Solution manual for finite mathematics 11th edition by sullivan

Solution Manual for Finite Mathematics 11th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter Linear Equations 1.1 Lines True 2 x + = 10 2x = x=2 True The equation of a vertical line with x-intercept at (–3, 0) is x = –3 If the slope of a line is undefined, the line is vertical 12 The set of points, (x, 3), where x is a real number, is a horizontal line passing through on the y-axis The equation of the line is y = The line y = –4x + has slope m = –4 and y-intercept (0, 6) The point-slope form of the equation of a line with slope m containing the point ( x1 , y1 ) is y − y1 = m ( x − x1 ) If the graph of a line slants downward from left to right, its slope m is negative A = (4, 2); B = (6, 2); C = (5, 3); D = (–2, 1) E = (–2, –3); F = (3, –2); G = (6, –2) H = (5, 0) 13 y = 2x + x −2 −2 y −4 12 −4 10 a 2nd quadrant b x-axis c 3rd quadrant d 1st quadrant e y-axis f 4th quadrant 14 y = –3x + x 2 −2 −4 y 0 12 −6 18 11 The set of points of the form, (2, y), where y is a real number, is a vertical line passing through on the x-axis The equation of the line is x = Full file at https://TestbankDirect.eu/ Solution Manual for Finite Mathematics 11th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter Linear Equations 15 2x – y = x −2 −4 y −6 −2 −10 −14 19 a The vertical line containing the point (–4, 1) is x = –4 b The horizontal line containing the point (–4, 1) is y = c y − = ( x + 4) y − = x + 20 −21 = x − y The line with a slope of containing the point (–4, 1) is 5x – y = –21 20 a The vertical line containing the point (–6, –3) is x = –6 b The horizontal line containing the point (–6, –3) is y = –3 16 x + 2y = x −2 −4 y c y + = ( x + 6) y + = x + 30 −27 = x − y The line with a slope of containing the point (–6, –3) is 5x – y = –27 21 a The vertical line containing the point (0, 3) is x = b The horizontal line containing the point (0, 3) is y = c y − = x −3 = x − y The line with a slope of containing the point (0, 3) is 5x – y = –3 17 a The vertical line containing the point (2, –3) is x = 22 a The vertical line containing the point (–6, 0) is x = –6 b The horizontal line containing the point (2, –3) is y = –3 b The horizontal line containing the point (–6, 0) is y = c y + = ( x − 2) y + = x − 10 13 = x − y The line with a slope of containing the point (2, –3) is 5x – y = 13 c y − = ( x + 6) y = x + 30 −30 = x − y The line with a slope of containing the point (–6, 0) is 5x – y = –30 18 a The vertical line containing the point (5, 4) is x = b The horizontal line containing the point (5, 4) is y = c y − = ( x − 5) y − = x − 25 21 = x − y The line with a slope of containing the point (5, 4) is 5x – y = 21 Full file at https://TestbankDirect.eu/ y − y1 − = = x2 − x1 − We interpret the slope to mean that for every unit change in x, y changes unit That is, for every units x increases, y increases by unit 23 m = y − y1 1− = =− x2 − x1 (−2) − We interpret the slope to mean that for every unit change in x, y changes –1 unit That is, for every units x increases, y decreases by unit 24 m = Solution Manual for Finite Mathematics 11th Edition by Sullivan Full file at https://TestbankDirect.eu/ 1.1 Lines y − y1 − = = −1 x − x1 −1 − We interpret the slope to mean that for every unit change in x, y changes by –1 unit That is, for every unit increase in x, y decreases by unit 25 m = 30 m = y − y1 −1 = = x − x1 − (−1) means that for every unit increase in x, y will increase units A slope of y − y1 −1 = = x2 − x1 − ( −1) We interpret the slope to mean that for every unit change in x, y changes by unit That is, for every units x increases, y increases by unit 26 m = y − y1 − = = =3 x2 − x1 − 1 A slope of means that for every unit change in x, y will change units 31 m = y − y1 − 2 = = =1 x2 − x1 − A slope of means that for every unit change in x, y will change unit 32 m = 27 m = y − y1 (−1) − (−1) = = =0 x2 − x1 − (−3) A slope of zero indicates that regardless of how x changes, y remains constant 28 m = y − y1 1− −2 = = =− 29 m = x − x1 − (−2) means that for every unit increase in x, y will decrease –1 unit A slope of − Full file at https://TestbankDirect.eu/ y − y1 2−2 = = =0 x − x1 (−5) − −9 A slope of zero indicates that regardless of how x changes, y remains constant y − y1 (−2) − −4 = = x2 − x1 (−1) − (−1) The slope is not defined 33 m = Solution Manual for Finite Mathematics 11th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter Linear Equations y − y1 − = = x − x1 − The slope is not defined 34 m = 39 40 35 41 36 42 37 38 Full file at https://TestbankDirect.eu/ 43 Use the points (0, 0) and (2, 1) to compute the slope of the line: y − y1 − m = = = x − x1 − Since the y-intercept, (0, 0), is given, use the slope-intercept form of the equation of the line: 1 y = x+0⇒ y = x 2 Then write the general form of the equation: x – 2y = Solution Manual for Finite Mathematics 11th Edition by Sullivan Full file at https://TestbankDirect.eu/ 1.1 Lines 44 Use the points (0, 0) and (–2, 1) to compute the slope of the line: y − y1 1− m = = =− x − x1 ( −2) − Since the y-intercept (0, 0) is given, use the slope-intercept form of the equation of the line: y = mx + b y =− x+0 y=− x 2y = −x x + 2y = This is the general form of the equation 45 Use the points (1, 1) and (–1, 3) to compute the slope of the line: y − y1 −1 m = = = = −1 x − x1 (−1) − −2 Now use the point (1, 1) and the slope m = –1 to write the point-slope form of the equation of the line: y − y1 = m( x − x1 ) y − = (−1)( x − 1) y −1 = −x +1 x+ y = This is the general form of the equation 46 Use the points (–1, 1) and (2, 2) to compute the slope of the line: y − y1 −1 m= = = x2 − x1 − (−1) 3 to write the point-slope form of the equation of the line y − y1 = m( x − x1 ) y − = ( x − (−1)) y − = ( x + 1) 3y − = x + x − y = −4 This is the general form of the equation Next use the point (–1, 1) and the slope m = 47 Since the slope and a point are given, use the point-slope form of the line: y − y1 = m ( x − x1 ) y − = ( x − ( −4)) y − = 2x + x − y = −9 This is the general form of the equation Full file at https://TestbankDirect.eu/ 48 Since the slope and a point are given, use the point-slope form of the line: y − y1 = m ( x − x1 ) y − = ( x − ( −3)) y − = 3x + 3x − y = −13 This is the general form of the equation 49 Since the slope and a point are given, use the point-slope form of the line: y − y1 = m ( x − x1 ) y − ( −1) = − ( x − 1) 3 y + = −2 ( x − 1) y + = −2 x + 2 x + y = −1 This is the general form of the equation 50 Since the slope and a point are given, use the point-slope form of the line: y − y1 = m ( x − x1 ) y − = ( x − 3) 2y − = x − x − 2y = This is the general form of the equation 51 Since we are given two points, (1, 3) and (–1, 2), first find the slope 3− m= = − (−1) Then use the slope, one of the points, (1, 3), and the point-slope form of the line: y − y1 = m( x − x1 ) y − = ( x − 1) 2y − = x −1 x − y = −5 This is the general form of the equation 52 Since we are given two points, (–3, 4) and (2, 5), first find the slope 4−5 m= = (−3) − Then use the slope, one of the points, (–3, 4), and the point-slope form of the line: y − y1 = m( x − x1 ) y − = ( x − ( −3)) 5 y − 20 = x + x − y = −23 This is the general form of the equation Solution Manual for Finite Mathematics 11th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter Linear Equations 53 Since we are given the slope m = –2 and the y-intercept (0, 3), we use the slope-intercept form of the line: y = mx + b y = −2 x + 2x + y = This is the general form of the equation 54 Since we are given the slope m = –3 and the y-intercept (0, –2), we use the slope-intercept form of the line: y = mx + b y = −3 x − x + y = −2 This is the general form of the equation 55 We are given the slope m =3 and the x-intercept (–4, 0), so we use the point-slope form of the line: y − y1 = m( x − x1 ) y − = ( x − ( −4)) y = 3x + 12 3x − y = −12 This is the general form of the equation 56 We are given the slope m = –4 and the x-intercept (2, 0) So we use the point-slope form of the line: y − y1 = m ( x − x1 ) y − = −4 ( x − ) y = −4 x + 4x + y = This is the general form of the equation and the point (0, 0), which is the y-intercept So, we use the slope-intercept form of the line: y = mx + b y = x+0 5 y = 4x 4x − y = This is the general form of the equation 57 We are given the slope m = and the point (0, 0), which is the y-intercept, we use the slope-intercept form of the line: y = mx + b y = x+0 3y = 7x 7x − 3y = This is the general form of the equation 58 Since we are given the slope m = Full file at https://TestbankDirect.eu/ 59 We are given two points, the x-intercept (2, 0) and the y-intercept (0, –1), so we need to find the slope and then use the slope-intercept form of the line to get the equation − (−1) slope = = 2−0 y = mx + b y = x −1 2y = x − x − 2y = This is the general form of the equation 60 We are given two points, the x-intercept (–4, 0) and the y-intercept (0, 4), so we need to find the slope and then use the slope-intercept form of the line to get the equation 4−0 slope = =1 − (−4) y = mx + b y = x+4 x − y = −4 This is the general form of the equation 61 Since the slope is undefined, the line is vertical The equation of the vertical line containing the point (1, 4) is x = 62 Since the slope is undefined, the line is vertical The equation of the vertical line containing the point (2, 1) is x = 63 Since the slope = 0, the line is horizontal The equation of the horizontal line containing the point (1, 4) is y = 64 Since the slope = 0, the line is horizontal The equation of the horizontal line containing the point (2, 1) is y = 65 y = 2x + slope: m = 2; y-intercept: (0, 3) Solution Manual for Finite Mathematics 11th Edition by Sullivan Full file at https://TestbankDirect.eu/ 1.1 Lines 66 y = –3x + slope: m = –3; y-intercept: (0, 4) 67 To obtain the slope and y-intercept, we transform the equation into its slope-intercept form by solving for y y = x −1 y = 2x − slope: m = 2; y-intercept: (0, –2) 68 To obtain the slope and y-intercept, we transform the equation into its slope-intercept form by solving for y x+ y =2 y =− x+2 slope: m = − ; y-intercept: (0, 2) Full file at https://TestbankDirect.eu/ 69 To obtain the slope and y-intercept, we transform the equation into its slope-intercept form by solving for y 2x − 3y = y = x−2 slope: m = ; y-intercept: (0, –2) 70 To obtain the slope and y-intercept, we transform the equation into its slope-intercept form by solving for y 3x + y = y = − x+3 slope: m = − ; y-intercept: (0, 3) 71 To obtain the slope and y-intercept, we transform the equation into its slope-intercept form by solving for y x + y =1 y = −x +1 slope: