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Grade 10 June 2006 Exam Solutions

Grade 10 June 2006 Exam Solutions. solutions prepared by Tammy the Tutor of the-mathroom.ca _____________________ Part A: questions 1 – 7 Part B: questions 8 – 16 Part C: questions 17 – 25. instructions. To animate the solutions and for the next question(s), click your mouse

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Grade 10 June 2006 Exam Solutions

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  1. Grade 10 June 2006 Exam Solutions solutions prepared byTammy the Tutorof the-mathroom.ca _____________________ Part A: questions 1 – 7 Part B: questions 8 – 16 Part C: questions 17 – 25

  2. instructions To animate the solutions and for the next question(s), click your mouse Check your work for formas well as precision. Questions you got wrongstudy the solution, thendo the question over again.

  3. Solution: use Cosine Law to find angle A: Question # 1 Part A:

  4. Question # 2 radii must be equal for equivalence.

  5. Question # 3 Solution: x held by arc and o angles ~ 8.6 cm, AB held by angles c and o ~ 8 cm x : 8.6 = 16 : 8 x = 2(8.6) = 17.2 cm.

  6. Solution: flip the fraction to get now factor the exponents: Question # 4

  7. Question # 5

  8. Question # 6 (u, v) t < 0 Solution: t is < 0 so 3t = p < 0. u is > 0 so –( u) = q < 0. v is > 0 so v + 2 = r > 0. p < 0, q < 0, and r > 0.

  9. Question # 7 Solution: sold more bikes, add 32 to y

  10. Question # 8 Solution: 500 people not enough too local as well.

  11. 1st quintile Question # 9 Solution: quintiles are marked. quintiles ordered backwards.

  12. Question # 10 Solution: x-int for line1 = w. x-int for line2 = w.

  13. Side, Angle, Side property of congruence. Question # 11 Part B:

  14. Question # 12 Solution: EF = 3/2 (BC) the triangle moved 2 units right and 1 unit down. 1st: dilatation (enlarged) by 3/2 (1.5) 2nd: translation (x, y)  (x + 2, y – 1)

  15. Question # 13 - 14 Solution: we have h, and k so we need a. f (x) = a (x – 3)² – 8 172 = a (– 7 – 3)² – 8 means a = 9/5 or 1.8 f (x) = 1.8 (x – 3)² – 8

  16. ? Question # 15 Solution: we need distance from a point to a line.

  17. Question # 16 Solution: we know points (15, 38) and (0, 20) on g (x) slope = 6/5, y-int = 20. g (x) = 6/5 x + 20

  18. Question # 17 Part C: Solution: We need the x-intercept of FE. BC // AF, so C is at (15, 14). slope CF = 2, so slope FE = – ½. Eq. of FE is (y – 8) = – ½ (x – 12). Set y = 0 to find x-intercept. – 8 = – ½ (x – 12)  x = 28.

  19. Solution: we need the coordinates of R. 17/85 = 1/5, so AR = 1/5 (AB). ER = 1/5 (DB) = 1/5 (40) = 8 AE = 1/5 (AD) = 1/5 (75) = 15. R is at (40, 48) = (32 + 8, 63 – 15) Distance CR = E 75 km 40 km D Question # 18

  20. (16, 196) (23, 175) 7 units Question # 19 Solution: Set y = y to get x² – 39x + 368 = 0. Factored form: (x – 16)(x – 23) = 0. point A is (16, 196), point B is (23, 175). AC = 196, BD = (23 – 16) = 7 use AC as base, BD as height to find Area = ½ (196)(7) = 686 units².

  21. = 9.4 110° 70° a = 10.78 55° Question # 20 Solution: angle GHF = 110° (180° – 2 × 35°). use Sine Law to find “a ”. a = 15.4 (sin 35°) ÷ (sin 110°) = 9.4 cm. angle FHI = 70° (180° – 110°). angle HFI = 55° = ½ (180° – 70°) use Sine Law to find “b ”. b = 9.4 (sin 70°)÷(sin 55°) = 10.78 cm. Perimeter = 2a + b + 2x + 15.4 52.2 = 2(9.4) + 10.78 + 2x + 15.4 x = 3.61 cm.

  22. ? sum to 180° Question # 21 Solution: angle EDG = ? (opposite angles) angle EDG = DGH = (2x – 19)° we need to find (2x – 19)°. angle H = x° = angle C (alt. int.) angle H + angle F = 180° (co-int.) x° + (3x + 20)° = 180°  x = 40°. (2x – 19)° = 61° angle EDG = angle BDI = 61°. 2x – 19 = 61° x°

  23. (25, 62) (x, 80) (50, 0) (0, 0) Question # 22 Solution: find x-value at D. Set g(x) = 80. 80 = 0.08(x – 25)² + 62  x = 10. Use zeros form to find f (x). f (x) = ax (x – 50)  80 = a (10)( – 40) a = – 1/5 f (x) = – 1/5 x (x – 50).

  24. 141st data Question # 23 Solution: 1st: find Myriam’s percentile. she’s 8th, so 7 ahead of her 31 – 7 = 24 < her mark. R100 = (24/31) × 100 = 77.42 = 78th. 2nd: find mark with 78th R100 . 0.78 × 170 = 132.6  132nd data. 132nd data = 86 3rd: find Dahlia’s mark =86 + 5 = 91. Find R100 for 91 = 141/170 = 0.829 Dahlia’s percentile is 83.

  25. x x A = 175 cm² Question # 24 Solution: solve for x: we know 175 = ½ x (x + 25)  x² + 25x – 350 = 0. (x + 35)(x – 10) = 0 so x = 10 similarity ratio k = 8/10 = 4/5, so k² = 16/25 Area of EFGH = 16/25 (175) = 112 cm².

  26. Question # 25 volume A volume A 8 (volume A) Solution: volume B = volume A (equivalence), volume C = 8 (volume A) so total volume of 960 cm³ = 10 (volume A) volume A = 96 cm³ so height = 96 cm³ ÷ 16 cm² = 6 cm. we know k ³ = 8, so k = 2 (k is ratio of similarity) height of C = 2 × height of A = 12 cm. use your BACK BUTTON to exit the Power Point if viewing on the web.

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