 24 16 Ratio 209 Continued Ratio Comparisons can also be made for three or more quantities. For example, the length of a rectangular solid is 75 centimeters, the width is 60 centimeters, and the height is 45 centimeters. The ratio of the length to the width is 75 : 60, and the ratio of the width to the height is 60 : 45. We can write these two ratios in an abbreviated form as the continued ratio 75 : 60 : 45. 45 cm 7 5 c m 60 cm A continued ratio is a comparison of three or more quantities in a definite order. Here, the ratio of the measures of the length, width, and height (in that order) of the rectangular solid is 75 : 60 : 45 or, in simplest form, 5 : 4 : 3. In general, the ratio of the numbers a, b, and c (b 0, c 0) is a : b : c. An oil tank with a capacity of 200 gallons contains 50 gallons of oil. a. Find the ratio of the number of gallons of oil in the tank to the capacity of the tank. b. What part of the tank is full? EXAMPLE 1 Solution a. Ratio number of gallons of oil in tank capacity of tank 5 50 200 5 1 4 . b. The tank is 1 4 full. Answers a. 1 4 b. 1 4 full EXAMPLE 2 Compute the ratio of 6.4 ounces to 1 pound. Solution First, express both quantities in the same unit of measure. Use the fact that 1 pound 16 ounces. 1 pound 5 6.4 ounces 6.4 ounces 16 ounces 5 6.4 16 5 6.4 16 3 10 10 5 64 160 5 64 4 32 160 4 32 5 2 5 210 Ratio and Proportion Calculator Solution On a calculator, divide 6.4 ounces by 16 ounces. ENTER: 6.4 16 ENTER DISPLAY: 6 . 4 / 1 6 . 4 Change the decimal in the display to a fraction. ENTER: DISPLAY: 2nd ANS MATH ENTER ENTER Answer The ratio is 2 : 5. EXAMPLE 3 Express the ratio 13 4 to 11 2 in simplest form. Solution Since a ratio is the quotient obtained when the first number is divided by the second, divide 13 4 by 11 . 2 13 4 4 11 14 2 12 5 7 6 Answer The ratio in simplest form is or 7 : 6. 7 6 EXERCISES Writing About Mathematics 1. Last week, Melanie answered 24 out of 30 questions correctly on a test. This week she answered 20 out of 24 questions correctly. On which test did Melanie have better results? Explain your answer. 2. Explain why the ratio 1.5 : 4.5 is not in simplest form. Developing Skills In 3–12, express each ratio in simplest form: a. as a fraction b. using a colon 3. 36 to 12 4. 48 to 24 5. 40 to 25 6. 12 to 3 7. 5 to 4 8. 8 to 32 9. 40 to 5 10. 0.2 to 8 11. 72 to 1.2 12. 3c to 5c Ratio 211 13. If the ratio of two numbers is 10 : 1, the larger number is how many times the smaller num- ber? 14. If the ratio of two numbers is 8 : 1, the smaller number is what fractional part of the larger number? In 15–19, express each ratio in simplest form. 15. 3 4 to 1 4 16. 11 8 to 3 8 17. 1.2 to 2.4 18. 0.75 to 0.25 19. 6 to 0.25 In 20–31, express each ratio in simplest form. 20. 80 m to 16 m 21. 75 g to 100 g 22. 36 cm to 72 cm 23. 54 g to 90 g 26. 11 2 hr to hr 1 2 29. 1 yd to 1 ft Applying Skills 24. 75 cm to 350 cm 25. 8 ounces to 1 pound 27. 3 in. to 1 2 in. 30. 1 hr to 15 min 28. 1 ft to 1 in. 31. 6 dollars to 50 cents 32. A baseball team played 162 games and won 90. a. What is the ratio of the number of games won to the number of games played? b. For every nine games played, how many games were won? 33. A student did six of ten problems correctly. a. What is the ratio of the number right to the number wrong? b. For every two answers that were wrong, how many answers were right? 34. A cake recipe calls for 11 4 cups of milk to 13 4 cups of flour. Write, in simplest form, the ratio of the number of cups of milk to the number of cups of flour in this recipe. 35. The perimeter of a rectangular garden is 30 feet, and the width is 5 feet. Find the ratio of the length of the rectangle to its width in simplest form. 36. In a freshman class, there are b boys and g girls. Express the ratio of the number of boys to the total number of pupils. 37. The length of a rectangular classroom is represented by 3x and its width by 2x. Find the ratio of the width of the classroom to its perimeter. 38. The ages of three teachers are 48, 28, and 24 years. Find, in simplest form, the continued ratio of these ages from oldest to youngest. 39. A woodworker is fashioning a base for a trophy. He starts with a block of wood whose length is twice its width and whose height is one-half its width. Write, in simplest form, the continued ratio of length to width to height. 40. Taya and Jed collect coins. The ratio of the number of coins in their collections, in some order, is 4 to 3. If Taya has 60 coins in her collection, how many coins could Jed have? 212 Ratio and Proportion 6-2 USING A RATIO TO EXPRESS A RATE When two quantities have the same unit of measure, their ratio has no unit of measure. A rate, like a ratio, is a comparison of two quantities, but the quantities may have different units of measures and their ratio has a unit of measure. For example, if a plane flies 1,920 kilometers in 3 hours, its rate of speed is a ratio that compares the distance traveled to the time that the plane was in flight. Rate 5 640 kilometers 1 hour The abbreviation
km/h is read “kilometers per hour.” 1,920 kilometers 3 hours time 5 distance 640 km/h A rate may be expressed in lowest terms when the numbers in its ratio are whole numbers with no common factor other than 1. However, a rate is most frequently written as a ratio with 1 as its second term. As shown in the example above, the second term may be omitted when it is 1. A rate that has a denominator of 1 is called a unit rate. A rate that identifies the cost of an item per unit is called the unit price. For example, $0.15 per ounce or $3.79 per pound are unit prices. Kareem scored 175 points in seven basketball games. Express, in lowest terms, the average rate of the number of points Kareem scored per game. EXAMPLE 1 Solution Rate 175 points 7 games 5 25 points 1 game 25 points per game Answer Kareem scored points at an average rate of 25 points per game. EXAMPLE 2 Solution There are 5 grams of salt in 100 cubic centimeters of a solution of salt and water. Express, in lowest terms, the ratio of the number of grams of salt per cubic centimeters in the solution. 5 g 100 cm3 5 1 g 20 cm3 5 1 20 g/cm3 Answer The solution contains 1 20 solution. grams or 0.05 grams of salt per cubic centimeter of Using a Ratio to Express a Rate 213 EXERCISES Writing About Mathematics 1. How is a rate a special kind of ratio? 2. How does the way in which a rate is usually expressed differ from a ratio in simplest form? Developing Skills In 3–8, express each rate in lowest terms. 3. The ratio of 36 apples to 18 people. 4. The ratio of 48 patients to 6 nurses. 5. The ratio of $1.50 to 3 liters. 6. The ratio of 96 cents to 16 grams. 7. The ratio of $2.25 to 6.75 ounces. 8. The ratio of 62 miles to 100 kilometers. Applying Skills In 9–12, in each case, find the average rate of speed, expressed in miles per hour. 9. A vacationer traveled 230 miles in 4 hours. 10. A post office truck delivered mail on a 9-mile route in 2 hours. 11. A commuter drove 48 miles to work in 11 2 hours. 12. A race-car driver traveled 31 miles in 15 minutes. (Use 15 minutes hour.) 1 4 13. If there are 240 tennis balls in 80 cans, how many tennis balls are in each can? 14. If an 11-ounce can of shaving cream costs 88 cents, what is the unit cost of the shaving cream in the can? 15. In a supermarket, the regular size of CleanRight cleanser contains 14 ounces and costs 49 cents. The giant size of CleanRight cleanser, which contains 20 ounces, costs 66 cents. a. Find, correct to the nearest tenth of a cent, the cost per ounce for the regular can. b. Find, correct to the nearest tenth of a cent, the cost per ounce for the giant can. c. Which is the better buy? 16. Johanna and Al use computers for word processing. Johanna can keyboard 920 words in 20 minutes, and Al can keyboard 1,290 words in 30 minutes. Who is faster at entering words on a keyboard? 17. Ronald runs 300 meters in 40 seconds. Carlos runs 200 meters in 30 seconds. Who is the faster runner for short races? 214 Ratio and Proportion 6-3 VERBAL PROBLEMS INVOLVING RATIO Any pair of numbers in the ratio 3 : 5 can be found by multiplying 3 and 5 by the same nonzero number. 3(3) 9 5(3) 15 3 : 5 9 : 15 3(7) 21 5(7) 35 3 : 5 21 : 35 3(0.3) 0.9 5(0.3) 1.5 3 : 5 0.9 : 1.5 3(x) 3x 5(x) 5x 3 : 5 3x : 5x Thus, for any nonzero number x, 3x : 5x 3 : 5. In general, when we know the ratio of two or more numbers, we can use the terms of the ratio and a nonzero variable, x, to express the numbers. Any two numbers in the ratio a : b can be written as ax and bx where x is a nonzero real number. EXAMPLE 1 The perimeter of a triangle is 60 feet. If the sides are in the ratio 3 : 4 : 5, find the length of each side of the triangle. Solution Let 3x the length of the first side, 4x the length of the second side, 5x the length of the third side. The perimeter of the triangle is 60 feet. 3x 4x 5x 60 12x 60 x 5 3x 3(5) 15 4x 4(5) 20 5x 5(5) 25 Check 15 : 20 : 25 3 : 4 : 5 ✔ 15 20 25 60 ✔ Answer The lengths of the sides are 15 feet, 20 feet, and 25 feet. EXAMPLE 2 Two numbers have the ratio 2 : 3. The larger is 30 more than of the smaller. Find the numbers. 1 2 Solution Let 2x the smaller number, 3x the larger number. Verbal Problems Involving Ratio 215 Check The ratio 30 : 45 in lowest terms is 2 : 3. ✔ One-half of the smaller number, 30, is 15. The larger number, 45, is 30 more than 15. ✔ The larger number is 30 more than of the smaller number. 1 2 1 2(2x) 1 30 3x 3x x 30 2x 30 x 15 2x 2(15) 30 3x 3(15) 45 Answer The numbers are 30 and 45. EXERCISES Writing About Mathematics 1. Two numbers in the ratio 2 : 3 can be written as 2x and 3x. Explain why x cannot equal zero. 2. The ratio of the length of a rectangle to its width is 7 : 4. Pete said that the ratio of the length to the perimeter is 7 : 11. Do you agree with Pete? Explain why or why not. Developing Skills 3. Two numbers are in the ratio 4 : 3. Their sum is 70. Find the numbers. 4. Find two numbers whose sum is 160 and that have the ratio 5 : 3. 5. Two numbers have the ratio 7 : 5. Their difference is 12. Find the numbers. 6. Find two numbers whose ratio is 4 : 1 and whose difference is 36. 7. The lengths of the sides of a triangle are in the ratio of 6 : 6 : 5. The perimeter of the triangle is 34 centimeters. Find the length of each side of the triangle. 8. The perimeter of a triangle is 48 centimeters. The lengths of the sides are in the ratio 3 : 4 : 5. Find the length of each side. 9. The perimeter of a rectangle is 360 centimeters. If the ratio of its length to its width is 11 : 4, find the dimensions of the rectangle. 10. The sum of the measures of two angles is 90°. The ratio of the measures of the angles is 2 : 3. Find the measure of each angle. 11. The sum of the measures of two angles is 180°. The ratio of the measures of the angles is 4 : 5. Find the measure of each angle. 216 Ratio and Proportion 12. The ratio of the measures of the three angles of a triangle is 2 : 2 : 5. Find the measures of each angle. 13. In a triangle, two sides have the same length. The ratio of each of these sides to the third side is 5 : 3. If the perimeter of the triangle is 65 inches, find the length of each side of the triangle. 14. Two positive numbers are in the ratio 3 : 7. The larger exceeds the smaller by 12. Find the numbers. 15. Two numbers are in the ratio 3 : 5. If 9 is added to their sum, the result is 41. Find the numbers. Applying Skills 16. A piece of wire 32 centimeters in length is divided into two parts that are in the ratio 3 : 5. Find the length of each part. 17. The ratio of the number of boys in a school to the number of girls is 11 to 10. If there are 525 pupils in the school, how many of them are boys? 18. The ratio of Carl’s money to Donald’s money is 7 : 3. If Carl gives Donald $20, the two then have equal amounts. Find the original amount that each one had. 19. In a basketball free-throw shooting contest, the points made by Sam and Wilbur were in the ratio 7 : 9. Wilbur made 6 more points than Sam. Find the number of points made by each. 20. A chemist wishes to make 121 2 liters of an acid solution by using water and acid in the ratio 3 : 2. How many liters of each should she use? 6-4 PROPORTION A proportion is an equation that states that two ratios are equal. Since the ratio 1 is equal to the ratio 1 : 5 or , we may write the proportion 4 : 20 or 5 4 20 4 : 20 1 : 5 or 4 20 5 1 5 Each of these proportions is read as “4 is to 20 as 1 is to 5.” The general form of a proportion may be written as: a : b c : d or b 5 c a d (b 0, d 0) Each of these proportions is read as “a is to b as c is to d.” There are four terms in this proportion, namely, a, b, c, and d. The outer terms, a and d, are called the extremes of the proportion. The inner terms, b and c, are the means. Proportion 217 means a : b c : d extremes or extreme mean ↓ ↓ a b 5 c d ↑ ↑ mean extreme In the proportion, 4 : 20 1 : 5, the product of the means, 20(1), is equal to the product of the extremes, 4(5). , the product of the means, 15(10), is equal to the In the proportion, 15 5 10 5 30 product of the extremes, 5(30). a b 5 c , we can show that the product of the means is equal In any proportion d a b 5 c to the product of the extremes, ad bc. Since is an equation, we can d multiply both members by bd, the least common denominator of the fractions in the equation. bd bd > b 5 c a d 5 bd ad bc d > B B Therefore, we have shown that the following statement is always true: In a proportion, the product of the means is equal to the product of the extremes. Notice that the end result, ad bc, is the result of multiplying the terms that are cross-wise from each other: ➤ ➤ a b 5 c d ➤ ➤ This is called cross-multiplying, which we have just shown to be valid. If the product of two cross-wise terms is called a cross product, then the fol- lowing is also true: In a proportion, the cross products are equal. If a, b, c, and d are nonzero numbers and b 5 c a d , then ad bc. There are three other proportions using a, b, c and d for which ad = bc. d c 5 b a is a proportion because 6(10) 4(15). For example, we know that 15 10 Therefore, each of the following is also a proportion. 6 15 5 4 10 4 5 15 10 6 10 15 5 4 6 218 Ratio and Proportion EXAMPLE 1 Show that 16 5 5 4 20 is a proportion. Solution Three methods are shown here. The first two use paper and pencil; the last makes use of a calculator. METHOD 1 Reduce each ratio to simplest form. 4 16 5 4 4 4 16 4 4 5 1 4 and 20 5 5 4 5 5 20 4 5 5 1 4 Since each ratio equals proportion. 1 4 , the ratios are equal and 16 5 5 4 20 is a METHOD 2 Show that the cross products are equal. ➤ ➤ ➤ ➤ 16 5 5 4 20 16 5 4 20 80 80 Therefore, 16 5 5 4 20 is a proportion. METHOD 3 Use a calculator. Enter the proportion. If the ratios are equal, then the calculator will display 1. If the ratios are not equal, the calculator will display 0. ENTER: 4 16 2nd TEST ENTER 5 20 ENTER DISPLAY Since the calculator displays 1, the statement is true. The ratios are equal and 16 5 5 4 20 is a proportion. Answer
Any one of the three methods shows that 16 5 5 4 20 is a proportion. Proportion 219 EXAMPLE 2 Solve the proportion 25 : q 5 : 2 for q. Solution Since 25 : q 5 : 2 is a proportion, the product of the means is equal to the product of the extremes. Therefore: Check Reduce each ratio to simplest form. 25 : q 5 : 2 25 : 10 5 : 2 5? 5 : 2 5 : 2 ✔ means 25 : q 5 : 2 extremes 5q 25(2) means extremes 5q 50 q 10 Answer q 10 Note: Example 2 could also have been solved by setting up the proportion q 5 5 25 and then using cross-multiplication to solve for the variable. 2 EXAMPLE 3 Solve for x: x 2 2 5 32 12 x 1 8 Solution Use the fact that the product of the means equals the product of the extremes (the cross products are equal). ➤ ➤ ➤ x 2 2 5 32 12 ➤ x 1 8 32(x – 2) 12(x 8) 32x 64 12x 96 12x 64 12x 64 160 20x 20 5 160 20x 20 x 8 Answer x 8 Check x 2 2 5 32 12 x 1 8 8 – 2 5? 32 12 8 1 8 6 5? 32 12 16 2 2 ✔ 220 Ratio and Proportion EXAMPLE 4 The denominator of a fraction exceeds the numerator by 7. If 3 is subtracted from the numerator of the fraction and the denominator is unchanged, the value of the resulting fraction becomes . Find the original fraction. 1 3 Solution Let x the numerator of original fraction, x 7 the denominator of the original fraction the original fraction. the new fraction. 1 The value of the new fraction is . (x 7) 3(x 3) x 7 3x 9 x 9 x 9 16 2x x 8 x 7 15 Answer The original fraction was 8 15 . Check The original fraction was 8 15 . The new fraction is 8 – 3 15 5 5 15 5 1 3 ✔ EXERCISES Writing About Mathematics 1. Jeremy said that if the means and the extremes of a proportion are interchanged, the result- ing ratios form a proportion. Do you agree with Jeremy? Explain why or why not. 2. Mike said that if the same number is added to each term of a proportion, the resulting ratios form a proportion. Do you agree with Mike? Explain why or why not. Developing Skills In 3–8, state, in each case, whether the given ratios may form a proportion. 3. 3 4 , 30 40 4. 2 3 , 10 5 5. 4 5 , 16 25 6. 2 5 , 5 2 7. 14 18 , 28 36 8. 36 30 , 18 15 In 9–16, find the missing term in each proportion. 2 5 ? 1 9. 8 13. 4 : ? 12 : 60 3 5 5 18 10. ? 14. ? : 9 35 : 63 4 5 6 1 11. ? 15. ? : 60 6 : 10 6 5 ? 4 12. 42 16. 16 : ? 12 : 9 In 17–25, solve each equation and check the solution. Proportion 221 17. 20. 23. 60 5 3 x 20 5 15 5 x x 1 8 3x 1 3 3 5 7x 2 1 5 18. 5 4 5 x 12 x 12 2 x 5 10 30 24. 12 : 15 x : 45 21. 19. 30 4x 5 10 24 16 8 5 21 2 x 25. 5 : x 2 4 : x 22. x In 26–28, in each case solve for x in terms of the other variables. 26. a : b c : x 27. 2r : s x : 3s 28. 2x : m 4r : s Applying Skills In 29–36, use a proportion to solve each problem. 29. The numerator of a fraction is 8 less than the denominator of the fraction. The value of the fraction is . Find the fraction. 3 5 30. The denominator of a fraction exceeds twice the numerator of the fraction by 10. The value of the fraction is . Find the fraction. 5 12 31. The denominator of a fraction is 30 more than the numerator of the fraction. If 10 is added to the numerator of the fraction and the denominator is unchanged, the value of the 3 resulting fraction becomes . Find the original fraction. 5 32. The numerator of a certain fraction is 3 times the denominator. If the numerator is decreased by 1 and the denominator is increased by 2, the value of the resulting fraction is . Find the original fraction. 5 2 33. What number must be added to both the numerator and denominator of the fraction make the resulting fraction equal to ? 3 4 7 19 to 34. The numerator of a fraction exceeds the denominator by 3. If 3 is added to the numerator and 3 is subtracted from the denominator, the resulting fraction is equal to . Find the original fraction. 5 2 35. The numerator of a fraction is 7 less than the denominator. If 3 is added to the numerator and 9 is subtracted from the denominator, the new fraction is equal to . Find the original fraction. 3 2 36. Slim Johnson was usually the best free-throw shooter on his basketball team. Early in the season, however, he had made only 9 of 20 shots. By the end of the season, he had made all the additional shots he had taken, thereby ending with a season record of 3 : 4. How many additional shots had he taken? 222 Ratio and Proportion 6-5 DIRECT VARIATION If the length of a side, s, of a square is 1 inch, then the perimeter, P, of the square is 4 inches. Also, if s is 2 inches, P is 8 inches; if s is 3 inches, P is 12 inches. These pairs of values are shown in the table at the right. s P 1 4 2 8 3 12 From the table, we observe that, as s varies, P also varies. Comparing each value of P to its corresponding value of s, we notice that all three sets of values result in the same ratio when reduced to lowest terms 12 3 If a relationship exists between two variables so that their ratio is a constant, that relationship between the variables is called a direct variation. In every direct variation, we say that one variable varies directly as the other, or that one variable is directly proportional to the other. The constant ratio is called a constant of variation. It is important to indicate the order in which the variables are being com- pared before stating the constant of variation. For example: • In comparing P to s, • In comparing s to P, . The constant of variation is 4. 1 . The constant of variation is . Note that each proportion, the perimeter of a square. and s P 5 1 4 , becomes P 4s, the formula for In a direct variation, the value of each term of the ratio increases when we multiply each variable by a factor greater than 1; the value of each term of the ratio decreases when we divide each variable by a factor greater than 1, as shown below 12 4 3 EXAMPLE 1 If x varies directly as y, and x 1.2 when y 7.2, find the constant of variation by comparing x to y. Solution Answer 1 6 Constant of variation x y 5 1.2 7.2 5 1.2 4 1.2 7.2 4 1.2 5 1 6 EXAMPLE 2 Direct Variation 223 The table gives pairs of values for the variables x and y. a. Show that one variable varies directly as the other. b. Find the constant of variation by comparing y to x. c. Express the relationship between the variables as a formula. d. Find the values missing in the table. x y 1 8 2 16 3 24 10 ? ? 1,600 Solution a. x y 5 2 16 5 2 4 2 x y 5 1 8 Since all the given pairs of values have the same ratio, x and y vary directly. 24 5 3 4 3 16 4 2 5 1 8 24 . Constant of variation 8. 8 1 c. Write a proportion that includes both variables and the constant of variation: x y 5 1 8 y 8x Cross multiply to obtain the formula. d. Substitute the known value in the equation written in b. Solve the equation. When x 10, find y. y 8x y 8(10) y 80 When y 1,600, find x. y 8x 1,600 8x 200 x Answers a. The variables vary directly because the ratio of each pair is the same constant. b. 8 c. y 8x d. When x 10, y 80; when y 1,600, x 200. 224 Ratio and Proportion EXAMPLE 3 There are about 90 calories in 20 grams of a cheese. Reggie ate 70 grams of this cheese. About how many calories were there in the cheese she ate if the number of calories varies directly as the weight of the cheese? Solution Let x number of calories in 70 grams of cheese. number of grams of cheese 5 90 20 70 5 90 x 20 20x 90(70) 20x 6,300 x 315 number of calories Answer There were about 315 calories in 70 grams of the cheese. EXERCISES Writing About Mathematics 1. On a cross-country trip, Natasha drives at an average speed of 65 miles per hour. She says that each day, her driving time and the distance that she travels are directly proportional. Do you agree with Natasha? Explain why or why not. 2. The cost of parking at the Center City Parking Garage is $5.50 for the first hour or part of an hour and $2.75 for each additional half hour or part of a half hour. The maximum cost for 24 hours is $50. Does the cost of parking vary directly as the number of hours? Explain your answer. Developing Skills In 3–11, in each case one value is given for each of two variables that vary directly. Find the constant of variation. 3. x 12, y 3 6. P 12.8, s 3.2 9. s 88, t 110 4. d 120, t 3 7. t 12, n 8 10. A 212, P 200 5. y 2, z 18 8. I 51, t 6 11. r 87, s 58 In 12–17, tell, in each case, whether one variable varies directly as the other. If it does, express the relation between the variables by means of a formula. 12. P s 3 1 6 2 9 3 13. n c 3 6 4 8 5 10 14. x y 4 6 5 8 6 10 15. t d 1 20 2 40 3 60 16. 2 3 x y 6 9 12 4 Direct Variation 225 17. x y 1 1 2 4 3 9 In 18–20, in each case one variable varies directly as the other. Write the formula that relates the variables and find the missing numbers. 18. h A 1 5 2 ? ? 25 19. h S 4 6 8 ? ? 15 20. l w 2 1 8 ? ? 7 In 21–24, state whether the relation between the variables in each equation is a direct variation. In each case, give a reason for your answer. 22. 15T D 21. R + T 80 25. C 7n is a formula for the cost of n articles that sell for $7 each. e 23. 20 i 24. bh 36 a. How do C and n vary? b. How will the cost of nine articles compare with the cost of three articles? c. If n is doubled, what change takes place in C? 26. A 12l is a formula for the area of any rectangle whose width is 12. a. Describe how A and l vary. b. How will the area of a rectangle whose length is 8 inches compare with the area of a rectangle whose length is 4 inches? c. If l is tripled, what change takes place in A? 27. The variable d varies directly as t. If d 520 when t 13, find d when t 9. 28. Y varies directly as x. If Y 35 when x 5, find Y when x 20. 29. A varies directly as h. A 48 when h 4. Find h when A 36. 30. N varies directly as d. N 10 when d 8. Find N when d 12. Applying Skills In 31–48, the quantities vary directly. Solve algebraically. 31. If 3 pounds of apples cost $0.89, what is the cost of 15 pounds of apples at the same rate? 32. If four tickets to a show cost $17.60, what is the cost of seven such tickets? 33. If pound of meat sells for $3.50, how much meat can be bought for $8.75? 1 2 34. Willis scores an average of 7
foul shots in every 10 attempts. At the same rate, how many shots would he score in 200 attempts? 35. There are about 60 calories in 30 grams of canned salmon. About how many calories are there in a 210-gram can? 226 Ratio and Proportion 36. There are 81 calories in a slice of bread that weighs 30 grams. How many calories are there in a loaf of this bread that weighs 600 grams? 37. There are about 17 calories in three medium-size shelled peanuts. Joan ate 30 such peanuts. How many calories were there in the peanuts she ate? 38. A train traveled 90 miles in 11 2 330 miles? hours. At the same rate, how long will the train take to travel 39. The weight of 20 meters of copper wire is 0.9 kilograms. Find the weight of 170 meters of the same wire. 40. A recipe calls for 11 2 cups of sugar for a 3-pound cake. How many cups of sugar should be used for a 5-pound cake? 41. In a certain concrete mixture, the ratio of cement to sand is 1 : 4. How many bags of cement would be used with 100 bags of sand? 42. The owner of a house that is assessed for $12,000 pays $960 in realty taxes. At the same rate, what should be the realty tax on a house assessed for $16,500? 43. The scale on a map is given as 5 centimeters to 3.5 kilometers. How far apart are two towns if the distance between these two towns on the map is 8 centimeters? 44. David received $8.75 in dividends on 25 shares of a stock. How much should Marie receive in dividends on 60 shares of the same stock? 45. A picture 21 8 inches. What will be the width of the enlarged picture? inches long and 31 4 61 2 inches wide is to be enlarged so that its length will become 46. An 11-pound turkey costs $9.79. At this rate, find: a. the cost of a 14.4-pound turkey, rounded to the nearest cent. b. the cost of a 17.5-pound turkey, rounded to the nearest cent. c. the price per pound at which the turkeys are sold. d. the largest size turkey, to the nearest tenth of a pound, that can be bought for $20 or less. 47. If a man can buy p kilograms of candy for d dollars, represent the cost of n kilograms of this candy. 48. If a family consumes q liters of milk in d days, represent the amount of milk consumed in h days. Percent and Percentage Problems 227 6-6 PERCENT AND PERCENTAGE PROBLEMS Base, Rate, and Percent Problems dealing with discounts, commissions, and taxes involve percents. A percent, which is a ratio of a number to 100, is also called a rate. Here, the word rate is treated as a comparison of a quantity to the whole. For example, 8% (read 8 as 8 percent) is the ratio of 8 to 100, or . A percent can be expressed as a frac100 tion or as a decimal: 8% 8 100 0.08 If an item is taxed at a rate of 8%, then a $50 pair of jeans will cost an addi- tional $4 for tax. Here, three quantities are involved. 1. The base, or the sum of money being taxed, is $50. 2. The rate, or the rate of tax, is 8% or 0.08 or 8 100 . 3. The percentage, or the amount of tax being charged, is $4. These three related terms may be written as a proportion or as a formula: As a proportion percentage rate base For example: 50 5 8 4 100 4 50 or 0.08 As a formula base rate percentage For example: 50 4 or 50 0.08 4 8 100 Just as we have seen two ways to look at this problem involving sales tax, we will see more than one approach to every percentage problem. Note that when we calculate using percent, we always use the fraction or decimal form of the percent. Percent of Error When we use a measuring device such as a ruler to obtain a measurement, the accuracy and precision of the measure is dependent on the type of instrument used and the care with which it is used. Error is the absolute value of the difference between a value found experimentally and the true theoretical value. For example, when the length and width of a rectangle are 13 inches and 84 inches, the true length of the diagonal, found by using the Pythagorean Theorem, is 85 inches. A student drew this rectangle and, using a ruler, found the inches. The error of measurement would be measure of the diagonal to be 847 8 228 Ratio and Proportion 85 2 847 1 8 8 value, written as a percent. or inches. The percent of error or is the ratio of the error to the true Percent of error zmeasured value2true valuez true value 3 100% In the example above, the percent of error is EXAMPLE 1 Solution 1 8 85 = 8 4 85 5 1 1 8 3 1 85 5 1 680 < 0.001470588 < 0.15% Note: The relative error is simply the percent of error written as a decimal. Find the amount of tax on a $60 radio when the tax rate is 8%. METHOD 1 Use the proportion: Let t the percentage or amount of tax. amount of tax base percentage base rate. 5 8 100 60 5 8 t 100 100t 480 t 4.80 The tax is $4.80. METHOD 2 Use the formula: base rate percentage. Let t percentage or amount of tax. Change 8% to a fraction. base rate percentage Change 8% to a decimal. base rate percentage 60 8% t 60 3 8 t 100 480 t 100 4.8 t 60 8% t 60 0.08 t 4.8 t Whether the fraction or the decimal form of 8% is used, the tax is $4.80. Answer The tax is $4.80. Percent and Percentage Problems 229 EXAMPLE 2 During a sale, a store offers a discount of 25% off any purchase. What is the regular price of a dress that a customer purchased for $73.50? Solution The rate of the discount is 25%. Therefore the customer paid (100 25)% or 75% of the regular price. The percentage is given as $73.50, and the base is not known. Let n the regular price, or base. METHOD 1 Use the proportion. percentage base 5 rate n 5 75 73.50 100 75n 7,350 n 98 Check If 25% of 98 is subtracted from 98 does the difference equal 73.50? 0.25 98 24.50 98 24.50 73.50 ✔ METHOD 2 Use the formula. base rate percentage n 75% 73.50 Use fractions n 3 75 n 3 75 73.50 100 100 3 100 75 5 73.50 3 100 75 n 98 Use decimals n 0.75 73.50 0.75 5 73.50 0.75n 0.75 n 98 Answer The regular price of the dress was $98. Alternative Solution Let n the regular price of the dress. Then, 0.25n the discount. The price of the dress minus the discount is the amount the customer paid. \|___________________________\| \|____________________\| ↓ ↓ 73.50 n ↓ ↓ \|___________\| ↓ 0.25n n 0.25n 73.50 1.00n 0.25n 73.50 0.75n 73.50 0.75 5 73.50 0.75n 0.75 n 98 The check is the same as that shown for Method 1. Answer The regular price of the dress was $98. 230 Ratio and Proportion Percent of Increase or Decrease A percent of increase or decrease gives the ratio of the amount of increase or decrease to the original amount. A sales tax is a percent of increase on the cost of a purchase. A discount is a percent of decrease on the regular price of a purchase. To find the percent of increase or decrease, find the difference between the original amount and the new amount. The original amount is the base, the absolute value of the difference is the percentage, and the percent of increase or decrease is the rate. Percent of increase or decrease zoriginal amount 2 new amountz original amount 100% EXAMPLE 3 Last year Marisa’s rent was $600 per month. This year, her rent increased to $630 per month. What was the percent of increase in her rent? Solution The original rent was $600. The new rent was $630. The amount of increase was \|$600 – $630\| $30. Percent of increase 30 600 0.05 1 20 Change 0.05 to a percent: 0.05 5% Answer The percent of increase is 5%. EXAMPLE 4 Solution A store reduced the price of a television from $840 to $504. What was the percent of decrease in the price of the television? Original price $840 New price $504 Amount of decrease \|$840 – $504\| $336 Percent of decrease 0.4 336 840 Change 0.4 to a percent: 0.4 40% Answer The percent of decrease was 40%. Percent and Percentage Problems 231 EXERCISES Writing About Mathematics 1. Callie said that two decimal places can be used in place of the percent sign. Therefore, 3.6% can be written as 0.36. Do you agree with Callie. Explain why or why not. 2. If Ms. Edwards salary was increased by 4%, her current salary is what percent of her salary before the increase? Explain your answer. Developing Skills In 3–11, find each indicated percentage. 3. 2% of 36 6. 2.5% of 400 9. 121 2% of 128 4. 6% of 150 7. 60% of 56 10. 331 3% of 72 In 12–19, find each number or base. 5. 15% of 48 8. 100% of 7.5 11. 150% of 18 12. 20 is 10% of what number? 13. 64 is 80% of what number? 14. 8% of what number is 16? 16. 125% of what number is 45? 18. 662 3% of what number is 54? 15. 72 is 100% of what number? 17. 371 2% of what number is 60? 19. 3% of what number is 1.86? In 20–27, find each percent. 20. 6 is what percent of 12? 22. What percent of 10 is 6? 24. 5 is what percent of 15? 26. 18 is what percent of 12? Applying Skills 21. 9 is what percent of 30? 23. What percent of 35 is 28? 25. 22 is what percent of 22? 27. 2 is what percent of 400? 28. A newspaper has 80 pages. If 20 of the 80 pages are devoted to advertising, what percent of the newspaper consists of advertising? 29. A test was passed by 90% of a class. If 27 students passed the test, how many students are in the class? 232 Ratio and Proportion 30. Marie bought a dress that was marked $24. The sales tax is 8%. a. Find the sales tax. b. Find the total amount Marie had to pay. 31. There were 120 planes on an airfield. If 75% of the planes took off for a flight, how many planes took off? 32. One year, the Ace Manufacturing Company made a profit of $480,000. This represented 6% of the volume of business for the year. What was the volume of business for the year? 33. The price of a new motorcycle that Mr. Klein bought was $5,430. Mr. Klein made a down payment of 15% of the price of the motorcycle and arranged to pay the rest in installments. How much was his down payment? 34. How much silver is in 75 kilograms of an alloy that is 8% silver? 35. In a factory, 54,650 parts were made. When they were tested, 4% were found to be defective. How many parts were good? 36. A baseball team won 9 games, which was 60% of the total number of games the team played. How many games did the team play? 37. The regular price of a sweater is $40. The sale price o
f the sweater is $34. What is the per- cent of decrease in the price? 38. A businessman is required to collect an 8% sales tax. One day, he collected $280 in taxes. Find the total amount of sales he made that day. 39. A merchant sold a stereo speaker for $150, which was 25% above its cost to her. Find the cost of the stereo speaker to the merchant. 40. Bill bought a wooden chess set at a sale. The original price was $120; the sale price was $90. What was the percent of decrease in the price? 41. If the sales tax on $150 is $7.50, what is the percent of the sales tax? 42. Mr. Taylor took a 2% discount on a bill. He paid the balance with a check for $76.44. What was the original amount of the bill? 43. Mrs. Sims bought some stock for $2,250 and sold the stock for $2,520. What was the percent increase in the value of the stock? 44. When Sharon sold a vacuum cleaner for $220, she received a commission of $17.60. What was the rate of commission? 45. On the first day of a sale, a camera was reduced by $8. This represented 10% of the original price. On the last day of the sale, the camera was sold for 75% of the original price. What was the final selling price of the camera? Percent and Percentage Problems 233 46. The regular ticketed prices of four items at Grumbell’s Clothier are as follows: coat, $139.99; blouse, $43.99; shoes, $89.99; jeans, $32.99. a. These four items were placed on sale at 20% off the regular price. Find, correct to the nearest cent, the sale price of each of these four items. b. Describe two different ways to find the sale prices. 47. At Relli’s Natural Goods, all items are being sold today at 30% off their regular prices. However, customers must still pay an 8% tax on these items. Edie, a good-natured owner, allows each customer to choose one of two plans at this sale: Plan 1. Deduct 30% of the cost of all items, then add 8% tax to the bill. Plan 2. Add 8% tax to the cost of all items, then deduct 30% of this total. Which plan if either, saves the customer more money? Explain why. 48. In early March, Phil Kalb bought shares of stocks in two different companies. Stock ABC rose 10% in value in March, then decreased 10% in April. Stock XYZ fell 10% in value in March, then rose 10% in April. What percent of its original price is each of these stocks now worth? 49. A dairy sells milk in gallon containers. The containers are filled by machine and the amount of milk may vary slightly. A quality control employee selects a container at random and makes an accurate measure of the amount of milk as 16.25 cups. Find the percent of error to the nearest tenth of a percent. 50. A carpenter measures the length of a board as 50.5 centimeters. The exact measure of the length was 50.1 centimeters. Find the percent of error in the carpenter’s measure to the nearest tenth of a percent. 51. A 5-pound weight is placed on a gymnasium scale. The scale dial displayed 51 2 pounds. If the scale is consistently off by the same percentage, how much does an athlete weigh, to the nearest tenth of a pound, if his weight displayed on this scale is 144 pounds? 52. Isaiah answered 80% of the questions correctly on the math midterm, and 90% of the questions correctly on the math final. Can you conclude that he answered 85% of all the questions correctly (the average of 80% and 90%)? Justify your answer or give a counterexample. 53. In January, Amy bought shares of stocks in two different companies. By the end of the year, shares of the first company had gone up by 12% while shares of the second company had gone up by 8%. Did Amy gain a total of 12% 8% 20% in her investments? Explain why or why not. 234 Ratio and Proportion 6-7 CHANGING UNITS OF MEASURE The weight and dimensions of a physical object are expressed in terms of units of measure. In applications, it is often necessary to change from one unit of measure to another by a process called dimensional analysis. To do this, we multiply by a fraction whose numerator and denominator are equal measures in two different units so that, in effect, we are multiplying by the identity element, 1. For example, since 100 centimeters and 1 meter are equal measures: 1 m 5 100 cm 100 cm 100 cm 5 1 1 m 100 cm 5 1 m 1 m 5 1 To change 4.25 meters to centimeters, . multiply by 100 cm 1 m 4.25 m 4.25 m 3 100 cm 1 m 425 cm To change 75 centimeters to meters, multiply by 1 m . 100 cm 75 cm 3 1 m 75 cm 100 cm 75 m 100 0.75 m Note that in each case, the fraction was chosen so that the given unit of measure occurred in the denominator and could be “cancelled” leaving just the unit of measure that we wanted in the result. Sometimes it is necessary to use more than one fraction to change to the required unit. For example, if we want to change 3.26 feet to centimeters and know that 1 foot 12 inches and that 1 inch 2.54 centimeters, it will be necessary to first use the fraction to change feet to inches. 12 in. 1 ft 3.26 ft 3.26 ft 3 12 in. 1 ft 39.12 1 in. 39.12 in. Then use the fraction to change inches to centimeters 2.54 cm 1 in. 39.12 in. 3 2.54 cm 1 in. 39.12 in. 99.3648 1 cm 99.3648 cm This answer, rounded to the nearest tenth, can be expressed as 99.4 centimeters. EXAMPLE 1 Solution If there are 5,280 feet in a mile, find, to the nearest hundredth, the number of miles in 1,200 feet. How to Proceed (1) Write a fraction equal to 1 with the required unit in the numerator and the given unit in the denominator: 1 mi 5,280 ft Changing Units of Measure 235 1,200 ft 1,200 ft 3 1 mi 5,280 ft mi 1,200 5,280 0.227 mi 1,200 ft 0.23 mi (2) Multiply the given measure by the fraction written in step 1: (3) Round the answer to the nearest hundredth: Answer 0.23 mi EXAMPLE 2 In France, apples cost 4.25 euros per kilogram. In the United States, apples cost $1.29 per pound. If the currency exchange rate is 0.95 euros for 1 dollar, in which country are apples more expensive? Solution Recall that “per” indicates division, that is, 4.25 euros per kilogram can be and $1.29 per pound as written as . 1.29 dollars 1 pound 4.25 euros 1 kilogram (1) Change euros in euros per kilogram to dollars. Use 1 dollar 0.95 euros , a fraction equal to 1. 4.25 euros 1 kilogram 3 1 dollar 0.95 euros 5 4.25 dollars 0.95 kilograms (2) Now change kilograms in dollars per kilogram to pounds. One pound equals 0.454 kilograms. Since kilograms is in the denominator, use the fraction with kilograms in the numerator. 4.25 dollars 0.95 kilograms 3 0.454 kilogram 1 pound 5 4.25(0.454) dollars 0.95(1) pounds (3) Use a calculator for the computation. (4) The number in the display, 2.031052632, is the cost of apples in France in dollars per pound. Round the number in the display to the nearest cent: $2.03 (5) Compare the cost of apples in France ($2.03 per pound) to the cost of apples in the United States ($1.29 per pound). Answer Apples are more expensive in France. 236 Ratio and Proportion EXAMPLE 3 Change 60 miles per hour to feet per second. Solution Use dimensional analysis to change the unit of measure in the given rate to the required unit of measure. (1) Write 60 miles per hour as a fraction: (2) Change miles to feet. Multiply by a ratio with miles in the denominator to cancel miles in the numerator: (3) Change hours to minutes. Multiply by a ratio with hours in the numerator to cancel hours in the denominator: (4) Change minutes to seconds. Multiply by a ratio with minutes in the numerator to cancel minutes in the denominator. 60 mi 1 hr 5 60 mi 1 hr 3 5,280 ft 1 mi 5 60(5,280) ft 1 hr 3 1 hr 60 min 5 60(5,280) ft 60 min 3 1 min 60 sec Alternative Solution (5) Compute and simplify: 60(60) sec 5 88 ft Write the ratios in one expression and compute on a calculator. 1 sec 5 60(5,280) ft 60 mi 1 hr 3 5,280 ft 1 mi 3 1 hr 60 min 3 1 min 60 sec 5 60(5,280) ft 60(60) sec Answer 60 miles per hour 88 feet per second. EXERCISES Writing About Mathematics 1. Sid cannot remember how many yards there are in a mile but knows that there are 5,280 feet in a mile and 3 feet in a yard. Explain how Sid can find the number of yards in a mile. 3 8 2. A recipe uses of a cup of butter. Abigail wants to use tablespoons to measure the butter and knows that 4 tablespoons equals cup. Explain how Abigail can find the number of tablespoons of butter needed for her recipe. 1 4 Changing Units of Measure 237 Developing Skills In 3–16: a. write, in each case, the fraction that can be used to change the given units of measure, b. find the indicated unit of measure. 3. Change 27 inches to feet. 4. Change 175 centimeters to meters. 5. Change 40 ounces to pounds. 6. Change 7,920 feet to miles. 7. Change 850 millimeters to centimeters. 8. Change 12 pints to gallons. 9. Change 10.5 yards to inches 4 11. Change yard to feet. 3 13. Change 1.2 pounds to ounces. 15. Change 44 centimeters to millimeters. Applying Skills 10. Change 31 2 feet to inches. 12. Change 1.5 meters to centimeters. 14. Change 2.5 miles to feet. 21 2 16. Change gallons to quarts. 17. Miranda needs boards 0.8 meters long for a building project. The boards available at the local lumberyard are 2 feet, 3 feet, and 4 feet long. a. Express the length, to the nearest hundredth of a foot, of the boards that Miranda needs to buy. b. Which size board should Miranda buy? Explain your answer. 18. Carlos needs 24 inches of fabric for a pillow that he is making. The fabric store has a piece of material of a yard long that is already cut that he can buy for $5.50. If he has the exact size piece he needs cut from a bolt of fabric, it will cost $8.98 a yard. 3 4 a. Is the piece of material that is already cut large enough for his pillow? b. What would be the cost of having exactly 24 inches of fabric cut? c. Which is the better buy for Carlos? 19. A highway sign in Canada gives the speed limit as 100 kilometers per hour. Tracy is driving at 62 miles per hour. One mile is approximately equal to 1.6 kilometers. a. Is Tracy exceeding the speed limit? b. What is the difference between the speed limit and Tracy’s speed in miles pe
