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Simplifying Radicals ALGEBRA 1 LESSON 11-1 (For help, go to Lessons 8-3 and 10-3.) Complete each equation. 1. a 3 = a2 • a 2. b 7 = b6 • b 3. c 6 = c3 • c 4. d 8 = d4 • d Find the value of each expression. 5. 4 8. 49 6. 169 11-1 7. 25 Simplifying Radicals ALGEBRA 1 LESSON 11-1 Solutions 1. a3 = a(2 + 1) = a2 • a1 2. b7 = b(6 + 1) = b6 • b1 3. c6 = c(3 + 3) = c3 • c3 4. d8 = d(4 + 4) = d4 • d4 5. 4=2 6. 169 = 13 7. 25 = 5 8. 49 = 7 11-1 Simplifying Radicals ALGEBRA 1 LESSON 11-1 Simplify 243 = = = 9 81 • 3 81 • 3 243. 81 is a perfect square and a factor of 243. 3 Use the Multiplication Property of Square Roots. Simplify 11-1 81. Simplifying Radicals ALGEBRA 1 LESSON 11-1 Simplify 28x7 = = 4x6 • 7x 4x6 • = 2x3 7x 28x7. 4x6 is a perfect square and a factor of 28x7. 7x Use the Multiplication Property of Square Roots. Simplify 11-1 4x6. Simplifying Radicals ALGEBRA 1 LESSON 11-1 Simplify each radical expression. a. 12 • 12 • 32 32 = 12 • 32 Use the Multiplication Property of Square Roots. = 384 Simplify under the radical. = 64 • 6 64 is a perfect square and a factor of 384. = 64 • = 8 6 6 Use the Multiplication Property of Square Roots. Simplify 11-1 64. Simplifying Radicals ALGEBRA 1 LESSON 11-1 (continued) b. 7 5x • 3 8x 7 5x • 3 8x = 21 40x2 Multiply the whole numbers and use the Multiplication Property of Square Roots. = 21 4x2 • 10 factor of 40x2. 4x2 is a perfect square and a = 21 4x2 • 10 Square Roots. Use the Multiplication Property of = 21 • 2x Simplify = 42x 10 10 Simplify. 11-1 4x2. Simplifying Radicals ALGEBRA 1 LESSON 11-1 Suppose you are looking out a fourth floor window 54 ft above the ground. Use the formula d = 1.5h to estimate the distance you can see to the horizon. d = 1.5h = 1.5 • 54 Substitute 54 for h. = 81 Multiply. =9 Simplify 81. The distance you can see is 9 miles. 11-1 Simplifying Radicals ALGEBRA 1 LESSON 11-1 Simplify each radical expression. a. b. 13 64 13 = 64 13 64 Use the Division Property of Square Roots. = 13 8 Simplify 49 = x4 49 x4 Use the Division Property of Square Roots. 64. 49 x4 7 = x2 Simplify 49 and 11-1 x4. Simplifying Radicals ALGEBRA 1 LESSON 11-1 Simplify each radical expression. a. 120 10 120 = 10 12 Divide. = 4•3 4 is a perfect square and a factor of 12. = 4• =2 3 3 Use the Multiplication Property of Square Roots. Simplify 11-1 4. Simplifying Radicals ALGEBRA 1 LESSON 11-1 (continued) b. 75x5 48x 75x5 = 48x 25x4 16 = 25x4 16 = = 25 • 16 5x2 4 Divide the numerator and denominator by 3x. Use the Division Property of Square Roots. x4 Use the Multiplication Property of Square Roots. Simplify 25, 11-1 x4, and 16. Simplifying Radicals ALGEBRA 1 LESSON 11-1 Simplify each radical expression. a. 3 7 3 = 7 = = 3 • 7 7 7 Multiply by 7 7 to make the denominator a perfect square. 3 7 49 3 7 7 Use the Multiplication Property of Square Roots. Simplify 11-1 49. Simplifying Radicals ALGEBRA 1 LESSON 11-1 (continued) Simplify the radical expression. b. 11 12x3 11 = 12x3 11 • 12x3 3x 3x = 33x 36x4 Use the Multiplication Property of Square Roots. = 33x 6x2 Simplify Multiply by 3x to make the denominator a 3x perfect square. 11-1 36x4. Simplifying Radicals ALGEBRA 1 LESSON 11-1 pages 581–583 Exercises 1. 10 2 3. 5 3 3 13. 20 2 2. 7 12. –4b2 27. 17 mi 14. 18 15. 11 2 16. 84 3 5 5. –6 30 18. –30 6. 40 5 19. 6n 2 7. 2n 7 20. 14t 2 21. 3x2 17 8. 6b2 9. 6x 10. 2n 11. 2a2 3 21 7 29. 3 3 2 30. 5 2 28. 4. –4 17. 7 26. 12 mi 3 3 31. 2 30 11 32. 5 3a 33. 7 4c 22. 80t 3 3 n 5a 23. 3a3 24. –6a2 25. 3 mi 11-1 2 2 34. 5 3a 7 Simplifying Radicals ALGEBRA 1 LESSON 11-1 35. 2n 2n 9 36. 5 54. Simplest form; radicand has no perfect-square factors other than 1. 21x 7x 47. 2 10n 5n 48. 9 2 4 46. 3 37. 3 2 38. 2 45. 3 39. –2 40. 2x 5 7 49. 2 55. Simplest form; radicand has no perfect-square factors other than 1. 18 • 10 = 180 = 36 • 5 = 6 5 b. Answers may vary. Sample: a = 36, b = 5; a = 9, b = 20 56. a. 3 41. s 50. 2b b 2 42. a 3 51. 55y 2y 43. 3 y 52. not simplest form; radical in the denominator of a fraction 44. 3 2 53. not simplest form; radical in the denominator of a fraction 3 2 2 11-1 57. 30 58. 2 13 59. 3 2 4 Simplifying Radicals ALGEBRA 1 LESSON 11-1 69. –3 ± 3 60. –2 a a2 61. 2 15 62. x y 70. 1 ± 3 64. 4 5 65. 2ab 66. ab2c 3m 67. 4m 68. 8 6a 3a 78. 5 3 50 = 25 • 2 = 25 • 2 = 5 2 b. The radicand has no perfect-square factors other than 1. 72. a. 5b abc seconds 79. C 71. 2 ± 10 y2 63. – 77. 30a4 2 73. Answers may vary. Sample: 12, 27, 48. 