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Chapter 2, continued Lesson 2.4
Chapter 2, continued Lesson 2.4 6. The line passes through (x1, y1) 5 (22, 5) and 1. The slope is m 5 3 and the y-intercept is b 5 1. Use the slope-intercept form to write an equation of the line. y 5 mx 1 b y 5 3x 1 1 2. The slope is m 5 22 and the y-intercept is b 5 24. Use the slope-intercept form to write an equation of the line. y 5 mx 1 b y 5 (22)x 1 (24) y 5 22x 2 4 3 7 3. The slope is m 5 2} and the y-intercept is b 5 }. 4 2 Use the slope-intercept form to write an equation of the line. y 5 mx 1 b 3 (x2, y2) 5 (4, 27). Find its slope. y2 2 y1 2 7 4. Because you know the slope and a point on the line, use point-slope form to write an equation of the line. Let (x1, y1) 5 (21, 6) and m 5 4. y 2 y1 5 m(x 2 x1) y 2 6 5 4(x 2 (21)) y 2 6 5 4(x 1 1) y 2 6 5 4x 1 4 y 5 4x 1 10 5. a. The given line has a slope of m1 5 3. So, a line parallel to it has a slope of m2 5 m1 5 3. You know the slope and a point on the line, so use the point-slope form with (x1, y1) 5 (4, 22) to write an equation of the line. y 2 y1 5 m2(x 2 x1) y 2 (22) 5 3(x 2 4) y 1 2 5 3(x 2 4) y 1 2 5 3x 2 12 y 5 3x 2 14 b. A line perpendicular to a line with slope m1 5 3 has a 1 1 . Use point-slope form slope of m2 5 2} m1 5 2} 3 with (x1, y1) 5 (4, 22). y 2 y1 5 m2(x 2 x1) 1 y 2 (22) 5 2}3 (x 2 4) 1 y 1 2 5 2}3 (x 2 4) 1 4 y 1 2 5 2}3 x 1 }3 1 2 y 5 2}3 x 2 }3 212 1 You know the slope and a point on the line, so use pointslope form with either given point to write an equation of the line. Choose (x1, y1) 5 (22, 5). y 2 y1 5 m(x 2 x1) y 2 5 5 22(x 2 (22)) y 2 5 5 22(x 1 2) y 2 5 5 22x 2 4 y 5 22x 1 1 7. The line passes through (x1, y1) 5 (6, 1) and (x2, y2) 5 (23, 28). Find its slope. y2 2 y1 28 2 1 29 m5} 5} 5} 51 23 2 6 29 x 2x 2 y 5 2}4 x 1 }2 27 2 5 m5} 5} 5} 5 22 x 2x 6 4 2 (22) 1 You know the slope and a point on the line, so use pointslope form with either given point to write an equation of the line. Choose (x1, y1) 5 (23, 28). y 2 y1 5 m(x 2 x1) y 2 (28) 5 11 x 2 (23) 2 y 1 8 5 1(x 1 3) y185x13 y5x25 8. The line passes through (x1, y1) 5 (21, 2) and (x2, y2) 5 (10, 0). Find its slope. y2 2 y1 022 22 2 m5} 5} 5} 5 2} x 2x 11 11 10 2 (21) 2 1 You know the slope and a point on the line, so use pointslope form with either given point to write an equation of the line. Choose (x1, y1) 5 (21, 2). y 2 y1 5 m(x 2 x1) 2 y 2 2 5 2} (x 2 (21)) 11 2 (x 1 1) y 2 2 5 2} 11 2 2 2 20 y 2 2 5 2} x2} 11 11 