r hour? 20. Taylor has a painting for which she paid 1 million yen when she was traveling in Japan. At that time, the exchange rate was 1 dollar for 126 yen. A friend has offered her $2,000 for the painting. a. Is the price offered larger or smaller than the purchase price? b. If she sells the painting, what will be her profit or loss, in dollars? c. Express the profit or loss as a percent of increase or decrease in the price of the painting. 238 Ratio and Proportion CHAPTER SUMMARY A ratio, which is a comparison of two numbers by division, is the quotient obtained when the first number is divided by a second, nonzero number. Quantities in a ratio are expressed in the same unit of measure before the quotient is found. Ratio of a to b: a b or a : b A rate is a comparison of two quantities that may have different units of measure, such as a rate of speed in miles per hour. A rate that has a denominator of 1 is called a unit rate. A proportion is an equation stating that two ratios are equal. Standard ways to write a proportion are shown below. In a proportion a : b c : d, the outer terms are called the extremes, and the inner terms are the means. Proportion: means a : b c : d extremes or extreme mean ↓ ↓ a b 5 c d ↑ ↑ mean extreme In a proportion, the product of the means is equal to the product of the extremes, or alternatively, the cross products are equal. This process is also called cross-multiplication. A direct variation is a relation between two variables such that their ratio is always the same value, called the constant of variation. For example, the diameter of a circle is always twice the radius, so 2 shows a direct variation between d and r with a constant of variation 2. d r A percent (%), which is a ratio of a number to 100, is also called a rate. Here, the word rate is treated as a comparison of a quantity to a whole. In basic formulas, such as those used with discounts and taxes, the base and percentage are numbers, and the rate is a percent. percentage base rate or base rate percentage The percent of error is the ratio of the absolute value of the difference between a measured value and a true value to the true value, expressed as a percent. Percent of error zmeasured value 2 true valuez true value 100% The relative error is the percent of error expressed as a decimal. Review Exercises 239 The percent of increase or decrease is the ratio of the absolute value of the difference between the original value and the new value to the original value. Percent of increase or decrease zoriginal value 2 new valuez original value 100% VOCABULARY 6-1 Ratio • Equivalent ratios • Simplest form • Continued ratio 6-2 Rate • Lowest terms • Unit rate • Unit price 6-4 Proportion • Extremes • Means • Cross-multiplying • Cross product 6-5 Direct variation • Directly proportional • Constant of variation 6-6 Percent • Base • Rate • Percentage • Error • Percent of error • Relative error • Percent increase • Percent decrease 6-7 Dimensional analysis REVIEW EXERCISES 1. Can an 8 inch by 12 inch photograph be reduced to a 3 inch by 5 inch pho- tograph? Explain why or why not? 2. Karen has a coupon for an additional 20% off the sale price of any dress. She wants to buy a dress that is on sale for 15% off of the original price. Will the original price of the dress be reduced by 35%? Explain why or why not. In 3–6, express each ratio in simplest form. 3. 30 : 35 4. 8w to 12w 6. 75 millimeters : 15 centimeters 5. 3 8 to 5 8 In 7–9, in each case solve for x and check. 7. 8 2x 5 12 9 8. x x 1 5 5 1 2 9. x 5 6 4 x 1 3 10. The ratio of two numbers is 1 : 4, and the sum of these numbers is 40. Find the numbers. 240 Ratio and Proportion In 11–13, in each case, select the numeral preceding the choice that makes the statement true. 11. In a class of 9 boys and 12 girls, the ratio of the number of girls to the number of students in the class is (1) 3 : 4 (2) 4 : 3 (3) 4 : 7 (4) 7 : 4 12. The perimeter of a triangle is 45 centimeters, and the lengths of its sides are in the ratio 2 : 3 : 4. The length of the longest side is (1) 5 cm (3) 20 cm (2) 10 cm 13. If a : x b : c, then x equals (4) 30 cm (1) ac b (2) bc a (3) ac – b (4) bc – a 14. Seven percent of what number is 21? 15. What percent of 36 is 45? 16. The sales tax collected on each sale varies directly as the amount of the sale. What is the constant of variation if a sales tax of $0.63 is collected on a sale of $9.00? 17. If 10 paper clips weigh 11 grams, what is the weight in grams of 150 paper clips? 18. Thelma can type 150 words in 3 minutes. At this rate, how many words can she type in 10 minutes? 19. What is the ratio of six nickels to four dimes? 20. On a stormy February day, 28% of the students enrolled at Southside High School were absent. How many students are enrolled at Southside High School if 476 students were absent? 21. After a 5-inch-by-7-inch photograph is enlarged, the shorter side of the enlargement measures 15 inches. Find the length in inches of its longer side. 22. A student who is 5 feet tall casts an 8-foot shadow. At the same time, a tree casts a 40-foot shadow. How many feet tall is the tree? 23. If four carpenters can build four tables in 4 days, how long will it take one carpenter to build one table? 24. How many girls would have to leave a room in which there are 99 girls and 1 boy in order that 98% of the remaining persons would be girls? 25. On an Australian highway, the speed limit was 110 kilometers per hour. A motorist was going 70 miles per hour. (Use 1.6 kilometers 1 mile) a. Should the motorist be stopped for speeding? b. How far over or under the speed limit was the motorist traveling? Review Exercises 241 26. The speed of light is 3.00 105 kilometers per second. Find the speed of light in miles per hour. Use 1.61 kilometers 1 mile. Write your answer in scientific notation with three significant digits. 27. Which offer, described in the chapter opener on page 207, is the better buy? Does the answer depend on the price of a package of frozen vegetables? Explain. 28. A proposal was made in the state senate to raise the minimum wage from $6.75 to $7.15 an hour. What is the proposed percent of increase to the nearest tenth of a percent? 29. In a chemistry lab, a student measured 1.0 cubic centimeters of acid to use in an experiment. The actual amount of acid that the student used was 0.95 cubic centimeters. What was the percent of error in the student’s measurement? Give your answer to the nearest tenth of a percent. Exploration a. Mark has two saving accounts at two different banks. In the first bank with a yearly interest rate of 2%, he invests $585. In the second bank with a yearly interest rate of 1.5%, he invests $360. Mark claims that at the end of one year, he will make a total of 2% 1.5% 3.5% on his investments. Is Mark correct? b. Two different clothing items, each costing $30, were on sale for 10% off the ticketed price. The manager of the store claims that if you buy both items, you will save a total of 20%. Is the manager correct? c. In a class of 505 graduating seniors, 59% were involved in some kind of after-school club and 41% played in a sport. The principal of the school claims that 59% 41% 100% of the graduating seniors were involved in some kind of after-school activity. Is the principal correct? d. A certain movie is shown in two versions, the original and the director’s cut. However, movie theatres can play only one of the versions. A journalist for XYZ News, reports that since 30% of theatres are showing the director’s cut and 60% are showing the original, the movie is playing in 90% of all movie theatres. Is the reporter correct? e. Based on your answers for parts a through d, write a rule stating when it makes sense to add percents. 242 Ratio and Proportion CUMULATIVE REVIEW CHAPTERS 1–6 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. When –4ab is subtracted from ab, the difference is (1) –3ab (2) –4 (3) 5ab (4) –5ab 2. If n – 4 represents an odd integer, the next larger odd integer is (1) n – 2 (2) n – 3 (3) n – 5 (4) n – 6 3. The expression 2.3 10–3 is equal to (1) 230 (2) 2,300 (3) 0.0023 (4) 0.023 4. The product 2x3y(–3xy4) is equal to (1) 6x3y4 (2) 6x4y5 (3) 6x4y5 (4) 6x3y4 5. What is the multiplicative inverse of ? 3 2 (1) 23 2 (2) 2 3 6. If 0.2x 4 x 0.6, then x equals (1) 46 (2) 460 7. The product 34 33 (3) 22 3 (4) 1 (3) 5.75 (4) 0.575 (1) 37 (2) 37 (3) 97 (4) 97 8. The diameter of a circle whose area is 144p square centimeters is (1) 24p cm (2) 6 cm (3) 12 cm (4) 24 cm 9. Jeannine paid $88 for a jacket that was on sale for 20% off the original price. The original price of the jacket was (1) $105.60 (2) $110.00 (3) $440.00 10. When a 2 and b 5, a2 ab equals (1) 14 (2) 14 (3) 6 Part II (4) $70.40 (4) 6 Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. Find two integers, x and x 1, whose squares differ by 25. 12. Solve and check: 4(2x 1) 5x 5 Cumulative Review 243 Part III Answer all questions in this part. Each correct answer will receive 3 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. Rectangle ABCD is separated into three squares as shown in the diagram at the right. The length of DA is 20 centimeters. a. Find the measure of AB . b. Find the ratio of the area of square AEFD to the area of rectangle ABCD. D 20 cm A F G E C H B 14. Sam drove a distance of 410 miles in 7 hours. For the first part of the trip his average speed was 40 miles per hour and for the remainder of the trip his average speed was 60 mile
s per hour. How long did he travel at each speed? Part IV Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. Three friends started a part-time business. They plan, each month, to share the profits in the ratio of the number of hours that each worked. During the first month, Rita worked 18 hours, Fred worked 30 hours, and Glen worked 12 hours. a. Express, in lowest terms, the ratio of the times that they worked during the first month. b. Find the amount each should receive if the profit for this first month was $540. c. Find the amount each should receive if the profit for this first month was $1,270. 244 Ratio and Proportion 16. A skating rink is in the form of a rectangle with a semicircle at each end as shown in the diagram. The rectangle is 150 feet long and 64 feet wide. Scott skates around the rink 2.5 feet from the edge. 64 ft 150 ft a. Scott skates once around the rink. Find, to the nearest ten feet, the dis- tance that he skated. b. Scott wants to skate at least 5 miles. What is the smallest number of complete trips around the rink that he must make? GEOMETRIC FIGURES, AREAS, AND VOLUMES A carpenter is building a deck on the back of a house. As he works, he follows a plan that he made in the form of a drawing or blueprint. His blueprint is a model of the deck that he is building. He begins by driving stakes into the ground to locate corners of the deck. Between each pair of stakes, he stretches a cord to indicate the edges of the deck. On the blueprint, the stakes are shown as points and the cords as segments of straight lines as shown in the sketch. • • • • • • At each corner, the edges of the deck meet at an angle that can be classified according to its size. Geometry combines points, lines, and planes to model the world in which we live. An understanding of geometry enables us to understand relationships involving the sizes of physical objects and the magnitude and direction of the forces that interact in daily life. In this chapter, you will review some of the information that you already know to describe angles and apply this information to learn more about geometry. CHAPTER 7 CHAPTER TABLE OF CONTENTS 7-1 Points, Lines, and Planes 7-2 Pairs of Angles 7-3 Angles and Parallel Lines 7-4 Triangles 7-5 Quadrilaterals 7-6 Areas of Irregular Polygons 7-7 Surface Areas of Solids 7-8 Volumes of Solids Chapter Summary Vocabulary Review Exercises Cumulative Review 245 246 Geometric Figures, Areas, and Volumes 7-1 POINTS, LINES, AND PLANES Undefined Terms We ordinarily define a word by using a simpler term. The simpler term can be defined by using one or more still simpler terms. But this process cannot go on endlessly; there comes a time when the definition must use a term whose meaning is assumed to be clear to all people. Because the meaning is accepted without definition, such a term is called an undefined term. In geometry, we use such ideas as point, line, and plane. Since we cannot give satisfactory definitions of these words by using simpler defined words, we will consider them to be undefined terms. Although point, line, and plane are undefined words, we must have a clear understanding of what they mean. Knowing the properties and characteristics they possess helps us to achieve this understanding. A point indicates a place or position. It has no length, width, or thickness. A point is usually indicated by a dot and named with a capital letter. For example, point A is shown on the left. A line is a set of points. The set of points may form a curved line, a broken line, or a straight line. A straight line is a line that is suggested by a stretched string but that extends without end in both directions. A Curved line Broken line A B l Straight line Unless otherwise stated, in this discussion the term line will mean straight line. A line is named by any two points on the line. For example, the straight line g BA shown above is line AB or line BA, usually written as . A line can also be named by one lowercase letter, for example, line l shown above. The arrows remind us that the line continues beyond what is drawn in the diagram. g AB or P A plane is a set of points suggested by a flat surface. A plane extends endlessly in all directions. A plane may be named by a single letter, as plane P shown on the right. A plane can also be named by three points of the plane, as plane ABC in the diagram on the right. A C B Facts About Straight Lines A statement that is accepted as true without proof is called an axiom or a postulate. If we examine the three accompanying figures pictured on the next page, we see that it is reasonable to accept the following three statements as postulates: S R Points, Lines, and Planes 247 1. In a plane, an infinite number of straight lines can be drawn through a given point. 2. One and only one straight line can be drawn that contains two given points. (Two points determine a straight line.) 3. In a plane, two different nonparallel straight lines will intersect at only one point. Line Segments The undefined terms, point, line, and plane, are used to define other geometric terms. A line segment or segment is a part of a line consisting of two points called endpoints and all points on the line between these endpoints. At the left is pictured a line segment whose endpoints are points R and S. We use these endpoints to name this segment as segment RS, which may be written as SR . RS or Recall that the measure of a line segment or the length of a line segment is the distance between its endpoints. We use a number line to associate a number with each endpoint. Since the coordinate of A is 0 and the coordinate of B is 5, the length of AB is 5 0 or AB 5 Note: The segment is written as endpoints. The length of the segment is written as AB, with no bar over the letters that name the endpoints. , with a bar over the letters that name the AB Half-Lines and Rays When we choose any point A on a line, the two sets of points that lie on opposite sides of A are called half-lines. Note that point A is not part of the half-line. In the diagram below, point A separates into two half-lines. g CD CB BD The two points B and D belong to the same half. Points B and C, howline since A is not a point of ever, do not belong to the same half-line since A is a point of . All points in the same half-line are said to be on the same side of A. We often talk about rays of sunlight, that is, the sun and the path that the sunlight travels to the earth. The ray of sunlight can be thought of as a point and a half-line. D B A C In geometry, a ray is a part of a line that consists of a point on the line, called an endpoint, and all the points on one side of the endpoint. To name a ray we use two capital letters and an arrow with one arrowhead. The first letter must be 248 Geometric Figures, Areas, and Volumes the letter that names the endpoint. The second letter is the name of any other point on the ray. The figure on the right shows ray AB, which is writ. This ray could also be called ray AC, written h AB ten as h AC . as Two rays are called opposite rays if they are rays of the same line that have a common endpoint but no other points in common. In the diagram on the left, h PQ are opposite rays. h PR and A C B Angles An angle is a set of points that is the union of two rays having the same endpoint. The common endpoint of the two rays is the vertex of the angle. The two rays forming the angle are also called the sides of the angle. In the diagram on the left, we can think of TOS as having been formed by h OT . The in a counterclockwise direction about O to the position rotating , that have the common endpoint O, is TOS. union of the two rays, Note that when three letters are used to name an angle, the letter that names the vertex is always in the middle. Since are the only rays in the diagram that have the common endpoint O, the angle could also have been called O. h OT h OT h OS h OS h OS and and Q P R S O T Measuring Angles To measure an angle means to determine the number of units of measure it contains. A common standard unit of measure of an angle is the degree; 1 degree is of the sum of all of the distinct angles about a point. written as 1°. A degree is In other words, if we think of an angle as having been formed by rotating a ray around its endpoint, then rotating a ray consecutively 360 times in 1-degree increments will result in one complete rotation. 1 360 A C E D Types of Angles Angles are classified according to their measures. g CD A right angle is an angle whose measure is 90°. In the diagram on the left, g AB intersect at E so that the four angles formed are right angles. The and measure of each angle is 90° and the sum of the angles about point E is 360°. We can say that B mAEC mCEB mBED mDEA 90. Note that the symbol mAEC is read “the measure of angle AEC.” In this book, the angle measure will always be given in degrees and the degree symbol will be omitted when using the symbol “m” to designate measure. I h H G Acute angle Points, Lines, and Planes 249 An acute angle is an angle whose measure is greater than 0° and less than 90°, that is, the measure of an acute angle is between 0° and 90°. Angle GHI, which can also be called h, is an acute angle (0 mh 90). An obtuse angle is an angle whose measure is greater than 90° but less than 180°; that is, its measure is between 90° and 180°. Angle LKN is an obtuse angle where 90 mLKN 180. A straight angle is an angle whose measure is 180°. Angle RST is a straight angle where mRST 180. A straight angle is the union of two opposite rays and forms a straight line. Here are three important facts about angles: N k K L Obtuse angle 180° T s S Straight angle R 1. The measure of an angle depends only on the amount of r
otation, not on the pictured lengths of the rays forming the angle. 2. Since every right angle measures 90°, all right angles are equal in measure. 3. Since every straight angle measures 180°, all straight angles are equal in measure. Perpendicularity Two lines are perpendicular if and only if the two lines intersect to form right angles. The symbol for perpendicularity is ⊥. g PR is perpendicular to g AB P bolized as In the diagram, g PR , sym— is used in a diagram to indicate that a right angle exists where the . The symbol g ' AB A B R perpendicular lines intersect. Segments of perpendicular lines that contain the point of intersection of the lines are also perpendicular. In the diagram on the left, g ST ' CD . S C T D EXERCISES Writing About Mathematics 1. Explain how the symbols AB, AB , and g AB differ in meaning. 2. Explain the difference between a half-line and a ray. 250 Geometric Figures, Areas, and Volumes Developing Skills In 3–8, tell whether each angle appears to be acute, right obtuse, or straight. 3. 6. 4. 7. 5. 8. 9. For the figure on the right: a. Name x by using three capital letters. b. Give the shorter name for COB. c. Name one acute angle. d. Name one obtuse angle. C B y x O A Applying Skills In 10–13, find the number of degrees in the angle formed by the hands of a clock at each given time. 10. 1 P.M. 11. 4 P.M. 12. 6 P.M. 13. 5:30 P.M. 14. At what time do the hands of the clock form an angle of 0°? 15. At what times do the hands of a clock form a right angle? 7-2 PAIRS OF ANGLES Adjacent Angles D B C A Adjacent angles are two angles in the same plane that have a common vertex and a common side but do not have any interior points in common. In the figure on the left, ABC and CBD are adjacent angles. Complementary Angles Two angles are complementary angles if and only if the sum of their measures is 90°. Each angle is the complement of the other. Pairs of Angles 251 In the figures shown below, because mCAB mFDE 25 65 90, CAB and FDE are complementary angles. Also, because mHGI mIGJ 53 37 90, HGI and IGJ are complementary angles. C 25° B A E F 65° D J I 37° 53° G H If the measure of an angle is 50°, the measure of its complement is (90 50)°, or 40°. In general, If the measure of an angle is x°, the measure of its complement is (90 x)°. Supplementary Angles Two angles are supplementary angles if and only if the sum of their measures is 180°. Each angle is the supplement of the other. As shown in the figures below, because mLKM mONP 50 130 180, LKM and ONP are supplementary angles. Also, RQS and SQT are supplementary angles because mRQS mSQT 115 65 180. M L P S 50° K 130° 65° 115° N O T Q R If the measure of an angle is 70°, the measure of its supplement is (180 70)°, or 110°. In general, If the measure of an angle is x°, the measure of its supplement is (180 x)°. 252 Geometric Figures, Areas, and Volumes Linear Pair DA . Draw and any Through two points, D and A, draw a line. Choose any point B on . The adjacent angles formed, DBC and CBA, point C not on are called a linear pair. A linear pair of angles are adjacent angles that are supplementary. The two sides that they do not share in common are opposite rays. The term linear tells us that a line is used to form this pair. Since the angles are supplementary, if mDBC x, then mCBA (180 x). h BC DA Vertical Angles g AB g CD intersect at E, then x and y share a If two straight lines such as common vertex at E but do not share a common side. Angles x and y are a pair of vertical angles. Vertical angles are two nonadjacent angles formed by two intersecting lines. In the diagram on the left, angles a and b are another pair of vertical angles. and If two lines intersect, four angles are formed that have no common interior intersect at E. There are four linear pairs of g CD g AB and point. In the diagram, angles: AED and DEB DEB and BEC BEC and CEA CEA and AED The angles of each linear pair are supplementary. • If mAED 130, then mDEB 180 130 50. • If mDEB 50, then mBEC 180 50 130. Therefore, mAED mBEC. A C 130° E D 50° 130° B DEFINITION When two angles have equal measures, they are congruent. We use the symbol to represent the phrase “is congruent to.” Since mAED mBEC, we can write AED BEC, read as “angle AED is congruent to angle BEC.” There are different correct ways to indicate angles with equal measures: 1. The angle measures are equal: mBEC mAED 2. The angles are congruent: BEC AED It would not be correct to say that the angles are equal, or that the angle measures are congruent. Notice that we have just shown that the two vertical angles BEC and AED are congruent. If we were to draw and measure additional pairs of vertical angles, we would find in each case that the vertical angles would be equal Pairs of Angles 253 in measure. No matter how many examples of a given situation we consider, however, we cannot assume that a conclusion that we draw from these examples will always be true. We must prove the conclusion. Statements that we prove are called theorems. We will use algebraic expressions and properties to write an informal proof of the following statement. If two lines intersect, the vertical angles formed are equal in measure; that is, they are congruent. (1) If g AB and g CD intersect at E, then AEB is a straight angle whose measure is 180°. Therefore, mAEC mCEB 180. (2) Let mAEC x. Then mCEB 180 x. (3) Likewise, CED is a straight angle whose measure is 180°. Therefore, mCEB mBED 180. (4) Since mCEB 180 x, then mBED 180 (180 x) x. C x 180 – x A E x B D (5) Since both mAEC x and mBED x, then mAEC mBED; that is, AEC BED. EXAMPLE 1 The measure of the complement of an angle is 4 times the measure of the angle. Find the measure of the angle. Solution Let x measure of angle. Then 4x measure of complement of angle. The sum of the measures of an angle and its complement is 90°. x 4x 90 5x 90 x 18 Check The measure of the first angle is 18°. The measure of the second angle is 4(18°) 72°. The sum of the measures of the angles is 18° 72° 90°. Thus, the angles are complementary. ✔ Answer The measure of the angle is 18°. 254 Geometric Figures, Areas, and Volumes Note: The unit of measure is very important in the solution of a problem. While it is not necessary to include the unit of measure in each step of the solution, each term in an equation must represent the same unit and the unit of measure must be included in the answer. EXAMPLE 2 The measure of an angle is 40° more than the measure of its supplement. Find the measure of the angle. Solution Let x the measure of the supplement of the angle. Then x 40 the measure of the angle. The sum of the measures of an angle and its supplement is 180°. x (x 40) 180 2x 40 180 2x 140 x 70 x 40 110 Check The sum of the measures is 110° 70° 180° and 110° is 40° more than 70°. ✔ Answer The measure of the angle is 110°. EXAMPLE 3 The algebraic expressions 5w 20 and 2w 16 represent the measures in degrees of a pair of vertical angles. a. Find the value of w. b. Find the measure of each angle. Solution a. Vertical angles are equal in measure. 5w 20 2w 16 3w 20 16 3w 36 w 12 b. 5w 20 5(12) 20 60 20 40 2w 16 2(12) 16 24 16 40 Check Since each angle has a measure of 40°, the vertical angles are equal in measure. ✔ Answers a. w 12 b. The measure of each angle is 40°. Pairs of Angles 255 EXERCISES Writing About Mathematics 1. Show that supplementary angles are always two right angles or an acute angle and an obtuse angle. 2. The measures of three angles are 15°, 26°, and 49°. Are these angles complementary? Explain why or why not. Developing Skills In 3–6, answer each of the following questions for an angle with the given measure. a. What is the measure of the complement of the angle? b. What is the measure of the supplement of the angle? c. The measure of the supplement of the angle is how much larger than the measure of its complement? 3. 15° 4. 37° 5. 67° 6. x° In 7–10, A and B are complementary. Find the measure of each angle if the measures of the two angles are represented by the given expressions. Solve the problem algebraically using an equation. 7. mA x, mB 5x 9. mA x, mB x 40 8. mA x, mB x 20 10. mA y, mB 2y 30 In 11–14, ABD and DBC are supplementary. Find the measure of each angle if the measures of the two angles are represented by the given expressions. Solve the problem algebraically using an equation. 11. mABD x, mDBC 3x 12. mABD x, mDBC x 80 13. mDBC x, mABD x 30 14. mDBC y, mABD 1 4y D A B C In 15–24, solve each problem algebraically using an equation. 15. Two angles are supplementary. The measure of one angle is twice as large as the measure of the other. Find the number of degrees in each angle. 16. The complement of an angle is 14 times as large as the angle. Find the measure of the com- plement. 17. The measure of the supplement of an angle is 40° more than the measure of the angle. Find the number of degrees in the supplement. 256 Geometric Figures, Areas, and Volumes 18. Two angles are complementary. One angle is twice as large as the other. Find the number of degrees in each angle. 19. The measure of the complement of an angle is one-ninth the measure of the angle. Find the measure of the angle. 20. Find the number of degrees in the measure of an angle that is 20° less than 4 times the mea- sure of its supplement. 21. The difference between the measures of two supplementary angles is 80°. Find the measure of the larger of the two angles. 22. The complement of an angle measures 20° more than the angle. Find the number of degrees in the angle. 23. Find the number of degrees in an angle that measures 10° more than its supplement. 24. Find the number of degrees in an angle that measures 8° less than its complement. 25. The supplement of the complement of an acute angle is always: (1) an acute angle (2) a right angle (3) an obtuse angle (4) a straight angle In 26–28, g MN and g RS intersect at T. 26. If mRTM 5x and mNTS 3x 10, find mRTM. 27. If mMTS 4x 60 and mNTR 2x, find mMTS. 28. If mRTM 7x 16 and
mNTS 3x 48, find mNTS. In 29–34, find the measure of each angle named, based on the given information. 29. Given: g EF h ' GH ; mEGI 62. H I 62° E G F a. Find mFGH. b. Find mHGI. 30. Given: g JK mNLM 48. h ' LM ; g NLO is a line; M N 48° J L K O a. Find mJLN. b. Find mMLK. c. Find mKLO. d. Find mJLO. 31. Given: GKH and HKI are a linear pair; h KH h ' KJ ; mIKJ 34. mPMN 40. Pairs of Angles 257 32. Given: h MO h ' MP g ; LMN is a line; H I G K 34° J a. Find mHKI. b. Find mHKG. c. Find mGKJ. 33. Given: g ABC ABD DBE. is a line; mEBC 40; D 40° A B a. Find mABD. b. Find mDBC. E C O P 40° L M N a. Find mPMO. b. Find mOML. 34. Given: g FI intersects g JH at K; mHKI 40; mFKG mFKJ. G H 40° F K I J a. Find mFKJ. b. Find mFKG. c. Find mGKH. d. Find mJKI. In 35–38, sketch and label a diagram in each case and find the measure of each angle named. 35. intersects g AB a. mCEB g CD at E; mAED 20. Find: b. mBED 36. PQR and RQS are complementary; mPQR 30; a. mRQS b. mSQT c. mCEA g RQT c. mPQT is a line. Find: 258 Geometric Figures, Areas, and Volumes at R. The measure of LRQ is 80 more than mLRP. Find: b. mLRQ c. mPRM at E. Point F is in the interior of CEB. The measure of CEF is 8 times the 37. 38. g PQ intersects g LM a. mLRP g g ' AB CD measure of FEB. Find: a. mFEB b. mCEF c. mAEF 39. The angles, ABD and DBC, form a linear pair and are congruent. What must be true about g ABC and g BD ? 7-3 ANGLES AND PARALLEL LINES A C B D In the figure on the left, g AB Not all lines in the same plane intersect. Two or more lines are called parallel lines if and only if the lines lie in the same plane but do not intersect. g g AB CD and is parallel to lie in the same plane but do not interg CD sect. Hence, we say that . Using the symbol \|\| for is parallel . When we speak of two parallel lines, we will mean two to, we write distinct lines. (In more advanced courses, you will see that a line is parallel to itself.) Line segments and rays are parallel if the lines that contain them are parlie in the same plane, they must be either allel. If two lines such as intersecting lines or parallel lines, as shown in the following figures. g i CD g AB g CD g AB and C A B D g g AB intersects CD. C D B A g g AB is parallel to CD. g AB g CD When two lines such as are parallel, they have no points in common. We can think of each line as a set of points. Hence, the intersection set of g AB and When two lines are cut by a third line, called a transversal, two sets of is the empty set symbolized as g d CD 5 . g CD g AB and 1 2 43 5 6 7 8 angles, each containing four angles, are formed. In the figure on the left: • Angles 3, 4, 5, 6 are called interior angles. • Angles 1, 2, 7, 8 are called exterior angles. • Angles 4 and 5 are interior angles on opposite sides of the transversal and do not have the same vertex. They are called alternate interior angles. Angles 3 and 6 are another pair of alternate interior angles. Angles and Parallel Lines 259 • Angles 1 and 8 are exterior angles on opposite sides of the transversal and do not have the same vertex. They are called alternate exterior angles. Angles 2 and 7 are another pair of alternate exterior angles. • Angles 4 and 6 are interior angles on the same side of the transversal. Angles 3 and 5 are another pair of interior angles on the same side of the transversal. • Angles 1 and 5 are on the same side of the transversal, one interior and one exterior, and at different vertices. They are called corresponding angles. Other pairs of corresponding angles are 2 and 6, 3 and 7, 4 and 8. Alternate Interior Angles and Parallel Lines In the figure, a transversal intersects two parallel lines, forming a pair of alternate interior angles, 3 and 6. If we measure 3 and 6 with a protractor, we will find that each angle measures 60°. Here, alternate interior angles 3 and 6 have equal measures, and 3 6. If we draw other pairs of parallel lines intersected by transversals, we will find again that pairs of alternate interior angles have equal measures. Yet, we would be hard-pressed to prove that this is always true. Therefore, we accept, without proof, the following statement: If two parallel lines are cut by a transversal, then the alternate interior angles that are formed have equal measures, that is, they are congruent. Note that 4 and 5 are another pair of alternate interior angles formed by a transversal that intersects the parallel lines. Therefore, 4 5. Corresponding Angles and Parallel Lines If two lines are cut by a transversal, four pairs of corresponding angles are formed. One such pair of corresponding angles is 2 and 6, as shown in the figure on the left. If the original two lines are parallel, do these corresponding angles have equal measures? We are ready to prove in an informal manner that they do. (1) Let m2 x. (2) If m2 x, then m3 x (because 2 and 3 are vertical angles, and vertical angles are congruent). (3) If m3 x, then m6 x (because 3 and 6 are alternate interior angles of parallel lines, and alternate interior angles of parallel lines are congruent). (4) Therefore m2 m6 (because the measure of each angle is x). 3 4 5 6 2 3 6 260 Geometric Figures, Areas, and Volumes The four steps on page 259 serve as a proof of the following theorem: Theorem. If two parallel lines are cut by a transversal, then the corresponding angles formed have equal measures, that is, they are congruent. Note that this theorem is true for each pair of corresponding angles: 1 5; 2 6; 3 7; 4 8 Alternate Exterior Angles and Parallel Lines If two parallel lines are cut by a transversal, we can prove informally that the alternate exterior angles formed have equal measures. One such pair of alternate exterior angles consists of 2 and 7, as shown in the figure on the left. (1) Let m2 x. (2) If m2 x, then m6 x (because 2 and 6 are corresponding angles of parallel lines, proved to have the same measure). (3) If m6 x, then m7 x (because 6 and 7 are vertical angles, previ- ously proved to have the same measure). (4) Therefore m2 m7 (because the measure of each angle is x). These four steps serve as a proof of the following theorem: Theorem. If two parallel lines are cut by a transversal, then the alternate exterior angles formed have equal measures, that is, they are congruent. Note that this theorem is true for each pair of alternate exterior angles: 1 8; 2 7. Interior Angles on the Same Side of the Transversal When two parallel lines are cut by a transversal, we can prove informally that the sum of the measures of the interior angles on the same side of the transversal is 180°. One such pair of interior angles on the same side of the transversal consists of 3 and 5, as shown in the figure on the left. (1) m5 m7 180 (5 and 7 are supplementary angles). (2) m7 m3 (7 and 3 are corresponding angles). (3) m5 m3 180 (by substituting m3 for its equal, m7). Angles and Parallel Lines 261 These three steps serve as a proof of the following theorem: Theorem. If two parallel lines are cut by a transversal, then the sum of the measures of the interior angles on the same side of the transversal is 180°. EXAMPLE 1 In the figure, the parallel lines are cut by a transversal. If ml (5x 10) and m2 (3x 60), find the measures of 1 and 2. Solution Since the lines are parallel, the alternate interior angles, 1 and 2, have equal measures. Write and solve an equation using the algebraic expressions for the measures of these angles: 5x 10 3x 60 5x 3x 70 2x 70 x 35 5x 10 5(35) 10 175 10 165 3x 60 3(35) 60 105 60 165 1 2 Answer m1 165 and m2 165. EXERCISES Writing About Mathematics 1. Give an example of a pair of lines that are neither parallel nor intersecting. 2. If a transversal is perpendicular to one of two parallel lines, is it perpendicular to the other? Prove your answer using definitions and theorems given in this chapter. Developing Skills In 3–10, the figure below shows two parallel lines cut by a transversal. For each given measure, find the measures of the other seven angles. 4. m6 150 6. m1 75 8. m4 10 10. m8 179 3. m3 80 5. m5 60 7. m2 55 9. m7 2 3 4 2 1 6 5 7 8 262 Geometric Figures, Areas, and Volumes In 11–15, the figure below shows two parallel lines cut by a transversal. In each exercise, find the measures of all eight angles under the given conditions. 11. m3 2x 40 and m7 3x 27 12. m4 4x 10 and m6 x 80 13. m4 3x 40 and m5 2x 14. m4 2x 10 and m2 x 60 15. 8 1 16. If g AB g i CD , m5 40, and m4 30, find the measures of the other angles in the figure 10 5 11 In 17–22, tell whether each statement is always, sometimes, or never true. 17. If two distinct lines intersect, then they are parallel. 18. If two distinct lines do not intersect, then they are parallel. 19. If two angles are alternate interior angles, then they are on opposite sides of the transversal. 20. If two parallel lines are cut by a transversal, then the alternate interior angles are congruent. 21. If two parallel lines are cut by a transversal, then the alternate interior angles are comple- mentary. 22. If two parallel lines are cut by a transversal, then the corresponding angles are supplemen- tary. 23. In the figure on the right, two parallel lines are cut by a transversal. Write an informal proof that 1 and 2 have equal measures. 1 2 7-4 TRIANGLES A polygon is a plane figure that consists of line segments joining three or more points. Each line segment is a side of the polygon and the endpoints of each side are called vertices. Each vertex is the endpoint of exactly two sides and no two sides have any other points in common. When three points that are not all on the same line are joined in pairs by line segments, the figure formed is a triangle. Each of the given points is the vertex of an angle of the triangle, and each line segment is a side of the triangle. There are many practical uses of the triangle, especially in construction work such as the building of bridges, radio towers, and airplane wings, because the tri- angle is a rigid figure. The shape of a triangle cannot be changed w