74. a. 2 6 in. b. 4.90 in. 80. F 81. B 82. I 83. [2] A = 96 ft2 s = 96 = 16 • 6 = 4 6 ft [1] correct answer, without work shown 84. quadratic; y = 0.2x2 75. 12x 85. exponential; y = 4(2.5)x 76. 10b2 86. linear; y = –4.2x + 7 11-1 Simplifying Radicals ALGEBRA 1 LESSON 11-1 87. 89. 88. 90. 3n2 + 5n + 5 91. 3v2 – v – 9 92. 5t 3 + 8t 2 – 14t – 11 93. –3b2 – 23b – 21 11-1 Simplifying Radicals ALGEBRA 1 LESSON 11-1 Simplify each radical expression. 1. 16 • 8 8 4. 2 a5 2 a a3 2 2. 4 5. 144 3x 15x3 11-1 48 5 5x 3. 12 36 3 3 The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 (For help, go to Lesson 10-4.) Simplify each expression. 2. 92 – 42 3. (3t)2 + (4t)2 4. c2 = 36 5. 24 + b2 = 49 6. a2 + 16 = 65 7. 12 + b2 = 32 8. 80 = c2 9. 100 = a2 + 52 1. 52 + 62 Solve each equation. 11-2 The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 Solutions 1. 52 + 62 = (5 • 5) + (6 • 6) = 25 + 36 = 61 2. 92 – 42 = (9 • 9) – (4 • 4) = 81 – 16 = 65 3. (3t)2 + (4t)2 = (32 • t2) + (42 • t2) = 9t2 + 16t2 = 25t 2 4. c2 = 36 c = ± 36 = ±6 5. 24 + b2 = 49 6. a2 + 16 = 65 b2 = 49 – 24 a2 = 65 – 16 b2 = 25 a2 = 49 b = ± 25 = ±5 a = ± 49 = ±7 7. 12 + b2 = 32 8. 80 = c2 b2 = 32 – 12 c = ± 80 = ± 16 • 5 = ± b2 = 20 b = ± 20 = ± 4 • 5 = ± 4 • 5 = ± 2 5 9. 100 = a2 + 52 100 – 52 = a2 48 = a2 a = ± 48 = ± 16 • 3 = ± 16 • 3 = ± 4 3 11-2 16 • 5 = ±4 5 The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 What is the length of the hypotenuse of this triangle? a2 + b2 = c2 Use the Pythagorean Theorem. 82 + 152 = c2 Substitute 8 for a and 15 for b. 64 + 225 = c2 289 = 17 = c c2 Simplify. Find the principal square root of each side. Simplify. The length of the hypotenuse is 17 m. 11-2 The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 A toy fire truck is near a toy building on a table such that the base of the ladder is 13 cm from the building. The ladder is extended 28 cm to the building. How high above the table is the top of the ladder? Define: Let b = height (in cm) of the ladder from a point 9 cm above the table. Relate: The triangle formed is a right triangle. Use the Pythagorean Theorem. 11-2 The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 (continued) Write: a2 + b2 = c2 132 + b2 = 282 169 + b2 = 784 Substitute. Simplify. b2 = 615 Subtract 169 from each side. b2 = Find the principal square root of each side. b 615 24.8 Use a calculator and round to the nearest tenth. The height to the top of the ladder is 9 cm higher than 24.8 cm, so it is about 33.8 cm from the table. 11-2 The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 Determine whether the given lengths are sides of a right triangle. a. 5 in., 5 in., and 7 in. 52 + 52 72 25 + 25 49 50 =/ 49 Determine whether a2 + b2 = c2, where c is the longest side. Simplify. This triangle is not a right triangle. b. 10 cm, 24 cm, and 26 cm 102 + 242 262 Determine whether a2 + b2 = c2, where c is the longest side. 100 + 576 676 Simplify. 676 = 676 This triangle is a right triangle. 11-2 The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 If two forces pull at right angles to each other, the resultant force is represented as the diagonal of a rectangle, as shown in the diagram. The diagonal forms a right triangle with two of the perpendicular sides of the rectangle. For a 50–lb force and a 120–lb force, the resultant force is 130 lb. Are the forces pulling at right angles to each other? 502 + 1202 2500 + 14,400 1302 Determine whether a2 + b2 = c2 where c is the greatest force. 16,900 16,900 = 16,900 The forces of 50 lb and 120 lb are pulling at right angles to each other. 11-2 The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 pages 587–590 Exercises 12. 0.6 24. no 1. 10 13. 1.2 m 25. yes 2. 25 14. about 15.5 ft 26. 1.5 3. 17 15. about 5.8 km 27. 4 or 0.3 4. 26 16. yes 28. 3 5. 2.5 17. no 29. 6 6. 1 18. no 30. 2.6 7. 4 19. yes 31. 7.0 8. 5 20. no 9. 12 21. yes 32. a. 6 5 ft b. 80.5 ft2 10. 7.1 11. 7.5 22. yes 23. no 11-2 15 33. yes 34. no The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 35. yes 36. yes 37. 4.2 cm 38. 1000 lb 39. 559.9 40. 9.0 44. a. 6.9 ft b. 89.2 ft2 c. 981 watts 45. 12.8 ft 46. a. Answers may vary. Sample: 5, 20, 5 b. 5 units2 47. a. 13.4 ft b. 17.0 ft These lengths could be c. 10.6 ft 2 legs or one leg and d. No; the hypotenuse d must the hypotenuse. be longer than each leg. about 12.8 in. or 6 in. 48. An integer has 2 as a factor; 62 + 82 = 36 + 64 = 100 = 102 the integer is even; if an integer is even, then it has 2 as a factor; true. 5; 12; 7; 41 Answers may vary. Sample: 10, 24, 26 41. 