x1} y 5 2} 11 11 9. Let x represent the time (in years) since 1993 and let y represent the number of participants (in millions). The initial value is 3.42. The rate of change is the slope m. Use (x1, y1) 5 (0, 3.42) and (x2, y2) 5 (10, 3.99). y2 2 y1 3.99 2 3.42 0.57 5} 5} 5 0.057 m5} x 2x 10 2 0 10 2 1 Participants 5 Initial 1 Rate of + change number y 5 3.42 1 0.057 + x In slope-intercept form, a linear model is y 5 0.057x 1 3.42 64 Algebra 2 Worked-Out Solution Key Years since 1993 Copyright © by McDougal Littell, a division of Houghton Mifflin Company. 2.4 Guided Practice (pp. 98–101) Chapter 2, 10. Company A song price continued Songs from + Company A Company B 1 song price + 10. Let (x1, y1) 5 (3, 21) and m 5 23. y 2 y1 5 m(x 2 x1) Songs from Company B 5 Your budget 0.69 + x 1 0.89 + y 5 30 An equation for this situation is 0.69x 1 0.89y 5 30. y 1 1 5 23(x 2 3) y 1 1 5 23x 1 9 y 5 23x 1 8 11. Let (x1, y1) 5 (24, 3) and m 5 2. 2.4 Exercises (pp. 101–104) y 2 y1 5 m(x 2 x1) Skill Practice y 2 3 5 2(x 2 (24)) 1. The linear equation 6x 1 8y 5 72 is written in standard form. 2. With two points on a line a slope can be calculated. Once the slope is calculated, simply substitute it in the point-slope form along with either one of the two points. 3. The slope is m 5 0 and the y-intercept is b 5 2. Use slope-intercept form to write an equation of the line. y 5 mx 1 b y 5 0x 1 2 y52 4. The slope is m 5 3 and the y-intercept is b 5 24. Use y 2 3 5 2(x 1 4) y 2 3 5 2x 1 8 y 5 2x 1 11 12. Let (x1, y1) 5 (25, 26) and m 5 0. y 2 y1 5 m(x 2 x1) y 2 (26) 5 0(x 2 (25)) y 1 6 5 0(x 1 5) y1650 y 5 26 13. Let (x1, y1) 5 (8, 13) and m 529. slope-intercept form to write in equation of the line. y 2 y1 5 m(x 2 x1) y 5 mx 1 b y 2 13 5 29(x 2 8) y 5 3x 1 (24) y 2 13 5 29x 1 72 y 5 3x 2 4 5. The slope is m 5 6 and the y-intercept is b 5 0. Use slope-intercept form to write an equation of the line. Copyright © by McDougal Littell, a division of Houghton Mifflin Company. y 2 (21) 5 23(x 2 3) y 5 29x 1 85 3 14. Let (x1, y1) 5 (12, 0) and m 5 }. 4 y 5 mx 1 b y 2 y1 5 m(x 2 x1) y 5 6x 1 0 y 2 0 5 }4 (x 2 12) y 5 6x 3 3 2 6. The slope is m 5 } and the y-intercept is b 5 4. Use 3 slope-intercept form to write an equation of the line. y 5 mx 1 b 2 y 5 }3 x 1 4 5 7. The slope is m 5 2} and the y-intercept is b 5 7. Use 4 slope-intercept form to write an equation of the line. y 5 mx 1 b 5 y 5 2}4 x 1 7 8. The slope is m 5 25 and the y-intercept is b 5 21. Use slope-intercept form to write an equation of the line. y 5 mx 1 b y 5 }4 x 2 9 4 15. Let (x1, y1) 5 (7, 23) and m 5 2}. 