F.2 Maths_6
2017/07/04 09:21
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F.2 Maths
發問:
[Quadratic Equations] The number of diagonals (D) od a polygon with n sides is given by the formula ; D= 1/2 n(n-3)(a)If the number of diagonals and the number of sides of a polygon are the same, how many sides does the polygon have? (b) If the number of sides of a polygon is less tahn than the number... 顯示更多 [Quadratic Equations] The number of diagonals (D) od a polygon with n sides is given by the formula ; D= 1/2 n(n-3) (a)If the number of diagonals and the number of sides of a polygon are the same, how many sides does the polygon have? (b) If the number of sides of a polygon is less tahn than the number od diagonals by 7, find the number of sides of the polygon. A circular ring of radius 10 cm is cut into two parts. Each part is then bent into a smaller ring. If the total aera bounded by the two smaller rings of radii r cm and R cm is 62.5*3.14cm^2 , find the values of r and R(assume R is greater than r.) The cost C, in thousand dollars, of producting n television sets per day in a factory is given by C=0.1n^2+10. The television sets produced are sold at $8 000 each. If the profit on a certain day is $150 000 , find the number of television sets produced on that day.
最佳解答:
Thenumber of diagonals (D) of a polygon with n sides is given by the formula: D=1/2 n(n-3) (a) Ifthe number of diagonals and the number of sides of a polygon are the same, howmany sides does the polygon have ? (b) If the number of sides of a polygon is less than the number ofdiagonals by 7, find the number of sides of the polygon. (a) D = (1/2)n(n - 3) ...... [1] D = n ...... [2] (1) = (2) : (1/2)n(n - 3) = n n2 - 5n = 0 n(n - 5) = 0 n= 0 (rejected) or n = 5 Hence, number of sides that the polygon has = 5 (b) D = (1/2)n(n - 3) ...... [3] D = n + 7 ...... [4] [3] = [4] : (1/2)n(n - 3) = n + 7 n2 - 5n - 14 = 0 (n + 2)(n - 7) = 0 n= -2 (rejected) or n = 7 Hence, number of sides that the polygon has = 7 = = = = = A circular ring of radius 10 cm is cut into twoparts. Each part is then bent into a smaller ring. If the total areabounded by the two smaller rings of radii r cm and R cm is 62.5*3.14cm^2 , findthe values of r and R (assume R is greater than r.) Take π = 3.14 2 ′ 3.14 ′ R+ 2 ′ 3.14 ′ r= 2 ′ 3.14 ′ 10...... [1] 3.14′ R2 + 3.14 ′ r2 = 62.5 ′ 3.14 ...... [2] From [1] : R + r = 10 r = 10 - R ...... [3] From [2] : R2 + r2 =62.5 ...... [4] Put [3] into [4] : R2 + (10 - R)2 =62.5 2R2 - 20R + 37.5 = 0 4R2 - 40R + 75 = 0 (2R - 15)(2R - 5) = 0 R = 7.5 or R = 2.5 When R = 7.5 : Put R = 7.5 into [1] : r = 10 - 7.5 r = 2.5 When R = 2.5 : Put R = 2.5 into [1] : r = 10 - 2.5 r = 7.5 (rejected) Hence, r = 2.5, R = 7.5 = = = = = The cost C, in thousand dollars, of producing ntelevision sets per day in a factory is given by C = 0.1n2 + 10.The television sets produced are sold at $8 000 each. If the profit on acertain day is $150 000, find the number of television sets produced on thatday. 8000n - (0.1n2 + 10) ′1000 = 150000 80n - (0.1n2 + 10) ′ 10= 1500 80n - n2 - 100 = 1500 n2 - 80n + 1600 = 0 (n - 40)2 = 0 n = 40 (double roots) The number of television sets produced on that day = 40
其他解答:
F.2 Maths
發問:
[Quadratic Equations] The number of diagonals (D) od a polygon with n sides is given by the formula ; D= 1/2 n(n-3)(a)If the number of diagonals and the number of sides of a polygon are the same, how many sides does the polygon have? (b) If the number of sides of a polygon is less tahn than the number... 顯示更多 [Quadratic Equations] The number of diagonals (D) od a polygon with n sides is given by the formula ; D= 1/2 n(n-3) (a)If the number of diagonals and the number of sides of a polygon are the same, how many sides does the polygon have? (b) If the number of sides of a polygon is less tahn than the number od diagonals by 7, find the number of sides of the polygon. A circular ring of radius 10 cm is cut into two parts. Each part is then bent into a smaller ring. If the total aera bounded by the two smaller rings of radii r cm and R cm is 62.5*3.14cm^2 , find the values of r and R(assume R is greater than r.) The cost C, in thousand dollars, of producting n television sets per day in a factory is given by C=0.1n^2+10. The television sets produced are sold at $8 000 each. If the profit on a certain day is $150 000 , find the number of television sets produced on that day.
最佳解答:
Thenumber of diagonals (D) of a polygon with n sides is given by the formula: D=1/2 n(n-3) (a) Ifthe number of diagonals and the number of sides of a polygon are the same, howmany sides does the polygon have ? (b) If the number of sides of a polygon is less than the number ofdiagonals by 7, find the number of sides of the polygon. (a) D = (1/2)n(n - 3) ...... [1] D = n ...... [2] (1) = (2) : (1/2)n(n - 3) = n n2 - 5n = 0 n(n - 5) = 0 n= 0 (rejected) or n = 5 Hence, number of sides that the polygon has = 5 (b) D = (1/2)n(n - 3) ...... [3] D = n + 7 ...... [4] [3] = [4] : (1/2)n(n - 3) = n + 7 n2 - 5n - 14 = 0 (n + 2)(n - 7) = 0 n= -2 (rejected) or n = 7 Hence, number of sides that the polygon has = 7 = = = = = A circular ring of radius 10 cm is cut into twoparts. Each part is then bent into a smaller ring. If the total areabounded by the two smaller rings of radii r cm and R cm is 62.5*3.14cm^2 , findthe values of r and R (assume R is greater than r.) Take π = 3.14 2 ′ 3.14 ′ R+ 2 ′ 3.14 ′ r= 2 ′ 3.14 ′ 10...... [1] 3.14′ R2 + 3.14 ′ r2 = 62.5 ′ 3.14 ...... [2] From [1] : R + r = 10 r = 10 - R ...... [3] From [2] : R2 + r2 =62.5 ...... [4] Put [3] into [4] : R2 + (10 - R)2 =62.5 2R2 - 20R + 37.5 = 0 4R2 - 40R + 75 = 0 (2R - 15)(2R - 5) = 0 R = 7.5 or R = 2.5 When R = 7.5 : Put R = 7.5 into [1] : r = 10 - 7.5 r = 2.5 When R = 2.5 : Put R = 2.5 into [1] : r = 10 - 2.5 r = 7.5 (rejected) Hence, r = 2.5, R = 7.5 = = = = = The cost C, in thousand dollars, of producing ntelevision sets per day in a factory is given by C = 0.1n2 + 10.The television sets produced are sold at $8 000 each. If the profit on acertain day is $150 000, find the number of television sets produced on thatday. 8000n - (0.1n2 + 10) ′1000 = 150000 80n - (0.1n2 + 10) ′ 10= 1500 80n - n2 - 100 = 1500 n2 - 80n + 1600 = 0 (n - 40)2 = 0 n = 40 (double roots) The number of television sets produced on that day = 40
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