9.7 42. a. b. 43. a. b. c. 11-2 The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 49. A figure is a square; the figure is a rectangle; if a figure is a rectangle then the figure is a square; false. 54. 10 50. You are in Brazil; you are south of the equator; if you are south of the equator you are in Brazil; false. 58. a. 51. An angle is a right angle; its measure is 90°; if the measure of an angle is 90°, then it is a right angle; true. 52. 52 55. 4 3 56. 5 57. n2 + (n + 1)2 = (n + 2)2; 3, 4, 5 b. 74 59. a. a2 + 2ab + b2 b. c2 c. ab 2 1 d. (a + b)2 = c2 + 4 ab ; a2 + 2ab + b2 = 2 2 2 2 2ab + c ; a + b = c2 e. This equation is the same as the Pythagorean Theorem. units2 53. 6 in. 11-2 The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 60. D 69. 2b2 10b 61. H 70. 63. C 2 2x2 71. 2 6v v4 64. A 72. 3 and 4 65. [2] It is a right triangle. Substitute 17, the longest side, for c and substitute the other lengths for a and b in the Pythagorean Theorem 82 + 152 172 –> 64 + 225 289 –> 289 = 289 [1] incorrect equation OR incorrect explanation 73. 8 and 9 62. B 66. 4 67. 3 75. 11 and 12 76. rational 77. irrational 78. irrational 6 3 68. 5 74. –8 and –7 79. rational 2 11-2 The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 80. 8x2 – 4x 81. 12a2 + 15a 82. 18t 3 – 6t 2 83. –10p4 + 26p3 84. 15b3 + 5b2 – 45b 85. –7v4 + 42v2 – 7v 11-2 The Pythagorean Theorem ALGEBRA 1 LESSON 11-2 1. Find the missing length to the nearest tenth. 2. Find the missing length to the nearest tenth. 16.6 5.7 3. A triangle has sides of lengths 12 in., 14 in., and 16 in. Is the triangle a right triangle? no 4. A triangular flag is attached to a post. The bottom of the flag is 48 in. above the ground. How far from the ground is the top of the flag? 57 in. 11-2 The Distance and Midpoint Formulas ALGEBRA 1 LESSON 11-3 (For help, go to Lessons 11-2 and 2-7.) Find the length of the hypotenuse with the given leg lengths. If necessary, round to the nearest tenth. 1. a = 3, b = 4 2. a = 2, b = 5 3. a = 3, b = 8 4. a = 7, b = 5 For each set of values, find the mean. 5. x1 = 6, x2 = 14 6. y1 = –4, y2 = 8 7. x1 = –5, x2 = –7 8. y1 = –10, y2 = –3 11-3 The Distance and Midpoint Formulas ALGEBRA 1 LESSON 11-3 Solutions 1. a2 + b2 = c2 2. 32 + 42 = c2 9 + 16 = c2 25 = c2 c = 25 = 5 The length of the hypotenuse is 5. 3. a2 + b2 = c2 4. 32 + 82 = c2 9 + 64 = c2 73 = c2 c = 73 8.5 The length of the hypotenuse is about 8.5. 6 + 14 20 = 2 = 10 2 6. mean = –4 + 8 = 4 = 2 2 2 5. mean = a2 + b2 = c2 22 + 52 = c2 4 + 25 = c2 29 = c2 c = 29 5.4 The length of the hypotenuse is about 5.4. a2 + b2 = c2 72 + 52 = c2 49 + 25 = c2 74 = c2 c = 74 8.6 The length of the hypotenuse is about 8.6. –12 –5 + (–7) = 2 = –6 2 8. mean = –10 + (–3) = –13 = –6.5 2 2 7. mean = 11-3 The Distance and Midpoint Formulas ALGEBRA 1 LESSON 11-3 Find the distance between F(6, –9) and G(9, –4). d= ( x2 – x1)2 + (y2 – y1)2 Use the distance formula. d= (9 – 6)2 + [–4 – (–9)]2 Substitute (9, –4) for (x2, y2) and (6, –9) for (x1, y1). d= 32 + 52 Simplify within parentheses. d= 34 Simplify to find the exact distance. d 5.8 Use a calculator. Round to the nearest tenth. The distance between F and G is about 5.8 units. 11-3 The Distance and Midpoint Formulas ALGEBRA 1 LESSON 11-3 Find the exact lengths of each side of quadrilateral EFGH. Then find the perimeter to the nearest tenth. The perimeter = EF = = = = [4 – (–1)]2 + (3 + 5)2 52 + (–2)2 25 + 4 29 FG = = = = (3 – 4)2 + (–2 – 3)2 (–1)2 + (–5)2 1 + 25 26 GH = |–2 – 3| = 5 EH = = = = [–2 – (–1)]2 + (–2 – 5)2 (–1)2 + (–7)2 1 + 49 50 29 + 26 + 5 + 50 11-3 22.6 units. The Distance and Midpoint Formulas ALGEBRA 1 LESSON 11-3 Find the midpoint of CD. x1+ x2 y1+ y2 2 , 2 (–3) + 5 7 + 2 = , 2 2 2 9 = 2,2 Substitute (–3, 7) for (x1, y1) and (5, 2) for (x2, y2). Simplify each numerator. 1 Write 9 as a mixed = 1, 42 2 number. 1 The midpoint of CD is M 1, 4 2 . 11-3 The Distance and Midpoint Formulas ALGEBRA 1 LESSON 11-3 A circle is drawn on a coordinate plane. The endpoints of the diameter are (–3, 5) and (4, –3). What are the coordinates of the center of the circle? x1+ x2 y1+ y2 2 , 2 = (–3) + 4 5 + (–3) , 2 2 1 2 = 2,2 1 = 2, 1 1 The center of the circle is at 2 , 1 . 11-3 Substitute (–3, 5) for (x1, y1) and (4, –3) for (x2, y2). The Distance and Midpoint Formulas ALGEBRA 1 LESSON 11-3 pages 594–597 Exercises 22. MN 1. 15 12. (–2, 3) 2. 14 13. 5, –5 1 3. 10 14. – 2 1 , –1 5. 2.8 23. JK = 4; KL 24. TU VT 2 2 4. 11.3 5.1; NP 15. (–4, 4) 2.2; LJ 7.6; UV 13.3 25. a. OR = 3.6; MP = 5 16.5; 29, ST = b. 5 ; 5 2 2 6. 8.1 16. 8 1 , –9 7. 16 17. 10.6 8. 21.6 18. 