7 y 2 y1 5 m(x 2 x1) 4 y 2 (23) 5 2}7 (x 2 7) 4 y 1 3 5 2}7 (x 2 7) 4 y 1 3 5 2}7 x 1 4 4 y 5 2}7 x 1 1 3 16. Let (x1, y1) 5 (24, 2) and m 5 }. 2 y 2 y1 5 m(x 2 x1) 3 y 5 25x 1 (21) y 2 2 5 }2 (x 2 (24)) y 5 25x 2 1 y 2 2 5 }2 (x 1 4) 9. Let (x1, y1) 5 (0, 22) and m 5 4. y 2 y1 5 m(x 2 x1) y 2 (22) 5 4(x 2 0) 3 3 y 2 2 5 }2 x 1 6 3 y 5 }2 x 1 8 y 1 2 5 4x y 5 4x 2 2 Algebra 2 Worked-Out Solution Key 65 continued 1 17. Let (x1, y1) 5 (9, 25) and m 5 2}. 3 y 2 y1 5 m(x 2 x1) 1 y 2 (25) 5 2}3 (x 2 9) 1 y 1 5 5 2}3 (x 2 9) 1 23. (4, 1); perpendicular to y 5 } x 1 3. 3 1 m1 5 }3 1 5 23 m2 5 2} m 1 Let (x1, y1) 5 (4, 1). 1 y 2 y1 5 m2(x 2 x1) 1 y 2 1 5 23x 1 12 y 1 5 5 2}3 x 1 3 y 5 2}3 x 2 2 18. The error was made when substituting for x1. It should be 24, not 4. y 2 y1 5 m(x 2 x1) y 2 2 5 3(x 2 (24)) y 2 2 5 3(x 1 4) y 2 2 5 3x 1 12 y 5 3x 1 14 19. The x1 and y1 values were transposed. y 2 y1 5 m(x 2 x1) y 2 1 5 22(x 2 5) y 2 1 5 22x 1 10 y 5 22x 1 11 20. (23, 25); parallel to y 5 24x 1 1. m1 5 24 m 2 5 m1 5 24 Let (x1, y1) 5 (23, 25). y 2 y1 5 m 2(x 2 x1) y 2 (25) 5 24(x 2 (23)) y 2 1 5 23(x 2 4) y 5 23x 1 13 24. (26, 2); perpendicular to y 5 22. m1 5 0 1 m2 5 2} 5 undefined m 1 Because the slope is undefined, you know the line is vertical. Because it passes through (26, 2), its equation is x 5 26. 25. (3, 21); perpendicular to y 5 4x 1 1. m1 5 4 1 1 Let (x1, y1) 5 (3, 21). y 2 y1 5 m2(x 2 x1) 1 y 2 (21) 5 2}4 (x 2 3) 1 y 1 1 5 2}4 (x 2 3) y 5 24x 2 17 21. (7, 1); parellel to y 5 2x 1 3. m1 5 21 m 2 5 m1 5 21 Let (x1, y1) 5 (7, 1). y 2 y1 5 m2(x 2 x1) y 2 1 5 21(x 2 7) y 2 1 5 2x 1 7 y 5 2x 1 8 22. (2, 8); parallel to y 5 3x 2 2. m1 5 3 m 2 5 m1 5 3 Let (x1, y1) 5 (2, 8). y 2 y1 5 m2(x 2 x1) y 2 8 5 3(x 2 2) y 2 8 5 3x 2 6 y 5 3x 1 2 66 Algebra 2 Worked-Out Solution Key 1 3 1 1 y 1 1 5 2}4 x 1 }4 y 1 5 5 24(x 1 3) y 1 5 5 24x 2 12 1 m2 5 2} 5 2}4 m y 5 2}4 x 2 }4 26. C; (1, 4); perpendicular to y 5 2x 2 3. m1 5 2 1 1 m2 5 2} 5 2}2 m 1 Let (x1, y1) 5 (1, 4). y 2 y1 5 m2(x 2 x1) 1 y 2 4 5 2}2 (x 2 1) 1 1 1 9 y 2 4 5 2}2 x 1 }2 y 5 2}2 x 1 }2 27. From the graph, you can see that the slope is m 5 22. Choose (x1, y1) 5 (3, 0). y 2 y1 5 m(x 2 x1) y 2 0 5 22(x 2 3) y 5 22x 1 6 Copyright © by McDougal Littell, a division of Houghton Mifflin Company. Chapter 2, Chapter 2, continued 28. From the graph, you can see that the slope is m 5 5. Choose (x1, y1) 5 (4, 4). y2 2 y1 527 2 22 y 2 y1 5 m(x 2 x1) 5} 5} 5 2}3 m5} x 2x 320 3 y 2 4 5 5(x 2 4) Choose (x1, y1) 5 (3, 5). 