ithout changing the length of at least one of its sides. Triangles 263 We begin our discussion of the triangle by considering triangle ABC, shown below, which is symbolized as ABC. In ABC, points A, B, and C are the verBC tices, and are the sides. Angles A, B, and C of the triangle are symbolized as A, B, and C. , and AB CA , We make the following observations: 1. Side AB 2. Side BC 3. Side CA is included between A and B. is included between B and C. is included between C and A. 4. Angle A is included between sides AB and CA . 5. Angle B is included between sides AB and BC . 6. Angle C is included between sides CA and BC . C A B Classifying Triangles According to Angles B F G J A C Acute triangle 90° E D Equiangular triangle H I Right triangle K L Obtuse triangle • An acute triangle has three acute angles. • An equiangular triangle has three angles equal in measure. • A right triangle has one right angle. • An obtuse triangle has one obtuse angle. In right triangle GHI above, the two sides of the triangle that form the right , are called the legs of the right triangle. The side opposite the GH angle, right angle, and GI HI , is called the hypotenuse. Sum of the Measures of the Angles of a Triangle When we change the shape of a triangle, changes take place also in the measures of its angles. Is there any relationship among the measures of the angles of a triangle that does not change? Let us see. 264 Geometric Figures, Areas, and Volumes Draw several triangles of different shapes. Measure the three angles of each triangle and find the sum of these measures. For example, in ABC on the right, mA + mB mC = 65 45 70 180. C 70° If you measured accurately, you should have found that in each triangle that you drew and measured, the sum of the measures of the three angles is 180°, regardless of its size or shape. We can write an informal algebraic proof of the following statement: A B 65° 45° The sum of the measures of the angles of a triangle is 180°. (1) In ABC, let mA x, mACB y, and mB z. g DCE (2) Let (3) Since DCE is a straight angle, be a line parallel to g AB . mDCE 180. C y x z D (4) mDCE mDCA mACB mBCE 180. x A (5) mDCA mA x. (6) mACB y. (7) mBCE mB z. (8) Substitute from statements (5), (6), and (7) in statement (4): x y z 180. E z B EXAMPLE 1 Find the measure of the third angle of a triangle if the measures of two of the angles are 72.6° and 84.2°. Solution Subtract the sum of the known measures from 180: 180 (72.6 84.2) 23.2 Answer 23.2° EXAMPLE 2 In ABC, the measure of B is twice the measure of A, and the measure of C is 3 times the measure of A. Find the number of degrees in each angle of the triangle. Triangles 265 Solution Let x the number of degrees in A. Then 2x the number of degrees in B. Then 3x the number of degrees in C. The sum of the measures of the angles of a triangle is 180°. B 2x 3x C x A x 2x 3x 180 6x 180 x 30 2x 60 3x 90 Check 60 2(30) 90 3(30) 30 60 90 180 ✔ Answer mA 30, mB 60, mC 90 Classifying Triangles According to Sides T Z N L Scalene triangle M R S Isosceles triangle X Equilateral triangle Y • A scalene triangle has no sides equal in length. • An isosceles triangle has two sides equal in length. • An equilateral triangle has three sides equal in length. C A B Isosceles Triangles In isosceles triangle ABC, shown on the left, the two sides that are equal in measure, , are called the legs. The third side, , is the base. and AB AC BC Two line segments that are equal in measure are said to be congruent. The angle formed by the two congruent sides, C, is called the vertex angle. The two angles at the endpoints of the base, A and B, are the base angles. In isosceles triangle ABC, if we measure the base angles, A and B, we find that each angle contains 65°. Therefore, mA mB. Similarly, if we measure the base angles in any other isosceles triangle, we find that they are equal in measure. Thus, we will accept the truth of the following statement: 266 Geometric Figures, Areas, and Volumes The base angles of an isosceles triangle are equal in measure; that is, they are congruent. This statement may be rephrased in a variety of ways. For example: 1. If a triangle is isosceles, then its base angles are congruent. 2. If two sides of a triangle are equal in measure, then the angles opposite these sides are equal in measure. The following statement is also true: If two angles of a triangle are equal in measure, then the sides opposite these angles are equal in measure and the triangle is an isosceles triangle. This statement may be rephrased as follows: 1. If two angles of a triangle are congruent, then the sides opposite these angles are congruent. 2. If two angles of a triangle have equal measures, then the sides opposite these angles have equal measures. Equilateral Triangles Triangle ABC is an equilateral triangle. Since AB BC, mC mA; also, since AC BC, mB mA. Therefore, mA mB mC. In an equilateral triangle, the measures of all of the angles are equal. In DEF, all of the angles are equal in measure. Since mD mE, EF DF; also, since mD mF, EF DE. Therefore, DE EF DF, and DEF is equilateral. If a triangle is equilateral, then it is equiangular. Properties of Special Triangles 1. If two sides of a triangle are equal in measure, the angles opposite these sides are also equal in measure. (The base angles of an isosceles triangle are equal in measure.) 2. If two angles of a triangle are equal in measure, the sides opposite these angles are also equal in measure. 3. All of the angles of an equilateral triangle are equal in measure. (An equi- lateral triangle is equiangular.) 4. If three angles of a triangle are equal in measure, the triangle is equilat- eral. (An equiangular triangle is equilateral.) C F B E A D Triangles 267 EXAMPLE 3 In isosceles triangle ABC, the measure of vertex angle C is 30° more than the measure of each base angle. Find the number of degrees in each angle of the triangle. Solution Let x number of degrees in one base angle, A. Then x number of degrees in the other base angle, B, and x 30 number of degrees in the vertex angle, C. C x + 30 The sum of the measures of the angles of a triangle is 180°. x A x B x x x 30 180 3x 30 180 3x 150 x 50 x 30 80 Check 50 50 80 180 ✔ Answer mA 50, mB 50, mC 80 EXERCISES Writing About Mathematics 1. Ayyam said that if the sum of the measures of two angles of a triangle is equal to the measure of the third angle, the triangle is a right triangle. Prove or disprove Ayyam’s statement. 2. Janice said that if two angles of a triangle each measure 60°, then the triangle is equilateral. Prove or disprove Janice’s statement. Developing Skills In 3–5, state, in each case, whether the angles with the given measures can be the three angles of the same triangle. 3. 30°, 70°, 80° 4. 70°, 80°, 90° 5. 30°, 110°, 40° In 6–9, find, in each case, the measure of the third angle of the triangle if the measures of two angles are: 6. 60°, 40° 7. 100°, 20° 8. 54.5°, 82.3° 9. 241 48 , 811 28 268 Geometric Figures, Areas, and Volumes 10. What is the measure of each angle of an equiangular triangle? 11. Can a triangle have: a. two right angles? obtuse angle? Explain why or why not. b. two obtuse angles? c. one right and one 12. What is the sum of the measures of the two acute angles of a right triangle? 13. In ABC, AC 4, CB 6, and AB 6. a. What type of triangle is ABC? b. Name two angles in ABC whose measures C are equal. B c. Why are these angles equal in measure? d. Name the legs, base, base angles and vertex angle of this triangle. 14. In RST, mR 70 and mT 40. a. Find the measure of S. b. Name two sides in RST that are congruent. c. Why are the two sides congruent? d. What type of triangle is RST? e. Name the legs, base, base angles, and vertex angle of this triangle. A T R S 15. Find the measure of the vertex angle of an isosceles triangle if the measure of each base angle is: a. 80° b. 55° c. 42° d. 221 28 e. 51.5° 16. Find the measure of each base angle of an isosceles triangle if the measure of the vertex angle is: a. 20° b. 50° c. 76° d. 100° e. 65° 17. What is the number of degrees in each acute angle of an isosceles right triangle? 18. If a triangle is equilateral, what is the measure of each angle? Applying Skills 19. The measure of each base angle of an isosceles triangle is 7 times the measure of the vertex angle. Find the measure of each angle of the triangle. 20. The measure of each of the congruent angles of an isosceles triangle is one-half of the mea- sure of the vertex angle. Find the measure of each angle of the triangle. 21. The measure of the vertex angle of an isosceles triangle is 3 times the measure of each base angle. Find the number of degrees in each angle of the triangle. 22. The measure of the vertex angle of an isosceles triangle is 15° more than the measure of each base angle. Find the number of degrees in each angle of the triangle. Triangles 269 23. The measure of each of the congruent angles of an isosceles triangle is 6° less than the mea- sure of the vertex angle. Find the measure of each angle of the triangle. 24. The measure of each of the congruent angles of an isosceles triangle is 9° less than 4 times the vertex angle. Find the measure of each angle of the triangle. 25. In ABC, mA x, mB x 45, and mC 3x 15. a. Find the measures of the three angles. b. What kind of triangle is ABC? 26. The measures of the three angles of DEF can be represented by (x 30)°, 2x°, and (4x 60)°. a. What is the measure of each angle? b. What kind of triangle is DEF? 27. In a triangle, the measure of the second angle is 3 times the measure of the first angle, and the measure of the third angle is 5 times the measure of the first angle. Find the number of degrees in each angle of the triangle. 28. In a triangle, the measure of the second angle is 4 times the measure of the first angle. The measure of the third angle is equal to the sum of the measures of the first two angles. Find the number of degrees in each angle of the triangle. What kind of triangle
is it? 29. In a triangle, the measure of the second angle is 30° more than the measure of the first angle, and the measure of the third angle is 45° more than the measure of the first angle. Find the number of degrees in each angle of the triangle. 30. In a triangle, the measure of the second angle is 5° more than twice the measure of the first angle. The measure of the third angle is 35° less than 3 times the measure of the first angle. Find the number of degrees in each angle of the triangle. 31. is a straight line, mAEG 130, and d–——S AEFB mBFG 140. a. Find mx, my, and mz. b. What kind of a triangle is EFG? 32. In RST, mR x, mS x 30, mT x 30. a. Find the measures of the three angles of the triangle. b. What kind of a triangle is RST? 33. In KLM, mK 2x, mL x 30, mM 3x 30. a. Find the measures of the three angles of the triangle. b. What kind of a triangle is KLM? G z x 130° E A y 140° F B 270 Geometric Figures, Areas, and Volumes Hands-On Activity 1: Constructing a Line Segment Congruent to a Given Segment To construct a geometric figure means that a specific design is accurately made by using only two instruments: a compass used to draw a complete circle or part of a circle and a straightedge used to draw a straight line. In this activity, you will learn how to construct a line segment congruent to a given segment, that is, construct a copy of a line segment. STEP 1. Use the straightedge to draw a line segment. Label the endpoints A and B. STEP 2. Use the straightedge to draw a ray and label the endpoint C. STEP 3. Place the compass so that the point of the compass is at A and the point of the pencil is at B. STEP 4. Keeping the opening of the compass unchanged, place the point at C and draw an arc that intersects the ray. Label this intersection D. Result: AB CD a. Now that you know how to construct congruent line segments, explain how you can construct a line segment that is three times the length of a given segment. b. Explain how to construct a line segment with length equal to the difference of two given seg- ments. c. Explain how to construct a line segment whose length is the sum of the lengths of two given line segments. Hands-On Activity 2: Constructing an Angle Congruent to a Given Angle In this activity, you will learn how to construct an angle congruent to a given angle, that is, construct a copy of an angle. STEP 1. Use the straightedge to draw an acute angle. Label the vertex S. STEP 2. Use the straightedge to draw a ray and label the endpoint M. STEP 3. With the point of the compass at S, draw an arc that intersects each ray of S. Label the point of intersection on one ray R and the point of intersection on the other ray T. STEP 4. Using the same opening of the compass as was used in step 3, place the point of the compass at M and draw an arc that intersects the ray and extends beyond the ray. (Draw at least half of a circle.) Label the point of intersection L. STEP 5. Place the point of the compass at R and the point of the pencil at T. STEP 6. Without changing the opening of the compass, place the point at L and draw an arc that intersects the arc drawn in step 4. Label this point of intersection N. h MN . STEP 7. Draw Result: RST LMN a. Now that you know how to construct congruent angles, explain how you can construct an angle that is three times the measure of a given angle. b. Explain how to construct an angle with a measure equal to the difference of two given angles. c. Explain how to construct a triangle congruent to a given triangle using two sides and the included angle. Triangles 271 Hands-On Activity 3: Constructing a Perpendicular Bisector In this activity, you will learn how to construct a perpendicular bisector of a line segment. A perpendicular bisector of a line segment is the line that divides a segment into two equal parts and is perpendicular to the segment. STEP 1. Use the straightedge to draw a line segment. Label one endpoint A and the other C. STEP 2. Open the compass so that the distance between the point and the pencil point is more than half of the length of AC . STEP 3. With the point of the compass at A, draw an arc above AC and an arc below AC . STEP 4. With the same opening of the compass and the point of the compass at C, draw an arc AC and an arc below above these intersections E and the other F. AC that intersect the arcs drawn in step 3. Call one of STEP 5. Use the straightedge to draw g EF , intersecting AC at B. g EF is perpendicular to Result: a. What is true about EAB and FAB? What is true about EAB and ECB? b. Explain how to construct an isosceles triangle with a vertex angle that is twice the measure of a , and B is the midpoint of AC AC . given angle. c. Explain how to construct an isosceles right triangle. Hands-On Activity 4: Constructing an Angle Bisector In this activity, you will learn how to construct an angle bisector. An angle bisector is the line that divides an angle into two congruent angles. STEP 1. Use the straightedge to draw an acute angle. Label the vertex S. STEP 2. With any convenient opening of the compass, place the point at S and draw an arc that intersects both rays of S. Call one of the intersections R and the other T. STEP 3. Place the point of the compass at R and draw an arc in the interior of S. STEP 4. With the same opening of the compass, place the point of the compass at T and draw an arc that intersects the arc drawn in step 3. Label the intersection of the arcs P. STEP 5. Draw h SP . Result: RSP PST; a. Now that you know how to construct an angle bisector, explain how you can construct an angle bisects angle S. h SP that is one and a half times the measure of a given angle. b. Is it possible to use an angle bisector to construct a 90° angle? Explain. c. Explain how to construct an isosceles triangle with a vertex angle that is congruent to a given angle. 272 Geometric Figures, Areas, and Volumes 7-5 QUADRILATERALS In your study of mathematics you have learned many facts about polygons that have more than three sides. In this text you have frequently solved problems using the formulas for the perimeter and area of a rectangle. In this section we will review what you already know and use that knowledge to demonstrate the truth of many of the properties of polygons. C B D A A quadrilateral is a polygon that has four sides. A point at which any two sides of the quadrilateral meet is a vertex of the quadrilateral. At each vertex, the two sides that meet form an angle of the quadrilateral. Thus, ABCD on the AB left is a quadrilateral whose sides are . Its vertices are A, , B, C, and D. Its angles are ABC, BCD, CDA, and DAB. , and BC , DA CD In a quadrilateral, two angles whose vertices are the endpoints of a side are called consecutive angles. For example, in quadrilateral ABCD, A and B are consecutive angles because their vertices are the endpoints of a side, . Other pairs of consecutive angles are B and C, C and D, and D and A. Two angles that are not consecutive angles are called opposite angles; A and C are opposite angles, and B and D are opposite angles. AB Special Quadrilaterals When we vary the shape of the quadrilateral by making some of its sides parallel, by making some of its sides equal in length, or by making its angles right angles, we get different members of the family of quadrilaterals, as shown and named below Trapezoid F E Parallelogram J K Q R W X Rectangle Rhombus Square A trapezoid is a quadrilateral in which two and only two opposite sides are are called the . Parallel sides AB i CD and CD AB parallel. In trapezoid ABCD, bases of the trapezoid. A parallelogram is a quadrilateral in which both pairs of opposite sides are . The symbol for a EH i FG EF i GH and parallel. In parallelogram EFGH, ~ . parallelogram is A rectangle is a parallelogram in which all four angles are right angles. Rectangle JKLM is a parallelogram in which J, K, L, and M are right angles. Quadrilaterals 273 A rhombus is a parallelogram in which all sides are of equal length. Rhombus QRST is a parallelogram in which QR = RS = ST TQ. A square is a rectangle in which all sides are of equal length. It is also correct to say that a square is a rhombus in which all angles are right angles. Therefore, square WXYZ is also a parallelogram in which W, X, Y, and Z are right angles, and WX XY YZ ZW. The Angles of a Quadrilateral C D Draw a large quadrilateral like the one shown at the left and measure each of its four angles. Is the sum 360°? It should be. If we do the same with several other quadrilaterals of different shapes and sizes, is the sum of the four measures 360° in each case? It should be. Relying on what you have just verified by experimentation, it seems reasonable to make the following statement: A B The sum of the measures of the angles of a quadrilateral is 360°. D AC To prove informally that this statement is true, we draw diagonal endpoints are the vertices of the two opposite angles, A and C. Then: , whose C (1) Diagonal ADC. AC divides quadrilateral ABCD into two triangles, ABC and (2) The sum of the measures of the angles of ABC is 180°, and the sum of the measures of the angles of ADC is 180°. (3) The sum of the measures of all the angles of ABC and ADC together A B is 360°. (4) Therefore, mA mB mC mD 360. The Family of Parallelograms Listed below are some relationships that are true for the family of parallelograms that includes rectangles, rhombuses, and squares. Parallelogram 1. All rectangles, rhombuses, and squares are parallelograms. Therefore any property of a parallelogram must also be a property of a rectangle, a rhombus, or a square. 2. A square is also a rectangle. Therefore, any property of a rectangle must also be a property of a square. 3. A square is also a rhombus. Therefore, any property of a rhombus must also be a property of a square. Rectangle Rhombus Square 274 Geometric Figures, Areas, and Volumes D C A B In parallelogram ABCD at the left, opposite sides are parallel.Thus, AD i BC . The following s
tatements are true for any parallelograms. and AB i DC Opposite sides of a parallelogram are congruent. Here, AB > DC and AD > BC . Therefore AB DC and AD BC. Opposite angles of a parallelogram are congruent. Here, A C and B D. Therefore mA mC and mB mD. Consecutive angles of a parallelogram are supplementary. Here, mA mB 180, mB mC 180, and so forth. Since rhombuses, rectangles, and squares are also parallelograms, these statements will be true for any rhombus, any rectangle, and any square. Informal Proofs for Statements About Angles in a Parallelogram The statements about angles formed by parallel lines cut by a transversal that were shown to be true in Section 3 of this chapter can now be used to establish the relationships among the measures of the angles of a parallelogram. (1) In parallelogram ABCD, AB and DC ments of parallel lines cut by transversal are segAD . (2) Let mA x. Then mD 180 x because A and D are interior angles on the same side of a transversal, and these angles have been shown to be supplementary. (3) In parallelogram ABCD, AD and BC ments of parallel lines cut by transversal are segAB . D 180 – x C x x A 180 – x B (4) Since mA x, mB 180 x because A and B are interior angles on the same side of a transversal, and these angles have been shown to be supplementary. (5) Similarly, AD and DC . Since mD 180 x, mC 180 (180 x) x because D and C are interior angles on the same side of the transversal. are segments of parallel lines cut by transversal BC Therefore, the consecutive angles of a parallelogram are supplementary: mA mD x (180 x) 180 mA mB x (180 x) 180 mC mB x (180 x) 180 mC mD x (180 x) 180 Quadrilaterals 275 Also, the opposite angles of a parallelogram have equal measures: mA x and mC x m B 180 x and mD 180 x or mA mC or mB mD Polygons and Angles D C We can use the sum of the measures of the interior angles of a triangle to find the sum of the measures of the interior angles of any polygon. Polygon ABCDE is a pentagon, a polygon with five sides. From vertex A, we draw diagonals to vertices C and D, the vertices that are not adjacent to A. These diagonals divide the pentagon into three triangles. The sum of the measures of the interior angles of ABCDE is the sum of the measures of ABC, ACD, and ADE. The sum of the measures of the interior angles of ABCDE 3(180°) 540°. We can use this same method to find the sum of the interior angles of any polygon of more than three sides. E A B Hexagon 4(180°) = 720° Octagon 6(180°) = 1,080° Decagon 8(180°) = 1,440° In each case, the number of triangles into which the polygon can be divided is 2 fewer than the number of sides of the polygon. In general: The sum of the measures of the interior angles of an n-sided polygon is 180(n 2). Trapezoids A trapezoid has one and only one pair of parallel lines. Each of the two parallel sides is called a base of the trapezoid. Therefore, we can use what we know about angles formed when parallel lines are cut by a transversal to demonstrate some facts about the angles of a trapezoid. Quadrilateral ABCD is a trapezoid with . Since DAB and AB i CD CDA are interior angles on the same side of transversal DA , they are supplementary. Also, CBA and DCB are interior angles on the same side of transversal , and they are supplementary. CB D A C B 276 Geometric Figures, Areas, and Volumes In a trapezoid, the parallel sides can never be congruent. But the nonparallel sides can be congruent. A trapezoid in which the nonparallel sides are congruent is called an isosceles trapezoid. In an isosceles trapezoid, the base angles, the two angles whose vertices are the endpoints of the same base, are congruent. To write a formula for the area of AC . Since AFCE is trapezoid ABCD, draw diagonal and altitudes and a rectangle AE CF. AE CF Let AB b1, CD b2, and AE CF h. b2 D E Area of triangle ABC Area of triangle ACD Area of trapezoid ABCD 1 2b1h 1 2b2h 1 2b1h 1 1 2b2h 1 2h(b1 1 b2) A B b1 C h F EXAMPLE 1 ABCD is a parallelogram where mA 2x 50 and mC 3x 40. D C 3x + 40 a. Find the value of x. b. Find the measure of each angle. 2x + 50 A B Solution a. In ~ABCD , mC mA because the opposite angles of a parallelogram have equal measure. Thus: b. By substitution: 3x 40 2x 50 x 10 mA 2x 50 2(10) 50 70 mB 180 mA 180 70 110 mC 3x 40 3(10) 40 70 mD 180 mC 180 70 110 Answer a. x 10 b. mA 70 mB 110 mC 70 mD 110 Quadrilaterals 277 EXERCISES Writing About Mathematics 1. Adam said that if a quadrilateral has four equal angles then the parallelogram is a rectangle. Do you agree with Adam? Explain why or why not. 2. Emmanuel said that if a parallelogram has one right angle then the parallelogram is a rec- tangle. Do you agree with Emmanuel? Explain why or why not. Developing Skills In 3–8, in each case, is the statement true or false? 3. If a polygon is a trapezoid, it is a quadrilateral. 4. If a polygon is a rectangle, it is a parallelogram. 5. If a polygon is a rhombus, it is a parallelogram. 6. If a polygon is a rhombus, it is a square. 7. If a polygon is a square, then it is a rhombus. 8. If two angles are opposite angles of a parallelogram, they are congruent. In 9–12, the angle measures are represented in each quadrilateral. In each case: a. Find the value of x. b. Find the measure of each angle of the quadrilateral. 9. D C 80° 120° 110° A x B 11. N M 2x 3x – 50 x K x – 10 L 10. G H x + 20 x – 20 x E x F 12. S 2x + 20 T 2x 3x – 20 R x + 10 Q 278 Geometric Figures, Areas, and Volumes In 13 and 14, polygon ABCD is a parallelogram. A D 13. AB 3x 8 and DC x 12. Find AB and DC. 14. mA 5x 40 and mC 3x 20. Find mA, mB, mC, and mD. B C In 15 and 16, polygon ABCD is a rectangle. 15. BC 4x 5, AD 2x 3. Find BC and AD. 16. mA 5x 10. Find the value of x. 17. ABCD is a square. If AB 8x 6 and BC 5x 12, find the length of each side of the square. D A 18. In rhombus KLMN, KL 3x and LM 2(x 3). Find the length of each side of the rhombus. Applying Skills 19. One side of a barn is in the shape of a seven-sided polygon called a heptagon. The heptagon, ABCDEFG, can be divided into rectangle ABFG, isosceles trapezoid BCEF, and isosceles triangle CDE, as shown in the diagram. a. Find the sum of the measures of the interior angles of ABCDEFG by using the sum of the measures of the angles of the two quadrilaterals and the triangle. b. Sketch the heptagon on your answer paper and show how it can be divided into triangles by drawing diagonals from A. Use these triangles to find the sum of the measures of the interior angles of ABCDEFG. If the measure of each lower base angle of trapezoid BCEF is 45° and the measure of each base angle of isosceles triangle CDE is 30°, find the measure of each angle of heptagon ABCDEFG. d. Find the sum of the measures of the interior angles of ABCDEFG by adding the angle measures found in c. 20. Cassandra had a piece of cardboard that was in the shape of an equilateral triangle. She cut an isosceles triangle from each vertex of the cardboard. The length of each leg of the triangles that she cut was one third of the length of a side of the original triangle. a. Show that each of the triangles that Cassandra cut off is an equilateral triangle. b. What is the measure of each angle of the remaining piece of cardboard? c. What is the shape of the remaining piece of cardboard? Areas of Irregular Polygons 279 21. A piece of land is bounded by two parallel roads and two roads that are not parallel forming a trapezoid. Along the parallel roads the land measures 1.3 miles and 1.7 miles. The distance between the parallel roads, measured perpendicular to the roads, is 2.82 miles. a. Find the area of the land. Express the area to the nearest tenth of a mile. b. An acre is a unit of area often used to measure land. There are 640 acres in a square mile. Express, to the nearest hundred acres, the area of the land. 22. If possible, draw the following quadrilaterals. If it is not possible, state why. a. A quadrilateral with four acute angles. b. A quadrilateral with four obtuse angles. c. A quadrilateral with one acute angle and three obtuse angles. d. A quadrilateral with three acute angles and one obtuse angle. e. A quadrilateral with exactly three right angles. 7-6 AREAS OF IRREGULAR POLYGONS Many polygons are irregular figures for which there is no formula for the area. However, the area of such figures can often be found by separating the figure into regions with known area formulas and adding or subtracting these areas to find the required area. EXAMPLE 1 and F is a point . If AB 12, FC 3, and DE 8, find the area ABCD is a square. E is a point on BC on of ABFE. DC Solution Since ABCD is a square and AB 12, then BC 12, CD 12, and DA 12. Therefore, CE CD DE 12 8 4. D 8 E C 3 F A 12 B Area of ABFE Area of ABCD Area of FCE Area of EDA 2(3)(4) 2 1 2(8)(12) 12(12) 2 1 144 6 48 90 Answer 90 square units 280 Geometric Figures, Areas, and Volumes EXAMPLE 2 ABCD is a quadrilateral. AB 5.30 centimeters, BC 5.5 centimeters, CD 3.2 centimeters, and DA 3.52 centimeters. Angle A and angle C are right angles. Find the area of ABCD to the correct number of significant digits. D 3.52 cm C c m 3 . 2 5.5 cm Solution Draw BD , separating the quadrilateral into two right triangles. In a right triangle, either leg is the base and the other leg is the altitude. A 5.30 cm B In ABD, b 5.30, and h 3.52: Area of ABD 1 2(5.30)(3.52) 9.328 In BCD, b 5.5, and h 3.2: Area of ABD 1 2(5.5)(3.2) 8.80 The first area should be given to three significant digits, the least number of significant digits involved in the calculation. Similarly, the second area should be given to two significant digits. However, the area of ABCD should be no more precise than the least precise measurement of the areas that are added. Since the least precise measurement is 8.8, given to the nearest tenth, the area of ABCD should also be written to the nearest tenth: Area of ABCD Area of ABD Area of BCD 9.328 8.80 18.128 18.1 Answer 18.1 square centimeters Note: When working with significant digits involving mult
iple operations, make sure to keep at least one extra digit for intermediate calculations to avoid roundoff error. Alternatively, when working with a calculator, you can round at the end of the entire calculation. EXERCISES Writing About Mathematics 1. ABCD is a trapezoid with AB i CD . The area of ABD is 35 square units. What is the area of ABC? Explain your answer. Areas of Irregular Polygons 281 2. ABCD is a quadrilateral. The area of ABD is 57 square inches, of BDC is 62 square inches, and of ABC is 43 square inches. What is the area of ADC? Explain your answer. Developing Skills In 3–10, find each measure to the correct number of significant digits. D C C 10.00 ft .00 ft E O 10.00 ft B A 10.0 cm B Ex. 3 Ex Ex. 5 3. ABCD is a square and arc BEC is a semicircle. If AB 10.0 cm, find the area of the figure. 4. ABCD is a trapezoid. The vertices of the trapezoid are on a circle whose center is at O. DE ' AB , OB DC = 10.00 ft, DE 8.66 ft, and AE 5.00 ft. Find the area of EBCD. 5. ABCDEF is a regular hexagon. The vertices of the hexagon are on a circle whose center is at O. OG ' BC , AB 15.0 m and OG 13.0 m. Find the area of ABCDEF. E 18.5 in. D 2.4 ft C 24.0 in. 9.60 in. C A 8.25 in. F B Ex. 6 D 2 . 4 ft ft A Ex. 7 D 24.0 cm C 7.00 cm E 5.00 cm A 12.0 cm B Ex. 8 6. ABCDE is a pentagon. DC ' ED , AC 24.0 in., and BF 8.25 in., find the area of ABCDE. ED i AC BF ' AC , and . If ED 18.5 in., DC 9.60 in., 7. ABCD is a quadrilateral. The diagonals of the quadrilateral are perpendicular to each other at E. If AE 8.7 ft, BE 5.6 ft, CE 2.4 ft, and DE 2.4 ft, find the area of ABCD. 8. ABCD is a trapezoid with BC ' AB BE 5.00 cm, and EC 7.00 cm. Find the area of triangle AED. and E a point on BC . AB 12.0 cm, DC = 24.0 cm, 282 Geometric Figures, Areas, and Volumes D C 15 yd 13 yd 18 yd B E 38 yd Ex. 9 A D 6.0 mm E 4.0 mm C 8.0 mm A Ex. 10 F 5.0 mm B 9. ABCD is a quadrilateral with DE ' AB BC 13 yd, and DE 15 yd. Find the area of ABCD. BC ' AB and . If AB 38 yd, EB 18 yd, 10. ABCD is a rectangle. AD 8 mm, DE 6 mm, EC 4 mm, and BF 5 mm. Find the area of ABFE. Applying Skills 11. Mason has a square of fabric that measures one yard on each side. He makes a straight cut from the center of one edge to the center of an adjacent edge. He now has one piece that is an isosceles triangle and another piece that is a pentagon. Find the area of each of these pieces in square inches. 12. A park is in the shape of isosceles trapezoid ABCD. The bases of the trapezoid, and measure 20.0 meters and 5.00 meters respectively. The measure of each base angle at is 60°, and the height of the trapezoid is 13.0 meters. A straight path from A to E, a CD AB point on trapezoid into two regions; a quadrilateral planted with grass and a triangle planted with shrubs and trees. Find the area of the region planted with grass to the nearest meter. and BE 10.0 meters. The path separates the , makes an angle of 30° with AB BC AB 7-7 SURFACE AREAS OF SOLIDS A right prism is a solid with congruent bases and with a height that is perpendicular to these bases. Some examples of right prisms, as seen in the diagrams, include solids whose bases are triangles, trapezoids, and rectangles. The prism is named for the shape of its base. The two bases may be any polygons, as long as they have the same size and shape. The remaining sides, or faces, are rectangles. B B h h Triangular right prism Trapezoidal right prism The surface area of a solid is the sum of the areas of the surfaces that make up the solid. The surface area of a right prism is the sum of the areas of the bases and the faces of the solid. The number of faces of a prism is equal to the number of sides of a base. Surface Areas of Solids 283 • If the base of the prism is a triangle, it has 2 bases 3 faces or 5 surfaces. • If the base of the prism is a quadrilateral, it has 2 bases 4 faces or 6 sur- faces. • If the base of the prism has n sides, it has 2 bases n faces or 2 n sur- faces. Two or more of the faces are congruent if and only if two or more of the sides of a base are equal in length. For example, if the bases are isosceles triangles, two of the faces will be congruent rectangles and if the bases are equilateral triangles, all three of the faces will be congruent rectangles. If the bases are squares, then the four faces will be congruent rectangles. The most common right prism is a rectangular solid that has rectangles as bases and as faces. Any two surfaces that have no edge in common can be the bases and the other four surfaces are the faces. If the dimensions of the rectangular solid are 3 by 5 by 4, then there are two rectangles that are 3 by 5, two that are 5 by 4 and two that are 3 by 4. The surface area of the rectangular solid is: 2(3)(5) 2(5)(4) 2(3)(4) 30 40 24 94 square units B h Rectangular solid (a prism) In general, when the dimensions of a rectangular solid are represented by l, w, and h, then the formula for the surface area is: Surface Area of a Rectangular Solid 2lw 2lh 2wh If the rectangular solid has six squares as bases and sides, the solid is a cube. If s represents the length of each side, then l s, w s, and h s. If we substitute in the formula for the area of a rectangular solid, then: S 2lw 2lh 2wh 2s(s) 2s(s) 2s(s) 2s2 2s2 2s2 6s2 Surface Area of a Cube 6s2 A right circular cylinder is a solid with two bases that are circles of the same size, and with a height that is perpendicular to these bases. Fruits and vegetables are often purchased in “cans” that are in the shape of a right circular cylinder. The label on the can corresponds to the surface area of the curved portion of the cylinder. That label is a rectangle whose length is the circumference of the can, 2pr, and whose width is the height of the can, h. Therefore, the area of the curved portion of the cylinder is 2prh and the area of each base is pr2. Surface Area of a Cylinder 2 the area of a base r r h B Right circular cylinder Surface Area of a Cylinder 2pr 2 2prh the area of the curved portion 284 Geometric Figures, Areas, and Volumes EXAMPLE 1 A rectangular solid has a square base. The height of the solid is 2 less than twice the length of one side of the square. The height of the solid is 22.2 centimeters. Find the surface area of the figure to the correct number of significant digits. Solution Let x the length of a side of the base. 2x 2 the height of the figure. 2x 2 22.2 2x 24.2 x 12.1 2x – 2 x x The surface area consists of two square bases with sides that measure 12.1 centimeters and four rectangular faces with width 12.1 and length 22.2. Surface area of the solid surface area of the bases surface area of the faces 2(12.1)2 4(12.1)(22.2) 292.82 1,074.48 1,367.3 1,370 The areas of the bases should be given to three significant digits, the least number of significant digits involved in the calculation. Similarly, the areas of the faces should be given to three significant digits. However, the surface area of the solid should be no more precise than the least precise measurement of the areas that are added. Since the least precise measurement is 1,074.48, given to the nearest ten, the surface area of the solid should also be written to the nearest ten: 1,370 square centimeters. Answer Surface area 1,370 cm2 Note: Recall that when working with significant digits involving multiple operations, you should keep at least one extra digit for intermediate calculations or wait until the end of the entire calculation to round properly. Surface Areas of Solids 285 EXERCISES Writing About Mathematics 1. A right prism has bases that are regular hexagons. The measure of each of the six sides of the hexagon is represented by a and the height of the solid by 2a. a. How many surfaces make up the solid? b. Describe the shape of each face c. Express the dimensions and the area of each face in terms of a. 2. A regular hexagon can be divided into six equilateral triangles. If the length of a side of an equilateral triangle is a, the height is 3 2 a " . For the rectangular solid described in exercise 1: a. Express the area of each base in terms of a. b. Express the surface area in terms of a. Developing Skills In 3–9, find the surface area of each rectangular prism or cylinder to the nearest tenth of a square unit. 3. Bases are circles with a radius of 18 inches. The height of the cylinder is 48 inches. 4. Bases are squares with sides that measure 27 inches. The height is 12 inches. 5. Bases are rectangles with dimensions of 8 feet by 12 feet. The height is 3 feet. 6. Bases are isosceles trapezoids with parallel sides that measure 15 centimeters and 25 cen- timeters. The distance between the parallel sides is 12 centimeters and the length of each of the equal sides is 13 centimeters. The height of the prism is 20 centimeters. 7. Bases are isosceles right triangles with legs that measure 5 centimeters. The height is 7 cen- timeters. 8. Bases are circles with a diameter of 42 millimeters. The height is 3.4 centimeters. 9. Bases are circles each with an area of 314.16 square feet. The height is 15 feet. Applying Skills 10. Agatha is using scraps of wallpaper to cover a box that is a rectangular solid whose base measures 8 inches by 5 inches and whose height is 3 inches. The box is open at the top. How many square inches of wallpaper does she need to cover the outside of the box? 11. Agatha wants to make a cardboard lid for the box described in exercise 10. Her lid will be a rectangular solid that is open at the top, with a base that is slightly larger than that of the box. She makes the base of the lid 8.1 inches by 5.1 inches with a height of 2.0 inches. To the nearest tenth of a square inch, how much wallpaper does she need to cover the outside of the lid? 286 Geometric Figures, Areas, and Volumes 12. Sandi wants to make a pillow in the shape of a right circular cylinder. The diameter of the circular ends is 10.0 inches and the length of the pillow (the height of the cylinder) is 16.0 inches. Find the number of square inches of fab
ric Sandi needs to make the pillow to the correct number of significant digits. 13. Mr. Breiner made a tree house for his son. The front and back walls of the house are trapezoids to allow for a slanted roof. The floor, roof, and remaining two sides are rectangles. The tree house is a rectangular solid. The front and back walls are the bases of this solid. The dimensions of the floor are 8 feet by 10 feet and the roof is 10 feet by 10 feet. The height of one side wall is 7 feet and the height of the other is 13 feet. The two side walls each contain a window measuring 3 feet by 3 feet. Mr. Breiner is going to buy paint for the floor and the exterior of the house, including the roof, walls, and the door. How many square feet must the paint cover, excluding the windows? 7-8 VOLUMES OF SOLIDS 10 ft 10 ft 13 ft 7 ft 3 ft 3 ft 10 ft 8 ft The volume of a solid is the number of unit cubes (or cubic units) that it contains. Both the volume V of a right prism and the volume V of a right circular cylinder can be found by multiplying the area B of the base by the height h. This formula is written as V Bh 2 4 3 V = Bh V = (lw)h = (4 · 2) · 3 = 8 · 3 = 24 cm3 To find the volume of a solid, all lengths must be expressed in the same unit of measure. The volume is then expressed in cubic units of this length. For example, in the rectangular solid or right prism shown at left, the base is a rectangle, 4 centimeters by 2 centimeters. The area, B, of this base is lw. Therefore, B 4(2) 8 cm2. Then, using a height h of 3 centimeters, the volume V of the rectangular solid Bh (8)(3) 24 cm3. For a rectangular solid, note that the volume formula can be written in two ways: V Bh or V lwh Volumes of Solids 287 To understand volume, count the cubes in the diagram on the previous page. There are 3 layers, each containing 8 cubes, for a total of 24 cubes. Note that 3 corresponds to the height, h, that 8 corresponds to the area of the base, B, and that 24 corresponds to the volume V in cubic units. A cube that measures 1 foot on each side represents 1 cubic foot. Each face of the cube is 1 square foot. Since each foot can be divided into 12 inches, the area of each face is 12 12 or 144 square inches and the volume of the cube is 12 12 12 or 1,728 cubic inches. 1 square foot 144 square inches 1 cubic foot 1,728 cubic inches A cube that measures 1 meter on each side represents 1 cubic meter. Each face of the cube is 1 square meter. Since each meter can be divided into 100 centimeters, the area of each face is 100 100 or 10,000 square centimeters and the volume of the cube is 100 100 100 or 1,000,000 cubic centimeters. 1 square meter 10,000 square centimeters 1 cubic meter 1,000,000 cubic centimeters EXAMPLE 1 A cylindrical can of soup has a radius of 1.5 inches and a height of 5 inches. Find the volume of this can: r = 1.5 in. a. in terms of p. b. to the nearest cubic inch. Solution This can is a right circular cylinder. Use the formula V Bh. Since the base is a circle whose area equals pr2, the area B of the base can be replaced by pr2. a. V Bh (pr2)h p(1.5)2(5) 11.25p O h = 5 in. b. When we use a calculator to evaluate 11.25p, the calculator gives 35.34291735 as a rational approximation. This answer rounded to the nearest integer is 35. Answers a. 11.25p cu in. b. 35 cu in. 288 Geometric Figures, Areas, and Volumes A pyramid is a solid figure with a base that is a polygon and triangular faces or sides that meet at a common point. The formula for the volume V of a pyramid is: V 5 1 3 Bh In the pyramid and prism shown here, the bases are the same in size and shape, and the heights are equal in measure. If the pyramid could be filled with water and that water poured into the prism, exactly three pyramids of water would be needed to fill the prism. In other words, the volume of a pyramid is one-third the volume of a right prism with the same base and same height as those of the pyramid. The volume of a cone is onethird of the volume of a right circular cylinder, when both the cone and the cylinder have circular bases and heights that are equal in measure. The formula for the volume of the cone is ) h ( t h g i e H Base (B) Base (B) Pyramid = prism Base (B) r Base (B) V 5 1 3Bh or V 5 1 3pr 2h As in the case of the pyramid and prism, if the cone could be filled with water and that water poured into the right circular cylinder, exactly three cones of water would be needed to fill the cylinder. C A B O Sphere A sphere is a solid figure whose points are all equally distant from one fixed point in space called its center. A sphere is not a circle, which is drawn on a flat surface or plane but similar terminology is used for a sphere. A line segment, , that joins the center O to any point on the sphere is called a radius such as of the sphere. A line segment, such as , that joins two points of the sphere and passes through its center is called a diameter of the sphere. AB OC The volume of a sphere with radius r is found by using the formula: V 5 4 3pr 3 Volumes of Solids 289 EXAMPLE 2 An ice cream cone that has a diameter of 6.4 centimeters at its top and a height of 12.2 centimeters is filled with ice cream and topped with a scoop of ice cream that is approximately in the shape of half of a sphere. Using the correct number of significant digits, how many cubic centimeters of ice cream are needed? Solution The amount of ice cream needed is approximately the volume of the cone plus one-half the volume of a sphere with a radius equal to the radius of the top of the cone. Volume of ice cream 4 3pr3 3pr2h 1 1 5 1 2 A 3p(3.2)2(12.2) 1 1 5 1 2 130.824 . . . 68.629 . . . 199.453 . . . 4 3 A B B p(3.2)3 The least precise volume is the volume of the cone, given to the nearest ten. Therefore, the answer should be rounded to the nearest ten. Note that in the answer, the zero in the tens place is significant. The zero in the ones place is not significant. Answer The volume of ice cream is 200 cubic centimeters. Error in Geometric Calculations When a linear measure is used to find area or volume, any error in the linear value will be increased in the higher-dimension calculations. EXAMPLE 3 The length of a side of a cube that is actually 10.0 centimeters is measured to be 10.5 centimeters. Find the percent error in: a. the linear measure. b. the surface area. c. the volume. d. compare the results in a–c above. Solution a. The true length is 10.0 centimeters. The measured value is 10.5 centimeters. Percent error \|measured value 2 true value\| true value 10.0 3 100% 5 0.5 0.05 100% 5% Answer \|10.5 – 10.0\| 100% 10.0 3 100% 290 Geometric Figures, Areas, and Volumes b. For a cube, S 6s2. Using the measured value: S 6(10.5)2 661.5 Using the true value: S 6(10.0)2 600.0 Percent error \|661.5 – 600.0\| 600.0 3 100% 5 61.5 600.0 3 100% 0.1025 100% 10.25% Answer c. For a cube, V s3. Using the measured value: V (10.5)3 1,157.625 Using the true value: V (10.0)3 1,000 Percent error \|1,157.625 – 1,000\| 1,000 3 100% 5 157.625 1,000 3 100% 0.157625 100% ≈ 15.76% Answer d. The error in the linear measure is 5%. The error increases to 10.25% when the surface area is calculated. The error increased even more to 15.76% when the volume is calculated. Answer EXERCISES Writing About Mathematics 1. A chef purchases thin squares of dough that she uses for the top crust of pies. The squares measure 8 inches on each side. From each square of dough she cuts either one circle with a diameter of 8 inches or four circles, each with a diameter of 4 inches. a. Compare the amount of dough left over when she makes one 8-inch circle with that left over when she makes four 4-inch circles. b. The chef uses the dough to form the top crust of pot pies that are 1.5 inches deep. Each pie is approximately a right circular cylinder. Compare the volume of one 8-inch pie to that of one 4-inch pie. 2. Tennis balls can be purchased in a cylindrical can in which three balls are stacked one above the other. How does the radius of each ball compare with the height of the can in which they are packaged? Developing Skills In 3–6, use the formula V lwh and the given dimensions to find the volume of each rectangular solid using the correct number of significant digits. 3. l 5.0 ft, w 4.0 ft, h 7.0 ft 31 5. l m, w cm, h 85 cm 4 21 2 4. l 8.5 cm, w 4.2 cm, h 6.0 cm 6. l 7.25 in., w 6.40 in., h 0.25 ft Volumes of Solids 291 7. Find the volume of a cube if each edge measures 8. The measure of each edge of a cube is represented by (2y 3) centimeters. Find the volume centimeters. 83 5 of the cube when y 7.25. 9. The base of a right prism is a triangle. One side of the triangular base measures 8 centime- ters and the altitude to that side measures 6 centimeters. The height h of the prism measures 35 millimeters. a. Find the area of the triangular base of the prism. b. Find the volume of the prism. 10. The base of a right prism is a trapezoid. This trapezoid has bases that measure 6 feet and 10 feet and an altitude that measures 4 feet. The height h of the prism is 2 yards. a. Find the area of the trapezoidal base of the prism. b. Find the volume of the prism. 11. The base of a pyramid has an area B of 48 square millimeters and a height h of 13 millime- ters. What is the volume of the pyramid? 12. The height of a pyramid is 4 inches, and the base is a rectangle 6 inches long and wide. What is the volume of the pyramid? 31 2 inches 13. A right circular cylinder has a base with a radius of 24.1 centimeters and a height of 17.3 centimeters. a. Express the volume of the cylinder in terms of p. b. Find the volume of the cylinder to the nearest hundred cubic centimeters. 14. A right circular cylinder has a base with a diameter of 25 meters and a height of 15 meters. a. Express the volume of the cylinder in terms of p. b. Find the volume of the cylinder to the nearest ten cubic meters. 15. The base of a cone has a radius of 7 inches. The height of the cone is 5 inches. a. Find the volume of the cone in terms of p. b. Find the volume o