9.4 9. (1, 6) 19. AB 4.1; BC 3.2; AC = 5 10. (–1, 12) 20. DE 6.1; EF 3.6; DF 5.7 11. (0, 0) 3.2; ST 5.7; RT 5.1 c. yes 2 21. RS 26. a. (20, 80), (–40, 30) b. 78.1 ft 11-3 3.6 c. (–10, 55) or 55L10 29 The Distance and Midpoint Formulas ALGEBRA 1 LESSON 11-3 27. Answers may vary. Sample: Suppose you have (1, 1) and (4, –3). To find the distance, square the difference between the x-coordinates. Square the difference between the y-coordinates. Find the sum and take the square root, so 9 + 16 = 5. To find the midpoint, add x-coordinates together and divide by 2. Repeat for y-coordinates. So, 1 + 4 1 – 3 5 , –1 , = 2 2 2 28. Check students’ work. 29. a. b. (0, – 1 ) 2 c. one half mile south of the substation 30. about 9.5 km apart 31. a. 38.1 mi b. 20 mi, 21.2 mi c. 15 min, 16 min 32. Yes; all sides are congruent. 11-3 The Distance and Midpoint Formulas ALGEBRA 1 LESSON 11-3 33. a. R(–27, –5) b. PR = 13 3.6 RQ = 13 3.6 34. a. about 4.3 mi b. about 17.4 mi 39. a. y = – 4 x + 1 3 b. 9 , – 7 5 5 c. 3 47. 11.7 48. 11.2 49. 10.2 50. 26 35. a. M(–0.5, 3); N(5.5, 3) b. They are equal. 40. Yes; the distance from each point to the center is 5. 36. (x, y) 41. 19.1 53. 4.1 37. They are opposites. 42. 5 54. –14, 14 38. a. 5 units b. Answers may vary. Sample: (5, 0), (0, 5), (–5, 0), (0, –5), (3, –4), (–3, 4), (–3, –4) c. circle 43. 22.9 55. –10, 10 44. –5 56. no solution 45. 3 57. –5, 5 46. –8 58. no solution 11-3 51. 3.5 52. 8 The Distance and Midpoint Formulas ALGEBRA 1 LESSON 11-3 59. – 2 , 2 3 3 60. k2 + 11k + 24 61. v2 + 2v – 35 62. 2p2 – 17p – 9 63. 8w4 + 19w2 + 11 64. 7t3 + 5t2 + 5t – 2 65. 6c3 – 27c2 + 33c + 24 11-3 The Distance and Midpoint Formulas ALGEBRA 1 LESSON 11-3 1. Find the distance between M(2, –1) and N(–4, 3) to the nearest tenth. 7.2 2. Find the distance between P(–2.5, 3.5) and R(–7.5, 8.5) to the nearest tenth. 7.1 3. Find the midpoint of AB, A(3, 6) and B(0, 2). (112 , 4) 4. Find the midpoint of CD, C(6, –4) and D(12, –2). (9, – 3) 5. Find the perimeter of triangle RST to the nearest tenth of a unit. 9.5 units 11-3 Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 (For help, go to Lesson 11-1.) Simplify each radical expression. 1. 52 2. 200 3. 4 54 Rationalize each denominator. 5. 3 11 6. 5 8 7. 11-4 15 2x 4. 125x2 Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 Solutions 1. 52 = 4 • 13 = 4 • 13 = 2 13 2. 200 = 100 • 2 = 100 • 2 = 10 2 3. 4 54 = 4 9 • 6 = 4 • 9 • 6 = 4 • 3 • 6 = 12 4. 125x2 = 5. 3 = 11 5 = 8 15 = 2x 6. 7. 25 • 5 • x2 = 3 • 11 5 • 8 15 • 2x 11 = 11 8 = 8 2x = 2x 25 • 5 • 3 • 11 = 33 11 • 11 11 5•8 = 40 = 8 8•8 15 • 2x = 30x 2x 2x • 2x 11-4 x2 = 5x 6 5 4 • 10 = 8 4• 8 10 = 2 10 = 8 10 4 Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 Simplify 4 4 3+ 3=4 3+1 = (4 + 1) =5 3 3+ 3 3 3. Both terms contain 3. Use the Distributive Property to combine like radicals. Simplify. 11-4 Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 5– Simplify 8 8 5– 45. 45 = 8 5+ 9•5 9 is a perfect square and a factor of 45. =8 5– 9• =8 5–3 Use the Multiplication Property of Square Roots. Simplify 9. = (8 – 3) =5 5 5 5 5 Use the Distributive Property to combine like terms. Simplify. 11-4 Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 Simplify 5( 8 + 9) = = =2 5( 40 + 9 4• 8 + 9). 5 10 + 9 10 + 9 Use the Distributive Property. 5 Use the Multiplication Property of Square Roots. 5 Simplify. 11-4 Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 Simplify ( ( 6–3 6 – 3 21)( 6 + 21) = 36 + 126 – 3 126 – 3 21)( 441 6+ 21). Use FOIL. =6–2 126 – 3(21) Combine like radicals and simplify 36 and 441. =6–2 9 • 14 – 63 9 is a perfect square factor of 126. =6–2 9• Use the Multiplication Property of Square Roots. =6–6 14 – 63 = –57 – 6 14 – 63 Simplify 14 Simplify. 11-4 9. Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 Simplify 8 7– = 3 7+ 7+ • 3 3 8 7– 3 . Multiply the numerator and denominator by the conjugate of the denominator. = 8( 7 + 3) 7–3 Multiply in the denominator. = 8( 7 + 4 Simplify the denominator. = 2( 7+ =2 7+2 3) 3) 3 Divide 8 and 4 by the common factor 4. Simplify the expression. 11-4 Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 A painting has a length : width ratio approximately equal to the golden ratio (1 + 5 ) : 2. The length of the painting is 51 in. Find the exact width of the painting in simplest radical form. Then find the approximate width to the nearest inch. Define: 51 = length of painting x = width of painting Relate: (1 + Write: 5) : 2 = length : width (1 + 5) = 51 x 2 x (1 + 5) = 102 x(1 + 5) = 102 (1 + 5) (1 + 5) Cross multiply. Solve for x. 11-4 Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 (continued) x= (1 – 102 • (1 – (1 + 5) x= 102(1 – 5) 1–5 x= 102(1 – –4 5) x = – 51(1 – 5) 2 x = 31.51973343 5) 5) Multiply the numerator and the denominator by the conjugate of the denominator. Multiply in the denominator. Simplify the denominator. Divide 102 and –4 by the common factor –2. Use a calculator. x 32 The exact width of the painting is – 51(1 – 2 5) inches. The approximate width of the painting is 32 inches. 