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BLM 1–9 Chapter 1 Test 1. B 2. C 3. B 4. D 5. B 6. There are 33 lockers. 7. Brittany travelled 1752 m. 8. 20 358 9. a) d = 6 b) t1 = 4 c) t100 = 598 10. a) r = 9 or –9 b) tn = 5(9)n – 1 or tn = 5(–9)n – 1 11. a) –3, –12, –48 b) The sequence is geometric. tn = –3(4)n – 1 12. a) Example, for the series 2 + 10 + 50 + ..., S10 = 4 882 812. b) Answers will vary. Students need to change the sign of the first term, while leaving the common ratio unchanged. In the example above, the series becomes –2 – 10 – 50 – ..., S10 = – 4 882 812. c) Answers will vary. Correct answers must have positive first term and negative common ratio. For example, 2 – 10 + 50 – …. 13. a) 6979, 7537, 8140 b) tn = 6462(1.08)n – 1 c) 27 888 d) Answers will vary. For example, we assume that population continues to grow at the same rate. 14. Answers will vary, however will all be in the form k, k, k, …, where k is a real number. a) Note that d = 0, so tn = k. b) Note that r = 1, so tn = k. c) There are infinitely many such sequences, but all sequences will have the same form. BLM 2–9 Chapter 2 Test 1. B 2. B 3. C 4. D 5. A 6. 9.5 yd 7. 21.9 cm 8. 14 cm 9. a) 3 b) – 3 2 c) 2 2 BLM 3–7 Chapter 3 Test 1. B 2. A 3. C 4. A 5. D 6. a) x b) y c) y 7. a) r and s b) t 8. a) f (x) = b) f (x) = – 1 (x + 3 x2 3 4)2 + 2 +5 9. a) y = (x – 2)2 + 8; vertex (2, 8) 1 4 b) y = – (x + 8)2 – 2; vertex(–8, –2) 10. vertex: (6, –21); axis of symmetry: x = 6; direction of opening: upward; domain: x R; range: y ≥ –21; x-intercepts: (6 42, 0); y-intercept: (0, –3) 11. a) R = –50x2 + 400x + 12 000 b) $80 c) $12 800 d) 40 seats BLM 4–9 Chapter 4 Test 1. A 2. B 3. D 4. B 5. A 6. 5 5 2 or 5.59 s 7. a) In line 2, –4 should be in brackets. 2 10 . 2 b) In step 3, each term should have been divided by 15. 3 39 15 . 8. a) x = 2 or 8; Example: Factoring, because the equation is easily factored to (x – 2)(x – 8). 2 3 b) x = –7 or x ; Example: Quadratic formula, because the equation is not readily factored. c) x 3 2; Example: Completing the square, because it is easy to find the perfect square. d) x = 1 or 5; Example: Determining square roots, because it is easy to find the roots for (x – 3)2 = 4 10. a) 20° b) 20°, 200°, 340° 9. x2 + 5x – 10 = 0; 11. a) 10. k b) 29.9°, 56.3°, 93.8° 5 2 11. 11.3 m by 9.3 m 12. 2 3 or 3 2 13. 2.57 s 12. 12.5 ft and 6.4 ft Copyright © 2011, McGraw-Hill Ryerson, ISBN: 9780070738836 5 65 2 c) 2 is not a solution because it is a non- BLM 5–7 Chapter 5 Test 1. B 2. D 3. A 4. C 5. B permissible value. d) 1 . 2 6. 6 2, 5 3, 2 19, 4 5, 9 7. False. 16 9 7 8. 7 6 units 9. 5 5 BLM 7–8 Chapter 7 Test 1. C 2. A 3. B 4. C 5. B 6. C 7. | 5 | 10. r 7, r 1 4 3 x 4, if x 3 8. y 3 x 4, if x 4 3 3 11. 6 5 13 12. a) No. Valerie’s error is in taking the square root of each term of the radicand. The equation v0 v 2 20h cannot be further simplified. b) 60 m 13. a) x 3 b) 3 is not a root of the equation. 2 2 9. Example: | x 5 | 3 10. x 3 11. x 3 and x 2 12. a) 14. a) y 3, y 7 b) y 5 3 BLM 6–9 Chapter 6 Test 1. D 2. D 3. B 4. B 5. B 6. C 7. D 2 ( x 3) 8. x 32 , 3; (2 x 3) 9. 