f the cone to the nearest cubic inch. 16. The base of a cone has a radius of 7 millimeters. The height of the cone is 2 centimeters. a. Find the volume of the cone in terms of p. b. Express the volume of the cone to the nearest cubic centimeter. 17. A sphere has a radius of 12.5 centimeters. a. Find the volume of the sphere in terms of p. b. Express the volume of the sphere to the nearest cubic centimeter. 18. A sphere has a diameter of 3 feet. a. Find the volume of the sphere in terms of p. b. Express the volume of the sphere to the nearest cubic foot. 292 Geometric Figures, Areas, and Volumes 19. The side of a cube that is actually 8 inches is measured to be 7.6 inches. Find, to the nearest tenth of a percent, the percent error in: a. the length of the side. b. the surface area. c. the volume. Applying Skills 20. An official handball has a diameter of 4.8 centimeters. Find its volume: a. to the nearest cubic centimeter b. to the nearest cubic inch. 21. A tank in the form of a right circular cylinder is used for storing water. It has a diameter of 12 feet and a height of 14 feet. How many gallons of water will it hold? (1 cubic foot contains 7.5 gallons.) 22. Four pieces of cardboard that are 8.0 inches by 12 inches and two pieces that are 12 inches by 12 inches are used to form a rectangular solid. a. What is the surface area of the rectangular solid formed by the six pieces of card- board? b. What is the volume of the rectangular solid in cubic inches? c. What is the volume of the rectangular solid in cubic feet? 23. A can of soda is almost in the shape of a cylinder with a diameter of 6.4 centimeters and a height of 12.3 centimeters. a. What is the volume of the can? b. If there are 1,000 cubic centimeters in a liter, find how many liters of soda the can holds. In 24–26, express each answer to the correct number of significant digits. 24. Cynthia used a shipping carton that is a rectangular solid measuring 12.0 inches by 15.0 inches by 3.20 inches. What is the volume of the carton? 25. The highway department stores sand in piles that are approximately the shape of a cone. What is the volume of a pile of sand if the diameter of the base is 7.0 yards and the height of the pile is 8.0 yards? 26. The largest pyramid in the world was built around 2500 B.C. by Khufu, or Cheops, a king of ancient Egypt. The pyramid had a square base 230˙ meters (756 feet) on each side, and a height of 147 meters (482 feet). (The length of a side of the base is given to the nearest meter. The zero in the ones place is significant.) Find the volume of Cheops’ pyramid using: a. cubic meters b. cubic feet CHAPTER SUMMARY Chapter Summary 293 Point, line, and plane are undefined terms that are used to define other terms. A line segment is a part of a line consisting of two points on the line, called endpoints, and all of the points on the line between the endpoints. A ray is a part of a line that consists of a point on the line and all of the points on one side of that point. An angle is the union of two rays with a common endpoint. Two angles are complementary if the sum of their measures is 90°. If the measure of an angle is x°, the measure of its complement is (90 x)°. Two angles are supplementary if the sum of their measures is 180°. If the measure of an angle is x°, the measure of its supplement is (180 x)°. A linear pair of angles are adjacent angles that are supplementary. Two angles are vertical angles if the sides of one are opposite rays of the sides of the other. Vertical angles are congruent. If two parallel lines are cut by a transversal, then: • The alternate interior angles are congruent. • The alternate exterior angles are congruent. • The corresponding angles are congruent. • Interior angles on the same side of the transversal are supplementary. The sum of the measures of the angles of a triangle is 180°. The base angles of an isosceles triangle are congruent. An equilateral triangle is equiangular. The sum of the measures of the angles of a quadrilateral is 360°. The sum of the measures of the angles of any polygon with n sides is 180(n 2). If the measures of the bases of a trapezoid are b1 and b2 and the measure of 1 2h(b11b2) . the altitude is h, then the formula for the area of the trapezoid is A Formulas for Surface Area Formulas for Volume Rectangular solid: S 2lw 2lh 2wh Cube: S 6s2 Cylinder: S 2pr2 2prh Any right prism: V Bh Rectangular solid: V lwh Right circular cylinder: V pr2h 1 3Bh 1 3pr2h Pyramid: V Cone: V Cube: V s3 Sphere: V 4 3pr3 294 Geometric Figures, Areas, and Volumes VOCABULARY 7-1 Undefined term • Point • Line • Straight line • Plane • Axiom (postulate) • Line segment (segment) • Endpoints of a segment • Measure of a line segment (length of a line segment) • Half-line • Ray • Endpoint of a ray • Opposite rays • Angle • Vertex of an angle • Sides an angle • Degree • Right angle • Acute angle • Obtuse angle • Straight angle • Perpendicular 7-2 Adjacent angles • Complementary angles • Complement • Supplementary angles • Linear pair of angles • Vertical angles • Congruent angles • Theorem 7-3 Parallel lines • Transversal • Interior angles • Alternate interior angles • Exterior angles • Alternate exterior angles • Interior angles on the same side of the transversal • Corresponding angles 7-4 Polygon • Sides of a polygon • Vertices • Triangle • Acute triangle • Equiangular triangle • Right triangle • Obtuse triangle • Legs of a right triangle • Hypotenuse • Scalene triangle • Isosceles triangle • Equilateral triangle • Legs of an isosceles triangle • Base of an isosceles triangle • Congruent line segments • Vertex angle of an isosceles triangle • Base angles of an isosceles triangle • Construction • Compass • Straightedge • Perpendicular bisector • Angle bisector 7-5 Quadrilateral • Consecutive angles • Opposite angles • Trapezoid • Bases of a trapezoid • Parallelogram • Rectangle • Rhombus • Square • Isosceles trapezoid 7-7 Right prism • Face • Surface area • Rectangular solid • Right circular cylinder 7-8 Volume • Pyramid • Cone • Sphere • Center of a sphere • Radius of a sphere • Diameter of a sphere REVIEW EXERCISES 1. If g AB g CD find mAEC. and intersect at E, mAEC x 10, and mDEB 2x 30, 2. If two angles of a triangle are complementary, what is the measure of the third angle? 3. The measure of the complement of an angle is 20° less than the measure of the angle. Find the number of degrees in the angle. 4. If each base angle of an isosceles triangle measures 55°, find the measure of the vertex angle of the triangle. 5. The measures of the angles of a triangle are consecutive even integers. What are the measures of the angles? Review Exercises 295 g CD , and these lines are cut by E In 6–8, g AB transversal is parallel to g EF at points G and H, respectively. 6. If mAGH 73, find mGHD. 7. If mEGB 70 and mGHD 3x 2, find x. 8. If mHGB 2x 10 and mGHD x 20, find mGHD. A G C H F B D 9. In ART, mA y 10, mR 2y, and mT 2y 30. a. Find the measure of each of the three angles. b. Is ART acute, right, or obtuse? c. Is ART scalene, isosceles but not equilateral or equiangular? 10. The measure of each base angle of an isosceles triangle is 15° more than the measure of the vertex angle. Find the measure of each angle. 11. The measure of an angle is 20° less than 3 times the measure of its supple- ment. What is the measure of the angle and its supplement? 12. In ABC, the measure of B is the measure of A, and the measure of C is 5 2 the measure of A. What are the measures of the three angles? 3 2 13. The measure of one angle is 3 times that of another angle, and the sum of these measures is 120°. What are the measures of the angles? 14. The measure of the smaller of two supplementary angles is of the mea- 4 5 sure of the larger. What are the measures of the angles? 15. The vertices of a trapezoid are A(3, 1), B(7, 1), C(5, 5), and D(7, 5). a. Draw ABCD on graph paper. b. E is the point on CD such that CD ' AE . What are the coordinates of E? c. Find AB, CD, and AE. d. Find the area of trapezoid ABCD. 16. In parallelogram ABCD, AB 3x 4 , BC 2x 5, and CD x 18. Find the measure of each side of the parallelogram. 17. A flowerpot in the shape of a right circular cylinder has a height of 4.5 inches. The diameter of the base of the pot is 4.1 inches. Find the volume of the pot to the nearest tenth. 18. Natali makes a mat in the shape of an octagon (an eight-sided polygon) by cutting four isosceles right triangles of equal size from the corners of a 9 by 15 inch rectangle. If the measure of each of the equal sides of the triangles is 2 inches, what is the area of the octagonal mat? 296 Geometric Figures, Areas, and Volumes 19. Marvin measures a block of wood and records the dimensions as 5.0 centimeters by 3.4 centimeters by 4.25 centimeters. He places the block of wood in a beaker that contains 245 milliliters of water. With the block completely submerged, the water level rises to 317 milliliters. a. Use the dimensions of the block to find the volume of the block of wood. b. What is the volume of the block of wood based on the change in the water level? c. Marvin knows that 1 ml 1 cm3. Can the answers to parts a and b both be correct? Explain your answer. 20. A watering trough for cattle is in the shape of a prism whose ends are the bases of the prism and whose length is the height of the prism. The ends of the trough are trapezoids with parallel edges 31 centimeters and 48 centimeters long and a height of 25 centimeters. The length of the trough is 496 centimeters. a. Find the surface area of one end of the trough. b. If the trough is filled to capacity, how many cubic centimeters of water does it hold? c. How many liters of water does the trough hold? (1 liter 1,000 cm3) Exploration A sector of a circle is a fractional part of the interior of a circle, determined by an angle whose vertex is at the center of the circle (a central angle). The area of a sector depends on the measure of its central angle. For example, • If
the central angle equals 90°, then the area of the sector is one-fourth the area of the circle, or 90 360 pr2. • If the central angle equals 180°, then the area of the sector is one-half the area of the circle, or 180 360 pr2. 0 r C The shaded region represents a sector with central angle u. • If the central angle equals 270°, then the area of the sector is three-fourths the area of the circle, or 270 360 pr2. In general, if the measure of the central angle is u (theta), then the area of the sector is Area of a sector u 360 pr 2 For this Exploration, use your knowledge of Geometry and the formula for the area of a sector to find the areas of the shaded regions. Express your answers in terms of p. Assume that all the arcs that are drawn are circular. Cumulative Review 297 30° 3 1 2 100° 1 2 2 a. c. e. 1 1 1 1 1 1 b. d. f. 1 1 1 45° 1 45° 45° 1 1 1 √3 2 60° 45° 1 1 1 √3 2 60° 1 1 CUMULATIVE REVIEW CHAPTERS 1–7 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. Which of the following inequalities is true? (1) 4 (3) 5 (2) 4 5 (3) (3) (3) 4 5 (4) 5 4 (3) 298 Geometric Figures, Areas, and Volumes 2. Which of the following is an example of the use of the associative property? (1) 2(x 5) 2(5 x) (2) 2(x 5) 2x 2(5) (3) 2 (x 5) 2 (5 x) (4) 2 (5 x) (2 5) x 3. Which of the following is not a rational number? (1) 0.09 " (2) 0.9 " 4. The product (a2b)(a3b) is equivalent to (3) 2–3 (4) 0.15 (1) a6b 5. The solution set of (1) {24} (2) a5b 3x 1 7 5 1 2 (2) {24} 6x 2 5 6. The sum of b2 7 and b2 3b is (3) a5b2 (4) a6b2 is (3) {6} (4) {6} (1) b4 4b (2) b4 3b 7 (3) 2b2 4b (4) 2b2 3b 7 7. Two angles are supplementary. If the measure of one angle is 85°, the mea- sure of the other is (1) 5° (2) 85° (3) 95° (4) 180° 8. In trapezoid ABCD, AB i CD . If mA is 75°, then mD is (1) 15° (2) 75° (3) 105° (4) 165° 9. The graph at the right is the solution set of (1) 2 x 3 (2) 2 x 3 (3) 2 x 3 (4) (2 x) or (x 3) –4 –3 –2 –1 0 1 2 3 4 10. When a 1.5, 3a2 a equals (1) 5.25 (2) 5.25 (3) 18.75 (4) 21.75 Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. The area of a rectangle is (x2 6x 8) square inches and its length is (x 4) inches. Express the width of the rectangle in terms of x. 12. Solve the following equation for x . Cumulative Review 299 Part III Answer all questions in this part. Each correct answer will receive 3 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. Mr. Popowich mailed two packages. The larger package weighed 12 ounces more than the smaller. If the total weight of the packages was 17 pounds, how much did each package weigh? 14. a. Solve for x in terms of a and b: ax 3b 7. b. Find, to the nearest hundredth, the value of x when a b . 5 " Part IV and 3 " Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. Calvin traveled 600 miles, averaging 40 miles per hour for part of the trip and 60 miles per hour for the remainder of the trip. The entire trip took 11 hours. How long did Calvin travel at each rate? 16. A box used for shipping is in the shape of a rectangular prism. The bases are right triangles. The lengths of the sides of the bases are 9.0, 12, and 15 feet. The height of the prism is 4.5 feet. a. Find the surface area of the prism. Express the answer using the correct number of significant digits. b. Find the volume of the prism. Express the answer using the correct number of significant digits. CHAPTER 8 CHAPTER TABLE OF CONTENTS 8-1 The Pythagorean Theorem 8-2 The Tangent Ratio 8-3 Applications of the Tangent Ratio 8-4 The Sine and Cosine Ratios 8-5 Applications of the Sine and Cosine Ratios 8-6 Solving Problems Using Trigonometric Ratios Chapter Summary Vocabulary Review Exercises Cumulative Review 300 TRIGONOMETRY OF THE RIGHT TRIANGLE The accurate measurement of land has been a critical challenge throughout the history of civilization. Today’s land measurement problems are not unlike those George Washington might have solved by using measurements made with a transit, but the modern surveyor has available a total workstation including EDM (electronic distance measuring) and a theodolite for angle measurement. Although modern instruments can perform many measurements and calculations, the surveyor needs to understand the principles of indirect measurement and trigonometry to correctly interpret and apply these results. In this chapter, we will begin the study of a branch of mathematics called trigonometry. The word trigonometry is Greek in origin and means “measurement of triangles.” Although the trigonometric functions have applications beyond the study of triangles, in this chapter we will limit the applications to the study of right triangles. 8-1 THE PYTHAGOREAN THEOREM The Pythagorean Theorem 301 The solutions of many problems require the measurement of line segments and angles. When we use a ruler or tape measure to determine the length of a segment, or a protractor to find the measure of an angle, we are taking a direct measurement of the segment or the angle. In many situations, however, it is inconvenient or impossible to make a measurement directly. For example, it is difficult to make the direct measurements needed to answer the following questions: What is the height of a 100-year-old oak tree? What is the width of a river? What is the distance to the sun? We can answer these questions by using methods that involve indirect measurement. Starting with some known lengths of segments or angle measures, we apply a formula or a mathematical relationship to indirectly find the measurement in question. Engineers, surveyors, physicists, and astronomers frequently use these trigonometric methods in their work. B The figure at the left represents a right triangle. Recall that such a triangle , which is contains one and only one right angle. In right triangle ABC, side opposite the right angle, is called the hypotenuse. AB The hypotenuse is the longest side of the triangle. The other two sides of the , form the right angle. They are called the legs of the right and AC BC triangle, triangle. C A More than 2,000 years ago, the Greek mathematician Pythagoras demonstrated the following property of the right triangle, which is called the Pythagorean Theorem: B a c C b A The Pythagorean Theorem In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. If we represent the length of the hypotenuse of right triangle ABC by c and the lengths of the other two sides by a and b, the Theorem of Pythagoras may be written as the following formula: c2 a2 b2 To show that this relationship is true for any right triangle ABC with the length of the hypotenuse represented by c and the lengths of the legs represented by a and b, consider a square with sides (a b). The area of the square is (a b)2. However, since it is divided into four triangles and one smaller square, its area can also be expressed as 302 Trigonometry of the Right Triangle Area of the square area of the four triangles area of the smaller square 4 1 2 ab c2 A Although the area is written in two different ways, both expressions are equal. B Thus, (a b)2 4 1 2 ab c2. A If we simplify, we obtain the relationship of the Pythagorean Theorema b)2 4 ab c2 1 2 A B a2 2ab b2 2ab c2 Expand the binomial term (a b)2. a2 2ab 2ab b2 2ab 2ab c2 Subtract 2ab from both sides of the equality. a2 b2 c2 Statements of the Pythagorean Theorem Two statements can be made for any right triangle where c represents the length of the hypotenuse (the longest side) and a and b represent the lengths of the other two sides. 1. If a triangle is a right triangle, then the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. If a triangle is a right triangle, then c2 a2 b2. 2. If the square of the length of the longest side of a triangle is equal to the sum of the squares of the lengths of the other two sides, the triangle is a right triangle. If c2 a2 b2 in a triangle, then the triangle is a right triangle. If we know the lengths of any two sides of a right triangle, we can find the length of the third side. For example, if the measures of the legs of a right triangle are 7 and 9, we can write: c2 a2 b2 c2 72 92 c2 49 81 c2 130 The Pythagorean Theorem 303 To solve this equation for c, we must do the opposite of squaring, that is, we must find the square root of 130. There are two square roots of 130, and 2 130 . 130 There are two things that we must consider here when finding the value of c. which we write as 130 " " " 1 1. Since c represents the length of a line segment, only the positive number is 1 an acceptable value. Therefore, c 130 . 2. There is no rational number that has a square of 130. The value of c is an irrational number. However, we usually use a calculator to find a rational approximation for the irrational number. " ENTER: 2nd ¯ 130 ENTER DISPLAY Therefore, to the nearest tenth, the length of the hypotenuse is 11.4. Note that the calculator gives only the positive rational approximation of the square root of 130. EXAMPLE 1 A ladder is placed 5 feet from the foot of a wall. The top of the ladder reaches a point 12 feet above the ground. Find the length of the ladder. Solution The ladder, the wall, and th
e ground form a right triangle. The length of the ladder is c, the length of the hypotenuse of the right triangle. The distance from the foot of the ladder to the wall is a 5, and the distance from the ground to the top of the ladder is b 12. Use the Theorem of Pythagoras. c2 a2 b2 c2 52 122 c2 25 144 c2 169 c 6 " 169 5 613 Reject the negative value. Note that in this case the exact value of c is a rational number because 169 is a perfect square. Answer The length of the ladder is 13 feet. c b = 12 a = 5 304 Trigonometry of the Right Triangle EXAMPLE 2 The hypotenuse of a right triangle is 36.0 centimeters long and one leg is 28.5 centimeters long. a. Find the length of the other leg to the nearest tenth of a centimeter. b. Find the area of the triangle using the correct number of significant digits. Solution a. The length of the hypotenuse is c = 36.0 and the length of one leg is a = 28.5. The length of the other leg is b. Substitute the known values in the Pythagorean Theorem. c2 a2 b2 36.02 28.52 b2 1,296 812.25 b2 483.75 b2 b 483.75 " 6 Reject the negative value. Use a calculator to find a rational approximation of the value of b. A calculator displays 21.99431745. Round the answer to the nearest tenth. Answer The length of the other leg is 22.0 centimeters. b. Area of ABC 1 2bh 1 2(28.5)(22.0) 313.5 Since the lengths are given to three significant digits, we will round the area to three significant digits. Answer The area of ABC is 314 square centimeters. EXAMPLE 3 Is a triangle whose sides measure 8 centimeters, 7 centimeters, and 4 centimeters a right triangle? Solution If the triangle is a right triangle, the longest side, whose measure is 8, must be the hypotenuse. Then: 82 c2 a2 b2 72 1 42 5? 5? 64 64 65 ✘ 49 1 16 Answer The triangle is not a right triangle. The Pythagorean Theorem 305 EXERCISES Writing About Mathematics 1. A Pythagorean triple is a set of three positive integers that make the equation c2 a2 b2 true. Luz said that 3, 4, and 5 is a Pythagorean triple, and, for any positive integer k, 3k, 4k, and 5k is also a Pythagorean triple. Do you agree with Luz? Explain why or why not. 2. Regina said that if n is a positive integer, 2n 1, 2n2 2n, and 2n2 2n 1 is a Pythagorean triple. Do you agree with Regina? Explain why or why not. Developing Skills In 3–11, c represents the length of the hypotenuse of a right triangle and a and b represent the lengths of the legs. For each right triangle, find the length of the side whose measure is not given. 3. a 3, b 4 6. c 13, a 12 9. a , b 2 " 2 " 4. a 8, b 15 7. c 17, b 15 10. a 1, b 3 " 5. c 10, a 6 8. c 25, b 20 11. a , c 3 8 " In 12–17, c represents the length of the hypotenuse of a right triangle and a and b represent the lengths of the legs. For each right triangle: a. Express the length of the third side in radical form. b. Express the length of the third side to the nearest hundredth. 12. a 2, b 3 15. a 7, b 2 13. a 3, b 3 16. b , c 3 " 14 " 14. a 4, c 8 17. a , c 6 7 " In 18–21, find x in each case and express irrational results in radical form. 18. 19. 20. 21. √ 2 0 0 4x 3x 90° x 90° 8 x 2x 9 x 90° x 6 90° x1 2 In 22–27, find, in each case, the length of the diagonal of a rectangle whose sides have the given measurements. 22. 7 inches by 24 inches 23. 9 centimeters by 40 centimeters 24. 28 feet by 45 feet 25. 17 meters by 144 meters 26. 15 yards by 20 yards 27. 18 millimeters by 24 millimeters 306 Trigonometry of the Right Triangle 28. The diagonal of a rectangle measures 65 centimeters. The length of the rectangle is 33 cen- timeters. Consider the measurements to be exact. a. Find the width of the rectangle. b. Find the area of the rectangle. 29. Approximate, to the nearest inch, the length of a rectangle whose diagonal measures 25.0 inches and whose width is 18.0 inches. 30. The altitude to the base of a triangle measures 17.6 meters. The altitude divides the base into two parts that are 12.3 meters and 15.6 meters long. What is the perimeter of the triangle to the nearest tenth of a meter? Applying Skills 31. A ladder 39 feet long leans against a building and reaches the ledge of a window. If the foot of the ladder is 15 feet from the foot of the building, how high is the window ledge above the ground to the nearest foot? 32. Mr. Rizzo placed a ladder so that it reached a window 15.0 feet above the ground when the foot of the ladder was 5.0 feet from the wall. Find the length of the ladder to the nearest tenth of a foot. 33. Mrs. Culkowski traveled 24.0 kilometers north and then 10.0 kilometers east. How far was she from her starting point? BC AB and 34. One day, Ronnie left his home at A and reached his , the sides school at C by walking along of a rectangular open field that was muddy. The dimensions of the field are 1,212 feet by 885 feet. When he was ready to return home, the field was dry and Ronnie decided to take a shortcut by walking diagonally across the field, along nearest whole foot, how much shorter was the trip home than the trip to school? . To the AC D A 885 ft C B 35. Corry and Torry have a two-way communication device that has a range of one-half mile (2,640 feet). Torry lives 3 blocks west and 2 blocks north of Corry. If the length of each block is 600 feet, can Corry and Torry communicate using this device when each is home? Explain your answer. 36. A baseball diamond has the shape of a square with the bases at the vertices of the square. If the distance from home plate to first base is 90.0 feet, approximate, to the nearest tenth of a foot, the distance from home plate to second base. 8-2 THE TANGENT RATIO Naming Sides The Tangent Ratio 307 In a right triangle, the hypotenuse, which is the longest side, is opposite the right angle. The other two sides in a right triangle are called the legs. However, in trigonometry of the right triangle, we call these legs the opposite side and the adjacent side to describe their relationship to one of the acute angles in the triangle. Notice that ABC is the same right triangle in both figures below, but the position names we apply to the legs change with respect to the angles. B B h y p o te n u se side opposite ∠A h y p o te n u se side adjacent to ∠B A side adjacent to ∠A C In ABC: BC AC is opposite A; is adjacent to A. A side opposite ∠B C In ABC: AC BC is opposite B; is adjacent to B. Similar Triangles Three right triangles are drawn to coincide at vertex A. Since each triangle contains a right angle as well as A, we know that the third angles of each triangle are congruent. When three angles of one triangle are congruent to the three angles of another, the triangles are similar. The corresponding sides of similar triangles are in proportion. Therefore: CB BA 5 ED DA 5 GF FA G E C A B D F 308 Trigonometry of the Right Triangle The similar triangles shown in the previous page, ABC, ADE, and AFG, are separated and shown below. G E C A B In ABC: CB BA is opposite A; is adjacent to A. A D A F In ADE: ED DA is opposite A; is adjacent to A. In AFG: GF FA is opposite A; is adjacent to A. CB BA 5 ED DA 5 GF FA a constant for ABC, Therefore, ADE, AFG and for any right triangle similar to these triangles, that is, for any right triangle with an acute angle congruent to A. This ratio is called the tangent of the angle. length of side opposite /A length of side adjacent to /A DEFINITION The tangent of an acute angle of a right triangle is the ratio of the length of the side opposite the acute angle to the length of the side adjacent to the acute angle. For right triangle ABC, with mC 90, the definition of the tangent of A is as follows: tangent A length of side opposite /A length of side adjacent to /A BC AC 5 a b By using “tan A” as an abbreviation for tangent A, “opp” as an abbreviation for the length of the leg opposite A, and “adj” as an abbreviation for the length of the leg adjacent to A, we can shorten the way we write the relationship given above as follows: B a c A b C tan A opp adj 5 BC AC 5 a b Finding Tangent Ratios on a Calculator The length of each side of equilateral triangle ABD is 2. The altitude B to hypotenuse rule to find BC. from forms two congruent right triangles with AC 1. We can use the AD BC (AC)2 (BC)2 (AB)2 12 (BC)2 22 1 (BC)2 4 (BC)2 3 BC 3 " The measure of each angle of an equilateral triangle is BC 60°. Therefore we can use the lengths of to find the exact value of the tangent of a 60° angle. and AC The Tangent Ratio 309 B 2 2 A 1 C 1 D tan 60° opp adj 5 BC 3 1 5 AC 5 " 3 " But how can we find the constant value of the tangent ratio when the right triangle has an angle of 40° or 76°? Since we want to work with the value of this ratio for any right triangle, no matter what the measures of the acute angles may be, mathematicians have compiled tables of the tangent values for angles with measures from 0° to 90°. Also, a calculator has the ability to display the value of this ratio for any angle. We will use a calculator to determine these values. The measure of an angle can be given in degrees or in radians. In this book, we will always express the measure of an angle in degrees. A graphing calculator can use either radians or degrees. To place the calculator in degree mode, , then use the down arrow and the right arrow keys to highlight press MODE Deg. Press and each time you turn it on. ENTER 2nd QUIT . Your calculator will be in degree mode CASE 1 Given an angle measure, find the tangent ratio. We saw that tan 60° is equal to 3 . The calculator will display this value as " an approximate decimal. To find tan 60°, enter the sequence of keys shown below. ENTER: TAN 60 ) ENTER DISPLAY , The value given in the calculator display is the rational approximation of " the value of tan 60° that we found using the ratio of the lengths of the legs of a right triangle with a 60° angle. Therefore, to the nearest ten-thousandth, tan 60° 1.7321. 310 Trigonometry of the Right Triangle CASE 2 Given a tangent ratio, find the angle measure. The value of the tangent ratio is different
for each different angle measure from 0° to 90°. Therefore, if we know the value of the tangent ratio, we can find the measure of the acute angle that has this tangent ratio.The calculator key used to do this is labeled . We can think of tan1 as “the angle whose tangent is.” Therefore, tan1 (0.9004) can be read as “the angle whose tangent is 0.9004.” and is accessed by first pressing TAN1 2nd To find the measure of A from the calculator, we use the following sequences of keys. ENTER: 2nd TAN1 0.9004 ) ENTER DISPLAY The measure of A to the nearest degree is 42°. EXAMPLE 1 In ABC, C is a right angle, BC = 3, AC = 4, and AB = 5. a. Find: (1) tan A (2) tan B (3) mA to the nearest degree (4) mB to the nearest degree b. Show that the acute angles of the triangle are complementary. Solution a. (1) tan A AC 5 3 4 BC 5 4 (2) tan B 3 Use a calculator to find the measures of A and B. opp adj 5 BC opp adj 5 AC Answer Answer (3) ENTER: 2nd TAN1 3 4 ) ENTER DISPLAY To the nearest degree, mA 37. Answer B 3 C 5 4 A The Tangent Ratio 311 (4) ENTER: 2nd TAN1 4 3 ) ENTER DISPLAY To the nearest degree, mB = 53. Answer b. mA mB 36.869889765 53.13010235 90.000000. Therefore, the acute angles of ABC are complementary. Answer Note: In a right triangle, the tangents of the two acute angles are reciprocals. EXERCISES Writing About Mathematics 1. Explain why the tangent of a 45° angle is 1. 2. Use one of the right triangles formed by drawing an altitude of an equilateral triangle to find tan 30°. Express the answer that you find to the nearest hundred-thousandth and compare this result to the valued obtained from a calculator. Developing Skills In 3–6, find: a. tan A b. tan B 3. √18 4. B 3 90° A 3 C A B 5 C 90° 13 12 5. B 6. 5 C 90° √61 6 A B A k 90° C p t 7. In ABC, mC 90, AC 6, and AB 10. Find tan A. 8. In RST, mT 90, RS 13, and ST 12. Find tan S. In 9–16, use a calculator to find each of the following to the nearest ten-thousandth: 9. tan 10° 13. tan 1° 10. tan 25° 14. tan 89° 11. tan 70° 15. tan 36° 12. tan 55° 16. tan 67° In 17–28, in each of the following, use a calculator to find the measure of A to the nearest degree. 17. tan A 0.0875 20. tan A 1.0000 18. tan A 0.3640 21. tan A 2.0503 19. tan A 0.5543 22. tan A 3.0777 312 Trigonometry of the Right Triangle 23. tan A 0.3754 26. tan A 0.3500 24. tan A 0.7654 27. tan A 0.1450 25. tan A 1.8000 28. tan A 2.9850 29. Does the tangent of an angle increase or decrease as the degree measure of the angle increases from 1° to 89°? 30. a. Use a calculator to find tan 20° and tan 40° to the nearest ten-thousandth. b. Is the tangent of the angle doubled when the measure of the angle is doubled? Applying Skills 31. In ABC, mC = 90, AC = 6, and BC = 6. a. Find tan A. b. Find the measure of A. 32. In ABC, mC = 90, BC = 4, and AC = 9. a. Find tan A. b. Find the measure of A to the nearest degree. c. Find tan B. d. Find the measure of B to the nearest degree. 33. In rectangle ABCD, AB 10 and BC 5. a. Find tan CAB. b. Find the measure of CAB to the nearest degree. c. Find tan CAD. d. Find the measure of CAD to the nearest degree. 34. In ABC, C is a right angle, mA = 45, AC = 4, BC = 4, and AB a. Using the given lengths, write the ratio for tan A. 4 " 2 . b. Use a calculator to find tan 45°. 35. In RST, T is a right angle and r, s, and t are lengths of sides. Using these lengths: a. Write the ratio for tan R. b. Write the ratio for tan S. c. Use parts a and b to find the numerical value of the product R (tan R)(tan S). B 4 C S r T 4√2 45° 4 A t s Applications of the Tangent Ratio 313 8-3 APPLICATIONS OF THE TANGENT RATIO The tangent ratio is often used to make indirect measurements when the measures of a leg and an acute angle of a right triangle are known. Angle of Elevation and Angle of Depression is the line of sight and When a telescope or some similar instrument is used to sight the top of a telephone pole, the instrument is elevated (tilted upward) from a horizontal position. Here, g OT is the horizontal line. The angle of elevation is the angle determined by the rays that are parts of the horizontal line and the line of sight when looking upward. Here, TOA is the angle of elevation. g OA When an instrument is used to sight a boat from a cliff, the instrument is depressed (tilted downward) g OB g OH from a horizontal position. Here, is the line of sight is the horizontal line. The angle of depression and is the angle determined by the rays that are parts of the horizontal line and of the line of sight when looking downward. Here, HOB is the angle of depression. T line of sight angle of elevation horizontal line A horizontal line O angle of depression line of sight O H B A g BA g BO Note that, if is a horizontal line and is the line of sight from the boat to the top of the cliff, ABO is called the angle of elevation from the boat to the top of the cliff. Since is a transversal, alternate interior and angles are congruent, namely, HOB ABO. Thus, the angle of elevation measured from B to O is congruent to the angle of depression measured from O to B. g HO g BA g OB 7 Using the Tangent Ratio to Solve Problems Procedure To solve a problem by using the tangent ratio: 1. For the given problem, make a diagram that includes a right triangle. Label the known measures of the sides and angles. Identify the unknown quantity by a variable. 2. If for the right triangle either (1) the lengths of two legs or (2) the length of one leg and the measure of one acute angle are known, write a formula for the tangent of an acute angle. 3. Substitute known values in the formula and solve the resulting equation for the unknown value. 314 Trigonometry of the Right Triangle EXAMPLE 1 Find to the nearest degree the measure of the angle of elevation of the sun when a vertical pole 6.5 meters high casts a shadow 8.3 meters long. Solution The angle of elevation of the sun is the same as A, the angle of elevation to the top of the pole from A, the tip of the shadow. Since the vertical pole and the shadow are the legs of a right triangle opposite and adjacent to A, use the tangent ratio. B 6.5 m tan A opp adj 5 BC AC 5 6.5 8.3 A 8.3 m C ENTER: 2nd TAN1 6.5 8.3 ) ENTER DISPLAY Answer To the nearest degree, the measure of the angle of elevation of the sun is 38°. EXAMPLE 2 At a point on the ground 39 meters from the foot of a tree, the measure of the angle of elevation of the top of the tree is 42°. Find the height of the tree to the nearest meter. Solution Let T be the top of the tree, A be the foot of the tree, and B be the point on the ground 39 meters from A. Draw ABT, and label the diagram: mB 42, AB 39. Let x height of tree (AT). The height of the tree is the length of the perpendicular from the top of the tree to the ground. Since the problem involves the measure of an acute angle and the measures of the legs of a right triangle, use the tangent ratio: T x tan B tan B tan 42° opp adj AT BA x 39 42° B 39 m A Substitute the given values. x 39 tan 42° Use a calculator for the computation: x 35.11575773 Solve for x. Answer To the nearest meter, the height of the tree is 35 meters. Applications of the Tangent Ratio 315 EXAMPLE 3 From the top of a lighthouse 165 feet above sea level, the measure of the angle of depression of a boat at sea is 35.0°. Find to the nearest foot the distance from the boat to the foot of the lighthouse. Solution Let L be the top of the lighthouse, LA be the length of the perpendicular from L to sea level, and B be the position of the boat. Draw right triangle ABL, and h LH draw , the horizontal line through L. Since HLB is the angle of depression, mHLB = 35.0, mLBA 35.0, and mBLA = 90 35.0 55.0. Let x distance from the boat to H B L 35.0° 55.0° 165 ft x A the foot of the lighthouse (BA). METHOD 1 METHOD 2 is the opposite is the adjacent side. BA Using BLA, LA side and Form the tangent ratio: tan BLA tan 55.0° BA LA x 165 x 165 tan 55.0° Use a calculator to perform the computation. The display will read 235.644421. is the opposite is the adjacent side. LA Using LBA, BA side and Form the tangent ratio: tan LBA tan 35.0° x tan 35.0° 165 LA BA 165 x x 165 tan 35.0o Use a calculator to perform the computation. The display will read 235.644421. Answer To the nearest foot, the boat is 236 feet from the foot of the lighthouse. EXERCISES Writing About Mathematics 1. Zack is solving a problem in which the measure of the angle of depression from the top of a building to a point 85 feet from the foot of the building is 64°. To find the height of the building, Zack draws the diagram shown at the right. Explain why Zack’s diagram is incorrect. 64° h 85 ft 316 Trigonometry of the Right Triangle 2. Explain why the angle of elevation from point A to point B is always congruent to the angle of depression from point B to point A. Developing Skills In 3–11, in each given triangle, find the length of the side marked x to the nearest foot or the measure of the angle marked x to the nearest degree. 3. 6. 9. x 42° 25 ft x 4. 7. 65° 13 ft 10 ft 50 ft 55° x 6.0 ft 9.0 ft x 10. 60° x 68° x 20 ft 5. 40° 18 ft x 8. 12 ft 24 ft x 11. 8.0 ft x 8.0 ft Applying Skills 12. At a point on the ground 52 meters from the foot of a tree, the measure of the angle of ele- vation of the top of the tree is 48°. Find the height of the tree to the nearest meter. 13. A ladder is leaning against a wall. The foot of the ladder is 6.25 feet from the wall. The ladder makes an angle of 74.5° with the level ground. How high on the wall does the ladder reach? Round the answer to the nearest tenth of a foot. The Sine and Cosine Ratios 317 14. From a point, A, on the ground that is 938 feet from the foot, C, of the Empire State Building, the angle of elevation of the top, B, of the building has a measure of 57.5°. Find the height of the building to the nearest ten feet. 15. Find to the nearest meter the height of a building if its shadow is 18 meters long when the angle of elevation of the sun has a measure of 38°. 16. From the top of a lighthouse 50.0 me
