Question： lim(x→∞) [x + √(x² + 7x)] = ? Solution：
二次不等式, 不難啊! x^2 - 4x - 21 ≦ 0 iff. (x-3)(x+7) ≦ 0 iff. -7 ≦ x ≦ 3 -9x^2 + 9x < 2 iff. 9x^2 - 9x + 2 > 0 iff. (3x-1)(3x-2) > 0 iff. 3x < 1 or 3x > 2 iff. x < 1/3 or x > 2/3 所以, x^2 - 4x - 21 ≦ 0 and -9x^2 + 9x < 2 iff. (-7 ≦ x ≦ 3) and (x...
設 X, Y 分別是第 1, 2 次抽取的硬幣, 則 X, Y 的聯合機率可以列表如下: X = x╲ Y = y │ y = 0.5 y = 1 y = 2 ——————┼———————————— x = 0.5 │ 20/132 15/132 20/132 x = 1 │ 15/132 6/132 12/132...
(a) C(7+13,1) × C(8+7,1) = 20 × 15 = 300 (b) Number of cases where all items are from boutique A = 7 × 8 = 56 The required number of cases that at least one item is from boutique B = 300 - 56 = 244
a) x(x-2)<80 and x-2>0 x²-2x-80<0 and x>2 (x-1O)(x+8)<0 and x>2 -8<x<1O and x>2 2<x<1O b) 2<x<1O and x is...
1. y=-3x^2+5x+4 y=2x-2 x=-1 y=-4 x=2 y=2 2 7x+2(2x+3)<-5 7x+4x+6<-5 11x<-11 x<-11 5-7x>-5x-13 18>2x 9>x so the ans is x<-11
cos(π/9)[sin(π/9)-sin(2π/9)-sin(3π/9)+sin(4π/9)] = (sin(2π/9))/2-(sin(3π/9)+sin(π/9))/2 -(sin(4π/9)+sin(2π/9))/2 + (sin(5π/9)+sin(3π/9))/2 = -(sin(π/9))/2 - (sin(4π/9))/2 + (sin(5π/9))/2 = -(sin(π/9))/2 (∵ sin(4π/9) = sin(π-5π/9) = sin(5π/9)) ∴ sin(π/9)-sin(2π/9)-sin(3π/9)+sin(4π/9) = [-(sin(π/9))/2]/cos(π/9...
(a) sin(3x) = sin(2x+x) = sin(2x)cos(x) + cos(2x)sin(x) = 2sin(x)cos^2(x) + (1-2sin^2(x))sin(x) = 2sin(x)(1-sin^2(x)) + (1-2sin^2(x))sin(x) =3sin(x) - 4sin^3(x) (b) csc(A) = 1/sin(A) csc(π/3-A) = 1/sin(π/3-A) = 1/(sin(π/3)cos(A) - cos(π/3)sin(A)) = 1/(√3 cos...
4sin(x)sin(3x)sin(5x) = 2(cos(3x-x)-cos(3x+x))sin(5x) = 2cos(2x)sin(5x) - 2cos(4x)sin(5x) = (sin(5x+2x)+sin(5x-2x)) - (sin(5x+4x)+sin(5x-4x)) = sin(7x) + sin(3x) - sin(9x) - sin(x) = -sin(x) + sin(3x) + sin(7x) - sin(9x) sin(75°)sin(45°) = (cos(75°-45°) - cos(75°+45°))/2 = (cos(30°) - cos(120°))/2 = (√3/2 + 1/2)/2 = (√3...
2021/01/13: 昨天看到你在回答區問的, 我就補充了. (1/2)tan(θ)(1+tan^2(θ/2))/(tan(θ)+tan(θ/2)) = (1/2)tan(θ)sec^2(θ/2)/(tan(θ)+tan(θ/2)) = (1/2)(1/cos^2(θ/2))/(1+tan(θ/2)/tan(θ)) = (1/2)/(cos^2(θ/2)+cos^2(θ/2)tan(θ/2)/tan(θ)) 請再看看還有哪裡寫得不夠清楚. ============================= 作 線段...