m = –1; y-intercept: (0, 1) Solution Manual for Finite Mathematics 11th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter Linear Equations 72 To obtain the slope and y-intercept, we transform the equation into its slope-intercept form by solving for y x− y =2 y = x−2 slope: m = 1; y-intercept: (0, –2) 73 The slope is not defined; there is no y- intercept So the graph is a vertical line 77 To obtain the slope and y-intercept, we transform the equation into its slope-intercept form by solving for y y−x=0 y=x slope: m = 1; y-intercept: (0, 0) 78 To obtain the slope and y-intercept, we transform the equation into its slope-intercept form by solving for y x+ y =0 y = −x slope: m = –1; y-intercept: (0, 0) 74 slope: m = 0; y-intercept: (0, –1) 75 slope: m = 0; y-intercept: (0, 5) 76 The slope is not defined; there is no y-intercept So the graph is a vertical line 79 To obtain the slope and y-intercept, we transform the equation into its slope-intercept form by solving for y y − 3x = y = 3x y= x slope: m = ; y-intercept: (0, 0) 80 To obtain the slope and y-intercept, we transform the equation into its slope-intercept form by solving for y 3x + y = y = −3 x y=− x slope: m = − ; y-intercept: (0, 0) (continued on next page) Full file at https://TestbankDirect.eu/ Solution Manual for Finite Mathematics 11th Edition by Sullivan Full file at https://TestbankDirect.eu/ 1.1 Lines (continued) Window: Xmin = –10; Xmax = 10 Ymin = –10; Ymax = 10 The x-intercept is (2.52, 0) The y-intercept is (0, –3.53) 81 To graph an equation on a graphing utility, first solve the equation for y 1.2 x + 0.8 y = 0.8 y = −1.2 x + y = −1.5 x + 2.5 Window: Xmin = –10; Xmax = 10 Ymin = –10; Ymax = 10 The x-intercept is (1.67, 0) The y-intercept is (0, 2.50) 82 To graph an equation on a graphing utility, first solve the equation for y −1.3 x + 2.7 y = 2.7 y = 1.3 x + 13 80 y= x+ 27 27 Window: Xmin = –10; Xmax = 10 Ymin = –10; Ymax = 10 84 To graph an equation on a graphing utility, first solve the equation for y x − y = 82 y = x − 82 82 y= x− 3 Window: Xmin = –10; Xmax = 30 Ymin = –35; Ymax = 10 The x-intercept is (16.4, 0) The y-intercept is (0, –27.33) 85 To graph an equation on a graphing utility, first solve the equation for y x+ y= 17 23 y = − x+ 23 17 23 ⎛ 2⎞ 46 23 y= x+ ⎜ − x + ⎟⎠ = − ⎝ 17 51 Window: Xmin = –10; Xmax = 10 Ymin = –10; Ymax = 10 The x-intercept is (–6.15, 0) The y-intercept is (0, 2.96) 83 To graph an equation on a graphing utility, first solve the equation for y 21x − 15 y = 53 15 y = 21x − 53 21 53 53 y= x− = x− 15 15 15 Full file at https://TestbankDirect.eu/ The x-intercept is (2.83, 0) The y-intercept is (0, 2.56) Solution Manual for Finite Mathematics 11th Edition by Sullivan Full file at https://TestbankDirect.eu/ 10 Chapter Linear Equations 86 To graph an equation on a graphing utility, first solve the equation for y x− y= 14 y= x− 14 8⎛ ⎞ 36 16 y = ⎜ x− ⎟ = x− ⎝ 14 ⎠ 21 21 12 16 y= x− 21 Window: Xmin = –10; Xmax = 10 Ymin = –10; Ymax = 10 The x-intercept is (0.44, 0) The y-intercept is (0, –0.76) 87 To graph an equation on a graphing utility, first solve the equation for y π x − 3y = 3y = π x − π y= x− Window: Xmin = –10; Xmax = 10 Ymin = –10; Ymax = 10 The x-intercept is (0.78, 0) The y-intercept is (0, –1.41) 88 To graph an equation on a graphing utility, first solve the equation for y x + π y = 15 π y = − x + 15 −x 15 y= + π π Window: Xmin = –10; Xmax = 10 Ymin = –10; Ymax = 10 The x-intercept is (3.87, 0) The y-intercept is (0, 1.23) Full file at https://TestbankDirect.eu/ 89 The graph passes through the points (0, 0) and (4, 8) We use the points to find the slope of the line: y − y1 − m= = = =2 x − x1 − The y-intercept (0, 0) is given, so we use the y-intercept and the slope m = 2, to obtain the slope-intercept form of the line y = mx + b y = 2x + ⇒ y = 2x This is choice (b) 90 The graph passes through the points (0, 0) and (8, 4) We use the points to find the slope of the line: y − y1 − m= = = = x2 − x1 − The y-intercept (0, 0) is given, so we use the y-intercept and the slope m = , to obtain the slope-intercept form of the line y = mx + b 1 y = x+0⇒ y = x 2 This is choice (c) 91 The graph passes through the points (0, 0) and (2, 8) We use the points to find the slope of the line: y − y1 − m= = = =4 x − x1 − The y-intercept (0, 0) is given, so we use the y-intercept and the slope m = 4, to obtain the slope-intercept form of the line y = mx + b y = 4x + ⇒ y = 4x This is choice (d) 92 The graph passes through the points (0, 0) and (2, 2) We use the points to find the slope of the line: y − y1 − m= = = =1 x2 − x1 − The y-intercept (0, 0) is given, so we use the y-intercept and the slope m = 1, to obtain the slope-intercept form of the line y = mx + b y = 1x + ⇒ y = x This is choice (a) Solution Manual for Finite Mathematics 11th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter Linear Equations 32 f f [–30, 10] by [0, 160] 16 a [–40, 10] by [–10, 35] 17 a b We select points (−30, 10) and (−14, 18) The slope of the line containing these points is: y − y1 18 − 10 m= = = = x − x1 (−14) − (−30) 16 The equation of the line is: y − y1 = m ( x − x1 ) y − 10 = ( x − (−30)) y − 10 = x + 15 y = x + 25 c The line on the scatter diagram will vary depending on the choice of points in part (b) b Answers will vary We select points (20, 16) and (50, 39) with numbers