ters high, the angle of depression of a boat out at sea has a measure of 15.0°. Find, to the nearest meter, the distance from the boat to the foot of the lighthouse, which is at sea level. 17. From the top of a school 61 feet high, the measure of the angle of depression to the road in front of the school is 38°. Find to the nearest foot the distance from the road to the school. 18. Find to the nearest degree the measure of the angle of elevation of the sun when a student 170 centimeters tall casts a shadow 170 centimeters long. 19. Find to the nearest degree the measure of the angle of elevation of the sun when a woman 150 centimeters tall casts a shadow 43 centimeters long. 20. A ladder leans against a building. The top of the ladder reaches a point on the building that is 18 feet above the ground. The foot of the ladder is 7.0 feet from the building. Find to the nearest degree the measure of the angle that the ladder makes with the level ground. 21. In any rhombus, the diagonals are perpendicular to each other and bisect each other. In BD rhombus ABCD, diagonals sure of each angle to the nearest degree. a. mBCM b. mMBC and AC meet at M. If BD 14 and AC 20, find the mea- c. mABC d. mBCD 8-4 THE SINE AND COSINE RATIOS Since the tangent is the ratio of the lengths of the two legs of a right triangle, it is not directly useful in solving problems in which the hypotenuse is involved. In trigonometry of the right triangle, two ratios that involve the hypotenuse are called the sine of an angle and the cosine of an angle. D F As in our discussion of the tangent ratio, we recognize that the figure at the right shows three similar triangles. Therefore, the ratios of corresponding sides are equal. B A C E G The Sine Ratio From the figure, we see that BC AB 5 DE This ratio is called the sine of A. AD 5 FG AF a constant 318 Trigonometry of the Right Triangle B a c A b C DEFINITION The sine of an acute angle of a right triangle is the ratio of the length of the side opposite the acute angle to the length of the hypotenuse. In right triangle ABC, with mC 90, the definition of the sine of A is: sine A length of side opposite /A length of hypotenuse BC AB 5 a c By using “sin A” as an abbreviation for sine A, “opp” as an abbreviation for the length of the leg opposite A, and “hyp” as an abbreviation for the length of the hypotenuse, we can shorten the way we write the definition of sine A as follows: sin A opp hyp BC AB 5 a c The Cosine Ratio From the preceding figure on page 317, which shows similar triangles, ABC, ADE, and AFG, we see that AC AB 5 AE a constant. AD 5 AG AF This ratio is called the cosine of /A. DEFINITION The cosine of an acute angle of a right triangle is the ratio of the length of the side adjacent to the acute angle to the length of the hypotenuse. In right triangle ABC, with mC 90, the definition of the cosine of A is: cosine A length of side adjacent to /A length of hypotenuse AC AB 5 b c By using “cos A” as an abbreviation for cosine A, “adj” as an abbreviation for the length of the leg adjacent to A, and “hyp” as an abbreviation for the length of the hypotenuse, we can shorten the way we write the definition of cosine A as follows: B a c A b C cos A adj hyp AC AB 5 b c The Sine and Cosine Ratios 319 Finding Sine and Cosine Ratios on a Calculator CASE 1 Given an angle measure, find the sine or cosine ratio. On a calculator we use the keys labeled to display the values of the sine and cosine of an angle. The sequence of keys that a calculator requires for tangent will be the same as the sequence for sine or cosine. For example, to find sin 50° and cos 50°, we use the following: and COS SIN ENTER: SIN 50 ) ENTER ENTER: COS 50 ) ENTER DISPLAY: s i n ( 5 0 ) DISPLAY CASE 2 Given a sine or cosine ratio, find the angle measure. A calculator will also find the measure of A when sin A or cos A is given. . These are the second func- To do this we use the keys labeled COS1 and SIN1 SIN COS and tions of . We can think of the meaning of sin1 as “the angle whose sine is.” Therefore, if sin A 0.2588, then sin1(0.2588) can be read as “the angle whose sine is 0.2588.” To find the measure of A from the calculator, we use the following and are accessed by first pressing 2nd sequences of keys: ENTER: 2nd SIN1 0.2588 ) ENTER DISPLAY The measure of A to the nearest degree is 15°. 320 Trigonometry of the Right Triangle EXAMPLE 1 In ABC, C is a right angle, BC 7, AC 24, and AB 25. Find: a. sin A b. cos A c. sin B d. cos B e. mB, to the nearest degree Solution a. sin A b. cos A c. sin B d. cos B Answers opp hyp 5 BC adj hyp 5 AC opp hyp 5 AC adj hyp 5 BC AB 5 7 25 AB 5 24 25 AB 5 24 25 AB 5 7 25 25 24 A B 7 C e. Use a calculator. Start with the ratio in part c and use SIN1 or start with the ratio in part d and use COS1 . METHOD 1 sin B = 24 25 METHOD 2 cos B = 7 25 ENTER: 2nd SIN1 24 25 ENTER: 2nd COS1 7 25 ) ENTER ) ENTER DISPLAY ) DISPLAY mB 74 to the nearest degree. Answer EXERCISES Writing About Mathematics 1. If A and B are the acute angles of right triangle ABC, show that sin A cos B. 2. If A is an acute angle of right triangle ABC, explain why it is always true that sin A 1 and cos A 1. The Sine and Cosine Ratios 321 Developing Skills In 3–6, find: a. sin A b. cos A c. sin B d. cos B 3. B 4. B 5. A 6. C 10 A 8 6 90° C 5 C 90° 13 12 29 A 20 A p B r 90° k 7. In ABC, mC 90, AC 4, and BC 3. Find sin A. 8. In RST, mS 90, RS 5, and ST 12. Find cos T. B 21 90° C In 9–20, for each of the following, use a calculator to find the trigonometric function value to the nearest ten-thousandth. 9. sin 18° 13. sin 1° 17. cos 40° 10. sin 42° 14. sin 89° 18. cos 59° 11. sin 58° 15. cos 21° 19. cos 74° 12. sin 76° 16. cos 35° 20. cos 88° In 21–38, for each of the following, use a calculator to find the measure of A to the nearest degree. 21. sin A 0.1908 24. cos A 0.9397 27. sin A 0.8910 30. sin A 0.1900 33. cos A 0.8545 36. cos A 0.2968 22. sin A 0.8387 25. cos A 0.0698 28. sin A 0.9986 31. cos A 0.9750 34. sin A 0.5800 37. sin A 0.1275 23. sin A 0.3420 26. cos A 0.8910 29. cos A 0.9986 32. sin A 0.8740 35. cos A 0.5934 38. cos A 0.8695 39. a. Use a calculator to find sin 25° and sin 50°. b. If the measure of an angle is doubled, is the sine of the angle also doubled? 40. a. Use a calculator to find cos 25° and cos 50°. b. If the measure of an angle is doubled, is the cosine of the angle also doubled? 41. As an angle increases in measure from 1° to 89°: a. Does the sine of the angle increase or decrease? b. Does the cosine of the angle increase or decrease? 322 Trigonometry of the Right Triangle In 42 and 43, complete each sentence by replacing ? with a degree measure that makes the sentence true. 42. a. sin 70° cos ? b. sin 23° cos ? c. sin 38° cos ? d. sin x° cos ? Applying Skills 43. a. cos 50° sin ? b. cos 17° sin ? c. cos 82° sin ? d. cos x° sin ? 44. In ABC, mC 90, BC = 20, and BA 40. a. Find sin A. b. Find the measure of A. 45. In ABC, mC 90, AC 40, and AB 80. a. Find cos A. b. Find the measure of A. 46. In ABC, C is a right angle, AC 8, BC 15, and AB 17. Find: a. sin A e. the measure of A to the nearest degree f. the measure of B to the nearest degree b. cos A c. sin B d. cos B 47. In RST, mT 90, ST 11, RT 60, and RS 61. Find: a. sin R e. the measure of R to the nearest degree f. the measure of S to the nearest degree b. cos R c. sin S d. cos S 48. In ABC, C is a right angle, AC 1.0, BC 2.4, and AB 2.6. Find: d. cos B c. sin B a. sin A e. the measure of A to the nearest degree. f. the measure of B to the nearest degree. b. cos A 49. In rectangle ABCD, AB 3.5 and CB 1.2. Find: a. sin ABD e. the measure of ABD to the nearest degree. f. the measure of CBD to the nearest degree. b. cos ABD c. sin CBD d. cos CBD 50. In right triangle ABC, C is the right angle, BC 1, AC 3 " a. Using the given lengths, write the ratios for sin A and cos A. and AB 2. b. Use a calculator to find sin 30° and cos 30°. c. What differences, if any, exist between the answers to parts a and b? 51. In ABC, mC 90 and sin A cos A. Find mA. 8-5 APPLICATIONS OF THE SINE AND COSINE RATIOS Applications of the Sine and Cosine Ratios 323 Since the sine and cosine ratios each have the length of the hypotenuse of a right triangle as the second term of the ratio, we can use these ratios to solve problems in the following cases: 1. We know the length of one leg and the measure of one acute angle and want to find the length of the hypotenuse. 2. We know the length of the hypotenuse and the measure of one acute angle and want to find the length of a leg. 3. We know the lengths of the hypotenuse and one leg and want to find the measure of an acute angle. EXAMPLE 1 While flying a kite, Betty lets out 322 feet of string. When the string is secured to the ground, it makes an angle of 38.0° with the ground. To the nearest foot, what is the height of the kite above the ground? (Assume that the string is stretched so that it is straight.) Solution Let K be the position of the kite in the air, B be the point on the ground at which the end of the string is secured, and G be the point on the ground directly below the kite, as shown in the diagram. The height of the kite is the length of the perpendicular from the ground to the kite. Therefore, mG 90, mB 38.0, and the length of the string BK 322 feet. K x G 322 ft 38.0° B Let x KG, the height of the kite. We know the length of the hypotenuse and the measure of one acute angle and want to find the length of a leg, KG. opp In Method 1 below, since leg KG is opposite B, we can use sin B . hyp In Method 2 below, since leg KG is adjacent to K, we can use cos K adj hyp with mK 90 38.0 52.0. METHOD 1 METHOD 2 sin B sin 38.0° KG BK x 322 cos K cos 52.0° KG BK x 322 x 322 sin 38.0° x 198.242995 x 322 cos 52.0° x 198.242995 Write the ratio: Substitute the given values: Solve for x: Compute using a calculator: Answer The height of the kite to the nearest foot is 198 feet. 324 Trigonometry of the Right Triangle EXAMPLE 2 A wire reaches from the to
p of a pole to a stake in the ground 3.5 meters from the foot of the pole. The wire makes an angle of 65° with the ground. Find to the nearest tenth of a meter the length of the wire. Solution In BTS, B is a right angle, BS 3.5, mS 65, and mT 90 65 25. Let x ST, the length of the wire. Since we know the length of one leg and the measure of one acute angle and want to find the length of the hypotenuse, we can use either the sine or the cosine ratio. METHOD 1 BS ST cos S adj hyp 3.5 cos 65° x x cos 65° 3.5 3.5 x cos 65o x 8.281705541 METHOD 2 sin T opp hyp 5 BS ST 3.5 sin 25° x x sin 25° 3.5 3.5 x sin 25o x 8.281705541 T x 65° 3.5 m S B Answer The wire is 8.3 meters long to the nearest tenth of a meter. EXAMPLE 3 A ladder 25 feet long leans against a building and reaches a point 23.5 feet above the ground. Find to the nearest degree the angle that the ladder makes with the ground. Solution In right triangle ABC, AB, the length of the hypotenuse is 25 feet and BC, the side opposite A, is 23.5 feet. Since the problem involves A, and (the hypotenuse), the sine ratio is used. opp hyp 5 23.5 25 sin A AB BC (its opposite side), B 25.0 ft 23.5 ft ENTER: 2nd SIN1 23.5 25 ) ENTER A C DISPLAY Answer To the nearest degree, the measure of the angle is 70°. Applications of the Sine and Cosine Ratios 325 EXERCISES Writing About Mathematics 1. Brittany said that for all acute angles, A, (tan A)(cos A) sin A. Do you agree with Brittany? Explain why or why not. 2. Pearl said that as the measure of an acute angle increases from 1° to 89°, the sine of the angle increases and the cosine of the angle decreases. Therefore, cos A is the reciprocal of sin A. Do you agree with Pearl? Explain why or why not. In 3–11, find to the nearest centimeter the length of the side marked x. x 4. x 124 cm 65.0° 5. x 55.0° 1 4 3 c m 7. 10. x 38° 45 cm 5 c m 2 x 71° 8. 11. 15 cm x 55° 61° 32 cm 40.0° x 15. 12 ft x 18 ft In 12–15, find to the nearest degree the measure of the angle marked x. 12. 13. 10.5 ft x 8.0 ft Applying Skills 24 ft x 12 ft 14. 15 ft x 21 ft 16. A wooden beam 6.0 meters long leans against a wall and makes an angle of 71° with the ground. Find to the nearest tenth of a meter how high up the wall the beam reaches. 3. 6. 9. 22 c m 40° 3 1 c m 38° x x 45° 15 cm 326 Trigonometry of the Right Triangle 17. A boy flying a kite lets out 392 feet of string, which makes an angle of 52° with the ground. Assuming that the string is tied to the ground, find to the nearest foot how high the kite is above the ground. 18. A ladder that leans against a building makes an angle of 75° with the ground and reaches a point on the building 9.7 meters above the ground. Find to the nearest meter the length of the ladder. 19. From an airplane that is flying at an altitude of 3,500 feet, the angle of depression of an airport ground signal measures 27°. Find to the nearest hundred feet the distance between the airplane and the airport signal. 20. A 22-foot pole that is leaning against a wall reaches a point that is 18 feet above the ground. Find to the nearest degree the measure of the angle that the pole makes with the ground. 21. To reach the top of a hill that is 1.0 kilometer high, one must travel 8.0 kilometers up a straight road that leads to the top. Find to the nearest degree the measure of the angle that the road makes with the horizontal. 22. A 25-foot ladder leans against a building and makes an angle of 72° with the ground. Find to the nearest foot the distance between the foot of the ladder and the building. 23. A wire 2.4 meters in length is attached from the top of a post to a stake in the ground. The measure of the angle that the wire makes with the ground is 35°. Find to the nearest tenth of a meter the distance from the stake to the foot of the post. 24. An airplane rises at an angle of 14° with the ground. Find to the nearest hundred feet the distance the airplane has flown when it has covered a horizontal distance of 1,500 feet. 25. A kite string makes an angle of 43° with the ground. The string is staked to a point 104 meters from a point on the ground directly below the kite. Find to the nearest meter the length of the kite string, which is stretched taut. 26. The top of a 43-foot ladder touches a point on the wall that is 36 feet above the ground. Find to the nearest degree the measure of the angle that the ladder makes with the wall. 27. In a park, a slide 9.1 feet long is perpendicular to the ladder to the top of the slide. The distance from the foot of the ladder to the bottom of the slide is 10.1 feet. Find to the nearest degree the measure of the angle that the slide makes with the horizontal. 9.1 ft 10.1 ft Solving Problems Using Trigonometric Ratios 327 28. A playground has the shape of an isosceles trapezoid ABCD. The length of the shorter base, CD , is 185 feet. The length of each of the equal sides is 115 feet and mA 65.0. a. Find DE, the length of the altitude from D, to the nearest foot. b. Find AE, to the nearest tenth of a foot. c. Find AB, to the nearest foot. d. What is the area of the playground to the nearest hundred square feet? e. What is the perimeter of the playground? 29. What is the area of a rhombus, to the nearest ten square feet, if the measure of one side is 43.7 centimeters and the measure of one angle is 78.0°? 30. A roofer wants to reach the roof of a house that is 21 feet above the ground. The measure of the steepest angle that a ladder can make with the house when it is placed directly under the roof is 27°. Find the length of the shortest ladder that can be used to reach the roof, to the nearest foot. 8-6 SOLVING PROBLEMS USING TRIGONOMETRIC RATIOS When the conditions of a problem can be modeled by a right triangle for which the measures of one side and an acute angle or of two sides are known, the trigonometric ratios can be used to find the measure of another side or of an acute angle. Procedure To solve a problem by using trigonometric ratios: 1. Draw the right triangle described in the problem. 2. Label the sides and angles with the given values. 3. Assign a variable to represent the measure to be determined. 4. Select the appropriate trigonometric ratio. 5. Substitute in the trigonometric ratio, and solve the resulting equation. sin A Given ABC with mC 90: opp hyp 5 BC adj hyp 5 AC opp adj 5 BC AB 5 a c AB 5 b c AC 5 a b tan A cos A c b A B a C sin B opp hyp 5 AC adj hyp 5 BC adj 5 AC AB 5 b c AB 5 a c BC 5 b a cos B tan B opp 328 Trigonometry of the Right Triangle EXAMPLE 1 Given: In isosceles triangle ABC, AC CB 20 and mA mB 68. is an altitude. C CD Find: a. Length of altitude CD to the nearest tenth. b. Length of AB to the nearest tenth. Solution a. In right BDC, sin B CD CB Let x CD. 20 x 20 68° A D 68° y B sin 68° x 20 x 20 sin 68° 18.54367709 18.5 b. Since the altitude drawn to the base of an isosceles triangle bisects the base, AB 2DB. Therefore, find DB in BDC and double it to find AB. In right BDC, cos B = Let y DB. DB CB . cos 68° y 20 y 20 cos 68° 7.492131868 AB 2y 2(7.492131868) 14.9843736 Answers a. CD 18.5, to the nearest tenth. b. AB 15.0, to the nearest tenth. EXERCISES Writing About Mathematics 1. If the measures of two sides of a right triangle are given, it is possible to find the measures of the third side and of the acute angles. Explain how you would find these measures. 2. If the measures of the acute angles of a right triangle are given, is it possible to find the measures of the sides? Explain why or why not. Solving Problems Using Trigonometric Ratios 329 Developing Skills In 3–10: In each given right triangle, find to the nearest foot the length of the side marked x; or find to the nearest degree the measure of the angle marked x. Assume that each measure is given to the nearest foot or to the nearest degree. 3. 7. x 37° 22 ft x 18 ft 12 ft 4. 8. x 33 ft 60° 12 ft x 15 ft 5. x 41° 25 ft 9. 8.0 ft 24 ft x 6. 10. 35 ft 65° x x 1 ° 5 3 9 ft BD is the altitude to AC . Find BD to the nearest tenth. 11. In ABC, mA 42, AB 14, and 12. In ABC, AC > BC , mA 50, and AB 30. Find to the nearest tenth the length of the altitude from vertex C. 13. The legs of a right triangle measure 84 and 13. Find to the nearest degree the measure of the smallest angle of this triangle. 14. The length of hypotenuse number of degrees in B. AB of right triangle ABC is twice the length of leg BC . Find the 15. The longer side of a rectangle measures 10, and a diagonal makes an angle of 27° with this side. Find to the nearest integer the length of the shorter side. 16. In rectangle ABCD, diagonal degree the measure of CAB. AC measures 11 and side AB measures 7. Find to the nearest 17. In right triangle ABC, CD is the altitude to hypotenuse AB , AB 25, and AC 20. Find lengths AD, DB, and CD to the nearest integer and the measure of B to the nearest degree. 18. The lengths of the diagonals of a rhombus are 10 and 24. a. Find the perimeter of the rhombus. b. Find to the nearest degree the measure of the angle that the longer diagonal makes with a side of the rhombus. 19. The altitude to the hypotenuse of a right triangle ABC divides the hypotenuse into segments whose measures are 9 and 4. The measure of the altitude is 6. Find to the nearest degree the measure of the smaller acute angle of ABC. 330 Trigonometry of the Right Triangle 20. In ABC, AB = 30, mB = 42, mC 36, and AD is an altitude. a. Find to the nearest integer the length of AD . b. Using the result of part a, find to the nearest integer the length of DC . 21. Angle D in quadrilateral ABCD is a right angle, is perpendicular to AC and diagonal mB 35, and mDAC = 65. a. Find AC to the nearest integer. BC , BC = 20, b. Using the result of part a, find DC to the nearest integer. A D 65° C 36° C A D 3 0 42° B B 35° 20 22. The diagonals of a rectangle each measure 198 and intersect at an angle whose measure is 110°. Find to the nearest integer the length and width of the rectangle. Hint: The diagonals of a rectangle bisect each other. 23. In rhombus ABC
D, the measure of diagonal AC is 80 and mBAC = 42. a. Find to the nearest integer the length of diagonal BD . b. Find to the nearest integer the length of a side of the rhombus. 24. In right triangle ABC, the length of hypotenuse AB is 100 and mA 18. a. Find AC and BC to the nearest integer. b. Show that the results of part a are approximately correct by using the relationship (AB)2 = (AC)2 + (BC)2. Applying Skills 25. Find to the nearest meter the height of a church spire that casts a shadow of 53.0 meters when the angle of elevation of the sun measures 68.0°. 26. From the top of a lighthouse 194 feet high, the angle of depression of a boat out at sea measures 34.0°. Find to the nearest foot the distance from the boat to the foot of the lighthouse. 27. A straight road to the top of a hill is 2,500 meters long and makes an angle of 12° with the horizontal. Find to the nearest ten meters the height of the hill. 28. A wire attached to the top of a pole reaches a stake in the ground 21 feet from the foot of the pole and makes an angle of 58° with the ground. Find to the nearest foot the length of the wire. 29. An airplane climbs at an angle of 11° with the ground. Find to the nearest hundred feet the distance the airplane has traveled when it has attained an altitude of 450 feet. 30. Find to the nearest degree the measure of the angle of elevation of the sun if a child 88 cen- timeters tall casts a shadow 180 centimeters long. 31. CD and AB represent cliffs on opposite sides of a river 125 meters wide. From B, the angle of elevation of D measures 20° and the angle of depression of C measures 25°. Find to the nearest meter: a. the height of the cliff represented by AB . b. the height of the cliff represented by CD . B A 20.0° 25.0° 125 m 32. Points A, B, and D are on level ground. CD represents the height of a building, BD = 86 feet, and mD 90. At B, the angle of elevation of the top of the building, CBD, measures 49°. At A, the angle of elevation of the top of the building, CAD, measures 26°. a. Find the height of the building, CD, to the nearest foot. b. Find AD to the nearest foot. 26° 49° A B 86 ft D Chapter Summary 331 D E C C CHAPTER SUMMARY The Pythagorean Theorem relates the lengths of the sides of a right triangle. If the lengths of the legs of a right triangle are a and b, and the length of the hypotenuse is c, then c2 a2 b2. The trigonometric functions associate the measure of each acute angle A with a number that is the ratio of the measures of two sides of a right triangle. The three most commonly used trigonometric functions are sine, cosine, and tangent. In the application of trigonometry to the right triangle, these ratios are defined as follows: sin A opp hyp cos A adj hyp tan A opp adj BC In right triangle ABC, with hypotenuse is opposite A and adjacent to B; is opposite B and adjacent to A. AC AB • • sin A cos A tan A BC AB AC AB BC AC BC cos B AB AC sin B AB tan B AC BC B C A 332 Trigonometry of the Right Triangle An angle of elevation, GDE in the diagram, is an angle between a horizontal line and a line of sight to an object at a higher elevation. An angle of depression, FED in the diagram, is an angle between a horizontal line and a line of sight to an object at a lower elevation. Angle of depression F D E G Angle of elevation VOCABULARY 8-1 Trigonometry • Direct measurement • Indirect measurement • Pythagorean Theorem • Pythagorean triple 8-2 Opposite side • Adjacent side • Similar • Tangent of an acute angle of a right triangle 8-3 Angle of elevation • Angle of depression 8-4 Sine of an acute angle of a right triangle • Cosine of an acute angle of a right triangle REVIEW EXERCISES 1. Talia’s calculator is not functioning properly and does not give the correct value when she uses the calculator are operating correctly. TAN key. Assume that all other keys of the a. Explain how Talia can find the measure of the leg ABC when BC 4.5 and mA 43. AC of right triangle b. Explain how Talia can use her calculator to find the tangent of any acute angle, given the measure of one side of a right triangle and the measure of an acute angle as in part a. 2. Jill made the following entry on her calculator: ENTER: 2nd SIN1 1.5 ) ENTER Explain why the calculator displayed an error message. In 3–8, refer to RST and express the value of each ratio as a fraction. 3. sin R 5. sin T 7. cos T 4. tan T 6. cos R 8. tan R 17 15 R T 8 S In 9–12: in each given triangle, find to the nearest centimeter the length of the side marked x. Assume that each given length is correct to the nearest centimeter. Review Exercises 333 40 c m 42° 9. 11. 50 cm x x 54° 18 cm 35° 10. x 12. 24° x 41 cm 13. If cos A sin 30° and 0° A 90°, what is the measure of A? 14. In right triangle ACB, mC 90, mA 66, and AC 100. Find BC to the nearest integer. 15. In right triangle ABC, mC 90, mB 28, and BC 30. Find AB to the nearest integer. 16. In ABC, mC 90, tan A 0.7, and AC 40. Find BC. 17. In ABC, mC 90, AB 30, and BC 15. What is the measure, in degrees, of A? 18. In ABC, mC 90, BC 5, and AC 9. Find to the nearest degree the measure of A. 19. Find to the nearest meter the height of a building if its shadow is 42 meters long when the angle of elevation of the sun measures 42°. 20. A 5-foot wire attached to the top of a tent pole reaches a stake in the ground 3 feet from the foot of the pole. Find to the nearest degree the measure of the angle made by the wire with the ground. 21. While flying a kite, Doris let out 425 feet of string. Assuming that the string is stretched taut and makes an angle of 48° with the ground, find to the nearest ten feet how high the kite is. 22. A rectangular field ABCD is crossed by a path from A to C. If mBAC 62 and BC 84 yards, find to the nearest yard: a. the width of the field, AB. b. the length of path, AC. 334 Trigonometry of the Right Triangle 23. Find the length of a leg of an isosceles right triangle if the length of the hypotenuse is 72 . " 24. The measure of each of the base angles of an isosceles triangle is 15 degrees more than twice the measure of the vertex angle. a. Find the measure of each angle of the triangle. b. Find to the nearest tenth of a centimeter the measure of each of the equal sides of the triangle if the measure of the altitude to the base is 88.0 centimeters. c. Find to the nearest tenth of a centimeter the measure of the base of the triangle. d. Find the perimeter of the triangle. e. Find the area of the triangle. 25. ABCD is a rectangle with E a point on BC . AB 12, BE 5, and EC 9. a. Find the perimeter of triangle AED. b. Find the area of triangle AED. /CDE . c. Find the measure of d. Find the measure of BAE. e. Find the measure of AED. Exploration A regular polygon with n sides can be divided into n congruent isosceles triangles. a. Express, in terms of n, the measure of the vertex angle of one of the isosceles triangles. b. Express, in terms of n, the measure of a base angle of one of the isosceles triangles. c. Let s be the measure of a side of the regular polygon. Express, in terms of n and s, the measure of the altitude to the base of one of the isosceles triangles. d. Express the area of one of the isosceles triangles in terms of n and s. e. Write a formula for the area of a regular polygon in terms of the measure of a side, s, and the number of sides, n. CUMULATIVE REVIEW CHAPTERS 1–8 Part I Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. Which of the following does not represent a real number when x 3? Cumulative Review 335 3 x (1) x 2 3 x 2. The coordinates of one point on the x-axis are (2) (3) x x (1) (1, 1) (2) (1, 1) (3) (1, 0) 3. The expression 0.2a2(10a3 2a) is equivalent to (1) 2a5 0.4a3 (2) 20a5 4a3 (3) 2a6 0.4a2 (4) 20a6 0.4a3 4. If 7 3, then x equals 1 4x 3 4x (1) 20 (2) 10 (3) 5 5. If sin A 0.3751, then, to the nearest degree, mA is (1) 21 (2) 22 (3) 68 6. Which of the following statements is false? (1) If a polygon is a square, then it is a parallelogram. (2) If a polygon is a square, then it is a rhombus. (3) If a polygon is a rectangle, then it is a parallelogram. (4) If a polygon is a rectangle, then it is a rhombus. (4) x x 2 3 (4) (0, 1) (4) 5 2 (4) 69 7. If the measures of two legs of a right triangle are 7.0 feet and 8.0 feet, then, to the nearest tenth of a foot, the length of the hypotenuse is (4) 48.9 (1) 10.6 (2) 15.0 (3) 41.2 8. The measure of the radius of a cylinder is 9.00 centimeters and the measure of its height is 24.00 centimeters. The surface area of the cylinder to the correct number of significant digits is (1) 1,610 square centimeters (2) 1,620 square centimeters (3) 1,860 square centimeters (4) 1,870 square centimeters 9. When 5b2 2b is subtracted from 8b the difference is (3) 5b2 10b (2) 5b2 6b (1) 6b 5b2 (4) 5b2 6b 10. When written in scientific notation, the fraction (1.2 3 1024) 3 (3.5 3 108) 8.4 3 1022 equals (1) 5.0 102 Part II (2) 5.0 101 (3) 5.0 105 (4) 5.0 106 Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 336 Trigonometry of the Right Triangle 11. The vertices of pentagon ABCDE are A(2, 2), B(7, 2), C(7, 5), D(0, 5), E(2, 0). a. Draw pentagon ABCDE on graph paper. b. Find the area of the pentagon. 12. In ABC, mC = 90, mB = 30, value of x. Part III AC 5 6x2 2 4x , and AB 5 2x . Find the Answer all questions in this part. Each correct answer will receive 3 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. Plank Road and Holt Road are perpendicular to each other. At a point 1.3 miles before the intersection of Plank and Holt, State Street cros
ses Plank at an angle of 57°. How far from the intersection of Plank and Holt will State Street intersect Holt? Write your answer to the nearest tenth of a mile. 14. Benny, Carlos, and Danny each play a different sport and have different career plans. Each of the four statements given below is true. The boy who plays baseball plans to be an engineer. Benny wants to be a lawyer. Carlos plays soccer. The boy who plans to be a doctor does not play basketball. What are the career plans of each boy and what sport does he play? Part IV Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 15. Bart wants to plant a garden around the base of a tree. To determine the amount of topsoil he will need to enrich his garden, he measured the circumference of the tree and found it to be 9.5 feet. His garden will be 2.0 feet wide, in the form of a ring around the tree. Find to the nearest square foot the surface area of the garden Bart intends to plant. 16. Samantha had a snapshot that is 3.75 inches wide and 6.5 inches high. She cut a strip off of the top of the snapshot so that an enlargement will fit into a frame that measures 5 inches by 8 inches. What were the dimensions of the strip that she cut off of the original snapshot? GRAPHING LINEAR FUNCTIONS AND RELATIONS The Tiny Tot Day Care Center charges $200 a week for children who stay at the center between 8:00 A.M. and 5:00 P.M. If a child is not picked up by 5:00 P.M., the center charges an additional $4.00 per hour or any part of an hour. Mr. Shubin often has to work late and is unable to pick up his daughter on time. For Mr. Shubin, the weekly cost of day care is a function of time; that is, his total cost depends on the time he arrives at the center. This function for the daily cost of day care can be expressed in terms of an equation in two variables: y 30 4x.The variable x represents the total time, in hours, after 5:00 P.M., that a child remains at the center, and the variable y represents the cost, in dollars, of day care for the day. CHAPTER 9 CHAPTER TABLE OF CONTENTS 9-1 Sets, Relations, and Functions 9-2 Graphing Linear Functions Using Their Solutions 9-3 Graphing a Line Parallel to an Axis 9-4 The Slope of a Line 9-5 The Slopes of Parallel and Perpendicular Lines 9-6 The Intercepts of a Line 9-7 Graphing Linear Functions Using Their Slopes 9-8 Graphing Direct Variation 9-9 Graphing First-Degree Inequalities in Two Variables 9-10 Graphs Involving Absolute Value 9-11 Graphs Involving Exponential Functions Chapter Summary Vocabulary Review Exercises Cumulative Review 337 338 Graphing Linear Functions and Relations 9-1 SETS, RELATIONS, AND FUNCTIONS Set-Builder Notation In Chapters 1 and 2, we used roster form to describe sets. In roster form, the elements of a set are enclosed by braces and listed once. Repeated elements are not allowed. For example, {. . ., 3, 2, 1, 0, 1, 2, 3, . . .} is the set of integers, in roster form. A second way to specify a set is to use set-builder notation. Set-builder notation is a mathematically concise way of describing a set without listing the elements of the set. For instance, using set-builder notation, the set of counting numbers from 1 to 100 is: {x x is an integer and 1 x 100} This reads as “the set of all x such that x is an integer and x is at least 1 and at most 100.” The vertical bar “” represents the phrase “such that,” and the description to the right of the bar is the rule which defines the set. Here are some other examples of set-builder notation: 1. {x x is an integer and x 6 } {7, 8, 9, 10, . . .} the set of integers greater than 6 2. {x 1 x 3} any real number in the interval [1, 3] 3. {2n 1 n is a whole number} {2(0) 1, 2(1) 1, 2(2) 1, 2(3) 1, . . .} {1, 3, 5, 7, . . .} the set of odd whole numbers Frequently used with set-builder notation is the symbol , which means “is an element of.” This symbol is used to indicate that an element is a member of a set. The symbol means “is not an element of,” and is used to indicate that an element is not a member of a set. For instance, 2 {x 1 x 3} 2 {x 1 x 3} since 2 is between 1 and 3. since 2 is not between 1 and 3. EXAMPLE 1 List the elements of each set or indicate that the set is the empty set. a. A {x x of the set of natural numbers and x 0} c. C {x x 3 5} b. B {2n n of the set of whole numbers} d. D {m m is a multiple of 5 and m 25} Sets, Relations, and Functions 339 Solution a. Since there are no natural numbers b. B {2(0), 2(1), 2(2), 2(3), . . .} less than 0, A 5 {0, 2, 4, 6, . . . } c. C represents the solution set of the equation x 3 5. Therefore, C {2} d. D consists of the set of multiples of 5 that are less than 25. D {. . ., 45, 40, 35, 30} Answers a. A 5 b. B {0, 2, 4, 6, . . .} c. C {2} d. D {. . ., 45, 40, 35, 30} Relations That are Finite Sets There are many instances in which one set of information is related to another. For example, we may identify the persons of a group who are 17, 18, or 19 to determine who is old enough to vote. This information can be shown in a diagram such as the one at the right, or as a set of ordered pairs. 17 18 19 Debbie Kurt Kim Eddie {(17, Debbie), (17, Kurt), (18, Kim), (19, Eddie)} DEFINITION A relation is a set of ordered pairs. The domain of a relation is the set of all first elements of the ordered pairs. For example, in the relation shown above, the domain is {17, 18, 19}, the set of ages. The range of a relation is the set of all second elements of the ordered pairs. In the relation above, the range is {Debbie, Kurt, Kim, Eddie}, the set of people. Let us consider the relation “is greater than,” using the set {1, 2, 3, 4} as both the domain and the range. Let (x, y) be an ordered pair of the relation. Then the relation can be shown in any of the following ways. 1. A Rule The relation can be described by the inequality x y. 2. Set of Ordered Pairs The relation can be listed in set notation as shown below: {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3)} 340 Graphing Linear Functions and Relations 3. Table of Values The ordered pairs shown in 2 can be displayed in a table. Graph In the coordinate plane, the domain is a subset of the numbers on the x-axis and the range is a subset of the numbers on the y-axis. The points that correspond to the ordered pairs of numbers from the domain and range are shown. Of the 16 ordered pairs shown, only six are enclosed to indicate the relation x y Relations That are Infinite Sets Charita wants to enclose a rectangular garden. How much fencing will she need if the width of the garden is to be 5 feet? The answer to this question depends on the length of the garden. Therefore, we say that the length of the garden, x, and the amount of fencing, y, form a set of ordered pairs or a relation. The formula for the perimeter of a rectangle can be used to express y in terms of x: P 2l 2w y 2x 2(5) y 2x 10 Some possible values for the amount of fencing can also be shown in a table. As the length of the garden changes, the amount of fencing changes, as the following table indicates. x 1 3 4.5 7.2 9.1 2x 10 2(1) 10 2(3) 10 2(4.5) 10 2(7.2) 10 2(9.1) 10 y 12 16 19 24.4 28.2 (x, y) (1, 12) (3, 16) (4.5, 19) (7.2, 24.4) (9.1, 28.2) Sets, Relations, and Functions 341 Each solution to y 2x 10 is a pair of numbers. In writing each pair, we place the value of x first and the value of y second. For example, when x 1, y 12, (1, 12) is a solution. When x 3, y 16, (3, 16) is another solution. It is not possible to list all ordered pairs in the solution set of the equation because the solution set is infinite. However, it is possible to determine whether a given ordered pair is a member of the solution set. Replace x with the first element of the pair (the x-coordinate) and y with the second element of the pair (the y-coordinate). If the result is a true statement, the ordered pair is a solution of the equation. • (1.5, 13) {(x, y) y 2x 10} because 13 2(1.5) 10 is true. • (4, 14) {(x, y) y 2x 10} because 14 2(4) 10 is false. Since in the equation y 2x 10, x represents the length of a garden, only positive numbers are acceptable replacements for x. Therefore, the domain of this relation is the set of positive real numbers. For every positive number x, 2x 10 will also be a positive number. Therefore, the range of this relation is the set of positive real numbers. Although a solution of y 2x 10 is (–2, 6), it is not possible for 2 to be the length of a garden; (2, 6) is not a pair of the relation. When we choose a positive real number to be the value of x, there is one and only one value of y that makes the equation true. Therefore, y 2x 10 is a special kind of relation called a function. DEFINITION A function is a relation in which no two ordered pairs have the same first element. or A function is a relation in which every element of the domain is paired with one and only one element of the range. Since a function is a special kind of relation, the domain of a function is the set of all first elements of the ordered pairs of the function. Similarly, the range of a function is the set of all second elements of the ordered pairs of the function. The notation f(a) b or “f of a equals b,” signifies that the value of the function, f, at a is equal to b. The independent variable is the variable that represents the first element of an ordered pair. The domain of a function is the set of all values that the independent variable is allowed to take. The dependent variable is the variable that represents the second element of an ordered pair. The range of a function is the set of all values that the dependent variable is allowed to take. 342 Graphing Linear Functions and Relations EXAMPLE 2 Find five members of the solution set of the sentence 3x y 7. Solution How to Proceed (1) T
ransform the equation into an equivalent equation with y alone as one member: (2) Choose any five values for x. Since no replacement set is given, any real numbers can be used: (3) For each selected value of x, determine y: 3x 3x y 7 3x y 3x 7 x 2 0 1 3 3 5 3x 7 3(2) 7 3(0) 7 A 1 1 7 23 3 B 3(3) 7 3(5) 7 y 13 7 6 2 8 Answer (–2, 13), (0, 7), , (3, –2), (5, –8) 1 3, 6 A B Note that many other solutions are also possible, and for every real number x, one and only one real number y will make the equation 3x y 7 true. Therefore, 3x y 7 defines a function. EXAMPLE 3 Determine whether each of the given ordered pairs is a solution of the inequality y 2x 4. a. (4, 0) b. (–1, 2) c. (0, 6) d. (0, 7) Solution a. y 2x 4 ? 0 2 2(4) $ 4 0 8 4 ✘ y 2x 4 c. ? 6 2 2(0) $ 4 6 0 4 ✔ b. y 2x 4 ? 2 2 2(21) $ 4 d. 2 2 4 ✔ y 2x 4 ? 7 2 2(0) $ 4 7 0 4 ✔ Answers a. Not a solution b. A solution c. A solution d. A solution EXAMPLE 4 Note that for the inequality y 2x 4, the pairs (0, 6) and (0, 7) have the same first element. Therefore y 2x 4 is a relation but not a function. Sets, Relations, and Functions 343 The cost of renting a car for 1 day is $64.00 plus $0.25 per mile. Let x represent the number of miles the car was driven, and let y represent the rental cost, in dollars, for a day. a. Write an equation for the rental cost of the car in terms of the number of miles driven. b. Find the missing member of each of the following ordered pairs, which are elements of the solution set of the equation written in part a, and explain the meaning of the pair. (1) (155, ?) (2) (?, 69) Solution a. Rental cost is $64.00 plus $0.25 times the number of miles. b. (1) For the pair (155, ?), x 155 and y is to be determined. Then: y 64.00 0.25x y 64.00 0.25x 64.00 0.25(155) 64.00 38.75 102.75 When the car was driven 155 miles, the rental cost was $102.75. (2) For the pair (?, 69), y 69 and x is to be determined. Then: y 64 0.25x 69 64 0.25x 64 64 5 0.25x 0.25x 0.25 5 0.25 20 x When the rental cost was $69, the car was driven 20 miles. Answers a. y 64.00 0.25x b. (1) (155, 102.75) The car was driven 155 miles, and the cost was $102.75. (2) (20, 69) The car was driven 20 miles, and the cost was $69.00. 344 Graphing Linear Functions and Relations EXAMPLE 5 Determine whether or not each set is a relation. If it is a relation, determine whether or not it is a function. a. {x 0 x 10} b. {(x, y) y 4 x} c. {(a, b) a 2 and b is a real number} d. {(x, y) y x2} Solution a. The set is not a set of ordered pairs. It is not a relation. Answer b. The set is a relation and a function. It is a set of ordered pairs in which every first element is paired with one and only one second element. Answer c. The set is a relation that is not a function. The same first element is paired with every second element. Answer d. The set is a relation and a function. Every first element is paired with one and only one second element, its square. Answer EXAMPLE 6 Jane would like to construct a rectangular pool having an area of 102 square feet. If l represents the length of the pool and w its height, express the set of all ordered pairs, (l, w), that represent the dimensions of the pool using set-builder notation. Solution Since the area must be 102 square units, the variables l and w are related by the area formula: A lw. Therefore, the set of ordered pairs representing the possible dimensions of the pool is: {(l, w) 102 lw, where l and w are both positive} Answer Note: Since the length and the width cannot be zero or negative, the rule must specify that the dimensions, l and w, of the pool are both positive. EXERCISES Writing About Mathematics 1. A function is a set of ordered pairs in which no two different pairs have the same first element. In Example 4, we can say that the cost of renting a car is a function of the number of miles driven. Explain how this example illustrates the definition of a function. Sets, Relations, and Functions 345 2. Usually when a car rental company charges for the number of miles driven, the number of miles is expressed as a whole number. For example, if the car was driven 145.3 miles, the driver would be charged for 146 miles. In Example 4, what would be the domain of the function? (Recall that the domain is the set of numbers that can replace the variable x.) Developing Skills In 3–7, find the missing member in each ordered pair if the second member of the pair is twice the first member. 3. (3, ?) 4. (0, ?) 5. (2, ?) 6. (?, 11) 7. (?, 8) In 8–12, find the missing member in each ordered pair if the first member of the pair is 4 more than the second member. 8. (?, 5) 9. 10. (?, 0) 11. 12. (8, ?) ?, 1 2 A B 91 4, ? A B In 13–27, state whether each given ordered pair of numbers is a solution of the equation or inequality. 13. y 5x; (3, 15) 16. 3x 2y 0; (3, 2) 19. 3y 2x l; (4, 3) 1 1 22. 4x 3y 2; 4, 3 25. y 6x; (1, 2) B A 14. y 4x; (16, 4) 17. y 4x; (2, 10) 20. 2x 3y 9; (0, 3) 23. 3x y 4; (7, 1) 26. 3x 4y; (5, 2) 15. y 3x 1; (7, 22) 18. y 2x 3; (0, 2) 21. x y 8; (4, 5) 24. x 2y 15; (1, 7) 27. 5x 2y 19; (3, 2) In 28–31, state which sets of ordered pairs represent functions. If the set is not a function, explain why. 28. {(1, 2), (2, 3), (3, 4), (5, 6)} 30. {(5, 5), (5, 5), (6, 6), (6, 6)} 29. {(1, 2), (2, 1), (3, 4), (4, 3)} 31. {(81, 9), (81, 9), (25, 5), (25, 5)} In 32–35, find the range of each function when the domain is: a. A {x 10 x 13 and x of the set of integers} b. B {6x x of the set of whole numbers} 32. y 2x 3 33. y 5 x 6 34. y x 2 35. y x 1 Applying Skills In 36–39, in each case: a. Write an equation or an inequality that expresses the relationship between x and y. b. Find two ordered pairs in the solution set of the equation that you wrote in part a. c. Is the relation determined by the equation or inequality a function? 36. The cost, y, of renting a bicycle is $5.00 plus $2.50 times the number of hours, x, that the bicycle is used. Assume that x can be a fractional part of an hour. 37. In an isosceles triangle whose perimeter is 54 centimeters, the length of the base in centime- ters is x and the length of each leg in centimeters is y. 346 Graphing Linear Functions and Relations 38. Jules has $265 at the beginning of the month. What is the maximum amount, y, that he has left after 30 days if he spends at least x dollars a day? 39. At the beginning of the day, the water in the swimming pool was 2 feet deep. Throughout the day, water was added so that the depth increased by less than 0.4 foot per hour. What was the depth, y, of the water in the pool after being filled for x hours? 40. A fence to enclose a rectangular space along a river is to be constructed using 176 feet of fencing. The two sides perpendicular to the river have length x. a. Complete five more rows of the table shown on the right. b. Is the area, A, a function of x? If so, write the function and determine its domain. Length (x) Width Area (A) 1 2 176 2(1) 176 2(2) 1[176 2(1)] 174 2[176 2(2)] 344 41. (1) At the Riverside Amusement Park, rides are paid for with tokens purchased at a central booth. Some rides require two tokens, and others one token. Tomas bought 10 tokens and spent them all on x two-token rides and y one-token rides. On how many two-token and how many one-token rides did Tomas go? (2) Minnie used 10 feet of fence to enclose three sides of a rectangular pen whose fourth side was the side of a garage. She used x feet of fence for each side perpendicular to the garage and y feet of fence for the side parallel to the garage. What were the dimensions of the pen that Minnie built? a. Write an equation that can be used to solve both problems. b. Write the six solutions for problem (1). c. Which of the six solutions for problem (1) are not solutions for problem (2)? d. Write two solutions for problem (2) that are not solutions for problem (1). e. There are only six solutions for problem (1) but infinitely many solutions for problem (2). Explain why. 42. A cylindrical wooden base for a trophy has radius r and height h. The volume of the base will be 102 cubic inches. Using set-builder notation, describe the set of ordered pairs, (r, h), that represent the possible dimensions of the base. 9-2 GRAPHING LINEAR FUNCTIONS USING THEIR SOLUTIONS Think of two numbers that add up to 6. If x and y represent these numbers, then x y 6 is an equation showing that their sum is 6. If the replacement set for both x and y is the set of real numbers, we can find infinitely many solutions for the equation x y 6. Some of the solutions are shown in the following table.5 1.5 4 2 3 3 2.2 3.8 2 4 0 6 2 8 Graphing Linear Functions Using Their Solutions 347 Each ordered pair of numbers, such as (8, –2), locates a point in the coordinate plane. When all the points that have the coordinates associated with the number pairs in the table are located, the points appear to lie on a straight line, as shown at the right. In fact, if the replacement set for both x and y is the set of real numbers, the following is true: All of the points whose coordinates are solutions of x y 6 lie on this same straight line, and all of the points whose coordinates are not solutions of x y 6 do not lie on this line2 –1 –1 – This line, which is the set of all the points and only the points whose coordinates make the equation x y 6 true, is called the graph of x y 6. DEFINITION A first-degree equation in standard form is written as Ax By C where A, B, and C are real numbers, with A and B not both 0. We call an equation that can be written in the form Ax + By C a linear equation since its graph is a straight line that contains all points and only those points whose coordinates make the equation true. The replacement set for both variables is the set of real numbers unless otherwise indicated. When we graph a linear equation, we can determine the line by plotting two points whose coordinates satisfy that equation. However, we usually plot a third point as a check on the first two. If the third point lies on the line determined b