11-4 Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 pages 603–606 Exercises 1. 5 12. 4 13. –2 6 2. 18 10 14. 3. –2 5 15. 8 4. 2 2 5 7 10 16. 4 – 4 7 5. 14 2 17. 9 + 6. –8 3 18. 6 – 2 2 3 3 7. yes 19. 3 5+2 8. yes 20. 3 2+6 21. 6 – 5 9. no 10. 4 11. –3 2 3 3 6 22. –9 – 14 23. 58 – 10 11-4 6 30 24. 11 – 4 7 25. 43 + 4 30 26. 32 + 9 11 27. 23 – 5 13 28. 2 7+2 29. –6 2 30. –4 6 – 12 31. 3( 10 + 5 3 2 5) 32. –5 11 – 5 3 33. 18 3+9 11 34. 10 2 + 10; 24.1 Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 35. – 4 ; –1.3 46. 4 36. 6 – 4 47. 3 2; 0.3 3+4 48. 8 38. 5 10 49. 10 + 10 39. 6 2+6 50. 6 41. 8 + 2 42. 13 + 65 + 130 + 5 2 8 14 44. –24 45. – 2 units 10 units 10 units 52. Answers may vary. Sample: 8 2 + 4 3, 2 7 + 9 3, 6 5+3 7 53. a. The student simplified 48 as 2 24 instead of 2 12 or 4 3. b. 2 6 + 4 3 54. a. 2 2 or 2.8 ft b. s 2 15 43. 15 + 4 2 units 51. 4x + x 3–6 40. 22 6+6 10 5 37. 7.4 ft 3 2+3 2 11-4 Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 55. 9.1% 63. 56. 12.8% b 2 65. 23 7 21 n 59. a+ 64. 9 2 57. 15.5% 58. a. x 2 n–1 b. x 2 a + b =/ x 66. 8 15 15 ab b 67. 2 60. about 251 years 68. 10 61. They are unlike radicals. 62. a. 1, 0, 1, 1; 4, 1, 5, 17; 5, 3, 8, 34; 8, 6, 14, 10; 10, 9, 19, 181 b. No; the only values it worked for were 0 and 1. 2 69. 70 70. 2 2– 6– 71. a. 2 6 b. 2 13 c. 2(p + q) 72. B 11-4 3+3 Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 73. I 74. [2] (3 5 – 2)( 5 + 5 2) = 3 25 + 15 10 – 10 – 5 4 = 3(5) + 15 10 – 10 – 5(2) = 15 + 15 10 – 10 – 10 = 5 + 14 10 [1] correct technique, but with a computational error 75. [4] 5 7+ 21 • 7– 7– 21 Multiply the numerator and 21 denominator by the conjugate of the denominator. Simplify the denominator. [3] correct steps but answer not completely simplified [2] correct technique, but with a computational error [1] correct answer but no work shown 11-4 Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 76. 9.2 units 88. 4p2 + 28p + 49 77. 6.7 units 89. 25g2 – 49 78. 26.2 units 90. 9x2 – 1 79. (3, 5) 91. 1 k2 – 81 80. (–2, 6.5) 92. d 2 – 2.2d + 1.21 9 81. 0, 7 82. –2, 9 83. –9, –3 84. –4, 6 85. –15, –2 86. –3, – 1 87. b2 2 + 22b + 121 11-4 Operations with Radical Expressions ALGEBRA 1 LESSON 11-4 Simplify each expression. 1. 12 16 – 2 40 4. ( 16 3–2 21)( –123 + 3 7 2. 3+3 20 – 4 –2 5 21) 5 5. 3. 16 5– –8 11-4 2( 2 + 3 2+3 6 7 5–8 7 3) Solving Radical Equations ALGEBRA 1 LESSON 11-5 (For help, go to Lesson 10-3.) Evaluate each expression for the given value. x – 3 for x = 16 1. 2. x + 7 for x = 9 3. 2 x + 3 for x = 1 Simplify each expression. 4. ( 3)2 5. ( x + 1) 2 11-5 6. ( 2x – 5) 2 Solving Radical Equations ALGEBRA 1 LESSON 11-5 Solutions 1. x – 3 for x = 16: 2. x + 7 for x = 9: 16 – 3 = 4 – 3 = 1 9+7= 3. 2 x + 3 for x = 1: 2 4. ( 3)2 = 3 5. ( x + 1)2 = x + 1 6. ( 2x – 5)2 = 2x – 5 16 = 4 1+3=2 4=2•2=4 11-5 Solving Radical Equations ALGEBRA 1 LESSON 11-5 Solve each equation. Check your answers. a. x–5 =4 x=9 ( x)2 = 92 Isolate the radical on the left side of the equation. Square each side. x = 81 Check: x–5 –5 9–5 = 4 4 4 Substitute 81 for x. 4=4 11-5 Solving Radical Equations ALGEBRA 1 LESSON 11-5 (continued) x–5 =4 b. ( x – 5)2 = 42 x–5=9 Square each side. Solve for x. x = 21 Check: x–5 21– 5 16 = 4 = 4 = 4 Substitute 21 for x. 4=4 11-5 Solving Radical Equations ALGEBRA 1 LESSON 11-5 On a roller coaster ride, your speed in a loop depends on the height of the hill you have just come down and the radius of the loop in feet. The equation v = 8 h – 2r gives the velocity v in feet per second of a car at the top of the loop. 11-5 Solving Radical Equations ALGEBRA 1 LESSON 11-5 (continued) The loop on a roller coaster ride has a radius of 18 ft. Your car has a velocity of 120 ft/s at the top of the loop. How high is the hill of the loop you have just come down before going into the loop? Solve v = 8 120 = 8 120 = 8 8 15 = h – 2r for h when v = 120 and r = 18. h – 2(18) Substitute 120 for v and 18 for r. h – 2(18) 8 h – 36 (15)2 = ( h – 36)2 225 = h – 36 261 = h The hill is 261 ft high. Divide each side by 8 to isolate the radical. Simplify. Square both sides. 