in thousands The slope of the line containing these points is: C − C1 39 − 16 23 m= = = I − I1 50 − 20 30 The equation of the line is: C − C1 = m ( I − I1 ) 23 C − 16 = ( I − 20) 30 23 46 C − 16 = I− 30 23 C= I+ 30 c The slope of this line indicates that a family will spend $23 of every extra $30 of disposable income d To find the consumption of a family whose disposable income is $42,000, substitute 42 for x in the equation from part (b) 23 C = ( 42) + 30 473 = ≈ 32.86667 15 The family will spend about $32,867 d e [–40, 10] by [–10, 35] e y = 0.75489x + 0.62663 y = 0.44207x + 23.45592 Full file at https://TestbankDirect.eu/ Solution Manual for Finite Mathematics 11th Edition by Sullivan Full file at https://TestbankDirect.eu/ 1.4 Scatter Diagrams; Linear Curve Fitting 18 a From the chart in Problem 17 we construct the savings chart and the scatter diagram Disposable Income I (thousands) Savings (thousands) S=I–C 20 20 18 27 36 37 11 45 50 11 33 e y = 0.24511x – 0.62663 19 a [–10, 110] by [50, 70] b y = 0.07818x + 59.0909 c b Answers will vary We select points (20, 4) and (50, 11) The slope of the line containing these points is: S − S1 11 − m= = = I − I1 50 − 20 30 The equation of the line is: S − S1 = m ( I − I1 ) S −4= ( I − 20) 30 14 S −4= I− 30 S= I− 30 [–10, 110] by [50, 70] d The slope indicates the apparent change in temperature in a 65ºF room for every percent increase in relative humidity e To determine the apparent temperature when the relative humidity is 75%, evaluate the equation of the line of best fit when x = 75 y = 0.07818 (75) + 59.0909 = 64.95 When the relative humidity is 75%, the temperature of the room will appear to be 65ºF 20 a c The slope indicates that a family of three will save $7.00 for every additional $30 of disposable income d To find the predicted amount of savings of a family with $42,000 of disposable income, evaluate S with I = 42 137 S= = 9.1333 (42) − = 30 15 A family of three with $42,000 of disposable income will save about $9,133.33 [–10, 110] by [60, 90] b y = 0.10818x + 68.68182 Full file at https://TestbankDirect.eu/ Solution Manual for Finite Mathematics 11th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter Linear Equations 34 c Chapter Review Exercises [–10, 110] by [60, 90] d The slope indicates the apparent change in temperature in a 75ºF room for every percent increase in relative humidity e To determine the apparent temperature when the relative humidity is 75%, evaluate the equation of the line of best fit when x = 75 y = 0.10818 (75) + 68.68182 = 76.80 When the relative humidity is 75%, the temperature of the room will appear to be 77ºF 21 a Using the LinReg function, the line of best fit is y = 0.0651t + 10.6049 The predicted energy use in 2015, t = 20, is y = 0.0651(20) + 10.6049 ≈ 11.91 quadrillion BTU The predicted energy use in 2030, t = 35, is y = 0.0651(35) + 10.6049 ≈ 12.88 quadrillion BTU The predicted energy use in 2035, t = 40, is y = 0.0651(40) + 10.6049 ≈ 13.21 quadrillion BTU b When we graph the scatterplot and the line of best fit, we see that the points are not strictly linear Therefore, estimates based on the line of best fit are approximations The line of best fit may differ depending on which points are chosen or a linear model may not have been used [0, 35] by [9, 13] Full file at https://TestbankDirect.eu/ a m = y − y1 4−2 = = =− x − x1 (−3) − −4 means that for every unit change in x, y will change (−1) unit That is, for every units x moves to the right, y will move down unit A slope of − b Use the point (1, 2) and the slope to get the point-slope form of the equation of the line: y − y1 = m ( x − x1 ) y − = − ( x − 1) (continued on next page) Solution Manual for Finite Mathematics 11th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter Review Exercises (continued) Simplifying and solving for y gives the slopeintercept form: 1 y−2=− x+ 2 y =− x+ 2 Rearranging terms gives the general form of the equation: x + y = 35 b Use the point (–1, 5) and the slope to get the point-slope form of the equation of the line: y − y1 = m ( x − x1 ) y − = ( x − (−1) ) Simplifying and solving for y gives the slopeintercept form: y − = 2x + y = 2x + Rearranging terms gives the general form of the equation: x − y = −7 c c y − y1 −1 = = = −1 x − x1 (−1) − −2 A slope of –1 means that for every one unit change in x, y changes (–1) unit That is, for every unit x moves to the right, y moves down unit a m = b Use the point (1, 1) and the slope to get the point-slope form of the equation of the line: y − y1 = m ( x − x1 ) y − = −1( x − 1) Simplifying and solving for y gives the slopeintercept form: y −1 = −x +1 y = −x + Rearranging terms gives the general form of the equation: x + y = c a m = means that for every unit change in x, y will change (–3) units That is, for every units x moves to the right, y will move down units A slope of − b Use the point (0, 0) and the slope to get the point-slope form of the equation of the line: y = mx + b 3 y = − x+0= − x 2 Rearranging terms gives the general form of the equation: x + y = c y − y1 5−3 = = =2 x − x1 (−1) − ( −2) A slope of means that for every one unit change in x, y changes by units That is, for every unit x moves to the right, y moves up units a m = Full file at https://TestbankDirect.eu/ y − y1 3−0 3 = = =− x − x1 (−2) − −2 Solution Manual for