y the first two points, we have probably made no error. Procedure To graph a linear equation by means of its solutions: 1. Transform the equation into an equivalent equation that is solved for y in terms of x. 2. Find three solutions of the equation by choosing values for x and finding corresponding values for y. 3. In the coordinate plane, graph the ordered pairs of numbers found in Step 2. 4. Draw the line that passes through the points graphed in Step 3. 348 Graphing Linear Functions and Relations EXAMPLE 1 Does the point (2, 3) lie on the graph of x 2y 4? Solution The point (2, 3) lies on the graph of x 2y 4 if and only if it is a solu- tion of x 2y 4. x 2y 4 2 2 2(23) 5? 2 1 6 5? 4 4 8 4 ✘ Answer Since 8 4 is not true, the point (2, 3) does not lie on the line x 2y 4. EXAMPLE 2 What must be the value of d if (d, 4) lies on line 3x y 10? Solution The coordinates (d, 4) must satisfy 3x y 10. Let x d and y 4. Then: 3d 4 10 3d 6 d 2 Answer The value of d is 2. EXAMPLE 3 a. Write the following verbal sentence as an equation: The sum of twice the x-coordinate of a point and the y-coordinate of that point is 4. b. Graph the equation written in part a. Solution a. Let x the x-coordinate of the point, and y the y-coordinate of the point. Then: 2x y 4 Answer b. How to Proceed (1) Transform the equation into an equivalent equation that has y alone as one member: 2x y 4 y 2x 4 Graphing Linear Functions Using Their Solutions 349 (2) Choose three values of x and find the corresponding values of y: (3) Plot the points that are associated with the three solutions: (4) Draw a line through the points that were plotted. Label the line with its equation: x 2x 4 0 2(0) 4 1 2(1) 4 2 2(23 –2 –1 –1 – To display the graph of a function on a calculator, follow the steps below. 1. Solve the equation for y and enter it as Y1. 2 X,T,,n ENTER: 4 DISPLAY: Y (-) ENTER . Use the standard viewing window to view the graph. The standard viewing window is a 20 by 20 graph centered at the origin. ENTER: ZOOM 6 ENTER DISPLAY: M e m o r y Zoom Tr i g 350 Graphing Linear Functions and Relations 3. Display the coordinates of points on the graph. ENTER: TRACE In the viewing window, the equation appears at the top left of the screen. The point at which the graph intersects the y-axis is marked with a star and the coordinates of that point appear at the bottom of the screen. Press the right and left arrow keys to move the star along the line to display other coordinates EXERCISES Writing About Mathematics 1. The points whose coordinates are (3, 1), (5, 1) and (7, 3) all lie on the same line. What could be the coordinates of another point that lies on that line? Explain how you found your answer. 2. Of the points (0, 5), (2, 4), (3, 3), and (6, 2), which one does not lie on the same line as the other three? Explain how you found your answer. Developing Skills In 3–5, state, in each case, whether the pair of values for x and y is a member of the solution set of the equation 2x y 6. 3. x 4, y 2 4. x 0, y 6 5. x 4, y 2 For 6–9, state in each case whether the point whose coordinates are given is on the graph of the given equation. 6. x y 7; (4, 3) 8. 3x 2y 8; (2, 1) 7. 2y x 7; (1, 3) 9. 2y 3x 5; (1, 4) In 10–13, find in each case the number or numbers that can replace k so that the resulting ordered number pair will be on the graph of the given equation. 10. x 2y 5; (k, 2) 12. x 3y 10; (13, k) 11. 3x 2y 22; (k, 5) 13. x y 0; (k, k) In 14–17, find in each case a value that can replace k so that the graph of the resulting equation will pass through the point whose coordinates are given. Graphing Linear Functions Using Their Solutions 351 14. x y k; (2, 5) 16. y 2x k; (2, 1) 15. x y k; (5, 3) 17. y x k; (5, 0) In 18–23, solve each equation for y in terms of x. 19. 4x y 6 22. 4x 2y 8 18. 3x y 1 21. 12x 3 2y 20. 2y 6x 23. 6x 3y 5 In 24–26: a. Find the missing values of the variable needed to complete each table. b. Plot the points described by the pairs of values in each completed table; then, draw a line through the points. 24. y 4x 25. y 3x 1 26. x 2y ? ? ? In 27–50, graph each equation. 27. y x 28. y 5x 29. y 3x 30. y x 1 2y 1 1 31. x 2y 3 35. y 3x 1 39. y x 0 43. 2x y 6 47. 3x 2y 4 51. a. Through points (0, 2) and (4, 0), draw a straight line. 32. x 36. y 2x 4 40. 3x y 12 44. 3x y 6 48. x 3y 9 33. y x 3 37. x y 8 41. x – 2y 0 45. x 3y 12 49. 2x y 4 34. y 2x 1 38. x y 5 42. y 3x –5 46. 2x 3y 6 50. 4x 3y b. Write the coordinates of two other points on the line drawn in part a. 52. a. Through points (2, 3) and (1, 3), draw a straight line. b. Does point (0, 1) lie on the line drawn in part a? Applying Skills In 53–60: a. Write an equation that can be used to represent each sentence. b. Graph the equation. 53. The length of a rectangle, y, is twice the width, x. 54. The distance to school, y, is 2 miles more than the distance to the library, x. 55. The cost of a loaf of bread, x, plus the cost of a pound of meat, y, is 6 dollars. 352 Graphing Linear Functions and Relations 56. The difference between Tim’s height, x, and Sarah’s height, y, is 1 foot. 57. The measure of each of three sides of the trapezoid is x, the measure of the remaining side is y, and the perimeter is 6. 58. The measure of each of the legs of an isosceles triangle is x, the measure of the base is y, and the perimeter is 9. 59. Bob’s age, y, is equal to five more than twice Alice’s present age, x. 60. The sum of the number of miles that Paul ran, x, and the number of miles that Sue ran, y, is 30 miles. 9-3 GRAPHING A LINE PARALLEL TO AN AXIS Lines Parallel to the x-Axis An equation such as y 2 can be graphed in the coordinate plane. Any pair of values whose y-coordinate is 2, no matter what the x-coordinate is, makes the equation y 2 true.Therefore, (3, 2), (2, 2), (1, 2), (0, 2), (1, 2), or any other pair (a, 2) for all values of a, are points on the graph of y 2. As shown at the right, the graph of y 2 is a horizontal line parallel to the x-axis and 2 units above it. The equation y 2 defines a function and can be displayed on a graphing calculator. If there are other equations entered in the before Y= menu, press entering the new equation. DISPLAY: ENTER: CLEAR 4 –3 –2 –1 –1 –2 –3 Y 2 GRAPH The equation of a line parallel to the x-axis and b units from the x-axis is y b. If b is positive, the line is above the x-axis, and if b is negative, the line is below the x-axis. Graphing a Line Parallel to an Axis 353 Lines Parallel to the y-Axis An equation such as x 3 can be graphed in the coordinate plane. Any pair of values whose x-coordinate is 3, no matter what the y-coordinate is, makes the equation x 3 true. Therefore, (3, 2), (–3, –1), (3, 0), (3, 1), (3, 2), or any other pair (3, b) for all values of b, are points on the graph of x 3. As shown at the right, the graph of x 3 is a vertical line parallel to the y-axis and 3 units to the left of it. y 3 2 1 O –4 –2 –1 1 2 3 –1 –2 –3 –4 3 – = x x The equation x 3 defines a relation that is not a function because it defines a set of ordered pairs but every ordered pair has the same first element. The method used to draw the graph of a function on a graphing calculator cannot be used to display this graph. The equation of a line parallel to the y-axis and a units from the y-axis is x a. If a is positive, the line is to the right of the y-axis, and if a is negative, the line is to the left of the x-axis. EXERCISES Writing About Mathematics 1. Mike said that the equation of the x-axis is x 0. Do you agree with Mike? Explain why or why not. 2. Does the line whose equation is y 4 intersect the x-axis? Determine the coordinates of the point of intersection if one exists or explain why a point of intersection does not exist. Developing Skills In 3–17, draw the graph of each equation. 3. x 6 8. y 4 x 5 1 2 13. 4. x 4 9. y 5 y 5 11 2 14. 5. x 0 10. y 0 15. y 2.5 6. x 3 11. y 4 x 5 23 2 16. 18. Write an equation of the line that is parallel to the x-axis 7. x 5 12. y 7 17. y 3.5 a. 1 unit above the axis. b. 5 units above the axis. c. 4 units below the axis. d. 8 units below the axis. e. 2.5 units above the axis. f. 3.5 units below the axis. 354 Graphing Linear Functions and Relations 19. Write an equation of the line that is parallel to the y-axis a. 3 units to the right of the axis b. 10 units to the right of the axis c. 41 2 units to the left of the axis d. 6 units to the left of the axis. e. 2.5 units to the right of the axis. f. 5.2 units to the left of the axis. 20. Which statement is true about the graph of the equation y 6? (1) It is parallel to the y-axis (2) It is parallel to the x-axis. (3) It is not parallel to either axis. (4) It goes through the origin. 21. Which statement is not true about the graph of the equation x 5? (3) It is a function. (1) It goes through (5, 0). (2) It is a vertical line. (4) It is parallel to the y-axis. 22. Which statement is true about the graph of the equation y x? (1) It is parallel to the y-axis. (2) It is parallel to the x-axis. (3) It goes through (2, 2). (4) It goes through the origin. Applying Skills 23. The cost of admission to an amusement park on Family Night is $25.00 for a family of any size. a. What is the cost of admission for the Gauger family of six persons? b. Write an equation for the cost of admission, y, for a family of x persons. 24. The coordinates of the vertices of rectangle ABCD are A(2, 1), B(4, 1), C(4, 5), and D(2, 5). a. Draw ABCD on graph paper. b. Write the equations of the lines g AB , g BC , g CD , and g DA . c. Draw a horizontal line and a vertical line that separate ABCD into four congruent rectangles. d. Write the equations of the lines drawn in part c. 25. During the years 2004, 2005, and 2006, there were 5,432 AnyClothes stores. a. Write a linear equation that gives the number of stores, y. b. Predict the number of stores for the years 2007 and 2008. 26. During the years 1999 through 2006, SmallTown News’ circulation was 74,000 each year. a. Write a linear equation that gives
the circulation, y. b. Predict the circulation of SmallTown News in the years 2007 and 2008. 9-4 THE SLOPE OF A LINE The Slope of a Line 355 Meaning of the Slope of a Line Easy Hill Tough Hill A 80 m B 20 m C D 80 m E 40 m F It is more difficult to hike up Tough Hill, shown above, than to hike up Easy Hill. Tough Hill rises 40 meters vertically over a horizontal distance of 80 meters, whereas Easy Hill rises only 20 meters vertically over the same horizontal distance of 80 meters. Therefore, Tough Hill is steeper than Easy Hill. To compare the steepness of roads , which lead up the two hills, we and compare their slopes. The slope of road is the ratio of the change in vertical distance, CB, to DE AB AB the change in horizontal distance, AC: slope of road AB 5 change in vertical distance, CB change in horizontal distance, AC 5 20 m 80 m 5 1 4 Also: slope of road DE 5 change in vertical distance, FE change in horizontal distance, DF 5 40 m 80 m 5 1 2 Note that the steeper hill has the larger slope. Finding the Slope of a Line To find the slope of the line determined by the two points (2, 3) and (5, 8), as shown at the right, we write the ratio of the difference in y-values to the difference in x-values as follows: slope 5 difference in y-values difference in x-values Suppose we change the order of the points (2, 3) and (5, 8) in performing the computation. We then have: slope 5 difference in y-values 23 5 5 difference in x-values 5 3 2 8 3 The result of both computations is the same. 2 2 5 5 25 5, 8) 5 = 3 – 8 (2, 3) 5 – 2 = 3 O –1 –1 1 2 3 4 5 6 7 x 356 Graphing Linear Functions and Relations When we compute the slope of a line that is determined by two points, it does not matter which point is considered as the first point and which the second. Also, when we find the slope of a line using two points on the line, it does not matter which two points on the line we use because all segments of a line have the same slope as the line. Procedure To find the slope of a line: 1. Select any two points on the line. 2. Find the vertical change, that is, the change in y-values by subtracting the y-coordinates in any order. 3. Find the horizontal change, that is, the change in x-values, by subtracting the x-coordinates in the same order as the y-coordinates. 4. Write the ratio of the vertical change to the horizontal change. We can use this procedure to find the , the lines shown in the g LM g RS and slopes of graph at the right. slope of g LM slope of g RS 5 5 or 2 5 5 6 – 2 vertical change horizontal change 2 5 2 3 – 1 5 4 1 vertical change horizontal change 4 – 1 5 22 24 2 (22) 3 5 22 3, 6) 4 O (1, 2) L –2 –1 1 2 3 4 5 6 x R Slope is a rate of change. It is a measure of the rate at which y changes compared to x. For g 2 LM whose slope is 2 or , y increases 2 units 1 when x increases 1 unit. When the second element of a rate is 1, the rate is a unit rate of change. If x is the independent variable and y is the dependent variable, then the slope is the rate of change in the dependent variable compared to the independent variable. –2 –3 –4 (4, –4) (1, –2) 2 – S 3 For g RS whose slope is , y decreases 2 units when x increases 3 units. We could also write this rate as to write the slope as a unit rate of change, that 22 3 2 3 1 is, y decreases of a unit when x increases 1 unit. 2 3 The Slope of a Line 357 In general: The slope, m, of a line that passes through any two points P1(x1, y1) and x2, is the ratio of the change or difference of the P2(x2, y2), where x1 y-values of these points to the change or difference of the corresponding x-values. Thus: slope of a line difference iny-values difference in x-values Therefore: slope of g P1P2 m 2 y y 1 2 x2 2 x1 x1, can The difference in x-values, x2 be represented by x, read as “delta x.” Similarly, the difference in y-values, y1, can be represented by y, read y2 as “delta y.” Therefore, we write: slope of a line m Dy Dx Positive Slopes g AB Examining from left to right and observing the path of a point, from C to D for example, we see that the line is rising. As the x-values increase, the y-values also increase. Between point C and point D, the change in y is 1, and the change in x is 2. Since both y and x are positive, the slope of must be positive. Thus: g AB slope of g AB m Dy Dx 5 1 2 y x2 – x1 1 y – 2 y O P1 (x1 , y1) P2 (x2 , y2) x x y A 1 –1 –1 B 2 1 D(3, 3) C(1, 2) O 1 Principle 1. As a point moves from left to right along a line that is rising, y increases as x increases and the slope of the line is positive. 358 Graphing Linear Functions and Relations Negative Slopes g EF from left to right and Now, examining observing the path of a point from C to D, we see that the line is falling. As the x-values increase, the y-values decrease. Between point C and point D, the change in y is 2, and the change in x is 3. Since y is negative and x is g EF positive, the slope of must be negative. slope of g EF m Dy Dx 5 22 3 5 22 3 y 1 –1 –1 E –2 O 1 C (2, 3) 3 D(5, 1) F x Principle 2. As a point moves from left to right along a line that is falling, y decreases as x increases and the slope of the line is negative. Zero Slope g GH g GH is parallel to the x-axis. We consider On the graph, a point moving along from left to right, for example from C to D. As the x-values increase, the y-values are unchanged. Between point C and point D, the change in y is 0 and the change in x is 3. Since y is 0 and x is 3, the slope of must be 0. Thus: g GH y 1 O –1 –1 G (–2, –2) C x H (1, –2) D slope of g GH m Dy Dx 5 0 3 5 0 Principle 3. If a line is parallel to the x-axis, its slope is 0. Note: The slope of the x-axis is also 0. No Slope g LM g LM is parallel to the y-axis. We consider On the graph, a point moving upward along , for example from C to D. The x-values are unchanged, but the y-values increase. Between point C and point D, the change in y is 3, and the change in x is 0. Since the slope is the change in y divided by the change in x and of has no defined a number cannot be divided by 0, slope. g LM g LM y M 1 O –1 –1 D(2, 1) x C(2, –2) L The Slope of a Line 359 slope of g MC m Dy Dx 5 1 2 (22) 0 5 undefined Principle 4. If a line is parallel to the y-axis, it has no defined slope. Note: The y-axis itself has no defined slope. EXAMPLE 1 Find the slope of the line that is determined by points (2, 4) and (4, 2). Solution Plot points (2, 4) and (4, 2). Let point (2, 4) be P1(x1, y1), and let point 2, y1 y 2 y 1 2 x2 2 x1 4, and y2 4, x2 2. y g slope of P1P2 (4, 2) be P2(x2, y2). Then, x1 Dy Dx 22 6 5 21 3 4 2 (22) Answer P1(–2, 4) –2 6 O 1 1 –1 –1 P2(4, 2) x EXAMPLE 2 3 Through point (2, 1), draw the line whose slope is . 2 Solution How to Proceed (1) Graph point A(2, 1): y (2) Note that since slope Dy Dx 5 3 2 , when y changes by 3, then x changes by 2. Start at point and move 3 units upward and 2 units to the right to locate point B: A(2, 21) (3) Start at B and repeat these movements to locate point C: (4) Draw a line that passes through points A, B, and C: C 2 B 3 2 1 –1–1 3 O 1 A(2, –1) x 360 Graphing Linear Functions and Relations KEEP IN MIND A fundamental property of a straight line is that its slope is constant. Therefore, any two points on a line may be used to compute the slope of the line. Applications of Slope In real-world contexts, the slope is either a ratio or a rate. When both the x and y variables have the same unit of measure, slope represents a ratio since it will have no unit of measure. However, when the x and y variables have different units of measure, slope represents a rate since it will have a unit of measure. In both cases, the slope represents a constant ratio or rate of change. EXAMPLE 3 The following are slopes of lines representing the daily sales, y, over time, x, for various sales representatives during the course of a year: (1) m 20 (3) m 39 (2) m 45 (4) m 7 a. Interpret the meaning of the slopes if sales are given in thousands of dollars and time is given in months. b. Which sales representative had the greatest monthly increase in sales? c. Is it possible to determine which sales representative had the greatest total sales at the end of the year? Solution a. The slopes represent the increase in sales per month: (1) $20,000 increase each month (2) $45,000 increase each month (3) $39,000 increase each month (4) $7,000 decrease each month b. At $45,000, (2) has the greatest monthly increase in sales. c. The slopes give the increase in sales each month, not the total monthly sales. With the given information, it is not possible to determine which sales representative had the greatest total sales at the end of the year. EXERCISES Writing About Mathematics 1. Regina said that in Example 1, the answer would be the same if the formula for slope had been written as . Do you agree with Regina? Explain why or why not. Use y1 y2 2 x1 2 x2 to y1 y2 2 x1 2 x2 find the answer to Example 1 to justify your response. The Slope of a Line 361 2. Explain why any two points that have the same x-coordinate lie on a line that has no slope. Developing Skills In 3–8: a. Tell whether each line has a positive slope, a negative slope, a slope of zero, or no slope. b. Find the slope of each line that has a slope. 3. 5. 7. y 1 –1 1 x y 1 –1 –1 1 y 1 –1 –1 1 x x 4. 6. 8. y 1 –1 1 x y 1 –1 –1 1 y 1 –1 –1 1 x x 362 Graphing Linear Functions and Relations In 9–17, in each case: a. Plot both points, and draw the line that they determine. b. Find the slope of this line. 9. (0, 0) and (4, 4) 12. (1, 5) and (3, 9) 15. (5, 2) and (7, 8) 10. (0, 0) and (4, 8) 13. (7, 3) and (1, 1) 16. (4, 2) and (8, 2) 11. (0, 0) and (3, 6) 14. (2, 4) and (0, 2) 17. (1, 3) and (2, 3) In 18–29, in each case, draw a line with the given slope, m, through the given point. m 5 2 3 18. (0, 0); m 2 21. (4, 5); 24. (1, 5); m 1 25 27. (1, 0); m 4 Applying Skills 19. (1, 3); m 3 22. (3, 1); m 0 25. (2, 4); m 28. (0, 2); m 23 2 22 3 20. (2, 5); m 4 23. (3, 4); m 2 26. (2, 3); m 29. (2, 0); m 21 3 1 2 30. Points A(2, 4), B(8,
4), and C(5, 1) are the vertices of ABC. Find the slope of each side of ABC. 31. Points A(3, 2), B(9, 2), C(7, 4), and D(1, 4) are the vertices of a quadrilateral. a. Graph the points and draw quadrilateral ABCD. b. What type of quadrilateral does ABCD appear to be? c. Compute the slope of and the slope of d. What is true of the slope of and the slope of g BC g BC g AD . g AD ? e. If two segments such as AD and BC , or two lines such as what appears to be true of their slopes? g AD and g BC , are parallel, f. Since AB and CD are parallel, what might be true of their slopes? g. Compute the slope of AB and the slope of CD . h. Is the slope of AB equal to the slope of CD ? 32. A path over a hill rises 100 feet vertically in a horizontal distance of 200 feet and then descends 100 feet vertically in a horizontal distance of 150 feet. B 100' a. Find the slope of the path up the hill when walking from point A to point B. A 200' C 150' b. Find the slope of the path down the hill when walking from point B to point C. c. What is the unit rate of change from point A to point B? d. What is the unit rate of change from point B to point C? The Slopes of Parallel and Perpendicular Lines 363 33. The amount that a certain internet phone company charges for international phone calls is represented by the equation C $0.02t $1.00, where C represents the total cost of the call and t represents time in minutes. Explain the meaning of the slope in terms of the information provided. 34. The monthly utility bill for a certain school can be represented by the equation C $0.075g $100, where C is the monthly bill and g is the amount of gas used by the hundred cubic feet (CCF). Explain the meaning of the slope in terms of the information provided. 35. A road that has an 8% grade rises 8 feet vertically for every 100 feet horizontally. a. Find the slope of the road. b. Explain the meaning of the slope in terms of the information provided. c. If you travel a horizontal distance of 200 feet on this road, what will be the amount of vertical change in your position? 9-5 THE SLOPES OF PARALLEL AND PERPENDICULAR LINES Parallel Lines /2 /1 and In the diagram, lines are parallel. Right triangle 1 has been drawn with one leg coinciding with the x-axis and with its hypotenuse coinciding with . Similarly, right triangle 2 has been drawn with one leg coinciding with the x-axis and with its hypotenuse coinciding with /1 /2 . Since triangles 1 and 2 are right triangles, tan a oppa adja tan b oppb adjb y oppa a adja Triangle 1 oppb x b adjb Triangle 2 ,1 ,2 The x-axis is a transversal that intersects parallel lines sponding angles, a and b, are congruent. Therefore, /1 and /2 . The corre- oppa adja oppb adjb Since oppa adja is the slope of /1 and oppb adj b two parallel lines have the same slope. is the slope of /2 , we have shown that If the slope of <1 is m1, the slope of <2 is m2, and <1 <2 , then m1 m2. 7 The following statement is also true: If the slope of <1 is m1, the slope of <2 is m2, and m1 m2, then <1 <2 . 7 364 Graphing Linear Functions and Relations Perpendicular Lines y a C a b ,2 ,1 ' B ' C /2 /1 /1 and . The slope of /1 are perpenIn the diagram, lines dicular. Right triangle ABC has been drawn with one leg parallel to the x-axis, the other leg parallel to the y-axis, and the hypotenuse is since y coinciding with increases by a units as x increases by b units. Triangle ABC is then rotated 90° counterclockwise about A so that the hypotenuse of ABC coincides with /2 . Notice that since each leg has been rotated by 90°, the two legs are still parallel to the axes. However, they are now parallel to different axes. Also, in the rotated triangle, the change in y is an increase of b units and the change in x is a decrease of a units, that is, a. Therefore, the slope b /2 /2 or of is the negative reciprocal of the 2a /1 slope of . We have thus shown that perpendicular lines have slopes that are negative reciprocals of each other. . In other words, the slope of 2b a is A B b b a x In general: If the slope of m1 ? m2 5 21 . <1 is m1, the slope of <2 is m2, and <1 ' <2 , then m1 2 1 m2 or The following statement is also true: If the slope of <1 ' <2 . then <1 is m1, the slope of <2 is m2, and m1 2 1 m2 or m1 ? m2 5 21 , EXAMPLE 1 The equation of g AB is y 2x 1. a. Find the coordinates of any two points on g AB . b. What is the slope of g AB ? c. What is the slope of a line that is parallel to g AB ? d. What is the slope of a line that is perpendicular to g AB ? Solution a. If x 0, y 2(0) 1 1. One point is (0, 1). If x 1, y 2(1) 1 1. Another point is (1, 1). b. Slope of g AB 1 2 (21) 0 2 1 5 2 21 5 22 . The Slopes of Parallel and Perpendicular Lines 365 c. The slope of a line parallel to g AB d. The slope of a line perpendicular to is equal to the slope of 2 1 g AB 22 5 1 2 is . g AB , –2. Answers a. (0, 1) and (1, 1) but many other answers are possible. b. 2 c. 2 d. 1 2 EXERCISES Writing About Mathematics 1. a. What is the slope of a line that is perpendicular to a line whose slope is 0? Explain your answer. b. What is the slope of a line that is perpendicular to a line that has no slope? Explain your answer. 2. What is the unit rate of change of y with respect to x if the slope of the graph of the equa- tion for y in terms of x is ? 1 4 Developing Skills In 3–11: a. Write the coordinates of two points on each line whose equation is given. b. Use the coordinates of the points found in part a to find the slope of the line. c. What is the slope of a line that is parallel to each line whose equation is given? d. What is the slope of a line that is perpendicular to the line whose equation is given? 4. y x 2 7. x y 4 10. x 4 5. y 3x 7 8. 2x 3y 6 11. y 5 3. y 2x 6 6. 3y x 9. x 2y 1 Applying Skills 12. A taxi driver charges $3.00 plus $0.75 per mile. a. What is the cost of a trip of 4 miles? b. What is the cost of a trip of 8 miles? c. Use the information from parts a and b to draw the graph of a linear function described by the taxi fares. d. Write an equation for y, the cost of a taxi ride, in terms of x, the length of the ride in miles. e. What is the change in the cost of a taxi ride when the length of the ride changes by 4 miles? f. What is the unit rate of change of the cost with respect to the length of the ride in dol- lars per mile? 366 Graphing Linear Functions and Relations 13. Mrs. Boyko is waiting for her oven to heat to the required temperature. When she first turns the oven on, the temperature registers 100°. She knows that the temperature will increase by 10° every 5 seconds. a. What is the temperature of the oven after 10 seconds? b. What is the temperature of the oven after 15 seconds? c. Use the information from parts a and b to draw the graph of the linear function described by the oven temperatures. d. Write an equation for y, the temperature of the oven, in terms of x, the number of sec- onds that it has been heating. e. What is the unit rate in degrees per second at which the oven heats? 9-6 THE INTERCEPTS OF A LINE The graph of any linear equation that is not parallel to an axis intersects both axes. The point at which the graph intersects the y-axis is a point whose x-coordinate is 0. The y-coordinate of this point is called the y-intercept of the equation. For example, for the equation 2x y 6, let x 0: 2x y 6 2(0 The graph of 2x y 6 intersects the y-axis at (0, 6) and the y-intercept is 6. The point at which the graph intersects the x-axis is a point whose y-coordinate is 0. The x-coordinate of this point is called the x-intercept of the equation. For example, for the equation 2x y 6, let y 0: 2x y 6 2x 0 6 2x 6 x 3 The graph of 2x y 6 intersects the x-axis at (3, 0) and the x-intercept is 3. Divide each term of the equation 2x y 6 by the constant term, 6 y 6 5 6 2x Note that x is divided by the x-intercept, 3, and y is divided by the y-intercept, –6. The Intercepts of a Line 367 In general, if a linear equation is written in the form y x b 5 1 a 1 y-intercept is b. , the x-intercept is a and the y x b 5 1 a 1 a y-intercept Q x-intercept An equation of the form x a 1 equation. Two points on the graph of these two points to find the slope of the line. a 2 0 5 2b 0 2 b slope a 5 2b a y b 5 1 is called the intercept form of a linear y x b 5 1 a 1 are (a, 0) and (0, b). We can use Slope and y-Intercept Consider the equation 2x y 6 from another point of view. Solve the equation for y: 2x y 6 y 2x 6 y 2x 6 There are two numbers in the equation that has been solved for y: the coefficient of x, 2, and the constant term, 6. Each of these numbers gives us important information about the graph. The y-intercept of a graph is the y-coordinate of the point at which the graph intersects the y-axis. For every point on the y-axis, x 0. When x 0, y 2(0) 6 0 6 6 Therefore, –6 is the y-intercept of the graph of the equation y 2x 6 and (0, 6) is a point on its graph. The y-intercept is the constant term of the equation when the equation is solved for y. Another point on the graph of y 2x 6 is (1, 4). Use the two points, (0, 6) and (1, 4), to find the slope of the line. slope of y 2x 6 26 2 (24) 0 2 1 5 22 21 5 2 The slope is the coefficient of x when the equation is solved for y. For any equation in the form y mx b: 1. The coefficient of x is the slope of the line. 2. The constant term is the y-intercept. 368 Graphing Linear Functions and Relations The equation we have been studying can be compared to a general equation of the same type: y 2x 3 Q slope a y-intercept y mx b Q slope a y-intercept The following statement is true for the general equation: If the equation of a line is written in the form y mx b, then the slope of the line is m and the y-intercept is b. An equation of the form y mx b is called the slope-intercept form of a linear equation. EXAMPLE 1 Find the intercepts of the graph of the equation 5x 2y 6. Solution To find the x intercept, let y 0: 5x 2y 6 5x 2(0) 6 5x 0 6 5x 6 x 6 5 To find the y intercept, let x 0: 5x 2y 6 5(0) 2y 6 0 2y
6 2y 6 y 3 6 Answer The x intercept is and the y intercept is 3. 5 EXAMPLE 2 Find the slope and the y-intercept of the line that is the graph of the equation 4x + 2y 10. Solution How to Proceed (1) Solve the given equation for y to obtain an equivalent equation of the form y mx b: (2) The coefficient of x is the slope: (3) The constant term is the y-intercept: 4x 2y 10 2y 4x 10 y 2x 5 slope 2 y-intercept 5 Answer slope 2, y-intercept 5 The Intercepts of a Line 369 EXERCISES Writing About Mathematics 1. Amanda said that the y-intercept of the line whose equation is 2y 3x 8 is 8 because the y-intercept is the constant term. Do you agree with Amanda? Explain why or why not. 2. Roberto said that the line whose equation is y x has no slope because x has no coefficient and no y-intercept because there is no constant term. Do you agree with Roberto? Explain why or why not. Developing Skills In 3–18, in each case, find the slope, y-intercept, and x-intercept of the line that is the graph of the equation. 3. y 3x 1 7. x 3 4. y 3 x 8. y 2 11. 2x y 5 12. 3y 6x 9 5. y 2x 22 9. y 3x 1 4 13. 2y 5x 4 y x 3 5 1 2 1 17. 6. y x 14. 10. y 3x 7 2x 1 3 4 5 1 1 3y 4x 5 2 2y 5 1 15. 4x 3y 0 19. What do the graphs of the equations y 4x, y 4x 2, and y 4x 2 all have in 16. 2x 5y 10 18. common? 20. What do the lines that are the graphs of the equations y 2x + 1, y 3x + 1, and y 5 24x 1 1 all have in common? 21. If two lines are parallel, how are their slopes related? 22. What is true of two lines whose slopes are equal? In 23–26, state in each case whether the lines are parallel, perpendicular, or neither. 23. y 3x 2, y 3x 5 25. y 4x 8, 4y x 3 27. Which of the following statements is true of the graph of y 3x? 24. y 2x 6, y 2x 6 26. y 2x, x 2y (1) It is parallel to the x-axis. (2) It is parallel to the y-axis. (3) Its slope is 3. (4) It does not have a y-intercept. 28. Which of the following statements is true of the graph of y 8? (1) It is parallel to the x-axis. (2) It is parallel to the y-axis. (3) It has no slope. (4) It goes through the origin. 370 Graphing Linear Functions and Relations 9-7 GRAPHING LINEAR FUNCTIONS USING THEIR SLOPES The slope and any one point can be used to draw the graph of a linear function. One convenient point is the point of intersection with the y-axis, that is, the point whose x-coordinate is 0 and whose y-coordinate is the y-intercept. EXAMPLE 1 Draw the graph of 2x 3y 9 using the slope and the y-intercept. Solution How to Proceed (1) Transform the equation into the form y 5 mx 1 b : (2) Find the slope of the line (the coefficient of x): (3) Find the y-intercept of the line (the constant): (4) On the y-axis, graph point A, whose y-coordinate is the y-intercept: (5) Use the slope to find two more points on the line. Since Dy Dx 5 22 slope , when y changes 3 by 2, x changes by 3. Therefore, start at point A and move 2 units down and 3 units to the right to locate point B. Then start at point B and repeat the procedure to locate point C: (6) Draw the line that passes through the three points: This line is the graph of 2x 3y 9. Check To check a graph, select two or more points on the line drawn and substitute their coordinates in the original equation. For example, check this graph using points (4.5, 0), (3, 1), and (6, 1). 2x 3y 9 3y 2x 9 y 22 3 x 1 3 slope 22 3 y-intercept 3 y 1 O y 1 O A(0, 3) 1 A(0, 3) 1 B B x C x C Graphing Linear Functions Using Their Slopes 371 2x 3y 9 2x 3y 9 2x 3y 9 2(4.5) 1 3(0) 5? 9 1 0 5? 9 9 9 9 ✔ 2(3) 1 3(1) 5? 6 1 3 5? 9 9 9 9 ✔ 2(6) 1 3(21) 5? 12 2 3 5? 9 9 9 9 ✔ In this example, we used the slope and the y-intercept to draw the graph. We could use any point on the graph of the equation as a starting point. EXAMPLE 2 Draw the graph of 3x 2y 9 using the slope and any point. Solution How to Proceed (1) Solve the equation for y. The slope 3x 2 2y 5 9 is and the y-intercept is 3 2 29 2 : (2) Since is not a convenient 0, 2 9 2 A B point to graph, choose another point. Let x 1. A convenient point on the graph is A(1, 3): (3) Graph point A whose coordinates are (1,23) : (4) Use the slope to find two more points on the line. Since slope when y changes by 3, x changes by 2. Dy Dx 5 3 2 , Therefore, start at point A and move 3 units up and 2 units to the right to locate point B. Then start at point B and repeat the procedure to locate point C: (5) Draw the line the passes through the three points. This is the graph of 3x 2y 9: 22y 5 23x 1 9 22 1 9 y 5 23x 22 2x 2 9 y 5 3 2 2(1 26 2 y 5 23 y 1 O C B x A(1, –3) Note that the graph intersects the x-axis at B(3, 0). The x-intercept is 3. 372 Graphing Linear Functions and Relations Translating, Reflecting, and Scaling Graphs of Linear Functions An alternative way of looking at the effects of changing the values of the y-intercept and slope is that of translating, reflecting, or scaling the graph of the linear function y x. For instance, the graph of y x 3 can be thought of as the graph of y x shifted 3 units up. Similarly, the graph of y x can be thought of the graph of y x reflected in the x-axis. On the other hand, the graph of y 4x can be thought of as the graph of stretched vertically by a factor is the graph of y x compressed vertically by a of 4, while the graph of 1 factor of . 4 y 5 1 4x = 4x Translation Rules for Linear Functions If c is positive: The graph of y x c is the graph of y x shifted c units up. The graph of y x c is the graph of y x shifted c units down. Reflection Rule for Linear Functions The graph of y x is the graph of y x reflected in the x-axis. Scaling Rules for Linear Functions When c 1, the graph of y cx is the graph of y x stretched vertically by a factor of c. When 0 c 1, the graph of y cx is the graph of y x compressed ver- tically by a factor of c. Graphing Linear Functions Using Their Slopes 373 EXAMPLE 3 In a–d, write an equation for the resulting function if the graph of y 5 x is: a. shifted 6 units down b. reflected in the x-axis and shifted 3 units up c. compressed vertically by a factor of 1 10 and shifted 1 unit up d. stretched vertically by a factor of 4 and shifted 1 unit down Solution a. y x 6 Answer b. First, reflect in the x-axis: Then, translate the resulting function 3 units up: c. First, compress vertically by a factor of 1 10 : Then, translate the resulting function 1 unit up: d. First, stretch vertically by a factor of 4: Then, translate the resulting function 1 unit down: y x y x 3 Answer y y 1 10x 1 10x 1 1 Answer y 4x y 4x 1 Answer EXERCISES Writing About Mathematics 1. Explain why (0, b) is always a point on the graph of y mx b. 2. Gunther said that in the first example in this section, since the slope of the line, , could have been written as , points on the line could have been located by moving up 2 units and to the left 3 units from the point (0, 3). Do you agree with Gunther? Explain why or why not. 2 23 22 3 3. Hypatia said that, for linear functions, vertical translations have the same effect as horizontal translations and that reflecting across the x-axis has the same effect as reflecting across the y-axis. Do you agree with Hypatia? Explain why or why not. Developing Skills In 4–15, graph each equation using the slope and the y-intercept. 5. y 2x 4. y 2x 3 6. y 2x 7. y 3x 2 374 Graphing Linear Functions and Relations 2 8. y 3x 1 2 12. y 2x 8 1 9. y 2x 2 1 13. 3x y 4 21 10. y 3x y x 3 5 1 2 1 14. 23 11. y 15. 4x 2y 4x 1 6 In 16–21, graph the equation using the slope and any point. 16. 3y 4x + 9 19. 2x 3y 1 17. 5x 2y 3 20. 3y x 5 18. 3x + 4y 7 21. 2x 3y 6 0 In 22–25, describe the translation, reflection, and/or scaling that must be applied to y x to obtain the graph of each given function. 23. y 25. y (x 2) 2 24. y 2x 1.5 22. y x 2 21 3x 2 2 Applying Skills 26. a. Draw the line through (2, 3) whose slope is 2. b. What appears to be the y-intercept of this line? c. Use the slope of the line and the answer to part b to write an equation of the line. d. Do the coordinates of point (2, 3) satisfy the equation written in part c? 2 27. a. Draw the line through (3, 5) whose slope is . 3 b. What appears to be the y-intercept of this line? c. Use the slope of the line and the answer to part b to write an equation of the line. d. Do the coordinates of point (3, 5) satisfy the equation written in part c? 28. a. Is (1, 1) a point on the graph of 3x 2y 1? b. What is the slope of 3x 2y 1? c. Draw the graph of 3x 2y 1 using point (1, 1) and the slope of the line. d. Why is it easier to use point (1, 1) rather than the y-intercept to draw this graph? 9-8 GRAPHING DIRECT VARIATION Recall that, when two variables represent quantities that are directly proportional or that vary directly, the ratio of the two variables is a constant. For example, when lemonade is made from a frozen concentrate, the amount of lemonade, y, that is obtained varies directly as the amount of concentrate, x, that is used. If we can make 3 cups of lemonade by using 1 cup of concentrate, the relationship can be expressed as y x 5 3 1 or y 3x Graphing Direct Variation 375 The constant of variation is 3. The equation y 3x can be represented by a line in the coordinate plane, as shown in the graph below. x 0 1 2 3 3x 3(0) 3(1) 3(2) 3(35 –5 x 5 Here, the replacement set for x and y is the set of positive numbers and 0. Thus, the graph includes only points in the first quadrant. In our example, if 0 cups of frozen concentrate are used, 0 cups of lemonade can be made. Thus, the ordered pair (0, 0) is a member of the solution set of y 3x. Using any two points from the table above, for example. (0, 0) and (1, 3), we can write: Dy Dx 5 3 2 0 1 2 0 5 3 Thus, we see that the slope of the line is also the constant of variation. An equation of a direct variation is a special case of an equation written in slopeand m the constant of variation, k. A direct variintercept form; that is, ation indicates that y is a multiple of x. Note that the graph of a direct variation is always a line through the origin. 1 5 3 b 5 0 The unit of measure
