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發問:
any 1991maths ce mc solution for Q.34, 44,35? thanks
最佳解答:
The 1991 CE Mathematics Multiple Choice solution for Q. 34, 35 and 44 are as follows: 34. logx : logy = m : n logx / logy = m / n logx = (m / n)logy x = y(m / n) So, the answer is D. 35. f(x) - f(1/x) = (x - 1/x) - [1/x - 1/(1/x)] = (x - 1/x) - (1/x - x) = 2x - 2/x = 2(x - 1/x) So, the answer is D. 44. 圖片參考：http://i295.photobucket.com/albums/mm158/Audrey_hepburn2008/A_Hepburn01Aug210928.jpg?t=1219282215 Join the line ABC to form a triangle, where A and C are the centres of the two circles respectively. AC = 1 + 1 = 2 cm AB = 3 - 1 - 1 = 1 cm So, ∠BAC = 60* We should calculate the value of BC BC = ACsin60* = √3 cm The two radii of the circles inside should also be calculated: So, Number of circles that can be cut = (40 - 1 - 1) / √3 = 21.9 = 22 (cor. to the nearest integer) So, the answer is C.
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1991maths ce mc solution發問:
any 1991maths ce mc solution for Q.34, 44,35? thanks
最佳解答:
The 1991 CE Mathematics Multiple Choice solution for Q. 34, 35 and 44 are as follows: 34. logx : logy = m : n logx / logy = m / n logx = (m / n)logy x = y(m / n) So, the answer is D. 35. f(x) - f(1/x) = (x - 1/x) - [1/x - 1/(1/x)] = (x - 1/x) - (1/x - x) = 2x - 2/x = 2(x - 1/x) So, the answer is D. 44. 圖片參考：http://i295.photobucket.com/albums/mm158/Audrey_hepburn2008/A_Hepburn01Aug210928.jpg?t=1219282215 Join the line ABC to form a triangle, where A and C are the centres of the two circles respectively. AC = 1 + 1 = 2 cm AB = 3 - 1 - 1 = 1 cm So, ∠BAC = 60* We should calculate the value of BC BC = ACsin60* = √3 cm The two radii of the circles inside should also be calculated: So, Number of circles that can be cut = (40 - 1 - 1) / √3 = 21.9 = 22 (cor. to the nearest integer) So, the answer is C.
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