Finite Mathematics 11th Edition by Sullivan Full file at https://TestbankDirect.eu/ 36 Chapter Linear Equations Since we are given the slope m = –3 and a point, we get the point-slope equation of the line: y − y1 = m ( x − x1 ) y − (−1) = −3 ( x − 2) Solving for y puts the equation into the slopeintercept form: y + = −3 x + y = −3 x + Rearranging terms gives the general form of the equation: 3x + y = 10 Since we are given the slope m = and a point, we get the point-slope equation of the line: y − y1 = m ( x − x1 ) y − (−3) = ( x − (−1) ) Solving for y puts the equation into the slopeintercept form: y + = 4x + y = 4x + Rearranging terms gives the general form of the equation: x − y = −1 11 Since we are given the slope m = and a point on the line We either use the point-slope formula or recognize that this is a horizontal line and the equation of a horizontal line y = b The general form of the equation is y = Full file at https://TestbankDirect.eu/ 12 Since we are told the slope is undefined, we know the line is vertical We use the point (–3, 4) to write the equation of a vertical line x = –3 This is the general form of the equation of a vertical line 13 We are told that the line is vertical, so the slope is not defined We also know the line contains the point (8, 5) The general equation of a vertical line is x = a, so the general form of this equation is x = 14 We are told that the line is horizontal, so the slope is We also know the line contains the point (5, 8) The general equation of a horizontal line is y = b, so the general form of this equation is y = Solution Manual for Finite Mathematics 11th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter Review Exercises 15 We are given the x-intercept and a point First we find the slope of the line containing the two y − y1 (−5) − −5 = = =− points m = x − x1 4−2 2 We then use the point (2, 0) and the slope to get the point-slope form of the equation of the line y − y1 = m ( x − x1 ) y − = − ( x − 2) To get the slope-intercept form, solve for y: y = − x+5 Rearrange the terms for the general form of the equation: x + y = 10 16 We are given two points, the y-intercept and another point First use the two points to find the slope of the line containing them y − y1 (−3) − (−2) −1 = = =− m= x − x1 5−0 5 Since one of the points is the y-intercept, use it and the slope to get the slope-intercept form of the equation: y = − x − Rearrange the terms for the general form of the equation: x + y = −10 Full file at https://TestbankDirect.eu/ 37 17 We are given the x-intercept and the y-intercept Use the two points to find the slope of the line y − y1 (−4) − −4 = = =− m= x − x1 − (−3) 3 Since one of the points is the y-intercept, use it and the slope to write the slope-intercept form of the equation: y = − x − Rearrange the terms for the general form of the equation: x + y = −12 18 We are given two points Use them to find the slope of the line: y − y1 − (−4) = = = −5 m= −1 x − x1 2−3 Use the point (2, 1) and the slope to write the point-slope form of the equation: y − y1 = m ( x − x1 ) y − = −5 ( x − ) Simplifying and solving for y, gives the slopeintercept form of the equation y − = −5 x + 10 y = −5 x + 11 Rearrange the terms for the general form of the equation: x + y = 11 Solution Manual for Finite Mathematics 11th Edition by Sullivan Full file at https://TestbankDirect.eu/ 38 Chapter Linear Equations 19 Since the line we are seeking is parallel to x + y = −4 , the slope of the two lines are the same Find the slope of the given line by putting it into slope-intercept form: x + y = −4 y = −2 x − 4 y=− x− 3 The slopes of the two lines are m = − Use the slope and the point (–5, 3) to write the pointslope form of the equation of the parallel line y − y1 = m ( x − x1 ) y − = − ( x − ( − )) y − = − ( x + 5) To put the equation into slope-intercept form, solve for y 10 y−3= − x− 3 y=− x− 3 Rearrange the terms to obtain the general form of the equation: x + y = −1 20 The line we are seeking is parallel to x + y = , so the slopes of both lines are the same To find the slope put the equation into slope-intercept form x+ y=2 y = −x + Use the slope m = –1 and the point (1, –3) to write the point-slope form of the equation of the parallel line y − y1 = m ( x − x1 ) y − (−3) = −1( x − 1) Solve for y to get the slope-intercept form: y = −x − Rearrange the terms to get the general form of the equation: x + y = −2 Full file at https://TestbankDirect.eu/ 21 To find the slope and y-intercept of the line, put the equation into the slope-intercept form x + y = 18 y = −9 x + 18 y = − x+9 The slope is − , and the y-intercept is (0, 9) 22 To find the slope and y-intercept of the line, put the equation into the slope-intercept form x + y = 20 y = −4 x + 20 y = − x+4 The slope is − , and the y-intercept is (0, 4) 23 To find the slope and y-intercept of the line, put the equation into the slope-intercept form 4x + y = y = −4 x + 9 y = −2 x + ⎛ 9⎞ The slope is –2, and the y-intercept is ⎜ 0, ⎟ ⎝ 2⎠ (continued on next page) Solution Manual for Finite Mathematics 11th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter Review Exercises (continued) 24 To find the slope and y-intercept of the line, put the equation into the slope-intercept form 3x + y = y = −3 x + y =− x+4 The slope is − , and the y-intercept is (0, 4) 39 26 To find the slope