for the lemonade and for the frozen concentrate is the same, namely, cups. Therefore, no unit of measure is associated with the ratio which, in this case, is or 3. or 6 cups 2 cups 5 3 1 lemonade concentrate 5 lemonade concentrate 5 9 cups 3 cups 5 3 1 There are many applications of direct variation in business and science, and it is important to recognize how the choice of unit can affect the constant of variation. For example, if a machine is used to pack boxes of cereal in cartons, the rate at which the machine works can be expressed in cartons per minute or in cartons per second. If the machine fills a carton every 6 seconds, it will fill 10 cartons in 1 minute. The rate can be expressed as: lemonade concentrate 5 or 3 1 3 cups 1 cup 5 3 1 1 carton 6 seconds 5 1 6 carton/second or 10 cartons 1 minute 10 cartons/minute Here each rate has been written as a unit rate. 376 Graphing Linear Functions and Relations As shown at right, each of these rates can be represented by a graph, where the rate, or is the constant of variation, slope of the line. The legend of the graph, that is, the units in which the rate is expressed, must be clearly stated if the graph is to be meaningful. y 10 10 Seconds 6 x 1 2 3 4 5 Minutes Cartons per second Cartons per minute EXAMPLE 1 The amount of flour needed to make a white sauce varies directly as the amount of milk used. To make a white sauce, a chef used 2 cups of flour and 8 cups of milk. Write an equation and draw the graph of the relationship. Solution Let x the number of cups of milk, and y the number of cups of flour. Then: y x 5 2 8 8y 2x y y 1 4 x 1 4(2) 1 4(4) 1 4(6) 2 8x 1 4x Milk EXERCISES Writing About Mathematics 1. A chef made the white sauce described in Example 1, measuring the flour and milk in ounces. Explain why the ratio of flour to milk should still be 1 : 4. 2. When Eduardo made white sauce, he used 1 cup of flour and 1 quart of milk. Is the ratio of flour to milk in Eduardo’s recipe the same as in the recipe used by the chef in the example? Explain why or why not. Developing Skills In 3–12, y varies directly as x. In each case: a. What is the constant of variation? b. Write an equation for y in terms of x. c. Using an appropriate scale, draw the graph of the equation written in part b. d. What is the slope of the line drawn in part c? 3. The perimeter of a square (y) is 12 centimeters when the length of a side of the square (x) is 3 centimeters. 4. Jeanne can type 90 words (y) in 2 minutes (x). 5. A printer can type 160 characters (y) in 10 seconds (x). Graphing Direct Variation 377 6. A cake recipe uses 2 cups of flour (y) to 11 2 7. The length of a photograph (y) is 12 centimeters when the length of the negative from cups of sugar (x). which it is developed (x) is 1.2 centimeters. 8. There are 20 slices (y) in 12 ounces of bread (x). 9. Three pounds of meat (y) will serve 15 people (x). 10. Twelve slices of cheese (y) weigh 8 ounces (x). 11. Willie averages 3 hits (y) for every 12 times at bat (x). 12. There are about 20 calories (y) in three crackers (x). Applying Skills 13. If a car travels at a constant rate of speed, the distance that it travels varies directly as time. If a car travels 75 miles in 2.5 hours, it will travel 110 feet in 2.5 seconds. a. Find the constant of variation in miles per hour. b. Draw a graph that compares the distance that the car travels in miles to the number of hours traveled. Let the horizontal axis represent hours and the vertical axis represent distance. c. Find the constant of variation in feet per second. d. Draw a graph that compares the distance the car travels in feet to the number of sec- onds traveled. 14. In typing class, Russ completed a speed test in which he typed 420 characters in 2 minutes. His teacher told him to let 5 characters equal 1 word. a. Find Russ’s rate on the test in characters per second. b. Draw a graph that compares the number of characters Russ typed to the number of minutes that he typed. (Let the horizontal axis represent minutes and the vertical axis represent characters.) c. Find Russ’s rate on the test in words per minute. d. Draw a graph that compares the number of words Russ typed to the number of min- utes that he typed. (Let the horizontal axis represent minutes and the vertical axis represent words.) In 15–20, determine if the two variables are directly proportional. If so, write the equation of variation. 15. Perimeter of a square (P) and the length of a side (s). 16. Volume of a sphere (V) and radius (r). 17. The total amount tucked away in a piggy bank (s) and weekly savings (w), starting with an initial balance of $50. 378 Graphing Linear Functions and Relations 18. Simple interest on an investment (I) in one year and the amount invested (P), at a rate of 2.5%. 19. Length measured in centimeters (c) and length measured in inches (i). 20. Total distance traveled (d) in one hour and average speed (s). 9-9 GRAPHING FIRST-DEGREE INEQUALITIES IN TWO VARIABLES When a line is graphed in the coordinate plane, the line is a plane divider because it separates the plane into two regions called half-planes. One of these regions is a half-plane on one side of the line; the other is a half-plane on the other side of the line. Let us consider, for example, the horizontal line y 3 as a plane divider. As shown in the graph below, the line y 3 and the two half-planes that it forms determine three sets of points: 1. The half-plane above the line y 3 is the set of all points whose y-coordinates are greater than 3, that is, y 3. For example, at point A, y 5; at point B, y 4. 2. The line y 3 is the set of all points whose y-coordinates are equal to 3. For example, y 3 at each point C(4, 3), D(0, 3), and G(6, 3). 3. The half-plane below the line y 3 is the set of all points whose y-coordinates are less than 3, that is, y 3. For example, at point E, y 1; at point F, y 2. y (–2, 4)B A(3, 5) y = 3 C(–4, 3) D(0, 3) (6, 3)G (–3, 1)E 1 –1 –1 O 1 F(0, –2) x Together, the three sets of points form the entire plane. To graph an inequality in the coordinate plane, we proceed as follows: 1. On the plane, represent the plane divider, for example, y 3, by a dashed line to show that this divider does not belong to the graph of the half-plane. y 3 y = 3 1 –1 –1 O 1 x Graphing First-Degree Inequalities in Two Variables 379 2. Shade the region of the half-plane whose points satisfy the inequality. To graph y 3, shade the region above the plane divider. 3. To graph y 3, shade the region below the plane divider. y 3 1 –1 –1 y > 3 y = 3 O 1 Graph of y > 3 y 3 1 –1 –1 y = 3 y < 3 O 1 Graph of y < 3 x x Let us consider another example, where the plane divider is not a horizontal line. To graph the inequality y 2x or the inequality y 2x, we use a dashed line to indicate that the line y 2x is not a part of the graph. This dashed line acts as a boundary line for the half-plane being graphed. y (1, 4)B y = 2x y > 2x O x y y = 2x y < 2x O x C (1, –1) Graph of y > 2x Graph of y < 2x The graph of y 2x is the shaded half-plane above the line y 2x. It is the set of all points in which the y-coordinate is greater than twice the x-coordinate. The graph of y 2x is the shaded half-plane below the line y 2x. It is the set of all points in which the y-coordinate is less than twice the x-coordinate. 380 Graphing Linear Functions and Relations An open sentence such as y 2x means y 2x or y 2x. The graph of y 2x is the union of two disjoint sets. It includes all of the points in the solution set of the inequality y 2x and all of the points in the solution set of the equality y 2x. To indicate that y 2x is part of the graph of y 2x, we draw the graph of y 2x as a solid line. Then we shade the region above the line to include the points for which y 2x. y y > 2x y = 2x O x Graph of y > 2x When the equation of a line is written in the form y mx b, the halfplane above the line is the graph of y mx b and the half-plane below the line is the graph of y , mx 1 b . To check whether the correct half-plane has been chosen as the graph of a linear inequality, we select any point in that half-plane. If the selected point satisfies the inequality, every point in that half-plane satisfies the inequality. On the other hand, if the point chosen does not satisfy the inequality, then the other half-plane is the graph of the inequality. EXAMPLE 1 Graph the inequality y 2x 2. Solution How to Proceed (1) Transform the inequality into one having y as the left member: (2) Graph the plane divider, y 2x 2, by using the y-intercept, 2, to locate the first point (0, 2) on the y-axis. Then use the slope, 2 or , to find other points by moving up 2 and to the right 1: 2 1 (3) Shade the half-plane above the line: This region and the line are the required graph. The half-plane is the graph of y 2 2x . 2 , and the line is the graph of y 2 2x 5 2 . Note that the line is now drawn solid to show that it is part of the graph. y 2x 2 y 2x 2 y 2 y – 2x = 2 O –2 2 x –2 y y – 2x > 2 2 y – 2x = 2 O –2 2 x –2 Graphing First-Degree Inequalities in Two Variables 381 (4) Check the solution. Choose any point in the half-plane selected as the solution to see whether it satisfies the original inequality, y 2x 2: Select point (0, 5) which is in the shaded region. y 2x 2 ? 5 2 2(0) $ 2 5 2 ✔ The above graph is the graph of y 2 2x $ 2 . EXAMPLE 2 Graph each of the following inequalities in the coordinate plane. a. x 1 b. x 1 c. y 1 d. y 1 Answers y 12 = x x > 1 x < 1 –2 O x 2 –2 y 2 O –2 y 2 1 = x x 2 –2 O –2 y 2 O –. x < 1 c. y > 1 d. y < 1 –2 a. x > 1 EXERCISES Writing About Mathematics 1. Brittany said that the graph of 2x y 5 is the region above the line that is the graph of 2x y 5. Do you agree with Brittany? Explain why or why not. 2. Brian said that the union of the graph of x 2 and the graph of x 2 consists of every point in the coordinate plane. Do you agree with Brian? Explain why or why not. Developing Skills In 3–8, transform each sentence into one whose left member is y. 3. y 2
x 0 6. 2x y 0 4. 5x 2y 7. 3x y 4 5. y x 3 8. 4y 3x 12 382 Graphing Linear Functions and Relations In 9–23, graph each sentence in the coordinate plane. 9. x 4 12. y 3 15. x y 3 18. 2y 6x 0 21. 9 x 3y 10. x 2 13. y x 2 16. x y 1 19. 2x 3y 6 y x 5 # 1 2 1 22. 11. y 5 y $ 1 2x 1 3 14. 17. x – 2y 4 2y . 1 1 20. 3x 1 1 23. 2x 2y 6 0 In 24–26: a. Write each verbal sentence as an open sentence. b. Graph each open sentence in the coordinate plane. 24. The y-coordinate of a point is equal to or greater than 3 more than the x-coordinate. 25. The sum of the x-coordinate and the y-coordinate of a point is less than or equal to 5. 26. The y-coordinate of a point decreased by 3 times the x-coordinate is greater than or equal to 2. Applying Skills In 27–31: a. Write each verbal sentence as an open sentence. b. Graph each open sentence in the c. Choose one pair of coordinates that could be reasonable values for x and y. coordinate plane. 27. The length of Mrs. Gauger’s garden (y) is greater than the width (x). 28. The cost of a shirt (y) is less than half the cost of a pair of shoes (x). 29. The distance to school (y) is at least 2 miles more than the distance to the library (x). 30. The height of the flagpole (y) is at most 4 feet more than the height of the oak tree (x) nearby. 31. At the water park, the cost of a hamburger (x) plus the cost of a can of soda (y) is greater than 5 dollars. 9-10 GRAPHS INVOLVING ABSOLUTE VALUE To draw the graph of the equation y \|x\|, we can choose values of x and then find the corresponding values of y. Let us consider the possible choices for x and the resulting y-values: 1. Choose x 0. Since the absolute value of 0 is 0, y will be 0. 2. Choose x as any positive number. Since the absolute value of any positive number is that positive number, y will have the same value as x. For example, if x 5, then y 5 5. Graphs Involving Absolute Value 383 3. Choose x as any negative number. Since the absolute value of any nega- tive number is positive, y will be the opposite of x. For example, if x 3, then y 3 3. Thus, we conclude that x can be 0, positive, or negative, but y will be only 0 or positive. Here is a table of values and the corresponding graphx (–5, 5) (5, 5) (–3, 3) (3, 3) (–1, 1) (1, 1) O 1 –1 –1 x Notice that for positive values of x, the graph of y x is the same as the graph of y x. For negative values of x, the graph of y x is the same as the graph of y x. y 1 y 1 y 1 –1 –1 O 1 x O 1 –1 –1 x O 1 –1 – It should be noted that for all values of x, y = x results in a negative value for y. Therefore, for positive values of x, the graph of y = x is the same as the graph of y x. For negative values of x, the graph of y x is the same as the graph of y x. 384 Graphing Linear Functions and Relations EXAMPLE 1 Draw the graph of y x 2. Solution (1) Make a table of values2 2 –2) Plot the points and draw rays to connect the points that were graphed: y (–4, 6) y = x + 2 (5, 7) (3, 5) (–2, 4) (–1, 3) (1, 3) (0, 2) x Calculator Solution (1) Enter the equation into Y1: (2) Graph to the standard window: ENTER: Y MATH ENTER ENTER: ZOOM 6 X,T,,n ) 2 DISPLAY DISPLAY: EXAMPLE 2 Draw the graph of x y 3. Solution By the definition of absolute value, x x and y y. • Since (1, 2) is a solution, (1, 2), (1, 2) and (1, 2) are solutions. • Since (2, 1) is a solution, (2, 1), (2, 1) and (2, 1) are solutions. • Since (0, 3) is a solution, (0, 3) is a solution. • Since (3, 0) is a solution, (3, 0) is a solution. Graphs Involving Absolute Value 385 y 1 –1 –1 O 1 x Plot the points that are solutions, and draw the line segments joining them. x + y = 3 Translating, Reflecting, and Scaling Graphs of Absolute Value Functions Just as linear functions can be translated, reflected, or scaled, graphs of absolute value functions can also be manipulated by working with the graph of the absolute value function y 5 ZxZ . For instance, the graph of y x 5 is the graph of y x shifted 5 units down. The graph of y –x is the graph of y x reflected in the x-axis. The graph of y 2x is the graph of y x stretched vertically by a factor of 2, is the graph of y x compressed vertically by a facwhile the graph of 1 tor of . 2 y 5 1 2 ZxZ = x1 2 x Translation Rules for Absolute Value Functions If c is positive: The graph of y x c is the graph of y x shifted c units up. The graph of y x c is the graph of y x shifted c units down. 386 Graphing Linear Functions and Relations For absolute value functions, there are two additional translations that can be done to the graph of y x, horizontal shifting to left or to the right. If c is positive: The graph of y x c is the graph of y x shifted c units to the left. The graph of y x c is the graph of y x shifted c units to the right. Reflection Rule for Absolute Value Functions The graph of y x is the graph of y x reflected across the x-axis. Scaling Rules for Absolute Value Functions When c 1, the graph of y cx is the graph of y x stretched vertically by a factor of c. When 0 c 1, the graph of y cx is the graph of y x compressed vertically by a factor of c. In a–e, write an equation for the resulting function if the graph of y 5 ZxZ is: a. shifted 2.5 units down b. shifted 6 units to the right c. stretched vertically by a factor of 3 and shifted 5 units up d. compressed vertically by a factor of and reflected in the x-axis 1 3 e. reflected in the x-axis, shifted 1 unit up, and shifted 1 unit to the left EXAMPLE 3 Solution a. y x 2.5 Answer b. y x 6 Answer c. First, stretch vertically by a factor of 3: Then, translate the resulting function 5 units up: 1 d. First, compress vertically by a factor of : 3 Then, reflect in the x-axis: e. First, reflect in the x-axis: Then, translate the resulting function 1 unit up: Finally, translate the resulting function 1 unit to the left: y 3x y 3x 5 Answer y Answer 1 3 ZxZ 21 y 3 ZxZ y x y x 1 y x 1 1 Answer Graphs Involving Exponential Functions 387 EXERCISES Writing About Mathematics 1. Charity said that the graph of y 2x 1 is the graph of y 2x 1 for x 0 and the graph of 2x 1 for x 0. Do you agree with Charity? Explain why or why not. 2. April said that x y 5 is a function. Do you agree with April? Explain why or why not. 3. Euclid said that, for positive values of c, the graph of y cx is the same as the graph of y cx. Do you agree with Euclid? If so, prove Euclid’s statement. Developing Skills In 4–15, graph each equation. 4. y x 1 8. y 2x 5. y x 3 9. y 2x 1 12. y x 13. x y 4 6. y x 1 10. x y 5 14. Zx Z 2 1 Zy Z 4 5 1 7. y x 3 11. x 2y 7 15. 2x 4y 6 0 In 16–19, describe the translation, reflection, and/or scaling that must be applied to y x to obtain the graph of each given function. 17. y 19. y x 1.5 4 18. y x 2 3 16. y –x 4 22ZxZ 1 2 9-11 GRAPHS INVOLVING EXPONENTIAL FUNCTIONS A piece of paper is one layer thick. If we place another piece of paper on top of it, the stack is two layers thick. If we place another piece of paper on top, the stack is now three layers thick. As the process continues, we can describe the number of layers, y, in terms of the number of sheets added, x, by the linear function y 1 x. For example, after we have added seven pieces of paper, the stack is eight layers thick. This is an example of linear growth because the change can be described by a linear function. Now consider another experiment. A piece of paper is one layer thick. If the paper is folded in half, the stack is two layers thick. If the stack is folded again, the stack is four layers thick. If the stack is folded a third time, it is eight layers thick. 388 Graphing Linear Functions and Relations 0 folds 1 fold 2 folds 3 folds 1 layer 2 layers 4 layers 8 layers Although we will reach a point at which it is impossible to fold the paper, imagine that this process can continue. We can describe the number of layers of paper, y, in terms of the number of folds, x, by an exponential function, y 2x. The table at the right shows some values for this function. Such functions are said to be nonlinear. After five folds, the stack is 32 layers thick. After ten folds, the stack would be 210 or 1,024 layers thick. In other words, it would be thinker than this book. This is an example of exponential growth because the growth is described by an exponential function y bx. When the base b is a positive number greater than 1 (b 1), y increases as x increases. x 0 1 2 3 4 5 2x 20 21 22 23 24 25 y 2x 1 2 4 8 16 32 We can draw the graph of the exponential function, adding zero and negative integral values to those given above. Locate the points on the coordinate plane and draw a smooth curve through them. When we draw the curve, we are assuming that the domain of the independent variable, x, is the set of real numbers. You will learn about powers that have exponents that are not integers in more advanced math courses. x 0 1 2 3 2x 20 21 22 23 y 2x 2, 4) (1, 2) 1 (0, 1) O 1 2 3 x (–3, )1 8 (–1, )1 2 (–2, )1 4 –3 –2 –1 There are many examples of exponential growth in the world around us. Over an interval, certain populations—for example of bacteria, or rabbits, or people—grow exponentially. Graphs Involving Exponential Functions 389 Compound interest is also an example of exponential growth. If a sum of money, called the principal, P, is invested at 4% interest, then after one year the value of the investment, A, is the principal plus the interest. Recall that I Prt. In this case, r 0.04 and t 5 1 . A P I = P Prt P(1 rt) P(1+ 0.04(1)) P(1.04) After 2 years, the new principal is (1.04)2P. Replace P by 1.04P in the formula A P(1.04): A 1.04P(1.04) = P(1.04)2 After 3 years, the new principal is (1.04)3P. Replace P by P(1.04)2 in the formula A 5 P(1.04) : A (1.04)2P(1.04) = P(1.04)3 In general, after n years, the value of an investment for which the rate of interest is r can be found by using the formula A P(1 r)n. Note that in this formula for exponential growth, P is the amount present at the beginning; r, a positive number, is the rate of growth; and n is the number of intervals of time during which the grow
th has been taking place. Since r is a positive number, the base, 1 r, is a number greater than 1. Compare the formula A P(1 r)n with the exponential function y 2x used to determine the number of layers in a stack after x folds. P 1 since we started with 1 layer. The base, (1 r) 2, so r 1 or 100%. Doubling the number of layers in the stack is an increase equal to the size of the stack, that is, an increase of 100%. An exponential change can be a decrease as well as an increase. Consider this example. Start with a large piece of paper. If we cut the paper into two equal parts, each part is one-half of the original piece. Now if we cut one of these pieces into two equal parts, each of the parts is one-fourth of the original piece. As this process continues, we can describe the part of the original piece that results from each cut in x terms of the number of cuts. The function B gives the part of the original piece, y, that is the result of x cuts. This is an example of exponential decay. For the exponential function y bx, when the base b is a positive number less than 1 (0 b 1), y decreases as x increases 16 1 32 Some examples of exponential decay include a decrease in population, a fund, or the value of a machine when that decrease can be represented by a constant rate at regular intervals. The radioactive decay of a chemical element such as carbon is also an example of exponential decay. 390 Graphing Linear Functions and Relations Problems of exponential decay can be solved using the same formula as that for exponential growth. In exponential decay, the rate of change is a negative number between 1 and 0 so that the base is a number between 0 and 1. This is illustrated in Example 3. EXAMPLE 1 Draw the graph of y 5 x . 2 3 A B Solution Make a table of values from x 3 to x 3. Plot the points and draw a smooth curve through them 23 5 22 5 21 27 27 y 5 4 3 (–3, )27 8 (–2, ) 9 4 2 (0, 1) 1 (–1, )3 2 (1, )2 3 (2, )4 9 (3, )8 27 –3 –2 –1 O 1 2 3 x Calculator Solution (1) Enter the equation into Y1: (2) Graph to the standard window: ENTER: Y ( 2 3 ) ENTER: ZOOM 6 ^ X,T,,n DISPLAY DISPLAY: Graphs Involving Exponential Functions 391 EXAMPLE 2 A bank is advertising a rate of 5% interest compounded annually. If $2,000 is invested in an account at that rate, find the amount of money in the account after 10 years. Solution This is an example of exponential growth. Use the formula A P (1 r)n with r 0.05, P the initial investment, and n 10 years. A 2,000(1 0.05)10 2,000(1.05)10 3,257.78954 Answer The value of the investment is $3,257.79 after 10 years. EXAMPLE 3 The population of a town is decreasing at the rate of 2.5% per year. If the population in the year 2000 was 28,000, what will be the expected population in 2015 if this rate of decrease continues? Give your answer to the nearest thousand. Solution Use the formula for exponential growth or decay. The rate of decrease is entered as a negative number so that the base is a number between 0 and 1. r 0.025 P the initial population n 15 years (x 0 corresponds to the year 2000) A P(1 r)n 28,000(1 (–0.025))15 28,000(0.975)15 19,152.5792 Answer The population will be about 19,000 in 2015. EXERCISES Writing About Mathematics 1. The equation y bx is an exponential function. If the graph of the function is a smooth curve, explain why the value of b cannot be negative. 2. The equation A P(1 r)n can be used for both exponential growth and exponential decay when r represents the percent of increase or decrease. a. How does the value of r that is used in this equation for exponential growth differ from that of exponential decay? b. If the equation represents exponential growth, can the rate be greater than 100%? Explain why or why not. c. If the equation represents exponential decay, can the rate be greater than 100%? Explain why or why not. 392 Graphing Linear Functions and Relations 3. Euler says that the graph of y 3x2 is the graph of y 3x shifted 2 units to the left and that the graph of y 3x 2 is the graph of y 3x shifted 2 units up. Do you agree with Euler? Explain why or why not. Developing Skills In 4–9, sketch each graph using as values of x the integers in the given interval. 4. y 3x, [3, 3] 5. y 4x, [2, 2] 7. y 2 5 x , [3, 3] 8. y x , [2, 2] 1 3 6. y 1.5x, [4, 4] 9. y 5 1.25x , [0, 10] A 10. Compare each of the exponential graphs drawn in exercises 4 through 9. A B B a. What point is common to all of the graphs? b. If the value of the base is greater than 1, in what direction does the graph curve? c. If the value of the base is between 0 and 1, in what direction does the graph curve? 11. a. Sketch the graph of y b. Sketch the graph of 1 2 B y 5 22xA x in the interval [3, 3]. in the interval [3, 3]. c. What do you observe about the graphs drawn in a and b? Applying Skills 12. In 2005, the population of a city was 25,000. The population increased by 20% in each of the next three years. If this rate of increase continues, what will be the population of the city in 2012? 13. Alberto invested $5,000 at 6% interest compounded annually. What will be the value of Alberto’s investment after 8 years? 14. Mrs. Boyko has a trust fund from which she withdraws 5% each year. If the fund has a value of $50,000 this year, what will be the value of the fund after 10 years? 15. Hailey has begun a fitness program. The first week she ran 1 mile every day. Each week she increases the amount that she runs each day by 20%. In week 10, how many miles does she run each day? Give your answer to the nearest mile. 16. Alex received $75 for his birthday. In the first week after his birthday, he spent one-third of the money. In the second week and each of the following weeks, he spent one-third of the money he had left. How much money will Alex have left after 5 weeks? 17. During your summer vacation, you are offered a job at which you can work as many days as you choose. If you work 1 day, you will be paid $0.01. If your work 2 days you will be paid a total of $0.02. If your work 3 days you will be paid a total of $0.04. If you continue to work, your total pay will continue to double each day. a. Would you accept this job if you planned to work 10 days? Explain why or why not. b. Would you accept this job if you planned to work 25 days? Explain why or why not. CHAPTER SUMMARY Chapter Summary 393 The solutions of equations or inequalities in two variables are ordered pairs of numbers. The set of points whose coordinates make an equation or inequality true is the graph of that equation or inequality. The graph of a linear function of the form Ax By C is a straight line. The domain of a function is the set of all values the independent variable, the x-coordinate, is allowed to take. This determines the range of a function, that is, the set of all values the dependent variable, the y-coordinate, will take. Set-builder notation provides a mathematically concise method of describing the elements of a set. Roster form lists every element of a set exactly once. A linear function can be written in the form y mx b, where m is the slope and b is the y-intercept of the line that is the graph of the function. A line parallel to the x-axis has a slope of 0, and a line parallel to the y-axis has no slope. The slope of a line is the ratio of the change in the vertical direction to the change in the horizontal direction. If (x1, y1) and (x2, y2) are two points on a line, the slope of the line is slope m y2 2 y1 x2 2 x2 . If y varies directly as x, the ratio of y to x is a constant. Direct variation can be represented by a line through the origin whose slope is the constant of variation. The graph of y mx b separates the plane into two half-planes. The halfplane above the graph of y mx b is the graph of y mx b, and the halfplane below the graph of y mx b is the graph of y mx b. The graph of y x is the union of the graph of y x for x 0 and the graph of y x for x 0. The equation y bx, b 0 and b 1, is an example of an exponential function. The exponential equation A P(1 r)n is a formula for exponential growth or decay. A is the amount present after n intervals of time, P is the amount present at time 0, and r is the rate of increase or decrease. In exponential growth, r is a positive number. In exponential decay, r is a negative number in the interval (1, 0). The function y x and y x can be translated, reflected, or scaled to graph other linear and absolute value functions. Vertical translations (c 0) Horizontal translations (c 0) Reflection across the x-axis Vertical stretching (c 1) Vertical compression (0 c 1) Linear Function Absolute Value y x c (up) y x c (down) y x c (left) y x c (right) y x y cx y cx y x c (up) y x c (down) y x c (left) y x c (right) y x y cx y cx 394 Graphing Linear Functions and Relations VOCABULARY 9-1 Roster form • Set-builder notation • Member of a set () • Not a member of a set () • Relation • Domain of a relation • Range of a relation • Function • Domain of a function • Range of a function • Independent variable • Dependent variable 9-2 Graph • Standard form • Linear equation 9-4 Slope • Unit rate of change 9-6 y-intercept • x-intercept • Intercept form a linear equation • Slope-intercept form of a linear equation 9-8 Equation of a direct variation 9-9 Plane divider • Half-plane 9-11 Linear growth • Exponential function • Nonlinear functions • Exponential growth • Exponential decay REVIEW EXERCISES 1. Determine a. the domain and b. the range of the relation shown in the graph to the right. c. Is the relation a function? Explain why or why not. y 1 O 1 x 2. Draw the graph of x y 5. Is the graph of x y 5 a square? Prove your answer. 3. A function is a set of ordered pairs in which no two ordered pairs have the same first element. Explain why {(x, y) y 2x 1} is not a function. 4. What is the slope of the graph of y 2x 5? 5. What is the slope of the line whose equation is 3x 2y 12? 6. What are the intercepts of the line whose equation is 3x 2y 12? 7. What is the slope of the line that passes thro
ugh points (4, 5) and (6, 1)? In 8–13, in each case: a. Graph the equation or inequality. b. Find the domain and range of the equation or inequality. 8. y x 2 9. y 3 10. y 2 3x Review Exercises 395 11. x 2y 8 12. y x 2 13. 2x y 4 In 14–22, refer to the coordinate graph to answer each question. 14. What is the slope of line k? 15. What is the x-intercept of line k? 16. What is the y-intercept of line k? y 1 O 1 k m x 17. What is the slope of a line that is parallel to line k? 18. What is the slope of a line that is perpendicular to line k? 19. What is the slope of line m? 20. What is the x-intercept of line m? 21. What is the y-intercept of line m? 22. What is the slope of a line that is perpendicular to line m? 23. If point (d, 3) lies on the graph of 3x y 9, what is the value of d? In 24–33, in each case select the numeral preceding the correct answer. 24. Which set of ordered pairs represents a function? (1) {(6, 1), (7, 1), (1, 7), (1, 6)} (2) {(0, 13), (1, 13), (6, 13), (112, 13)} (3) {(3, 3), (3, 4), (3, 4), (3, 3)} (4) {(1, 12), (11, 1), (112, 11), (11, 112)} 25. Which point does not lie on the graph of 3x y 9? (2) (2, 3) (1) (1, 6) (3) (3, 0) (4) (0, 9) 26. Which ordered pair is in the solution set of y 2x 4? (1) (0, 5) (2) (2, 0) (3) (3, 3) (4) (0, 2) 27. Which equation has a graph parallel to the graph of y 5x 2? (l) y 5x (2) y 5x + 3 (3) y 2x (4) y 2x 5 28. The graph of 2x y 8 intersects the x-axis at point (2) (8, 0) 29. What is the slope of the graph of the equation y 4? (3) (0, 4) (1) (0, 8) (4) (4, 0) (1) The line has no slope. (3) 4 (2) 0 (4) 4 30. In which ordered pair is the x-coordinate 3 more than the y-coordinate? (1) (1, 4) (2) (1, 3) (3) (3, 1) (4) (4, 1) 396 Graphing Linear Functions and Relations 31. Which of the following is not a graph of a function? y 1 O (1) (2) (3) (4 32. Using the domain D {4x x the set of integers and 10 4x 20}, what is the range of the function y 2 x 2? (1) {10, 14, 18} (2) {12, 16, 20} (3) {10, 14, 16, 20} (4) {4, 14, 24, 34} 33. Which equation best describes the graph of y \|x\| reflected across the x-axis, shifted 9 units up, and shifted 2 units to the left? (1) y x 2 9 (2) y = x + 2 9 (3) y x 2 9 (4) y x 2 9 34. a. Plot points A(5, 2), B(3, 3), C(3, 3), and D(5, 2). b. Draw polygon ABCD. c. What kind of polygon is ABCD? d. Find DA and BC. e. Find the length of the altitude from C to DA . f. Find the area of ABCD. Review Exercises 397 In 35–38, graph each equation. 36. y x 2 35. y x 2 37. x 2y 6 38. y 2.5x 39. Tiny Tot Day Care Center has changed its rates. It now charges $350 a week for children who stay at the center between 8:00 A.M. and 5:00 P.M. The center charges an additional $2.00 for each day that the child is not picked up by 5:00 P.M. a. Write an equation for the cost of day care for the week, y, in terms of x, the number of days that the child stays beyond 5:00 P.M. b. Define the slope of the equation found in part a in terms of the infor- mation provided. c. What is the charge for 1 week for a child who was picked up at the fol- lowing times: Monday at 5:00, Tuesday at 5:10, Wednesday at 6:00, Thursday at 5:00, and Friday at 5:25? 40. Each time Raphael put gasoline into his car, he recorded the number of gallons of gas he needed to fill the tank and the number of miles driven since the last fill-up. The chart below shows his record for one month. Gallons of Gasoline 12 8.5 10 4.5 Number of Miles 370 260 310 140 13 420 a. Plot points on a graph to represent the information in the chart. Let the horizontal axis represent the number of gallons of gasoline and the vertical axis the number of miles driven. b. Find the average number of gallons of gasoline per fill-up. c. Find the average number of miles driven between fill-ups. d. Locate a point on the graph whose x-coordinate represents the average number of gallons of gasoline per fill-up and whose y-coordinate is the average number of miles driven between fill-ups. e. It is reasonable that (0, 0) is a point on the graph? Draw, through the point that you plotted in d and the point (0, 0), a line that could represent the information in the chart. f. Raphael drove 200 miles since his last fill-up. How many gallons of gasoline should he expect to need to fill the tank based on the line drawn in e? 41. Mandy and Jim are standing 20 feet apart. Each second, they decrease the distance between each other by one-half. a. Write an equation to show their distance apart, D, at s seconds. b. How far apart will they be after 5 seconds? c. According to your equation, will Mandy and Jim ever meet? 398 Graphing Linear Functions and Relations 42. According to the curator of zoology at the Rochester Museum and Science Center, if you count the number of chirps of a tree cricket in 15 seconds and add 40, you will have a close approximation of the actual air temperature in degrees Fahrenheit. To test this statement, Alexa recorded, on several summer evenings, the temperature in degrees Fahrenheit and the number of chirps of a tree cricket in 1 minute. She obtained the following results: Chirps/minute 150 170 140 125 108 118 145 Temperature 80 85 76 71 65 70 76 68 56 95 63 110 67 a. Write an equation that states the relationship between the number of chirps per minute of the tree cricket, c, to the temperature in degrees Fahrenheit, F. (To change chirps per minute to chirps in 15 seconds, divide c by 4.) b. Draw a graph of the data given in the table. Record the number of chirps per minute on the horizontal axis, and the temperature on the vertical axis. c. Draw the graph of the equation that you wrote in part a on the same set of axes that you used for part b. d. Can the data that Alexa recorded be represented by the equation that you wrote for part a? Explain your answer. 43. In order to control the deer population in a local park, an environmental group plans to reduce the number of deer by 5% each year. If the deer population is now estimated to be 4,000, how many deer will there be after 8 years? Exploration STEP 1. Draw the graph of y x 2. STEP 2. Let A be the point at which the graph intersects the x-axis, B be any point on the line that has an x-coordinate greater than A, and C be the point at which a vertical line from B intersects the x-axis. STEP 3. Compare the slope of y x 2 with tan BAC. STEP 4. Use a calculator to find mBAC. STEP 5. Repeat Steps 2 through 4 for three other lines that have a positive slope. STEP 6. Draw the graph of y x 2 and repeat Step 2. Let the measure of an acute angle between the x-axis and a line that slants upward be positive and the measure of an acute angle between the x-axis and a line that slants downward be negative. STEP 7. Repeat Steps 3 and 4 for y x 2. STEP 8. Repeat Steps 2 through 4 for three other lines that have a negative slope. STEP 9. What conclusion can you draw? CUMULATIVE REVIEW Part I Cumulative Review 399 CHAPTERS 1–9 Answer all questions in this part. Each correct answer will receive 2 credits. No partial credit will be allowed. 1. The product of 5a3 and 3a4 is (1) 15a7 (2) 15a12 (3) 15a7 (4) 15a12 2. A basketball team won b games and lost 4. The ratio of games won to games played is (1) b 4 (2) b b 1 4 (3) 4 b (4) 4 b 1 4 3. In decimal notation, 8.72 10–2 is (1) 87,200 (2) 872 (3) 0.0872 (4) 0.00872 4. Which equation is not an example of direct variation? y x 5 2 (1) y 2x y 5 2 x (2) (3) (4) y 5 x 2 5. The graph of y 2x 4 is parallel to the graph of (1) y 2x 5 (2) 2x y 7 (3) y 2x 3 (4) 2x y 0 6. The area of a triangle is 48.6 square centimeters. The length of the base of the triangle is 3.00 centimeters. The length of the altitude of the triangle is (1) 32.4 centimeters (2) 16.2 centimeters (3) 8.10 centimeters (4) 3.24 centimeters 7. The measure of the larger acute angle of a right triangle is 15 more than twice the measure of the smaller. What is the measure of the larger acute angle? (1) 25 (2) 37.5 (3) 55 (4) 65 8. Solve for x: 1 2(x 4) x. (1) 21 2 (2) 2 (3) 3 (4) 4.5 9. The measure of one leg of a right triangle is 9 and the measure of the hypotenuse is 41. The measure of the other leg is (1) 32 (2) 40 (3) 41.98 (4) 1,762 10. Which of the following geometric figures does not always have a pair of " congruent angles? (1) a parallelogram (2) an isosceles triangle (3) a rhombus (4) a trapezoid 400 Graphing Linear Functions and Relations Part II Answer all questions in this part. Each correct answer will receive 2 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 11. Cory has planted a rectangular garden. The ratio of the length to the width of the garden is 5 : 7. Cory bought 100 feet of fencing. After enclosing the garden with the fence, he still had 4 feet of fence left. What are the dimensions of his garden? 12. A square, ABCD, has a vertex at A(4, 2). Side AB is parallel to the x-axis and AB 7. What could be the coordinates of the other three vertices? Explain how you know that for the coordinates you selected ABCD is a square. Part III Answer all questions in this part. Each correct answer will receive 3 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a correct numerical answer with no work shown will receive only 1 credit. 13. A straight line with a slope of 2 contains the point (2, 4). Find the y-coordinate of a point on this line whose x-coordinate is 5. 14. Mrs. Gantrish paid $42 for 8 boxes of file folders. She bought some on sale for $4 a box and the rest at a later time for $6 a box. How many boxes of file folders did she buy on sale? Part IV Answer all questions in this part. Each correct answer will receive 4 credits. Clearly indicate the necessary steps, including appropriate formula substitutions, diagrams, graphs, charts, etc. For all questions in this part, a co
rrect numerical answer with no work shown will receive only 1 credit. 15. Concentric circles (circles that have the same center) are used to divide the circular face of a dartboard into 4 regions of equal area. If the radius of the board is 12.0 inches, what is the radius of each of the concentric circles? Express your answers to the nearest tenth of an inch. 16. A ramp is to be built from the ground to a doorway that is 4.5 feet above the ground. What will be the length of the ramp if it makes an angle of 22° with the ground? Express your answer to the nearest foot. WRITING AND SOLVING SYSTEMS OF LINEAR FUNCTIONS Architects often add an outdoor stairway to a building as a design feature or as an approach to an entrance above ground level. But stairways are an obstacle to persons with disabilities, and most buildings are now approached by means of ramps in addition to or in place of stairways. In designing a ramp, the architect must keep the slant or slope gradual enough to easily accommodate a wheelchair. If the slant is too gradual, however, the ramp may become inconveniently long or may require turns to fit it into the available space.The architect will also want to include design features that harmonize with the rest of the building and its surroundings. Solving a problem such as the design of a ramp often involves writing and determining the solution of several equations or inequalities at the same time. CHAPTER 10 CHAPTER TABLE OF CONTENTS 10-1 Writing an Equation Given Slope and One Point 10-2 Writing an Equation Given Two Points 10-3 Writing an Equation Given the Intercepts 10-4 Using a Graph to Solve a System of Linear Equations 10-5 Using Addition to Solve a System of Linear Equations 10-6 Using Substitution to Solve a System of Linear Equations 10-7 Using Systems of Equations to Solve Verbal Problems 10-8 Graphing the Solution Set of a System of Inequalities Chapter Summary Vocabulary Review Exercises Cumulative Review 401 402 Writing and Solving Systems of Linear Functions 10-1 WRITING AN EQUATION GIVEN SLOPE AND ONE POINT You have learned to draw a line using the slope and the coordinates of one point on the line. With this same information, we can also write an equation of a line. EXAMPLE 1 Write an equation of the line that has a slope of 4 and that passes through point (3, 5). Solution METHOD 1. Use the definition of slope. How to Proceed (1) Write the definition of slope: (2) Let (x1, y1) (3, 5) and (x2, y2) (x, y). Substitute these values in the definition of slope: (3) Solve the equation for y in terms of x: slope 4 y2 y1 2 x2 2 x1 y 2 5 x 2 3 y 5 4(x 3) y 5 4x 12 y 4x 7 METHOD 2. Use the slope-intercept form of an equation. How to Proceed (1) In the equation of a line, y = mx b, replace m by the given slope, 4: (2) Since the given point, (3, 5), is on the line, its coordinates satisfy the equation y 4x b. Replace x by 3 and y by 5: (3) Solve the resulting equation to find the value of b, the y-intercept: (4) In y 4x b, replace b by 7: y mx b y 4x b 5 4(3) b 5 12 b –7 b y 4x 7 Answer y 4x 7 In standard form (Ax By C), the equation of the line is Note: 4x y 7. Writing an Equation Given Slope and One Point 403 EXERCISES Writing About Mathematics 1. Micha says that there is another set of information that can be used to find the equation of a line: the y-intercept and the slope of the line. Jen says that that is the same information as the coordinates of one point and the slope of the line. Do you agree with Jen? Explain why or why not. 2. In Method 1 used to find the equation of a line with a slope of 4 and that passes through point (3, 5): a. How can the slope, 4, be written as a ratio? b. What are the principles used in each step of the solution? Developing Skills In 3–6, in each case, write an equation of the line that has the given slope, m, and that passes through the given point. 3. m 2; (1, 4) 4. m 2; (3, 4) 5. m 3; (2, 1) 6. m 5 25 3 ; (3, 0) 1 2 23 4 7. Write an equation of the line that has a slope of and a y-intercept of 2. and a y-intercept of 0. 8. Write an equation of the line that has a slope of 9. Write an equation of the line, in the form y mx b, that is: a. parallel to the line y 2x 4, and has a y-intercept of 1. b. perpendicular to the line y 2x 4, and has a y-intercept of 1. c. parallel to the line y 3x 6, and has a y-intercept of 2. d. perpendicular to the line y 3x 6, and has a y-intercept of 2. e. parallel to the line 2x 3y 12, and passes through the origin. f. perpendicular to the line 2x 3y 12, and passes through the origin. 10. Write an equation of the line, in the form Ax By C, that is: a. parallel to the line y 4x 1, and passes through point (2, 3). b. perpendicular to the line y 4x 1, and passes through point (2, 3). c. parallel to the line 2y 6x 9, and passes through point (2, 1). d. perpendicular to the line 2y 6x 9, and passes through point (2, 1). e. parallel to the line y 4x 3, and has the same y-intercept as the line y 5x 3. f. perpendicular to the line y 4x 3, and has the same y-intercept as the line y 5x 3. 404 Writing and Solving Systems of Linear Functions 11. For a–c, write an equation, in the form y mx b, that describes each graph. a. y 1 1 O 1 Applying Skills y = x + 1 (2, 3) x + y = 5 b. (–4, 1) c. O–1 –1 (1, 2) O 12. Follow the directions below to draw a map in the coordinate plane. a. The map begins at the point A(0, 3) with a straight line segment that has a slope of 1. Write an equation for this segment. b. Point B has an x-coordinate of 4 and is a point on the line whose equation you wrote in a. Find the coordinates of B and draw AB . c. Next, write an equation for a segment from point B that has a slope of 21 2 . d. Point C has an x-coordinate of 6 and is a point on the line whose equation you wrote in c. Find the coordinates of C and draw BC . 1 e. Finally, write an equation for a line through C that has a slope of . 2 f. Point D lies on the line whose equation you wrote in e and on the y-axis. Does point D coincide with point A? 13. If distance is represented by y and time by x, then the rate at which a car travels along a straight road can be represented as slope. Tom leaves his home and drives for 12 miles to the thruway entrance. On the thruway, he travels 65 miles per hour. a. Write an equation for Tom’s distance from his home in terms of the number of hours that he traveled on the thruway. (Hint: Tom enters the thruway at time 0 when he is at a distance of 12 miles from his home.) b. How far from home is Tom after 3 hours? c. How many hours has Tom driven on the thruway when he is 285 miles from home? 10-2 WRITING AN EQUATION GIVEN TWO POINTS You have learned to draw a line using two points on the line. With this same information, we can also write an equation of a line. Writing an Equation Given Two Points 405 EXAMPLE 1 Write an equation of the line that passes through points (2, 5) and (4, 11). Solution METHOD 1. Use the definition of slope and the coordinates of the two points to find the slope of the line. Then find the equation of the line using the slope and the coordinates of one point. How to Proceed (1) Find the slope of the line that passes through the two given points, (2, 5) and (4, 11): (2) In y = mx b, replace m by the slope, 3: (3) Select one point that is on the line, for example, (2, 5). Its coordinates must satisfy the equation y 3x b. Replace x by 2 and y by 5: 2 and y1 4 and y2 5] 11] (2, 5). (4, 11). [x1 [x2 4 2 2 5 6 5 11 2 5 2 5 3 Let P1 Let P2 2 y y 1 2 m 5 x2 2 x1 y mx b y 3x b 5 3(2) b (4) Solve the resulting equation to find the value of b, the y-intercept: 1 b 5 6 b (5) In y 3x b, replace b by 1: y 3x 1 METHOD 2. Let A(2, 5) and B(4, 11) be the two given points on the line and P(x, y) be any point on the line. Use the fact that the slope of PA to write an equation. equals the slope of AB How to Proceed (1) Set slope of PA equal to slope of AB : (2) Solve the resulting equation for y 11 26 22 (x 2 2) y 2 5 5 3x 2 6 y 5 3x 2 1 Check Do the coordinates of the second point, (4, 11), satisfy the equation y 3x 1? 3(4) 1 5? 11 11 11 ✔ Answer y 3x 1 406 Writing and Solving Systems of Linear Functions EXERCISES Writing About Mathematics 1. In step 3 of Method 1, the coordinates (2, 5) were substituted into the equation. Could the coordinates (4, 11), the coordinates of the other point on the line, be substituted instead? Explain your answer and show that you are correct. 2. Name the principle used in each step of the solution of the equation in Method 2. Developing Skills In 3–6, in each case write an equation of the line, in the form y mx b, that passes through the given points. 3. (0, 5), (2, 0) 4. (0, 3), (1, 1) 5. (1, 4), (3, 8) 6. (3, 1), (9, 7) In 7–10, in each case write an equation of the line, in the form Ax By C, that passes through the given points. 10. (0, 0), (3, 5) 7. (1, 2), (10, 14) 11. A triangle is determined by the three points A(3, 5), B(6, 4), and C(1, 1). Write the equa- 9. (2, 5), (1, 2) 8. (0, 1), (6, 8) tion of each line in the form y mx b: g CA g AB g BC b. a. c. d. Is triangle ABC a right triangle? Explain your answer. 1 12. A quadrilateral is determined by the four points W(1, 1), X(4, 6), Y , 2 101 2 , Z B A 53 5 , 53 5 . Is B A the quadrilateral a trapezoid? Explain your answer. Applying Skills 13. Latonya uploads her digital photos to an internet service that archives them onto CDs for a fee per CD plus a fixed amount for postage and handling; that is, the amount for postage and handling is the same no matter how many archive CDs she purchases. Last month Latonya paid $7.00 for two archive CDs, and this month she paid $13.00 for five archive CDs. a. Write two ordered pairs such that the x-coordinate is the number of archive CDs pur- chased and the y-coordinate is the total cost of archiving the photos. b. Write an equation for the total cost, y, for x archive CDs. c. What is the domain of the equation found in b? Write your answer in set-builder notation. d. What is the range of the equat