11-5 Solving Radical Equations ALGEBRA 1 LESSON 11-5 Solve ( 3x – 4 = 3x – 4)2 = ( 2x + 3)2 3x – 4 = 2x + 3 3x = 2x + 7 x=7 Check: 3x – 4 = 3(7) – 4 17 = 2x + 3. Square both sides. Simplify. Add 4 to each side. Subtract 2x from each side. 2x + 3 2(7) + 3 17 Substitute 7 for x. The solution is 7. 11-5 Solving Radical Equations ALGEBRA 1 LESSON 11-5 Solve x = x + 12. (x)2 = ( x + 12)2 Square both sides. x2 = x + 12 x2 – x – 12 = 0 Simplify. (x – 4)(x + 3) = 0 Solve the quadratic equation by factoring. (x – 4) = 0 or (x + 3) = 0 x = 4 or x = –3 Use the Zero–Product Property. Solve for x. Check: x = x + 12 4 4 + 12 4 = 4 –3 –3 =/ –3 + 12 3 The solution to the original equation is 4. The value –3 is an extraneous solution. 11-5 Solving Radical Equations ALGEBRA 1 LESSON 11-5 Solve 3x + 8 = 2. 3x = –6 ( 3x)2 = (–6)2 Square both sides. 3x = 36 x = 12 Check: 3x 3(12) 36 6 +8=2 +8 2 +8 2 + 8 =/ 2 Substitute 12 for x. x = 12 does not solve the original equation. 3x + 8 = 2 has no solution. 11-5 Solving Radical Equations ALGEBRA 1 LESSON 11-5 12. –2 24. 2 1. 4 13. 4 25. no solution 2. 49 14. 2.5 26. 4 3. 36 15. 2 27. 1.25 or 5 4. 137 16. –4 5. 15 17. none 28. 1 , 1 4 6. 16 18. – 1 7. 576 ft 19. –7 8. 602 watts 20. none 9. 4.5 21. 3 10. 3 22. 5 11. 7 23. no solution pages 610–612 Exercises 4 2 11-5 29. a. 25 b. 11.25 30. about 2.5 in. 31. An extraneous solution is a solution of a new equation that does not satisfy the original equation. 32. Answers may vary. Sample: x – 2 = 7 – 2x , 3x = 3 Solving Radical Equations ALGEBRA 1 LESSON 11-5 33. 1600 ft 34. 3 35. no solution 36. no solution 37. 1, 6 38. 1.5 39. 11 44. a. 68 ft b. 20.5 mi/h c. As radius increases, velocity decreases. As height decreases, velocity decreases. d. Velocity depends upon the difference of the height and the radius. 45. a. 40. 0, 12 41. 3, 6 42. 44 43. no solution b. approximately (6, 3.6) c. 6; it is the x-coordinate of the point of intersection. 46. a. V = 10x2 b. x = c. 2, 3, 4, 5, 6, 7 47. a. – 7, 7 b. 49 c. In both cases 3 is added to each side. To solve the first equation you find the square roots of each side, and in the second equation you find the square of each side. 48. –2, 8 49. 0 11-5 V 10 Solving Radical Equations ALGEBRA 1 LESSON 11-5 51. –1 52. Subtract 2x from each side. Square both sides. Solve for x. Check the solution if there is one. 60. C 15 – 5x = 4x – 3 15 – 5x = 4x – 3 –9x = –18 x =2 Check: 15 – 5(2) 4(2) – 3 5= 5 The solution is 2. [1] correct technique with a minor error OR correct answer, no work shown 61. [2] 53. The square of x – 1 will have only 2 terms while x – 1 squared will have 3 terms. 54. a. about 2.0 m b. about 32.4 m 55. C 56. G 57. B 58. A 62. 5 5 63. 3 2+4 64. 65. 32 59. B 11-5 2 3 Solving Radical Equations ALGEBRA 1 LESSON 11-5 66. 4( 5– 67. 54 2 3) 78. 3(d – 1)(d + 5) 79. (4v – 5)(v – 5) 68. 2.5, –4.5 69. 8.4, –0.4 70. –0.2, –4.8 71. –10.7, 0.7 72. –11.7, 1.7 73. –1.6, 3.1 74. (x + 12)(x – 2) 75. (m – 13)(m – 1) 76. (b + 18)(b – 2) 77. (2p + 1)(p + 7) 11-5 Solving Radical Equations ALGEBRA 1 LESSON 11-5 Solve each radical equation. 25 1. 7x – 3 = 4 3. 2x + 7 = 5. 3x + 4 + 5 = 3 2. 7 5x – 8 5 3x – 2 = 4. x = no solution 11-5 2x + 8 x+2 4 2 Graphing Square Root Functions ALGEBRA 1 LESSON 11-6 (For help, go to Lessons 10-1 and 10-3.) Graph each pair of quadratic functions on the same graph. 1. y = x2, y = x2 + 3 2. y = x2, y = x2 – 4 Evaluate each expression for the given value of x. 3. x for x = 4 4. x + 7 – 3 for x = 2 5. 3 x + 2 for x = 9 11-6 Graphing Square Root Functions ALGEBRA 1 LESSON 11-6 Solutions 1. y = x2, y = x2 + 3 2. y = x2, y = x2 – 4 3. x for x = 4: 4 = 2 4. x – 7 – 3 for x = 2: 2 + 7 – 3 = – 3 = 9 – 3 – 3 = 0 5. 3 x + 2 for x = 9: 3 9 + 2 = 3(3) + 2 = 9 + 2 = 11 11-6 Graphing Square Root Functions ALGEBRA 1 LESSON 11-6 Find the domain of each function. a. y = x+5 x+5> –0 Make the radicand > – 0. x> – –5 The domain is the set of all numbers greater than or equal to –5. b. y = 6 4x – 12 4x – 12 > –0 Make the radicand > – 0. 4x > – 12 x> – 3 The domain is the set of all numbers greater than or equal to 3. 11-6 Graphing Square Root Functions ALGEBRA 1 LESSON 11-6 The size of a television screen is the length of the screen’s diagonal d in inches.The equation d = 2A estimates the length of a diagonal of a television with screen area A. Graph the function. Domain 2A > –0 A> –0 Screen Area (sq. in.) 0 50 100 200 300 400 Length of Diagonal (in.) 0 10 14.1 20 24.5 28.3 11-6 Graphing Square Root Functions ALGEBRA 1 LESSON 11-6 Graph y = y= x + 4 by translating the graph of x. For the graph y = the graph of y = 11-6 x + 4, x is shifted 4 units up. Graphing Square Root Functions ALGEBRA 1 LESSON 11-6 Graph ƒ(x) = y= x + 3 by translating the graph of x. For the graph ƒ(x) = the graph of y = 11-6 x + 3, x is shifted to the left 3 units. Graphing Square Root Functions ALGEBRA 1 LESSON 11-6 pages 616–619 Exercises 1. x> –2 2. 