and y-intercept of the line, put the equation into the slope-intercept form 1 x− y= 12 1 − y = − x+ 12 y = x− 4 5⎞ ⎛ The slope is , and the y-intercept is ⎜ 0, − ⎟ ⎝ 4⎠ 27 Put each equation into slope-intercept form: 3x − y = −12 x − y = −9 −4 y = −3x − 12 −8 y = −6 x − 9 y = x+3 y = x+ 8 y = x+ Since both lines have the same slope but different y-intercepts, the lines are parallel 25 To find the slope and y-intercept of the line, put the equation into the slope-intercept form 1 x+ y= 1 y=− x+ y =− x+ 2 ⎛ 1⎞ The slope is − , and the y-intercept is ⎜ 0, ⎟ ⎝ 2⎠ 28 Put each equation into slope-intercept form: x + y = −5 x + y = −10 y = −2 x − y = −4 x − 10 10 y=− x− y=− x− 3 6 y = − x− 3 Since both lines have the same slope and the same y-intercept, the lines are coincident 29 Put each equation into slope-intercept form: x − y = −2 3x − y = −12 − y = −x − −4 y = −3x − 12 y = x+2 y = x+3 Since the lines have different slopes, they intersect 30 Put each equation into slope-intercept form: x + y = 12 2x + 3y = y = − x + 12 y = −2 x + 5 y =− x+ 3 Since the lines have different slopes, they intersect Full file at https://TestbankDirect.eu/ Solution Manual for Finite Mathematics 11th Edition by Sullivan Full file at https://TestbankDirect.eu/ 40 Chapter Linear Equations 31 Put each equation into slope-intercept form: x + y = −12 x + y = −6 y = −4 x − 12 y = −2 x − 2 y =− x−2 y =− x−2 3 Since both lines have the same slope and the same y-intercept, the lines are coincident 32 Put each equation into slope-intercept form: −3 x + y = x − y = −5 y = 3x −2 y = −6 x − 5 y = 3x + Since both lines have the same slope but different y-intercepts, the lines are parallel 33 To find the point of intersection of two lines, first put the lines in slope-intercept form: L: x − y = M: x + 2y = y= x−4 y=− x+ 2 Since the point of intersection, (x0, y0), must be on both L and M, we set the two equations equal to each other and solve for x0 Then we substitute the value of x0 into the equation of one of the lines to find y0 y = x0 − x0 − = − x0 + 2 y0 = − x0 − = − x0 + y0 = 3x0 = 15 x0 = 34 To find the point of intersection of two lines, first put the lines in slope-intercept form: L: x + y = M: x − 2y = y = −x + 1 y= x− 2 Since the point of intersection, (x0, y0), must be on both L and M, we set the two equations equal to each other and solve for x0 Then we substitute the value of x0 into the equation of one of the lines to find y0 1 y = − x0 + − x + = x0 − 2 y = −3 + −2 x0 + = x0 − y0 = 3x0 = x0 = The point of intersection is (3, 1) Full file at https://TestbankDirect.eu/ 35 To find the point of intersection of two lines, first put the lines in slope-intercept form: M: x + 2y = L: x − y = −2 y= x+2 y=− x+ 2 Since the point of intersection, (x0, y0), must be on both L and M, we set the two equations equal to each other and solve for x0 Then we substitute the value of x0 into the equation of one of the lines to find y0 y = x0 + x0 + 2 y0 = + 2 x0 + = − x0 + y0 = 3 x0 = x0 = The point of intersection is (1, 3) x0 + = − 36 To find the point of intersection of two lines, first put the lines in slope-intercept form: L : 2x + y = M : 2x − y = y = −2 x + − y = −2 x + 1 y = − x +1 y = x−2 2 Since the point of intersection, (x0, y0), must be on both L and M, we set the two equations equal to each other and solve for x0 Then we substitute the value of x0 into the equation of one of the lines to find y0 1 − x0 + = x0 − y = x0 − 2 2 − x0 + = x0 − y0 = ⋅ − −2 x = − x0 = y0 = − 1⎞ ⎛ The point of intersection is ⎜ 3, − ⎟ ⎝ 2⎠ Solution Manual for Finite Mathematics 11th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter Review Exercises 41 1⎞ ⎛4 The point of intersection is ⎜ , − ⎟ ⎝5 10 ⎠ 37 To find the point of intersection of two lines, first put the lines in slope-intercept form: L: x − y = −8 M : 3x + y = −4 y = −2 x − y = −3 x 1 y= x+2 y=− x 2 Since the point of intersection, (x0, y0), must be on both L and M, we set the two equations equal to each other and solve for x0 Then we substitute the value of x0 into the equation of one of the lines to find y0 1 y = − x0 x + = − x0 2 x0 + = − x0 y = − ⋅ (−2) x = −4 y0 = x = −2 The point of intersection is (–2, 1) 39 Use a table to organize the information Amount Invested B-Bonds Bank Total Interest Interest Earned Rate x 0.12 0.12x y = 90,000 – x 0.05 0.05(90,000 – x) x + y = 90,000 10,000 The last column gives the information needed for the equation since the sum of the interest earned on the individual investments must equal the total interest earned 0.12 x + 0.05 (90, 000 − x ) = 10, 000 0.12 x + 4500 − 0.05 x = 10, 000 0.07 x = 5, 500 x ≈ 78,571.43 y = 90, 000 − 78,571.43 = 11, 428.57 Karen should invest $78,571.43 in B-rated bonds and $11,428.57 in the well-known bank in order to achieve their investment goals 40 Use a table to organize the information 38 To find the point of intersection of two lines, first put the lines in slope-intercept form: L : 3x + y = M: x − 2y = y = −3 x + − 2y = −x +1 1 y = − x+ y= x− 2 Since the point of intersection, (x0, y0), must be on both L and M, we set the two equations equal to each other and solve for x0 Then we substitute the value of x0 into the equation of one of the lines to find y0 1 1 y = x0 − − x0 + = x0 − 2 2 −3 x + = x − y0 = ⋅ − −5 x = − 2 1 y0 = − = − x0 = 10 Full file at https://TestbankDirect.eu/ Amount Mixed Percent Acid Total Acid Solution x 0.20 0.20x Solution y = 100 – x 0.12 0.12(100 – x) Mixture x + y = 100 0.15 0.15(100) The last column contains the information needed to solve the problem because the sum of the concentrations of the individual solutions must equal the concentration in the mixture 0.20 x + 0.12 (100 − x ) = 15 0.20 x + 12 − 0.12 x = 15 0.08 x = x = 37.5 y = 100 − 37.5 = 62.5 (continued on next page) Solution Manual for Finite Mathematics 11th Edition by Sullivan Full file at https://TestbankDirect.eu/ 42 Chapter Linear Equations (continued) So, 37.5 cubic centimeters of 20% HCL acid should be mixed with 62.5 cubic centimeters of 12% HCL acid to obtain 100 cubic centimeters of a solution with a 15% HCL acid 41 a The break-even point is the point where the cost equals the revenue, or when the profit is zero Before we can find the break-even point we need the equation that describes cost We are told the fixed costs, the band and the advertising, and the variable costs If we let x denote the number of tickets sold, the cost of the dance is described by the equation: C = 500 + 100 + x = x + 600 The revenue is given by the equation, R = 10x, since each ticket costs $10 Setting C = R, and solving for x, will tell how many tickets must be sold to break even x + 600 = 10 x 600 = x x = 120 So, 120 tickets must be sold for the group to break even b Profit is the difference between the revenue and the cost To determine the number of tickets that need to be sold to clear a profit of $900, we will solve the equation: P = R−C 900 = 10 x − (5 x + 600) 900 = x − 600 1500 = x x = 300 The church group must sell 300 tickets to realize a profit of $900 c If tickets cost $12, the break-even point will come from the equation R=C 12 x = x + 600 x = 600 x = 85.71 To break even, 86 tickets must be sold To find the number of ticket sales needed to have a $900 profit, solve the equation: P = R−C 900 = 12 x − (5 x + 600) 900 = x − 600 1500 = x x = 214.29 So the church group needs to sell 215 tickets at $12 each to realize a profit of $900 Full file at https://TestbankDirect.eu/ 42 Use a table to organize the information Type of Coffee Amount of Coffee Price per Pound Total Cost of Coffee Type x 5x Type y = 100 – x 7.5 7.5(100 – x) Blend x + y = 100 6(100) The last column provides the information necessary to solve the problem, since the sum of the costs of the individual coffees must equal the cost of the blend x + 7.5 (100 − x ) = 600 x + 750 − 7.5 x = 600 −2.5 x = −150 x = 60 The manufacturer should mix 60 pounds of type coffee and 40 pounds of type coffee to get 100 pounds of a blend worth $6.00 per pound 43 This relation does not appear to be linear 44 This relation appears to be linear 45 a The market price is the price for which supply equals demand To find market price set S = D and solve for p 0.8 p + 0.2 = −0.4 p + 1.8 1.2 p = 1.6 p = 1.33 The market price is $1.33 per bushel b When p = 1.33, S = 0.8(1.33) + 0.2 = 1.264 There will be 1.264 million bushels of corn supplied at the market price of $1.33 Solution Manual for Finite Mathematics 11th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter Review Exercises 43 h The average annual decrease in the concentration of carbon monoxide is not constant The slope depends on the points used to calculate it c i The average level of carbon monoxide is decreasing 47 a d At a price of $1.33 per bushel the supply and the demand for corn are equal 46 a p − p1 228, 700 − 181, 900 = x − x1 2002 − 1998 46,800 = = 11, 700 b m = y − y1 4.03 − 5.47 = x2 − x1 1998 − 1992 −1.44 = = − 0.24 b m = c The slope indicates the average annual decrease in concentration of carbon monoxide in the air between 1992 and 1998 y − y1 2.57 − 3.51 = x − x1 2008 − 2000 −0.94 = = − 0.1175 d m = c The mean price of houses sold in the United States increased by an average of $11,700 per year from 1998 through 2002 d m = p − p1 = 292, 600 − 228, 700 x − x1 2008 − 2002 63,900 = = 10, 650 e The mean price of houses sold in the United States increased by an average of $10,650 per year from 2002 through 2008 f e The slope indicates the average annual decrease in concentration of carbon monoxide in the air between 2000 and 2008 f The slope of the line of best fit is a = 13,960 The slope of the line of best fit is a = –0.2377 g The slope indicates the average annual decrease in the concentration of carbon monoxide in the air between 1992 and 2008 g The mean price of houses sold in the United States increased by an average of $13,960 per year from 1998 through 2008 h The average annual increase in mean price of houses sold in the United States is not constant The slope depends on the points used to calculate it i The average price of houses sold in the United States increased from 1998 though 2007 and then started to decline Full file at https://TestbankDirect.eu/ Solution Manual for Finite Mathematics 11th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter Linear Equations 44 50 Answers will vary 48 a 51 a This is the equation of a vertical line that includes the point (0, 0) Its graph is the yaxis b The graph of y = is the x-axis It is a horizontal line and has a slope of zero b Yes, the data are roughly linear v2 − v1 135.15 − 102.67 = x2 − x1 2007 − 2003 32.48 = = 8.12 c m = d The slope is the average annual rate of change