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F3 Maths Problems Plz quick
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1) In the figure, ABCD is a rectangle and AECF is a rhombus. AB=8 and BC=5. Find AE.Figure:http://postimg.org/image/6upr28r0v/2)(a) Factorize a^2-14a-51(b) Using (a), or otherwise, factorize a^3+a^2-14a-243) The area of a rectangle is 100 cm^2. The length of the rectangle is increased by x% while the... 顯示更多 1) In the figure, ABCD is a rectangle and AECF is a rhombus. AB=8 and BC=5. Find AE. Figure:http://postimg.org/image/6upr28r0v/ 2)(a) Factorize a^2-14a-51 (b) Using (a), or otherwise, factorize a^3+a^2-14a-24 3) The area of a rectangle is 100 cm^2. The length of the rectangle is increased by x% while the width is decreased by 2%. (a) Show that the new area is (100-x-x^2/50) cm^2. (b) If the new area is 52 cm^2, find x. 4)In the figure, AB//EG//DC and AD is perpendicular with DC. AE=4, FC=9, ED=6 and BG=8. AED, AFC and BGC are straight lines. (a) Find AF and DC. (b) Find AB. Figure:http://postimg.org/image/6kdevpov7/ 5)(a) Show that 1/(x+ √7)=(x- √7)/(x^2-7). (b) Hence, rationalize 1/(1+ √(6)+ √(7)). Need Step, Plz!
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F3 Maths Problems Plz quick
發問:
1) In the figure, ABCD is a rectangle and AECF is a rhombus. AB=8 and BC=5. Find AE.Figure:http://postimg.org/image/6upr28r0v/2)(a) Factorize a^2-14a-51(b) Using (a), or otherwise, factorize a^3+a^2-14a-243) The area of a rectangle is 100 cm^2. The length of the rectangle is increased by x% while the... 顯示更多 1) In the figure, ABCD is a rectangle and AECF is a rhombus. AB=8 and BC=5. Find AE. Figure:http://postimg.org/image/6upr28r0v/ 2)(a) Factorize a^2-14a-51 (b) Using (a), or otherwise, factorize a^3+a^2-14a-24 3) The area of a rectangle is 100 cm^2. The length of the rectangle is increased by x% while the width is decreased by 2%. (a) Show that the new area is (100-x-x^2/50) cm^2. (b) If the new area is 52 cm^2, find x. 4)In the figure, AB//EG//DC and AD is perpendicular with DC. AE=4, FC=9, ED=6 and BG=8. AED, AFC and BGC are straight lines. (a) Find AF and DC. (b) Find AB. Figure:http://postimg.org/image/6kdevpov7/ 5)(a) Show that 1/(x+ √7)=(x- √7)/(x^2-7). (b) Hence, rationalize 1/(1+ √(6)+ √(7)). Need Step, Plz!
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1. Let AE = u Then, BE = 8 - u In ΔBCE : CE2 = BC2 +BE2 CE2 = 52 + (8- u)2 AECF is a rhombus. Then, CE = AE CE = AE 52 + (8 - u)2 = u2 25 + 64 - 16u + u2 = u2 16u = 89 u = 89/16 AE = 89/16 ==== 2. (a) a2 - 14a - 51 = (a - 17)(a + 3) (b) a3 + a2 -14a - 24 = a3 + (a2 -14a) - 24 = a3 + (a2 -14a - 51) + 51 - 24 = (a3 + 27) + (a2 -14a - 51) = (a3 + 33) +(a2 - 14a - 51) = (a + 3)(a2 - 3a + 9) + (a - 17)(a + 3) = (a + 3) [(a2 - 3a + 9) + (a - 17)] = (a + 3)(a2 - 2a - 8) ==== 3. (a) "2%" should be "2x%" instead. Let a cm and b cm be the original length and width respectively. ab = 100 New area = [a(1 + x%)] [b(1 - 2x%)] cm2 = ab(1 + x%)(1 - 2x%) cm2 = 100 [1 + (x/100)] [1 - (2x/100)] cm2 = 100 [1 - (x/100) - (x2/5000)] cm2 = [100 - x - (x2/50)] cm2 (b) [100 - x - (x2/50)] = 52 50[100 - x - (x2/50)] = 50(52) 5000 - 50x - x2 = 2600 x2 + 50x - 2400 = 0 (x - 30)(x + 80) = 0 x = 30 or x = -80 (rejected) Hence, x = 30 ==== 4. (a) EF//DC, and thus ΔAEF ~ ΔADC AE/AD = AF/AC 4/(4 + 6) = AF/(AF + 9) 4/6 = AF/9 6 AF = 36 AF = 6 In ΔADC, by Pythagorean theorem : AD2 + DC2 =AC2 (4 + 6)2 + DC2 = (6+ 9)2 100 + DC2 = 225 DC2 = 125 DC = 5√5 DC ≈ 11.2 (to 3 sig. fig.) (b) FG//AB, and thus ΔCFG ~ ΔCAB CF/CA = CG/(CG + 8) 9/(9 + 6) = CG/(CG + 8) 9/6 = CG/8 6 CG = 72 CG = 12 Draw CH perpendicular to AB, and CH cuts AB at H. In ΔCHB, by Pythagorean theorem : CH2 + HB2 = BC2 (6 + 4)2 + HB2 = (12+ 8)2 HB2 = 300 HB = 10√3 AB = AH + HB = 5√5 + 10√3 ≈ 28.5 (to 3 sig. fig.) ==== 5. (a) 1 / (x + √7) = (x - √7) / (x + √7)(x - √7) = (x - √7) / [x2 - (√7)2] = (x - √7) / (x2 - 7) (b) Put x = 1 + √6 1 / (1 + √6 + √7) = 1 / (x + √7) = (x - √7) / (x2 - 7) = [(1 + √6) - √7] / [(1 + √6)2 - 7] = [1 + √6 - √7] / [1 + 2√6 + (√6)2 - 7] = [1 + √6 - √7] / 2√6 = [1 + √6 - √7](√6) / 2(√6)2 = (6 + √6 - √42) / 12其他解答:
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