3 x> – 3. x> –0 4. x> – –7 5. x> – –3 6. x> –5 7. 5 x> –– 8. x> – –2 9. 4 x> – 13. x 0 3 5.3 y 0 3 4 11. x ƒ(x) 0 0 1 2 4 4 14. x 0 1 4 ƒ(x) 0 3 6 12. x 2 3 6 15. x 0 1 4 y 0 –3 –6 10. x 0 2 4.5 4 y 0 2 3 3 3 y 0 2 4 11-6 Graphing Square Root Functions ALGEBRA 1 LESSON 11-6 16. h 0 1 4 v 0 8 16 22. 25. 23. 26. 24. 27. 17. D 18. A 19. C 20. B 21. 11-6 Graphing Square Root Functions ALGEBRA 1 LESSON 11-6 28. 33. a-d. a. b. c. d. 29. 30. x > – 4; y 31. x < – 4; y Answers may vary. Samples: y= x+2 y= x+2 y=2 x Check students’ work. 34. Translate the graph of y = x 8 units to the left. > –0 > –0 32. Form an inequality setting the radicand > – 0. Solve for x. Answers may vary. Sample: y= x–2 Domain: x – 2 > –0 x> –2 35. Translate the graph of y = x 10 units down. 36. Translate the graph of y = x 12 units up. 37. Translate the graph of y = x 9 units right. 11-6 Graphing Square Root Functions ALGEBRA 1 LESSON 11-6 38. x 2.5 3.5 6.5 y 0 1 2 39. x 0 1 2 4 ƒ(x) 0 4 5.7 8 40. x –6 –5 –2 0 y 0 1 2 2.4 41. x 0 2 4 8 y 0 1 1.4 2 42. x 2 3 6 y 3 4 5 43. x –2 –1 2 ƒ(x) –4 –3 –2 11-6 Graphing Square Root Functions ALGEBRA 1 LESSON 11-6 44. x 0 1 2 3 y 3 4.4 5 5.4 47. B 48. D 49. A 50. C 45. x –3 –2 –1 0 y 1 2.4 3 3.4 46. x 1 2 3 4 y –2 –0.3 0.4 1 51. a. p > 0 b. c. about 45 lb/in.2 52. a. no b. Answers may vary. Sample: The graph of y = x is the first quadrant of the graph of x = y2. c. y = – x 53. y = 3 x rises more steeply because 3 x > 3x for all positive values of x. 54. False; x must equal 81. 55. False; only combine like terms. 56. true 57. False; x = –1. 11-6 Graphing Square Root Functions ALGEBRA 1 LESSON 11-6 58. a. about 213 cameras b. month 4 61. a. i. ii. 59. a. iii. iv. b. y = |x| + 5 b. The greater the absolute value of n, 60. Translate the graph of the steeper the graph. If n < 0, then the y = x right 2 units and up 3 units. graph lies in Quadrant II. If n > 0, the graph lies in Quadrant I. 11-6 Graphing Square Root Functions ALGEBRA 1 LESSON 11-6 62. Check students’ work. 70. 16 63. B 71. 7 64. H 72. 169 65. C 73. 14.76 66. H 74. no solution 67. B 75. 2 3 68. H 69. [2] x y 6 0 7 1 8 1.4 9 1.7 10 2 [1] incorrect coordinates on graph 11-6 76. –2 – 3 2 , –2 + 3 2 2 2 77. 4 – 39, 4 + 39 78. no solution 11, –1 + 11 3 3 79. –1 – Graphing Square Root Functions ALGEBRA 1 LESSON 11-6 80. no solution 421 , –13 + 421 18 18 81. –13 – 82. (2x + 1)(x – 4) 83. (3x – 5)(x + 2) 84. (2x + 1)(2x + 9) 85. 2(x – 8)(x + 3) 86. 4(x2 – x – 15) 87. x(x – 13)(x + 1) 11-6 Graphing Square Root Functions ALGEBRA 1 LESSON 11-6 1. Find the domain of the function ƒ(x) = 2. Graph y = 3 3. Graph y = 2x – 4. x. x – 3. 4. Describe how to translate the graph of y = x to obtain the graph of the function y = x – 15. Shift the graph to the right 15 units. 11-6 x>2 Trigonometric Ratios ALGEBRA 1 LESSON 11-7 (For help, go to Lessons 1-6 and 4-2.) Let c = A , s = O, t = O . Calculate c, s, and t for the given values. H H A 1. A = 3, O = 4, H = 5 2. A = 5, O = 12, H = 13 Solve each equation. 3. 15 0.75 = 1 x 5. 0.84 = 21 1 x x 0.34 4. 20 = 1 6. x 14 = 0.52 1 11-7 Trigonometric Ratios ALGEBRA 1 LESSON 11-7 Solutions 1. For A = 3, O = 4, H = 5: c= A 3 O 4 O 3 = ,s= = ,t= = H 5 H 5 A 4 2. For A = 5, O = 12, H = 13: c = A = 5 , s = O = 12, t = O = 12 H 13 15 = 0.75 x 1 3. H 13 A 5 4. x = 0.34 1 20 0.75x = 15 x = 20(0.34) x = 20 5. 0.84 = 21 1 x x = 6.8 6. x 0.52 0.84 = 21 = 14 1 x = 14(0.52) x = 25 x = 7.28 11-7 Trigonometric Ratios ALGEBRA 1 LESSON 11-7 Use the triangle. Find sin A, cos A, and tan A. opposite leg 6 3 sin A = hypotenuse = = 10 5 adjacent leg 8 4 cos A = hypotenuse = = 10 5 opposite leg tan A = adjacent leg = 6 = 3 8 4 11-7 Trigonometric Ratios ALGEBRA 1 LESSON 11-7 Find sin 40° by using a calculator. To find sin 40°, press Use degree mode when finding trigonometric ratios. Rounded to the nearest ten-thousandth, the sin 40° is 0.6428. 11-7 40 . Trigonometric Ratios ALGEBRA 1 LESSON 11-7 Find the value of x in the triangle. Step 1: Decide which trigonometric ratio to use. You know the angle and the length of the hypotenuse. You are trying to find the adjacent side. Use the cosine. Step 2: Write an equation and solve. adjacent leg hypotenuse x cos 30° = 15 cos 30° = x = 15(cos 30°) 15 30 12.99038106 x 13.0 The value of x is about 13.0. Substitute x for adjacent leg and 15 for hypotenuse. Solve for x. Use a calculator. Round to the nearest tenth. 