2 y 2 4 5 5x 2 20 1 29. From the graph, you can see that the slope is m 5 2}. 4 Choose (x1, y1) 5 (21, 5). 2 2 1 y 5 2}3 x 1 7 y 2 5 5 2}4 (x 2 (21)) 34. Let (x1, y1) 5 (21, 2) and (x2, y2) 5 (3, 24). 1 y 2 5 5 2}4 (x 1 1) y2 2 y1 2 6 y 2 2 5 2}2 (x 1 1) 1 Choose (x1, y1) 5 (2, 9). y 2 9 5 2(x 2 2) 3 3 1 y 5 2}2 x 1 }2 y 2 9 5 2x 2 4 35. Let (x1, y1) 5 (25, 22) and (x2, y2) 5 (23, 8). y 5 2x 1 5 31. Let (x1, y1) 5 (4, 21) and (x2, y2) 5 (6, 27). 26 5} 5} 5 23 m5} x 2x 624 2 2 3 y 2 2 = 2}2 x 2 }2 y 2 y1 5 m(x 2 x1) 27 2 (21) 3 y 2 2 5 2}2 (x 2 (21)) 3 5} 5 }3 5 2 m5} x 2x 2 2 (21) y2 2 y1 1 Choose (x1, y1) 5 (4, 21). y 2 y1 5 m(x 2 x1) y 2 (21) 5 23(x 2 4) y 1 1 5 23(x 2 4) y 1 1 5 23x 1 12 y 5 23x 1 11 y2 2 y1 8 2 (22) 2 1 Choose (x1, y1) 5 (25, 22). y 2 y1 5 m(x 2 x1) y 2 (22) 5 5(x 2 (25)) y 1 2 5 5(x 1 5) y 1 2 5 5x 1 25 y 5 5x 1 23 36. Let (x1, y1) 5 (15, 20) and (x2, y2) 5 (212, 29). 29 2 20 2 Choose (x1, y1) 5 (2, 21). y 2 20 5 2}3 (x 2 15) 2 1 y 2 y1 5 m(x 2 x1) 1 y 2 (21) 5 }2 (x 2 2) 1 y 1 1 5 }2 (x 2 2) 1 1 1 Choose (x1, y1) 5 (15, 20). y 2 y1 5 m(x 2 x1) 2 9 5} 5} 5 2}3 m5} x 2x 212 2 15 227 5} 5 }4 5 }2 m5} x 2x 2 2 (22) 21 2 (23) 10 5} 5} 55 m 5} x 2x 2 23 2 (25) y2 2 y1 32. Let (x1, y1) 5 (22, 23) and (x2, y2) 5 (2, 21). y2 2 y1 3 y 2 y1 5 m(x 2 x1) 30. Let (x1, y1) 5 (21, 3) and (x2, y2) 5 (2, 9). 2 26 1 Choose (x1, y1) 5 (21, 2). 19 y 5 2}4 x 1 } 4 923 24 2 2 5} 5} 5 2}2 m5} x 2x 4 3 2 (21) 1 1 y 2 5 5 2}4 x 2 }4 y2 2 y1 2 y 2 5 5 2}3 (x 2 3) y 2 5 5 2}3 x 1 2 y 2 y1 5 m(x 2 x1) 1 1 y 2 y1 5 m(x 2 x1) y 5 5x 2 16 Copyright © by McDougal Littell, a division of Houghton Mifflin Company. 33. Let (x1, y1) 5 (0, 7) and (x2, y2) 5 (3, 5). 1 1 y 2 20 5 2}3 x 1 5 1 y 5 2}3 x 1 25 1 y 1 1 5 }2 x 2 1 1 y 5 }2 x 2 2 Algebra 2 Worked-Out Solution Key 67 continued 37. Let (x1, y1) 5 (3.5, 7) and (x2, y2) 5 (21, 20.5). y2 2 y1 20.5 2 7 13.5 5} 5} 5 23 m5} x 2x 21 2 3.5 24.5 2 1 Choose (x1, y1) 5 (21, 20.5). y 2 y1 5 m(x 2 x1) 4 43. Using m 5 } and (x1, y1) 5 (2, 3): 5 y 2 y1 5 m(x 2 x1) 4 y 2 3 5 }5 (x 2 2) y 2 20.5 5 23(x 2 (21)) 22.6 2 0.9 23.5 m5} 5} 5 21.25 3.4 2 0.6 2.8 Choose (x1, y1) 5 (0.6, 0.9). y 2 y1 5 m(x 2 x1) y 2 0.9 5 21.25(x 2 0.6) y 2 0.9 5 21.25x 1 0.75 y 5 21.25x 1 1.65 39. C; Let (x1, y1) 5 (9, 25) and m 5 26. y 2 y1 5 m(x 2 x1) y 2 (25) 5 26(x 2 9) y 1 5 5 26(x 2 9) y 1 5 5 26x 1 54 y 5 26x 1 49 So, an equation