ion found in b? Write your answer in set-builder notation. Writing an Equation Given the Intercepts 407 14. Every repair bill at Chickie’s Service Spot includes a fixed charge for an estimate of the repairs plus an hourly fee for labor. Jack paid $123 for a TV repair that required 3 hours of labor. Nina paid $65 for a DVD player repair that required 1 hour of labor. a. Write two ordered pairs (x, y), where x represents the number of hours of labor and y is the total cost for repairs. b. Write an equation in the form y = mx b that expresses the value of y, the total cost of repairs, in terms of x, the number of hours of labor. c. What is the fixed charge for an estimate at Chickie’s Service Spot? d. What is the hourly fee for labor at Chickie’s? 15. A copying service charges a uniform rate for the first one hundred copies or less and a fee for each additional copy. Nancy Taylor paid $7.00 to make 200 copies and Rosie Barbi paid $9.20 to make 310 copies. a. Write two ordered pairs (x, y), where x represents the number of copies over one hun- dred and y represents the cost of the copies. b. Write an equation in the form y = mx b that expresses the value of y, the total cost of the copies, in terms of x, the number of copies over one hundred. c. What is the cost of the first one hundred copies? d. What is the cost of each additional copy? 10-3 WRITING AN EQUATION GIVEN THE INTERCEPTS The x-intercept and the y-intercept are two points on the graph of a line. If we know these two points, we can graph the line and we can write the equation of the line. For example, to write the equation of the line whose x-intercept is 5 and whose y-intercept is 3, we can use two points. • If the x-intercept is 5, one point is (5, 0). • If the y-intercept is 3, one point is (0, 3). Let (x, y) be any other point on the line. Set the slope of the line from (x, y) to (5, 0) equal to the slope of the line from (5, 0) to (0, 3). 0 2 (23 Solve the resulting equation for y. 5y 3(x 5) 5y 3x 15 y 5 3 5x 2 3 408 Writing and Solving Systems of Linear Functions The slope-intercept form of the equation shows that the y-intercept is 3. Recall from Section 9-6 that we can write this equation in intercept form to show both the x-intercept and the y-intercept. Start with the equation: Add 3x to each side: Divide each term by the constant term: 5y 3x 15 3x 5y 15 5y 215 5 215 23x 215 1 215 y x 23 5 1 5 1 When the equation is in this form, x is divided by the x-intercept and y is divided by the y-intercept. In other words, when the equation of a line is written in the form , the x-intercept is a and the y-intercept is b. y x b 5 1 a 1 EXAMPLE 1 Write an equation of the line whose x-intercept is 1 and whose y-intercept is 2 . Write the answer in standard form (Ax By C). 3 Solution How to Proceed (1) In y x b 5 1 a 1 2 , replace a by 1 and b by : 3 (2) Write this equation in simpler form: x 21 21 1 3y x 2 5 1 21 1 2 3 Note: In the original form of the equation, y is divided by . To divide by the same as to multiply by the reciprocal, . 3 2 2 3 is (3) Write the equation with integral coefficients by multiplying each term of the equation by the product of the denominators, 2: 22 x 21 A B 3y 2 2 2 5 22(1) A B 2x 2 3y 5 22 Answer 2x 3y 2 EXAMPLE 2 Find the x-intercept and y-intercept of the equation x 5y 2. Writing an Equation Given the Intercepts 409 Solution Write the equation in the form tion by 2, the constant term: y x b 5 1 a 1 . Divide each member of the equa- x 1 5y 5 2 5y x 2 5 2 2 1 2 5y x 2 1 2 5 1 The number that divides x, 2, is the x-intercept. The variable y is multiplied by 5 , which is equivalent to saying that it is divided by the reciprocal, , so the 2 2 y-intercept is . 5 2 5 2 Answer The x-intercept is 2 and the y-intercept is . 5 EXERCISES Writing About Mathematics 1. a. Can the equation y 4 be put into the form cept of the line? Explain your answer. y x b 5 1 a 1 b. Can the equation x 4 be put into the form intercept of the line? Explain your answer. y x b 5 1 a 1 to find the x-intercept and y-inter- to find the x-intercept and y- 2. Can the equation x y 0 be put into the form intercept of the line? Explain your answer. y x b 5 1 a 1 to find the x-intercept and y- Developing Skills In 3–10, find the intercepts of each equation if they exist. 3. x y 5 7. 3x 4y 8 4. x 3y 6 8. 7 x 2y y x b 5 1 a 1 11. Express the slope of in terms of a and b. 5. 5y x 10 9. y 3x 1 6. 2x y 2 0 10. y 4 12. For a–c, use the x- and y-intercepts to write an equation for each graph. a. b. c. O 2 –1.5 O –1 –1 2 31 O 1 1 3 410 Writing and Solving Systems of Linear Functions Applying Skills 13. Triangle ABC is drawn on the coordinate plane. Point A is at (4, 0), point B is at (0, 3), and point C is at the origin. a. What is the equation of g AB written in the form y x b 5 1 a 1 ? b. What is the length of AB ? 14. a. Isosceles right triangle ABC is drawn on the coordinate plane. Write the equation of each side of the triangle if C is the right angle at the origin and the length of each leg is 7. b. Is more than one answer to a possible? Explain your answer. 10-4 USING A GRAPH TO SOLVE A SYSTEM OF LINEAR EQUATIONS Consistent Equations The perimeter of a rectangle is 10 feet. When we let x represent the width of the rectangle and y represent the length, the equation 2x 2y 10 expresses the perimeter of the rectangle. This equation, which can be simplified to x y 5, has infinitely many solutions. Let x the width of the rectangle, and y the length of the rectangle. Perimeter: 2x 2y 10 x y 5 (Simplified) Measures of the sides: y x 1 If we also know that the length of the rectangle is 1 foot more than the width, the dimensions of the rectangle can be represented by the equation y x 1. We want both of the equations, x y 5 and y = x 1, to be true for the same pair of numbers. The two equations are called a system of simultaneous equations or a linear system. System of simultaneous equations: x y 5 y x 1 The solution of a system of simultaneous equations in two variables is an ordered pair of numbers that satisfies both equations. The graphs of x y 5 and y x 1, drawn using the same set of axes in a coordinate plane, are shown at the right. The possible solutions of x y 5 are all ordered pairs that are coordinates of the points on the line x y 5. The possible solutions of y x 1 are all ordered pairs that are coordinates of the points on the line y x 1. The coordinates of the point of intersection, (2, 3), are a solution of both equations. The ordered pair (2, 3) is a solution of the system of simultaneous equations. y y = x + 1 (2, 3) x 1 –1 –1 O 1 + y = 5 x Using a Graph to Solve a System of Linear Equations 411 Check: Substitute (2, 3) in both equations: x y 5 2 1 3 5? 3 3 ✔ Since two straight lines can intersect in no more than one point, there is no other ordered pair that is a solution of this system. Therefore x 2 and y 3, or (2, 3), is the solution of the system of equations. The width of the rectangle is 2 feet, and the length of the rectangle is 3 feet. When two lines are graphed in the same coordinate plane on the same set of axes, one and only one of the following three possibilities can occur. The pair of lines will: 1. intersect in one point and have one ordered number pair in common; 2. be parallel and have no ordered number pairs in common; 3. coincide, that is, be the same line with an infinite number of ordered num- ber pairs in common. If a system of linear equations such as x y 5 and y x 1 has at least one common solution, it is called a system of consistent equations. If a system has exactly one solution, it is a system of independent equations. Inconsistent Equations Sometimes, as shown at the right, when two linear equations are graphed in a coordinate plane using the same set of axes, the lines are parallel and fail to intersect, as in the case of x y 2 and x y 4. There is no common solution for the system of equations x y 2 and x y 4. It is obvious that there can be no ordered number pair (x, y) such that the sum of those numbers, x y, is both 2 and 4. Since the solution set of the system has no members, it is the empty set. y 1 –1 – If a system of linear equations such as x y 2 and x y 4 has no common solution, it is called a system of inconsistent equations. The graphs of two inconsistent linear equations are lines that have equal slopes or lines that have no slopes. Such lines are parallel. 412 Writing and Solving Systems of Linear Functions Dependent Equations Sometimes, as shown at the right, when two linear equations are graphed in a coordinate plane using the same set of axes, the graphs turn out to be the same line; that is, they coincide. This happens in the case of the equations x y 2 and 2x 2y 4. Every one of the infinite number of solutions of x y 2 is also a solution of 2x 2y 4. Thus, we see that 2x 2y 4 and x y 2 are equivalent equations with identical solutions. We note that, when both sides of the equation 2x 2y 4 are divided by 2, the result is x y 2. y 1 –1 – If a system of two linear equations, for example, x y 2 and 2x 2y 4, is such that every solution of one of the equations is also a solution of the other, it is called a system of dependent equations. The graphs of two dependent linear equations are the same line. Note that a system of dependent equations is considered consistent because there is at least one solution. Procedure To solve a pair of linear equations graphically: 1. Graph one equation in a coordinate plane. 2. Graph the other equation using the same set of coordinate axes. 3. One of three relationships will apply: a. If the graphs intersect in one point, the common solution is the ordered pair of numbers that are the coordinates of the point of intersection of the two graphs.The equations are independent and consistent. b. If the graphs have no points in common, that is, the graphs are parallel, there is no solution.The equations are inconsistent. c. If the graphs have all points in common, that is, the graphs are the
same line, every point on the line is a solution.The equations are consistent and dependent. 4. Check any solution by verifying that the ordered pair satisfies both equations. EXAMPLE 1 Solve graphically and check: 2x y 8 y x 2 Using a Graph to Solve a System of Linear Equations 413 Solution (1) Graph the first equation in a coordinate plane. Solve the first equation for y: 2x y 8 The y-intercept is 8 and the slope is 2. Start at the point (0, 8) and move down 2 units and to the right 1 unit to determine two other points. Or, choose three values of x and find the corresponding values of y to determine three points. Draw the line that is the graph of 2x y 8: y 2x 8 y (0, 81 O –1 1 (2) Graph the second equation using the same set of coordinate axes. Solve the second equation for y: The y-intercept is 2 and the slope is 1. Start at the point (0, 2) and move up 1 unit and to the right 1 unit to determine two other points. Or, choose three values of x and find the corresponding values of y to determine three points. Draw the line that is the graph of y – x 2 on the same set of coordinate axes: y x 2 y x 2 y (0, 8) P (2, 4) 2 x + y = 8 x (0, 2) y – x = 2 1 –1 O –1 1 (3) The graphs intersect at one point, P(2, 4). (4) Check: Substitute (2, 4) in each equation: 2x y 8 2(2) 1 4 5? 2 2 2 ✔ 414 Writing and Solving Systems of Linear Functions Calculator Solution (1) Solve the equations for y: 2x y 8 y x 2 y 2x 8 y x 2 (2) Enter the equations in the Y menu. ENTER: Y X,T,,n (-) 2 X,T,,n 8 ENTER 2 DISPLAY+ 2 (3) Use ZStandard to display the equations. ENTER: ZOOM 6 DISPLAY: (4) Use TRACE to determine the coordinates of the point of intersection. Use the right and left arrow keys to move along the first equation to what appears to be the point of intersection and note the coordinates, (2, 4). Use the up arrow key to change to the second equation. Again note the coordinates of the point of intersection, (2, 4). ENTER: TRACE DISPLAY: Y1=–2X+8 * X=2 Y=4 Answer (2, 4), or x 2, y 4, or the solution set {(2, 4)} Using a Graph to Solve a System of Linear Equations 415 EXERCISES Writing About Mathematics 1. Are there any ordered pairs that satisfy both the equations 2x y 7 and 2x 5 y? Explain your answer. 2. Are there any ordered pairs that satisfy the equation y x 4 but do not satisfy the equa- tion 2y 8 2x? Explain your answer. Developing Skills In 3–22, solve each system of equations graphically, and check. 3. y 2x 5 y x 4 6. x y 5 x 3y 9 9. y 3x 2x y 10 12. 3x y 6 y 3 1 15. y 3 3x 2x y 8 18. 5x 3y 9 5y 13 x 21. y 2x 6 0 y x 4. y 2x 3 1 y 3 2x 7. y 2x 1 x 2y 7 10. x 3y 9 x 3 13. y 2x 4 x y 5 16. 3x y 13 x 6y 7 19. x 0 y 0 22. 7x 4y 7 0 3x 5y 3 0 5. x y 1 x y 7 8. x 2y 12 y 2x 6 11. y x 2 x 2y 4 14. 2x – y 1 x y 1 17. 2x y 9 6x 3y 15 20. x y 2 0 x y 8 In 23–28, in each case: a. Graph both equations. b. State whether the system is consistent and independent, consistent and dependent, or inconsistent. 24. x y 5 23. x y 1 x y 3 26. 2x y 1 2y 4x 2 2x 2y 10 27. y 3x 2 y 3x 2 25. y 2x 1 y 3x 3 28. x 4y 6 x 2 Applying Skills In 29–32, in each case: a. Write a system of two first-degree equations involving the variables x and y that represent the conditions stated in the problem. b. Solve the system graphically. 29. The sum of two numbers is 8. The difference of these numbers is 2. Find the numbers. 416 Writing and Solving Systems of Linear Functions 30. The sum of two numbers is 5. The larger number is 7 more than the smaller number. Find the numbers. 31. The perimeter of a rectangle is 12 meters. Its length is twice its width. Find the dimensions of the rectangle. 32. The perimeter of a rectangle is 14 centimeters. Its length is 3 centimeters more than its width. Find the length and the width. 33. a. The U-Drive-It car rental agency rents cars for $50 a day with unlimited free mileage. Write an equation to show the cost of renting a car from U-Drive-It for one day, y, if the car is driven for x miles. b. The Safe Travel car rental agency rents cars for $30 a day plus $0.20 a mile. Write an equation to show the cost of renting a car from Safe Travel for one day, y, if the car is driven for x miles. c. Draw, on the same set of axes, the graphs of the equations written in parts a and b. d. If Greg will drive the car he rents for 200 miles, which agency offers the less expensive car? e. If Sarah will drive the car she rents for 50 miles, which agency offers the less expensive car? f. If Philip finds that the price for both agencies will be the same, how far is he planning to drive the car? 10-5 USING ADDITION TO SOLVE A SYSTEM OF LINEAR EQUATIONS In the preceding section, graphs were used to find solutions of systems of simultaneous equations. Since most of the solutions were integers, the values of x and y were easily read from the graphs. However, it is not always possible to read values accurately from a graph. For example, the graphs of the system of equations 2x y 2 and x y 2 are shown at the right. The solution of this system of equations is not a pair of integers. We could approximate the solution and then determine whether our approximation was correct by checking. However, there are other, more direct methods of solution. Algebraic methods can be used to solve a system of linear equations in two variables. Solutions by these methods often take less time and lead to more accurate results than the graphic method used in Section 4 of this chapter. y x + y = 2 2x – y = 2 1 O –1–1 1 x Using Addition to Solve a System of Linear Equations 417 To solve a system of linear equations such as 2x y 2 and x y 2, we make use of the properties of equality to obtain an equation in one variable. When the coefficient of one of the variables in the first equation is the additive inverse of the coefficient of the same variable in the second, that variable can be eliminated by adding corresponding members of the two equations. The system 2x y 2 and x y 2 can be solved by the addition method as follows: Solve for x (1) Since the coefficients of y in the two equations are additive inverses, add the equations to obtain an equation that has only one variable: (2) Solve the resulting equation for x: 2x y 2 x y 2 4 3x x 5 4 3 (3) Replace x by its value in either of the given equations: Solve for y (4) Solve the resulting equation for y 24 Check: Substitute for x and that these values make the given equations true: 2 3 4 3 for y in each of the given equations, and show 2 A 2x 2 y 5 2 4 2 2 3 5? 2 6 3 5? 2 6 3 5? 2 2 5 2 ✔ We were able to add the equations in two variables to obtain an equation in one variable because the coefficients of one of the variables were additive inverses. If the coefficients of neither variable are additive inverses, we can multiply one or both equations by a convenient constant or constants. For clarity, it is often helpful to label the equations in a system. The procedure is shown in the following examples. KEEP IN MIND When solving a system of linear equations for a variable using the addition method, if the result is: • the equation 0 0, then the system is dependent; • a false statement (such as 0 3), then the system is inconsistent. 418 Writing and Solving Systems of Linear Functions EXAMPLE 1 Solve the system of equations and check: x 3y 13 x y 5 [A] [B] Solution How to Proceed (1) Since the coefficients of the variable x are the same in both equations, write an equation equivalent to equation [B] by multiplying both sides of equation [B] by 1. Now, since the coefficients of x are additive inverses, add the two equations so that the resulting equation involves one variable, y: (2) Solve the resulting equation for the variable y: (3) Replace y by its value in either of the given equations: (4) Solve the resulting equation for the remaining variable, x: x 3y 13 [A] x y 5 1[B] 2y B] Check Substitute 1 for x and 4 for y in each of the given equations to verify the solution: x 3y 13 1 1 3(4) 5? 13 1 1 12 5? 13 13 13 ✔ x y 5 1 1 4 5? 5 5 5 ✔ Answer x 1, y 4 or (1, 4) Note: If equation [A] in Example 1 was x y 13, then the system of equations would be inconsistent. x y 13 [A] x y 5 [B] 0 8 ✘ Using Addition to Solve a System of Linear Equations 419 EXAMPLE 2 Solve the system of equations and check: 5a b 13 4a 3b 18 [A] [B] Solution How to Proceed (1) Multiply both sides of equation [A] by 3 to obtain an equivalent equation, 3[A]: 5a b 13 15a 3b 39 [A] 3[A] Note that the coefficient of b in 3[A] is the additive inverse of the coefficient of b in [B]. (2) Add the corresponding members of equations 3[A] and [B] to eliminate the variable b: (3) Solve the resulting equation for a: (4) Replace a by its value in either of the given equations and solve for b: 3[A] [B] [A] 15a 3b 39 4a 3b 18 57 19a a 3 5a b 13 5(3) b 13 15 b 13 b 2 Check Substitute 3 for a and 2 for b in the given equations: 5a b 13 4a 3b 18 5(3) 1 (22) 5? 13 15 1 (22) 5? 13 13 13 ✔ 4(3) 2 3(22) 5? 18 12 1 6 5? 18 18 18 ✔ Answer a 3, b 2, or (a, b) (3, 2) Order is critical when expressing solutions. While (a, b) (3, 2) is the solution to the above system of equations, (a, b) (2, 3) is not. Be sure that all variables correspond to their correct values in your answers. 420 Writing and Solving Systems of Linear Functions EXAMPLE 3 Solve and check: 7x 5 2y 3y 16 2x [A] [B] Solution How to Proceed (1) Transform each of the given equations into equivalent equations in which the terms containing the variables appear on one side and the constant appears on the other side: (2) To eliminate y, find the least common multiple of the coefficients of y in equations [A] and [B]. That least common multiple is 6. We want to write one equation in which the coefficient of y is 6 and the other in which the coefficient of y is 6. Multiply both sides of equation [A] by 3, and multiply both sides of equation [B] by 2 so that the new coefficients of y will be additive inverses, 6 and 6: (3) Add the corresponding members of this last pair of equations to eliminate variable y: (4) Solve the resulting eq
uation for variable x: (5) Replace x by its value in any equation containing both variables: (6) Solve the resulting equation for the remaining variable, y: 7x 5 2y 7x 2y 5 3y 16 2x 2x 3y 16 [A] [B] 3(7x 2y 5) 2(2x 3y 16) 3[A] 2[B] 3[A] 2[B] [B] 21x 6y 15 4x 6y 32 17 17x x 1 3y 16 2x 3y 16 2(1) 3y 16 2 3y 18 y 6 Check Substitute 1 for x and 6 for y in each of the original equations to verify the answer. The check is left to you. Answer x 1, y 6 or (1, 6) Using Addition to Solve a System of Linear Equations 421 EXERCISES Writing About Mathematics 1. Raphael said that the solution to Example 3 could have been found by multiplying equation [A] by 2 and equation [B] by 7. Do you agree with Raphael? Explain why or why not. 2. Fernando said that the solution to Example 3 could have been found by multiplying equation [A] by 3 and equation [B] by 2. Do you agree with Fernando? Explain why or why not. Developing Skills In 3–26, solve each system of equations by using addition to eliminate one of the variables. Check your solution. 3. x y 12 x y 4 6. c 2d 14 c 3d 9 9. –2m 4n 13 6m 4n 9 12. 5x 2y 20 2x 3y 27 15. 5r 2s 8 3r 7s 1 18. 4a 6b 15 6a – 4b 10 21. 3x 4y 2 x 2(7 y) 2a 1 1 1 3 2a 2 4 3b 5 8 3b 5 24 24. 4. a b 13 a b 5 7. a 4b 8 a 2b 0 10. 4x y 10 2x 3y 12 13. 2x y 26 3x 2y 42 16. 3x 7y 2 2x 3y 3 19. 2x y 17 5x 25 y 22. 3x 5(y 2) 1 8y 3x 25. c 2d 1 2 3c 5d 26 5. 3x y 16 2x y 11 8. 8a 5b 9 2a 5b 4 11. 5x 8y 1 3x 4y 1 14. 2x 3y 6 3x 5y 15 17. 4x 3y 1 5x 4y 1 20. 5r 3s 30 2r 12 3s 3x 1 1 1 3x 2 1 1 26. 2a 3b 3a 2 1 2 4y 5 10 2y 5 4 2b 5 2 23. Applying Skills 27. Pepe invested x dollars in a savings account and y dollars in a certificate of deposit. His total investment was $500. After 1 year he received 4% interest on the money in his savings account and 6% interest on the certificate of deposit. His total interest was $26. To find how much he invested at each rate, solve the following system of equations: x y 500 0.04x 0.06y 26 422 Writing and Solving Systems of Linear Functions 28. Mrs. Briggs deposited a total of $400 in two different banks. One bank paid 3% interest and the other paid 5%. In one year, the total interest was $17. Let x be the amount invested at 3% and y be the amount invested at 5%. To find the amount invested in each bank, solve the following system of equations: x y 400 0.03x 0.05y 17 29. Heather deposited a total of $600 in two different banks. One bank paid 3% interest and the other 6%. The interest on the account that pays 3% was $9 more than the interest on the account that pays 6%. Let x be the amount invested at 3% and y be the amount invested at 6%. To find the amount Heather invested in each account, solve the following system of equations: x y 600 0.03x 0.06y 9 30. Greta is twice as old as Robin. In 3 years, Robin will be 4 years younger than Greta is now. Let g represent Greta’s current age, and let r represent Robin’s current age. To find their current ages, solve the following system of equations: g 2r r 3 g 4 31. At the grocery store, Keith buys 3 kiwis and 4 zucchinis for a total of $4.95. A kiwi costs $0.25 more than a zucchini. Let k represent the cost of a kiwi, and let z represent the cost of a zucchini. To find the cost of a kiwi and the cost of a zucchini, solve the following system of equations: 3k 4z 4.95 k z 0.25 32. A 4,000 gallon oil truck is loaded with gasoline and kerosene. The profit on one gallon of gasoline is $0.10 and $0.13 for a gallon of kerosene. Let g represent the number of gallons of gasoline loaded onto the truck and k the number of gallons of kerosene. To find the number of gallons of each fuel that were loaded into the truck when the profit is $430, solve the following system of equations: g k 4,000 0.10g 0.13k 430 10-6 USING SUBSTITUTION TO SOLVE A SYSTEM OF LINEAR EQUATIONS Another algebraic method, called the substitution method, can be used to eliminate one of the variables when solving a system of equations. When we use this method, we apply the substitution principle to transform one of the equations of the system into an equivalent equation that involves only one variable. Using Substitution to Solve a System of Linear Equations 423 Substitution Principle: In any statement of equality, a quantity may be substituted for its equal. To use the substitution method, we must express one variable in terms of the other. Often one of the given equations already expresses one of the variables in terms of the other, as seen in Example 1. EXAMPLE 1 Solve the system of equations and check: 4x 3y 27 y 2x 1 [A] [B] Solution How to Proceed (1) Since in equation [B], both y and 2x 1 name the same number, eliminate y in equation [A] by replacing it with 2x 1: (2) Solve the resulting equation for x: (3) Replace x with its value in any equation involving both variables: (4) Solve the resulting equation for y: 4x 3y 27 y 2x 1 4x 3(2x 1) 27 4x 6x 3 27 10x 3 27 10x 30 x 3 y 2x 1 y 2(3) 1 y 6 1 y 5 [A] [B] [B→A] [B] Check Substitute 3 for x and 5 for y in each of the given equations to verify that the resulting sentences are true. 4x 3y 27 4(3) 1 3(5) 5? 27 12 1 15 5? 27 27 27 ✔ y 2x 1 5 5? 2(3) 2 1 5 5? 6 2 1 5 5 ✔ Answer x 3, y 5 or (3, 5) 424 Writing and Solving Systems of Linear Functions EXAMPLE 2 Solve the system of equations and check: 3x 4y 26 x 2y 2 [A] [B] Solution In neither equation is one of the variables expressed in terms of the other. We will use equation [B] in which the coefficient of x is 1 to solve for x in terms of y. How to Proceed (1) Transform one of the equations into an equivalent equation in which one of the variables is expressed in terms of the other. In equation [B], solve for x in terms of y: (2) Eliminate x from equation [A] by replacing it with its equal, 2 2y, from step (1): (3) Solve the resulting equation for y: (4) Replace y by its value in any equation that has both variables: (5) Solve the resulting equation for x: x 2y 2 x 2 2y [B] 3x 4y 26 3(2 2y) 4y 26 [A] [B→A] 6 6y 4y 26 6 10y 26 10y 20 y 2 x 2 2y x 2 2(2) x 2 4 x 6 [B] Check Substitute 6 for x and 2 for y in each of the given equations to verify that the resulting sentences are true. This check is left to you. Answer x 6, y 2 or (6, 2) EXERCISES Writing About Mathematics 1. In Example 2, the system of equations could have been solved by first solving the equation 3x 4y 26 for y. Explain why the method used in Example 2 was easier. 2. Try to solve the system of equations x 2y 5 and y in the first equation. What conclusion can you draw? 1 2x 1 1 by substituting 1 2x 1 1 for y Using Substitution to Solve a System of Linear Equations 425 Developing Skills In 3–14, solve each system of equations by using substitution to eliminate one of the variables. Check. 3. y x x y 14 6. y x 1 x y 9 9. a 2b 2 2a b 5 12. 2x 3y 4x 3y 12 4. y 2x x y 21 7. a 3b 1 5b 2a 1 10. 7x 3y 23 x 2y 13 13. 4y 3x 5x 8y 4 5. a 2b 5a 3b 13 8. a b 11 3a 2b 8 11. 4d 3h 25 3d 12h 9 14. 2x 3y 7 4x 5y 25 In 15–26, solve each system of equations by using any convenient algebraic method. Check. 15. s r 0 r s 6 18. 3x 8y 16 5x 10y 25 21. 3(y 6) 2x 3x 5y 11 24. 3x 4y 3x 1 8 5 5 3y 2 1 2 Applying Skills 16. 3a b 13 2a 3b 16 19. y 3x 3x 1 1 1 2y 5 11 22. x y 300 25. 0.1x 0.3y 78 a 3 5 a 1 b 6 5a 2b 49 17. y x 2 20. 3x y 16 a 2 2 3b 5 4 3 5a 1 b 5 15 23. 3d 13 2c 3c 1 d 2 8 26. a 3(b 1) 0 2(a 1) 2b 16 27. The length of the base of an isosceles triangle is 6 centimeters less than the sum of the lengths of the two congruent sides. The perimeter of the triangle is 78 centimeters. a. If x represents the length of each of the congruent sides and y represents the length of the base, express y in terms of x. b. Write an equation that represents the perimeter in terms of x and y. c. Solve the equations that you wrote in a and b to find the lengths of the sides of the triangle. 28. A package of batteries costs $1.16 more than a roll of film. Martina paid $11.48 for 3 rolls of film and a package of batteries. a. Express the cost of a package of batteries, y, in terms of the cost of a roll of film, x. b. Write an equation that expresses the amount that Martina paid for the film and batter- ies in terms of x and y. c. Solve the equations that you wrote in a and b to find what Martina paid for a roll of film and of a package of batteries. 426 Writing and Solving Systems of Linear Functions 29. The cost of the hotdog was $0.30 less than twice the cost of the cola. Jules paid $3.90 for a hotdog and a cola. a. Express the cost of a hotdog, y, in terms of the cost of a cola, x. b. Express what Jules paid for a hotdog and a cola in terms of x and y. c. Solve the equations that you wrote in a and b to find what Jules paid for a hotdog and for a cola. 30. Terri is 12 years older than Jessica. The sum of their ages is 54. a. Express Terri’s age, y, in terms of Jessica’s age, x. b. Express the sum of their ages in terms of x and y. c. Solve the equations that you wrote in a and b to find Terri’s age and Jessica’s age. 10-7 USING SYSTEMS OF EQUATIONS TO SOLVE VERBAL PROBLEMS You have previously learned how to solve word problems by using one variable. Frequently, however, a problem can be solved more easily by using two variables rather than one variable. For example, we can use two variables to solve the following problem: The sum of two numbers is 8.6. Three times the larger number decreased by twice the smaller is 6.3. What are the numbers? First, we will represent each number by a different variable: Let x the larger number and y the smaller number. Now use the conditions of the problem to write two equations: x y 8.6 The sum of the numbers is 8.6: Three times the larger decreased by twice the smaller is 6.3: 3x 2y 6.3 Solve the system of equations to find the numbers: x y 8.6 → 2(x y) 2(8.6) 2x 2y 17.2 3x 2y 6.3 23.5 5x x 4.7 The two numbers are 4.7 and 3.9. x y 8.6 4.7 y 8.6 y 3.9 Using Systems of Equations to Solve Verbal Problems 427 Procedure To solve a word problem by using a system of two equations involving two variables: 1. Use two di
fferent variables to represent the different unknown quantities in the problem. 2. Use formulas or information given in the problem to write a system of two equations in two variables. 3. Solve the system of equations. 4. Use the solution to determine the answer(s) to the problem. 5. Check the answer(s) in the original problem. EXAMPLE 1 The owner of a men’s clothing store bought six belts and eight hats for $140. A week later, at the same prices, he bought nine belts and six hats for $132. Find the price of a belt and the price of a hat. Solution (1) Let b the price, in dollars, of a belt, and h the price, in dollars, of a hat. (2) 6 belts and 8 hats \|_____\| \|______\| ↓ ↓ ↓ 8h 6b cost $140. ↓ ↓ 140 9 belts and 6 hats \|_____\| \|______\| ↓ ↓ ↓ 6h 9b cost $132. ↓ ↓ 132 (3) The least common multiple of the coefficients of b is 18. To eliminate b, write equivalent equations with 18 and 18 as the coefficients of b. Obtain these equations by multiplying both sides of the first equation by 3 and both members of the second equation by 2. Then add the equations and solve for h. 3(6b 8h) 3(140) → 18b 24h 420 2(9b 6h) 2(132) → 18b 12h 264 12h 156 h 13 428 Writing and Solving Systems of Linear Functions Substitute 13 for h in any equation containing both variables. 6b 8h 140 6b 8(13) 140 6b 104 140 6b 36 b 6 (4) Check: 6 belts and 8 hats cost 6($6) 8($13) $36 $104 $140. ✔ 9 belts and 6 hats cost 9($6) 6($13) $54 $78 $132. ✔ Answer A belt costs $6; a hat costs $13. EXAMPLE 2 When Angelo cashed a check for $170, the bank teller gave him 12 bills, some $20 bills and the rest $10 bills. How many bills of each denomination did Angelo receive? Solution (1) Represent the unknowns using two variables: Let x number of $10 bills, and y number of $20 bills. In this problem, part of the information is in terms of the number of bills (the bank teller gave Angelo 12 bills) and part of the information is in terms of the value of the bills (the check was for $170). (2) Write one equation using the number of bills: Write a second equation using the value of the bills. The value of x $10 bills is l0x, the value of y $20 bills is 20y, and the total value is $170: (3) Solve the system of equations: x y 12 10x 20y 170 10x 20y 170 10(x y) 10(12) → –10x 10y 120 10y 50 5 y x y 12 x 5 12 x 7 (4) Check the number of bills: 7 ten-dollar bills and 5 twenty-dollar bills 12 bills ✔ Check the value of the bills: 7 ten-dollar bills are worth $70 and 5 twenty-dollar bills are worth $100. Total value $70 $100 $170 ✔ Answer Angelo received 7 ten-dollar bills and 5 twenty-dollar bills. Using Systems of Equations to Solve Verbal Problems 429 EXERCISES Writing About Mathematics 1. Midori solved the equations in Example 2 by using the substitution method. Which equation do you think she would have solved for one of the variables? Explain your answer. 2. The following system can be solved by drawing a graph, by using the addition method, or by using the substitution method. x y 1 5x 10y 8 a. Which method do you think is the more efficient way of solving this system of equa- tions? Explain why you chose this method. b. Which method do you think is the less efficient way of solving this system of equa- tions? Explain why you chose this method. Developing Skills In 3–9, solve each problem algebraically, using two variables. 3. The sum of two numbers is 36. Their difference is 24. Find the numbers. 4. The sum of two numbers is 74. The larger number is 3 more than the smaller number. Find the numbers. 5. The sum of two numbers is 104. The larger number is 1 less than twice the smaller number. Find the numbers. 6. The difference between two numbers is 25. The larger exceeds 3 times the smaller by 4. Find the numbers. 7. If 5 times the smaller of two numbers is subtracted from twice the larger, the result is 16. If the larger is increased by 3 times the smaller, the result is 63. Find the numbers. 8. One number is 15 more than another. The sum of twice the larger and 3 times the smaller is 182. Find the numbers. 9. The sum of two numbers is 900. When 4% of the larger is added to 7% of the smaller, the sum is 48. Find the numbers. Applying Skills In 10–31, solve each problem algebraically using two variables. 10. The perimeter of a rectangle is 50 centimeters. The length is 9 centimeters more than the width. Find the length and the width of the rectangle. 11. A rectangle has a perimeter of 38 feet. The length is 1 foot less than 3 times the width. Find the dimensions of the rectangle. 12. Two angles are supplementary. The larger angle measures 120° more than the smaller. Find the degree measure of each angle. 430 Writing and Solving Systems of Linear Functions 13. Two angles are supplementary. The larger angle measures 15° less than twice the smaller. Find the degree measure of each angle. 14. Two angles are complementary. The measure of the larger angle is 30° more than the mea- sure of the smaller angle. Find the degree measure of each angle. 15. The measure of the larger of two complementary angles is 6° less than twice the measure of the smaller angle. Find the degree measure of each angle. 16. In an isosceles triangle, each base angle measures 30° more than the vertex angle. Find the degree measures of the three angles of the triangle. 17. At a snack bar, 3 pretzels and 1 can of soda cost $2.75. Two pretzels and 1 can of soda cost $2.00. Find the cost of a pretzel and the cost of a can of soda. 18. On one day, 4 gardeners and 4 helpers earned $360. On another day, working the same number of hours and at the same rate of pay, 5 gardeners and 6 helpers earned $480. Each gardener receives the same pay for a day’s work and each helper receives the same pay for a day’s work. How much does a gardener and how much does a helper earn each day? 19. A baseball manager bought 4 bats and 9 balls for $76.50. On another day, she bought 3 bats and 1 dozen balls at the same prices and paid $81.00. How much did she pay for each bat and each ball? 20. Mrs. Black bought 2 pounds of veal and 3 pounds of pork, for which she paid $20.00. Mr. Cook, paying the same prices, paid $11.25 for 1 pound of veal and 2 pounds of pork. Find the price of a pound of veal and the price of a pound of pork. 21. One day, Mrs. Rubero paid $18.70 for 4 kilograms of brown rice and 3 kilograms of basmati rice. Next day, Mrs. Leung paid $13.30 for 3 kilograms of brown rice and 2 kilograms of basmati rice. If the prices were the same on each day, find the price per kilogram for each type of rice. 22. Tickets for a high school dance cost $10 each if purchased in advance of the dance, but $15 each if bought at the door. If 100 tickets were sold and $1,200 was collected, how many tickets were sold in advance and how many were sold at the door? 23. A dealer sold 200 tennis racquets. Some were sold for $33 each, and the rest were sold on sale for $18 each. The total receipts from these sales were $4,800. How many racquets did the dealer sell at $18 each? 24. Mrs. Rinaldo changed a $100 bill in a bank. She received $20 bills and $10 bills. The number of $20 bills was 2 more than the number of $10 bills. How many bills of each kind did she receive? 25. Linda spent $4.50 for stamps to mail packages. Some were 39-cent stamps and the rest were 24-cent stamps. The number of 39-cent stamps was 3 less than the number of 24-cent stamps. How many stamps of each kind did Linda buy? 26. At the Savemore Supermarket, 3 pounds of squash and 2 pounds of eggplant cost $2.85. The cost of 4 pounds of squash and 5 pounds of eggplant is $5.41. What is the cost of one pound of squash, and what is the cost of one pound of eggplant? 27. One year, Roger Jackson and his wife Wilma together earned $67,000. If Roger earned $4,000 more than Wilma earned that year, how much did each earn? Graphing the Solution Set of a System of Inequalities 431 28. Mrs. Moto invested $1,400, part at 5% and part at 8%. Her total annual income from both investments was $100. Find the amount she invested at each rate. 29. Mr. Stein invested a sum of money in certificates of deposit yielding 4% a year and another sum in bonds yielding 6% a year. In all, he invested $4,000. If his total annual income from the two investments was $188, how much did he invest at each rate? 30. Mr. May invested $21,000, part at 8% and the rest at 6%. If the annual incomes from both investments were equal, find the amount invested at each rate. 31. A dealer has some hard candy worth $2.00 a pound and some worth $3.00 a pound. He makes a mixture of these candies worth $45.00. If he used 10 pounds more of the less expensive candy than he used of the more expensive candy, how many pounds of each kind did he use? 10-8 GRAPHING THE SOLUTION SET OF A SYSTEM OF INEQUALITIES To find the solution set of a system of inequalities, we must find the ordered pairs that satisfy the open sentences of the system. We do this by a graphic procedure that is similar to the method used in finding the solution set of a system of equations. EXAMPLE 1 Graph the solution set of this system: x 2 y 2 Solution (1) Graph x 2 by first graphing the plane divider x 2, which, in the figure at the right, is represented by the dashed line labeled l. The half-plane to the right of this line is the graph of the solution set of x 2. (2) Using the same set of axes, graph y 2 by first graphing the plane divider y 2, which, in the figure at the right, is represented by the dashed line labeled m. The half-plane below this line is the graph of the solution set of y 2. m y l 1 1O 3) The solution set of the system x 2 and y 2 consists of the intersection of the solution sets of x 2 and y 2. Therefore, the dark colored region in the lower figure, which is the intersection of the graphs made in steps (1) and (2), is the graph of the solution set of the system x 2 and y 2 432 Writing and Solving Systems of Linear Functions From the graph on page 431, all points in the solution region, and no others, satisfy both inequalities of the sy

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