2 x 3x 3 ( x 6)( x 3) 11. Example: x 1 x3 10. 8 does not belong because it 13. a) | A –16 | is not a rational expression. 12. a) x 5, 4, 0, 6 b) 13. a) 15 x b) 15 x x ( x 5)( x 6) x( x 5) c) 15 x 1.25; x 60 buckets 14. No. Example: Restrictions on a variable in an expression come from the denominator of the factored form before any of the common factors in the numerator and denominator are simplified. 15. Example: Mary cannot cancel 3y from the denominator and from one term in the numerator. However, she can divide each term by 3y, which results in 5y 1 , 1 b) (5, 0) c) domain: {x | x R}; range: {y | y 0, y R} 3 4 b) 15 years and 3 months, or 16 years and 9 months c) Alain could be older or younger than the average age. 14. a) x 4 and x 1.5 b) or 5y 1. 16. 2; Example: I first found the LCD on each side of the equation and then performed the required operations: addition on the left side and subtraction on the right. I then used multiplication to eliminate the denominator on each side of the equation. Finally, I isolated the variable, and solved. 17. a) x 1, 2, 3 1 2 b) , 2 BLM 8–6 Chapter 8 Test Copyright © 2011, McGraw-Hill Ryerson, ISBN: 9780070738836 1. A 2. B 3. B 4. D 5. A 6.{(5, 8), (0, 3)} 7. Example: An object is released from a launcher on the ground, and a person standing on a platform throws a ball, trying to hit the object with the ball. 8. Example: ay a(x2 6x 5), a R 9. a) LS 2x2 x 7 RS y 2(3)2 3 7 14 14 LS RS LS 3x y 23 RS 0 3(3) 14 23 0 LS RS b) (5, 38) solution half-plane. 10. 52 , 2 , 12 , 1 11. a) m 5, k 2 b) k 8, m 2 12. a) two b) k 4 or k 0 c) (5.43, 1.08) or (1.43, 1.08) 13. a) {(4.3, 6.8), (21.7, 177.3)} b) The coordinates represent where the two streams of water meet. However, only the (4.3, 6.8) solution makes sense because the distance cannot be negative in this context. 14. a) perimeter: 2y 4x 26; area: 3y 9 x2 13x 36 b) x 7 and y 1, or x 12 and y 11 c) Substituting 7 results in a negative dimension, so x must be 12. The dimensions are 8 units and 3 units. d) perimeter: 22 units; area: 24 square units BLM 9–7 Chapter 9 Test 1. B 2. D 3. D 4. C 5. A 6. Example: When the point is substituted into the inequality, it makes a true statement. Shade the half-plane containing the point. 7. Example: The point (0, 0) lies on the boundary. It cannot be used to determine the Copyright © 2011, McGraw-Hill Ryerson, ISBN: 9780070738836 10. {x 2 3 x 2 3, x R} 11. 14. {P 30 P 70, P R} Example: Test point (0, 3): 1 2 3 (0)2 3(0) 1 31 The test point satisfies the inequality, so the region above the parabola should be shaded. 12. a) 13.95x 9.50y 150, where x is the number of tickets for adults and y is the number of tickets for children b) c) Example: The number of tickets must be a whole number. The number of tickets can be various combinations of 0 to 10 adult tickets and 0 to 15 children’s tickets, where the total cost does not go over $150. 13. a) Example: x2 x 2 0 b) Example: x2 x 12 0 Copyright © 2011, McGraw-Hill Ryerson, ISBN: 9780070738836