in the value of a share of the Vanguard 500 Index Fund The positive value indicates that the value of a share is increasing e c This is an equation of a line with a slope of –1 and a y-intercept of (0, 0) Its graph is a negatively sloped diagonal line through the origin Chapter Project Since Metro PCS charges a flat rate that does not depend on the number of minutes, x, the equation is that of a horizontal line M = 60 If x ≤ 450, the cost of the Verizon plan is constant, so V = 39.99 If x > 450, then V = 0.45(450 – x) + 39.99 3, The slope of the line of best fit is a = 8.391 f Evaluate the line of best fit at x = 2008 y = 8.391x − 16, 705 y = 8.391( 2008) − 16, 705 ≈ 144.13 According to the model, the value of a share of Vanguard 500 Index Fund in 2008 was worth 144.13 49 a Set V (for x ≥ 450) = M to find the intersection of the two graphs 0.45 ( x − 450) + 39.99 = 60 0.45 x − 162.51 = 60 0.45 x = 222.51 x ≈ 494.5 From the graph, we can see that if you talk more than 494.5 minutes per month, the Metro PCS plan is more economical If x ≤ 650, the cost of the T-Mobile plan is constant, so T = 49.99 If x > 650, then T = 0.50(650 – x) + 49.99 b The data not appear to be linearly related c Since the data are not linearly related, the prediction in Problem 48 is not valid d Answers will vary Full file at https://TestbankDirect.eu/ Solution Manual for Finite Mathematics 11th Edition by Sullivan Full file at https://TestbankDirect.eu/ Chapter Mathematical Questions from Professional Exams Set T (for x ≥ 650) = M to find the intersection of the two graphs 0.50 ( x − 650) + 49.99 = 60 0.50 x − 275.01 = 60 0.50 x = 335.01 x ≈ 670 Set V (for x ≥ 450) = T = 49.99 to find the intersection of the two graphs 0.45 ( x − 450) + 39.99 = 49.99 0.45 x − 162.51 = 49.99 0.45 x = 212.5 x ≈ 472.2 45 Verizon: 2000 minutes: 0.45 (2000 − 450) + 39.99 = 737.49 2400 minutes: 0.45 (2400 − 450) + 39.99 = 917.49 Find when the costs of the two plans are equal 0.50 ( x − 650) + 49.99 = 0.45 ( x − 450) + 39.99 0.50 x − 275.01 = 0.45 x − 162.51 0.05 x = 112.5 x = 2250 If you talk between 2000 and 2250 minutes, then the T-Mobile plan is less expensive If you talk more than 2250 minutes, then the Verizon plan is less expensive Both plans are expensive If you talk 480 minutes a month, then the T-Mobile plan is the cheapest T-Mobile is cheapest if you talk between 472 and 670 minutes per month Metro PCS is cheapest if you talk more than 670 minutes per month 10 0.50 ( x − 650) + 49.99 = 55 0.50 x − 275.01 = 55 0.50 x = 330.01 x ≈ 660 If you want to spend only $55 a month using T-Mobile, then you are limited to about 660 minutes per month 11 The monthly charge for Metro PCS is $70 including the phone 0.45 ( x − 450) + 39.99 = 70 0.45 x − 162.51 = 70 0.45 x = 232.51 x ≈ 516.7 Verizon is the better deal if you talk less than 517 minutes per month 12 T-Mobile: 2000 minutes: 0.50 (2000 − 650) + 49.99 = 724.99 2400 minutes: 0.50 (2400 − 650) + 49.99 = 924.99 Mathematical Questions from Professional Exams The break-even point is the value of x for which the revenue equals cost If x units are sold at price of $2.00 each, the revenue is R = 2x Cost is the total of the fixed and variable costs We are told that the fixed costs are $6000, and that the variable cost per item is 40% of the price So the cost is given by the equation: C = (0.40)(2) x + 6000 = 0.8 x + 6000 Setting R = C and solving for x yields R=C x = 0.8 x + 6000 1.2 x = 6000 x = 5000 The answer is (b) Profit is defined as revenue minus cost If x rodaks are sold for $6.00 each, the revenue equation is R = 6x We are told that to manufacture rodaks costs $2.00 per unit and $37,500 in fixed costs So the cost equation for producing x rodaks is C = 2x + 37,500 The Breiden Company wants to realize a before tax profit equal to 15% of sales, which is P = 0.15R = (0.15)(6x) = 0.9x (continued on next page) Full file at https://TestbankDirect.eu/ Solution Manual for Finite Mathematics 11th Edition by Sullivan Full file at https://TestbankDirect.eu/ 46 Chapter Linear Equations (continued) To find the number of rodaks that need to be sold to meet Breiden’s goal, we solve the equation P = R−C 0.9 x = x − (2 x + 37, 500) 0.9 x = x − 37,500 3.1x = 37, 500 x = 12, 096.77 The answer is (d) The break-even point is the number of units that must be sold for revenue to equal cost Using the notation given, we have R = SPx and C = VCx + FC To find the sales level necessary to break even, we set R = C and solve for x R=C SPx = VCx + FC SPx − VCx = FC ( SP − VC ) x = FC FC x= SP − VC The answer is (d) Full file at https://TestbankDirect.eu/ At the break-even point of x = 400 units sold, cost equals revenue We are told cost, C = $400 + 200 = 600, so revenue R = $600 Since R = price · quantity, the price of each item $600 is = $1.50 The variable cost per unit is 400 $1.00, so the 401st unit contributes $1.50 – $1.00 = $0.50 to profits The answer is (b) Since straight-line depreciation remains the same over the life of the property, its expense over time will be a horizontal line Sum-ofyear’s-digits depreciation expense decreases as time increases The answer is (c) The answer is (c) Y = $1000 + $2X is a linear relationship The answer is (b) Y is an estimate of total factory overhead The answer is (b) In the equation $2 is the estimate of variable cost per direct labor hour