of the line is y 5 26x 1 49. The point (7, 7) is a solution of this equation, so it lies on the line. 40. When m 5 23 and b 5 5: y 5 mx 1 b y 5 23x 1 5 7 24x 1 5y 5 7 44. Let (x1, y1) 5 (21, 3) and (x2, y2) 5 (26, 27). y2 2 y1 27 2 3 210 m5} 5} 5} 52 25 x2 2 x1 26 2 (21) Choose (x1, y1) 5 (21, 3). y 2 y1 5 m(x 2 x1) y 2 3 5 2(x 2 (21)) y 2 3 5 2(x 1 1) y 2 3 5 2x 1 2 y 5 2x 1 5 22x 1 y 5 5 45. Let (x1, y1) 5 (2, 8) and (x2, y2) 5 (24, 16). y2 2 y1 16 2 8 8 4 m5} 5} 5} 5 2}3 x 2x 24 2 2 26 2 1 Choose (x1, y1) 5 (24, 16). y 2 y1 5 m(x 2 x1) 4 y 2 16 5 2}3 (x 2 (24)) 4 y 2 16 5 2}3 (x 1 4) 4 16 4 32 y 2 16 5 2}3 x 2 } 3 y 5 2}3 x 1 } 3 3x 1 y 5 5 41. When m 5 4 and b 5 23: y 5 mx 1 b y 5 4x 1 (23) y 5 4x 2 3 24x 1 y 5 23 3 42. Using m 5 2} and (x1, y1) 5 (4, 27): 2 y 2 y1 5 m(x 2 x1) 3 y 2 (27) 5 2}2 (x 2 4) 3 y 1 7 5 2}2 (x 2 4) 3 y 1 7 5 2}2 x 1 6 3 y 5 2}2 x 2 1 2y 5 23x 2 2 3x 1 2y 5 22 68 4 5y 5 4x 1 7 y 2 20.5 5 23x 2 3 38. Let (x1, y1) 5 (0.6, 0.9) and (x2, y2) 5 (3.4, 22.6). 8 y 5 }5 x 1 }5 y 2 20.5 5 23(x 1 1) y 5 23x 1 17.5 4 y 2 3 5 }5 x 2 }5 Algebra 2 Worked-Out Solution Key 3y 5 24x 1 32 4x 1 3y 5 32 46. a. The line y 5 22 is horizontal. So, a line parallel to it is also horizontal. Because it passes through (3, 4), its equation is y 5 4. b. The line y 5 22 is horizontal. So, a line perpendicular to it is vertical. Because it passes through (3, 4), its equation is x 5 3. c. The line x 5 22 is vertical. So a line parallel to it is also vertical. Because it passes through (3, 4), its equation is x 5 3. d. The line x 5 22 is vertical. So, a line perpendicular to it is horizontal. Because it passes through (3, 4), its equation is y 5 4. Copyright © by McDougal Littell, a division of Houghton Mifflin Company. Chapter 2, Chapter 2, continued 47. Sample answer: 49. The line has an x-intercept at point (a, 0) and a y 5 23x 1 5; m1 5 23, b1 5 5 y-intercept at point (0, b). y 5 2x 1 1; m2 5 2, b2 5 1 Let (x1, y1) 5 (a, 0) and (x2, y2) 5 (0, b). 1 y2 2 y1 Because m2 Þ 2} m , the two lines are not perpendicular. To form a right triangle line * must be perpendicular to either y 5 23x 1 5 or y 5 2x 1 1. Because the triangle can be any size, line * can be placed anywhere except the point of intersection of y 5 23x 1 5 and y 5 2x 1 1. One possible line * could be perpendicular to y 5 23x 1 5 through the point (0, 0). 