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Final Exam Option 1 Answers Multiple Choice and Numerical Response Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. C 68° C A B D A D B B C D B C 165 ft2 B C B 1620 in.3 D C 14.9° B C B $9.09 17.2 m D A B 13.8 mm C C B B A 24 A B D 3124 D D 3 C 21 B D 30 29.5° A B D A D B D A 0.5 A Written Response Answers ________ 1. a) 4 √302 + 202 = 144.222 204 The amount of cable needed is approximately 144 m. 30 b) tan θ = __ 20 ( ) θ = 56.309 932… θ = 56.3 The angle is approximately 56.3°. -30 c) ____ = -1.5 20 d) y = -1.5x + 30; in general form: 5x = 2y – 60 = 0 2. a) V = π (302)(80) V = 226 194.690 265… The volume of the barrel is approximately 226 195 cm3. b) V = π (32)(12) V = 339.292 035… The volume of the spigot is approximately 339 cm2. Copyright © 2010, McGraw-Hill Ryerson Limited, a subsidiary of the McGraw-Hill Companies. All rights reserved. This page may be reproduced for classroom use by the purchaser of this book without written permission of the publisher. Name: ____________________________________________ V c) Date: _______________________________ 3. a) Diagrams may vary. Example: Rate of Water Flow 200 000 Volume (cm3) 150 000 b) 2x2 – 5x – 3 c) Answers will vary. Students may represent a linear or a non-linear relation. d) Answers will vary. Check that the graph passes the vertical line test. e) Answers will vary. Check that the graph does not pass the vertical line test. 100 000 50 000 200 0 400 600 800 Time (min) 1000 1200 t 1 d) __4 (226 195) = 56 548.75; 56 549 _____ 200 = 282.745 It would take approximately 283 s or 4.7 min before the barrel was three quarters full. Alternatively, students may use the graph in part c) to determine the time. e) Bryan is correct. The slope is decreasing (or negative). As time passes, the level of the water in the barrel decreases. 4 f) Vball = __3 π(6)3 Vball = 904.778 761 061 947… The volume of the ball is approximately 905 cm3. V πr 905 h = _____ 900π h = ___2 905 h = ____________ 2827.433 628… h = 0.320 078… The level of the water rose about 0.32 cm after the ball sank. Copyright © 2010, McGraw-Hill Ryerson Limited, a subsidiary of the McGraw-Hill Companies. All rights reserved. This page may be reproduced for classroom use by the purchaser of this book without written permission of the publisher.