11-7 Trigonometric Ratios ALGEBRA 1 LESSON 11-7 Suppose the angle of elevation from a rowboat to the top of a lighthouse is 708. You know that the lighthouse is 70 ft tall. How far from the lighthouse is the rowboat? Round your answer to the nearest foot. Draw a diagram. Define: Let x = the distance from the boat to the lighthouse. Relate: You know the angle of elevation and the opposite side. You are trying to find the adjacent side. Use the tangent. 11-7 Trigonometric Ratios ALGEBRA 1 LESSON 11-7 (continued) Write: opposite leg tan A = adjacent leg 70 tan 70° = X x(tan 70°) = 70 Substitute for the angle and the sides. Multiply each side by x. x= 70 tan 70 Divide each side by tan 70°. x 25.4779164 Use a calculator. x 25 Round to the nearest unit. The rowboat is about 25 feet from the lighthouse. 11-7 Trigonometric Ratios ALGEBRA 1 LESSON 11-7 A pilot is flying a plane 15,000 ft above the ground. The pilot begins a 3° descent to an airport runway. How far is the airplane from the start of the runway (in ground distance)? Draw a diagram. Define: Let x = the ground distance from the start of the runway. Relate: You know the angle of depression and the opposite side. You are trying to find the adjacent side. Use the tangent. 11-7 Trigonometric Ratios ALGEBRA 1 LESSON 11-7 (continued) Write: opposite leg tan A = adjacent leg tan 3° = 15,000 x x(tan 3°) = 15,000 Substitute for the angle and the sides. Multiply each side by x. x= 15,000 tan 3 Divide each side by tan 3°. x 286217.05 Use a calculator. x 290,000 Round to the nearest 10,000 feet. The airplane is about 290,000 feet (or about 55 miles) from the start of the runway. 11-7 Trigonometric Ratios ALGEBRA 1 LESSON 11-7 pages 625–627 Exercises 1. 2. 3. 4. 5. 6. 3 5 4 5 3 4 4 5 3 5 4 3 7. 0.5299 8. 0.5736 9. 1.2799 10. 0.9962 11. 0.9659 23. b = 15; sin A = 8 ; 12. 5.5 17 15 cos A = ; tan A = 17 24. a = 24; sin A = 12 ; 13 5 cos A = ; tan A = 13 13. 10.4 14. 19.2 15. 38.1 16. 66.0 25. AC 6; AB 17. 21.1 26. AC 36; BC 18. about 2.0 mi 27. BC 6; AB 18 19. about 172 ft 28. AC 4; AB 50 20. about 816 ft 29. about 55 m 21. about 0.4 mi 30. about 4.4 m 22. c = 29; sin A = 21 ; 31. a. 1,720,000 ft b. 326 mi cos A = 20 ; tan A = 21 29 29 11-7 20 8 22 8 15 12 5 Trigonometric Ratios ALGEBRA 1 LESSON 11-7 40. Answers may vary. Sample: about 8.2 cm 32. 12 m 33. about 6.8 m 34. 514.3 35. 4.5 36. 78.4 37. q 6.1; r 7.9 38. a. about 177 ft b. about 86 ft 39. a. about 252 ft b. about 377 ft 49. [2] sin A 0.6271, cos A 0.7881, tan A 0.7957 [1] at least one correct equation 50. [4] 850 41. about 203 ft 42. about 5938 m 43. about 57 ft tan 56° = x x 573.3 573 ft 44. 550 ft 45. 0.25 46. a. b. 47. C about 229 km about 4 km 48. G 11-7 [3] correct equation, but minor computational error [2] incorrect equation used [1] no work shown Trigonometric Ratios ALGEBRA 1 LESSON 11-7 51. 54. 2 55. 0 56. 1 57. (n – 20)(n + 20) 52. 58. (x – 15)2 59. (10p – 7)(10p + 7) 60. 1 d – 3 4 53. 1d + 3 4 2 2 61. 2(7w – 8)(7w + 8) 62. (x + 13)2 11-7 Trigonometric Ratios ALGEBRA 1 LESSON 11-7 1. Use the figure to find sin A, cos A, and tan A. 21 20 21 , , 29 29 20 2. Find the value of x to the nearest tenth. 6.9 3. A group of skateboarders wants to build a ramp with an angle of incline of 14°. What should the rise be for every 10 meters of run? about 2.5 m 4. A park ranger on a 220 ft tower spots a fire at an angle of depression of 4°. How far is the fire from the base of the tower? about 3146 ft 5. A wheelchair ramp is to have an angle of 3.5° with the ground. The deck at the top of the ramp is 18 in. above the ground. How long should the ramp be? about 294.3 in. 11-7 Radical Expressions and Equations ALGEBRA 1 CHAPTER 11 1. yes 2. no 12. – 1 , – 2 13. 3. no 4. yes 5. 57.8 cm 6. 9.8 units 7. 4.1 units 8. 19.8 units 9. 7.6 units 10. 21, 2 2 11. 1 1 ,–3 2 2 2 – 1 , 5 (– 1 , 5) 2 2 14. 29.5 15. 27.0 23. 11 3 24. 35 5 – 15 25. 8 2–8 26. 5 3 27. 2 16. 13 ft 10 – 7 3 6 17. 15 mi 28. Answers may vary. Sample: 2 5+4 5=6 5 18. 29. B 2 19. 3 30. 4 5 20. 4 3 31. 13 21. 4 6 32. 6 22. 17 2 33. 7 11-A Radical Expressions and Equations ALGEBRA 1 CHAPTER 11 5 35. 36. 2 41. 24 cm 38. x > – 0; 34. 9 42. The graph of y = x is shifted 3 units down. 10.65 5 ft 43. r = 37. x > – 0; V h 44. about 32 ft 45. about 8.1 39. x > – 4; 46. about 12.1 47. about 1.106 48. about 0.7431 40. x > – –9; 49. BC 6.2, AC 50. about 5 ft 11-A 3.3