1 1 2 b y 2 0 5 2} (x 2 a) a b y 5 2}a x 1 b bx a y x }1}51 b a }1y5b 1 1 y 2 0 5 }3 (x 2 0) 1 Problem Solving y 5 }3 x y 5 23x 1 5 50. Let x represent the time (in months) since buying the car y and let y represent the total cost (in dollars). Total Initial Rate of Months Cost 5 number 1 change + from now y 5 2x 1 1 y5 1 x 3 b 1 Choose (x1, y1) 5 (a, 0). m1 5 23, m2 5 2} 5 2} 5 }3 m 23 1 b20 5} 5 2}a m5} x 2x 02a 1 1 y x 1 5 6500 + 350 1 x In slope-intercept form, a linear model is y 5 350x 1 6500. 48. Begin by finding the slope of each line. A1x 1 B1 y 5 C1 A2 x 1 B2 y 5 C2 B1y 5 C1 2 A1x B2 y 5 C2 2 A2x C1 A1 C2 A2 1 1 2 2 y5} 2} x B B Copyright © by McDougal Littell, a division of Houghton Mifflin Company. 51. A1 1 x2} y 5 2} B A 1 n 1 2 A1 A2 1 2 2 A2 m1 5 2} B m2 5 2} B 1 50 1 15 + t 52. Area of plot 5 25 + 16 5 400 square feet. C2 y 5 2} x2} B A 2 5 In slope-intercept form, a linear model is n 5 15t 1 50. y5} 2} x B B C1 Number Initial Rate of Years of houses 5 number 1 change + from now Number of Space of Space of tomato plant + tomato plants 1 pepper plant + 2 a. If A1 x 1 B1 y 5 C1 and A2 x 1 B2 y 5 C2 are parallel, then m2 5 m1. Using the slopes you found above, substitute for m1 and m2 as shown. m1 5 m2 A1 A2 1 2 2} 5 2} B B 2A1B2 5 2A2 B1 A1B2 5 A2B1 b. If A1x 1 B1 y 5 C1 and A2 x 1 B2 y 5 C2 are 1 . Using the slopes you perpendicular, then m2 5 2} m1 found above, substitute for m1 and m2 as shown. Number of Area of pepper plants 5 plot 8 + x 1 5 + y 5 400 An equation for this situation is 8x 1 5y 5 400. Let x 5 15. 8(15) 1 5y 5 400 120 1 5y 5 400 5y 5 280 y 5 56 If you grow 15 tomato plants, you can grow 56 pepper plants. 21 m2 5 } m 1 A2 21 2} 5} B A 2 1 2}B 2 1 1 A2 B1 2 1 2} 5} B A 2A1A2 5 B1B2 0 5 A1A2 1 B1B2 Algebra 2 Worked-Out Solution Key 69 Chapter 2, Monthly Initial Rate of Years 5 number 1 change + since 1994 cost General Student General admission + tickets 1 admission + Student tickets y 5 21.62 1 1.66x In 2010, x 5 16. Total in 5 ticket sales y 5 21.62 1 1.66(16) y 5 21.62 1 26.56 15 + x 1 9 + y 5 4500 y 5 48.18 An equation that models this situation is 15x 1 9y 5 4500. Start at 200 on the genral admission axis and move up until you reach the graph. Then find the point that it corresponds to on the student admission axis. If 200 general admission tickets were sold, about 167 student tickets were sold. The predicted average monthly cost for basic cable in 2010 is $48.18. 56. Let y represent the tire’s pressure (in pounds per square inch) and let x represent the air temperature (8F). 