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Please help me! The first problem is implicit function and another one is about matrics. I tried but I can't find out these answers. 1)Find the slope of the tangent line to the curve sqrt(1x+3y)+sqrt(2xy)=9.72at the point (3,5)2)Find a and b such that[-34,7,-27]=a[1,-1,3]+b[11,-2,8]a=?b=? 顯示更多 Please help me! The first problem is implicit function and another one is about matrics. I tried but I can't find out these answers. 1)Find the slope of the tangent line to the curve sqrt(1x+3y)+sqrt(2xy)=9.72 at the point (3,5) 2)Find a and b such that [-34,7,-27]=a[1,-1,3]+b[11,-2,8] a=? b=?
最佳解答:
Answer for Q1: -1.14 Solution: sqrt(x + 3y) + sqrt(2xy) = 9.72 d[sqrt(x + 3y) + sqrt(2xy)] / dx = d(9.72) / dx [1 + 3(dy/dx)] / [2sqrt(x + 3y)] + [y + x(dy/dx)] / sqrt(2xy) = 0 [1 + 3(dy/dx)][sqrt(2xy)] + [y + x(dy/dx)][2sqrt(x + 3y)] = 0 sub. x = 3, y = 5, we have [1 + 3(dy/dx)]{sqrt[2(3)(5)]} + [5 + 3(dy/dx)][2sqrt(3 + 3(5))] = 0 [1 + 3(dy/dx)](sqrt30) + [5 + 3(dy/dx)](2sqrt18) = 0 sqrt30 + 3(sqrt30)(dy/dx) + 5(2sqrt18) + 3(2sqrt18)(dy/dx) = 0 sqrt30 + 3(sqrt30)(dy/dx) + 30sqrt2 + 18sqrt2(dy/dx) = 0 3(sqrt30)(dy/dx) + 18sqrt2(dy/dx) = - (sqrt30) - 30sqrt2 [3(sqrt30) + 18sqrt2](dy/dx) = - (sqrt30) - 30sqrt2 dy/dx = [- (sqrt30) - 30sqrt2] / [3(sqrt30) + 18sqrt2] = -1.14 Answer for Q2: a = -1, b = -3 Solution: [-34, 7, -27] = a[1, -1, 3] + b[11, -2, 8] [-34, 7, -27] = [a, -a, 3a] + [11b, -2b, 8b] [-34, 7, -27] = [a + 11b, -a - 2b, 3a + 8b] thus, we have a + 11b = -34 ------- (i) -a - 2b = 7 ----------- (ii) 3a + 8b = -27 ------- (iii) (i) + (ii) => 9b = -27 => b = -3 sub b = -3 into (i) => a + 11(-3) = -34 => a = -1 sub a = -1 and b = -3 into (iii) for checking, LHS = 3(-1) + 8(-3) = -27 = RHS therefore, a = -1, b = -3
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Pre Calculus發問:
Please help me! The first problem is implicit function and another one is about matrics. I tried but I can't find out these answers. 1)Find the slope of the tangent line to the curve sqrt(1x+3y)+sqrt(2xy)=9.72at the point (3,5)2)Find a and b such that[-34,7,-27]=a[1,-1,3]+b[11,-2,8]a=?b=? 顯示更多 Please help me! The first problem is implicit function and another one is about matrics. I tried but I can't find out these answers. 1)Find the slope of the tangent line to the curve sqrt(1x+3y)+sqrt(2xy)=9.72 at the point (3,5) 2)Find a and b such that [-34,7,-27]=a[1,-1,3]+b[11,-2,8] a=? b=?
最佳解答:
Answer for Q1: -1.14 Solution: sqrt(x + 3y) + sqrt(2xy) = 9.72 d[sqrt(x + 3y) + sqrt(2xy)] / dx = d(9.72) / dx [1 + 3(dy/dx)] / [2sqrt(x + 3y)] + [y + x(dy/dx)] / sqrt(2xy) = 0 [1 + 3(dy/dx)][sqrt(2xy)] + [y + x(dy/dx)][2sqrt(x + 3y)] = 0 sub. x = 3, y = 5, we have [1 + 3(dy/dx)]{sqrt[2(3)(5)]} + [5 + 3(dy/dx)][2sqrt(3 + 3(5))] = 0 [1 + 3(dy/dx)](sqrt30) + [5 + 3(dy/dx)](2sqrt18) = 0 sqrt30 + 3(sqrt30)(dy/dx) + 5(2sqrt18) + 3(2sqrt18)(dy/dx) = 0 sqrt30 + 3(sqrt30)(dy/dx) + 30sqrt2 + 18sqrt2(dy/dx) = 0 3(sqrt30)(dy/dx) + 18sqrt2(dy/dx) = - (sqrt30) - 30sqrt2 [3(sqrt30) + 18sqrt2](dy/dx) = - (sqrt30) - 30sqrt2 dy/dx = [- (sqrt30) - 30sqrt2] / [3(sqrt30) + 18sqrt2] = -1.14 Answer for Q2: a = -1, b = -3 Solution: [-34, 7, -27] = a[1, -1, 3] + b[11, -2, 8] [-34, 7, -27] = [a, -a, 3a] + [11b, -2b, 8b] [-34, 7, -27] = [a + 11b, -a - 2b, 3a + 8b] thus, we have a + 11b = -34 ------- (i) -a - 2b = 7 ----------- (ii) 3a + 8b = -27 ------- (iii) (i) + (ii) => 9b = -27 => b = -3 sub b = -3 into (i) => a + 11(-3) = -34 => a = -1 sub a = -1 and b = -3 into (iii) for checking, LHS = 3(-1) + 8(-3) = -27 = RHS therefore, a = -1, b = -3
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