1 psi 108F Rate of change 5 } m 5 0.1 y 500 Let (x1, y1) 5 (55, 30). y 2 y1 5 m(x 2 x1) 400 y 2 30 5 0.1(x 2 55) 300 y 2 30 5 0.1x 2 5.5 200 y 5 0.1x 1 24.5 Width of 1 rectangle rectangle 100 0 54. a. 57. a. Length of 0 50 100 150 200 x * 1 w 5 12 b. w 5 2* 1 12 w 5 Budget 16 21.75 + x 1 17 + y 5 86,000 12 21.75x 1 17y 5 86,000 8 b. Let y 5 2500. 21.75y 1 17(2500) 5 86,000 21.75x 1 42,500 5 86,000 21.75y 5 43,500 x 5 2000 2000 square feet can be rented in the first building. c. Let x 5 y. 21.75y 1 17y 5 86,000 38.75y 5 86,000 y ø 2219 x 1 y ø 4438 The total number of square feet that can be rented is about 4438 square feet. 55. Let y represent the average monthly cost (in dollars), and let x represent the number of years since 1994. Use (x1, y1) 5 (0, 21.62) and (x2, y2) 5 (10, 38.23). Rate of change: y2 2 y1 38.23 2 21.62 10 2 0 16.61 10 m 5 } 5 }} 5 } ø 1.66 x2 2 x1 70 Perimeter 2* 1 2w 5 24 First Number of Second building rate + square feet 1 building rate + Number of square feet 5 Algebra 2 Worked-Out Solution Key 4 0 c. 58. 0 4 8 12 16 w 6 5 4 3 2 * 6 7 8 9 10 Total Money raised 5 Total Total Hours fixed 1 amount + danced amount per hour y 5 (15 1 35 1 20) 1 (4 1 8 1 3) + x y 5 70 1 15x Mixed Review for TAKS 59. C; x 2 30 1 125 2 22 5 180 x 5 107 John had $107 in his bank account at the beginning of the week. Copyright © by McDougal Littell, a division of Houghton Mifflin Company. 53. continued Chapter 2, continued 60. G; Using slope-intercept form, use the point (6, 13). measure of interior angles 5 180(n 2 2) Problem Solving Workshop 2.4 (p. 105) 1. Find the slope of the line. 16 5 b Choose the point (2, 15). Substitute m 5 4 and (x, y) 5 (2, 5) into the slope-intercept form and solve for b. y 5 mx 1 b 15 5 4(2) 1 b 15 5 8 1 b 75b Substitute m and b into the slope-intercept form. y 5 4x 1 7 change in calories per hour 2. Average rate of change 5 }} change in pounds 600 2 420 5} 172 2 120 180 52 5} 45 calories per hour per pound 5} 13 Using point-slope form, let (x1, y1) 5 (120, 420). y 2 y1 5 m(x 2 x1) 45 y 2 420 5 } (x 2 120) 13 Copyright © by McDougal Littell, a division of Houghton Mifflin Company. 5400 x2} y 2 420 5 } 13 13 45 1 13 5 2}2 (6) 1 b 13 5 23 1 b 35 2 15 20 m5} 5} 54 5 722 45 y 5 mx 1 b 60 1 y 5 2}2 x 1 16 9.5 2 8 1.5 4. m 5 } 5 } 5 0.75 624 2 Using point-slope form, choose (x1, y1) 5 (4, 8). y 2 y1 5 m(x 2 x1) y 2 8 5 0.75(x 2 4) y 2 8 5 0.75x 2 3 y 5 0.75x 1 5 Using slope-intercept form, use the point (6, 9.5). y 5 mx 1 b 9.5 5 0.75(6) 1 b 9.5 5 4.5 1 b 55b y 5 0.75x 1 5 3500 2 2600 900 225 5. m 5 } 5 } 5 } 7 54 2 26 28 Using point-slope form, choose (x1, y1) 5 (54, 3500). y 2 y1 5 m(x 2 x1) 225 y 2 3500 5 } (x 2 54) 7 x1} y5} 13 13 Using slope-intercept form, use the point (172, 600). y 5 mx 1 b 45 (172) 1 b 600 5 } 13 7740 1b 600 5 } 13 60 13 }5b 225 12,150 225 12,350 x2} y 2 3500 5 } 7 7 x1} y5} 7 7 Using slope-intercept form, use the point (26, 2600). y 5 mx 1 b 225 (26) 1 b 2600 5 } 7 5850 1b 2600 5 } 7 12,350 7 }5b 45 60 x1} y5} 13 13 3. (6, 13), (20, 6) 6 2 13 27 1 m5} 5} 5 2}2 20 2 6 14 Using point-slope form, choose (x1, y1) 5 (20, 6). y 2 y1 5 m(x 2 x1) 225 12,350 x1} y5} 7 7 6. y 5 mx 1 b y1 5 mx1 1 b y1 2 mx1 5 b b 5 y1 2 mx1 1 y 2 6 5 2}2 (x 2 20) 1 y 2 6 5 2}2 x 1 10 1 y 5 2}2 x 1 16 Algebra 2 Worked-Out Solution Key 71 Chapter 2, continued Mixed Review for TEKS (p.106) 3. y 5 ax 1. B; v 5 50,000 1 1200t 4. y 5 ax 3 5 a(5) 22 5 a(6) 3 5 2}3 5 a 1 }5a 2. J; 3 1 So, y 5 2}3x. So, y 5 }5x. 1 4 } y 2 3x 5 5 y 1 4 } y 5 3x 1 5 y 1 y 5 12x 1 20 1 x 22 x 21 The slope is 12. 3. A; 5. If the radius is 0.6 inches, then the diameter is x 1 3y 5 12 2(0.6) 5 1.2 inches. When d 5 1.2: 3y 5 2x 1 12 d 5 0.0625t 1 y 5 2}3 x 1 4 1.2 5 0.0625t 19.2 5 t 1 The slope must be 2}3 and the intercept must be negative. 1 y 5 2}3x 2 4 4. H; 227,818 2 219,478 2000 2 1990 }} 5 834 Tooth length, t (cm) 1.8 2.4 Body mass, m (kg) 80 220 375 730 1690 3195 5. D; } ø 92 730 3.6 } ø 360 } ø 203 7g 1 4s 5 11,200 3.6 4.7 5.8 220 2.4 } ø 129 375 2.9 1690 4.7 } ø 551 3195 5.8 8x 1 4y 5 512 2.5 Exercises (pp. 109 –111) 3 023 7. slope of given line: } 5 2} 5 520 Skill Practice 5 slope of > line: }3 5 1.67 1. The constant of variation is the nonzero constant a in the equation y 5 ax, for two variables x and y that vary directly. Lesson 2.5 2. In a table of ordered pairs (x, y), find the ratio of y to x 2.5 Guided Practice (pp. 107–109) 2. y 5 ax for each term in the table. If the ratios are approximately equal, the data show direct variation. y 5 ax 4 5 a(27) 29 5 a(3) 3. y 5 ax 4 2}7 5 a 23 5 a 4. 6 5 a(2) 12 5 a(23) 35a 4 So, y 5 23x. So, y 5 2}7x. y y y 5 ax 24 5 a So, y 5 3x. So, y 5 24x. y 3 y 2 22 21 72 Algebra 2 Worked-Out Solution Key x 1 x 21 x 21 x Copyright © by McDougal Littell, a division of Houghton Mifflin Company. Because the ratios are not approximately equal, tooth length and body mass do not show direct variation. 6. F; 3 2.9 80 1.8 } ø 44 The population increases by about 834 people per year. 1. The hailstone has been forming for 19.2 minutes. 6.