MaxWong

I haven't been active here for a long time, so I will post an answer :)

\(\begin{array}{rcl} f(x) &=& xe^{-x} + 3x\\ f'(x) &=& x\dfrac{d}{dx} \left(e^{-x}\right)+e^{-x}\dfrac{d}{dx} \left(x\right) + 3\dfrac{d}{dx} x\\ &=& x\left(-e^{-x}\right) + e^{-x} + 3\\ &=& (1 - x)e^{-x} + 3 \end{array}\)

\(\text{Let }\angle A = x, \angle PBC = \angle PBF = y, \angle PCB = \angle PCE = z\\ \begin{cases} x + y +z = 111^{\circ}\\ x+2y+2z = 180^{\circ} \end{cases} \implies y + z = 69^{\circ}\\ x + 69^{\circ} = 111^{\circ}\\ x = 42^{\circ}\\ \angle A = 42^{\circ}\)

1)

\(f(x) = q(x) d(x) + r(x)\\ \text{If }\text{deg}\;d \le 3, \text{then deg } r < \text{deg }d \le 3 \implies \text{Contradiction!}\\ \therefore \text{deg }d \ge 4\\ \text{Minimum possible deg }d = 4\\ \implies \text{Maximum possible deg }q = \text{deg }f - \text{deg }d = \boxed5\)

Your expansion is correct! But you don't have any term that is independent of x, so the constant term is 0.

1. 

Parent function: \(f(x) = \left(\dfrac{4}5{}\right)^x\\\)

Transformation: Vertical compression with a factor of 16/25.

2.

Parent function: \(f(x) = 2^x\)

Transformation: Reflect along x-axis, vertical stretch with a factor of 3/2, then translation upwards by 7 units.

\(\phantom{=} \displaystyle\sum_{x = 0^{\circ}}^{90^{\circ}} \cos^2 x\\ = \cos^2 0^{\circ} + \cos^2 1^{\circ} + \cdots + \cos^2 90^{\circ}\\ = 1 + (\cos^2 1^{\circ} + \cos^2 2^{\circ} + \cos^2 3^{\circ} + \cdots + \cos^2 89^{\circ}) + 0\\ = 1 + (\cos^2 1^{\circ} + \cos^2 89^{\circ}) + (\cos^2 2^{\circ} + \cos^2 88^{\circ}) + \cdots + (\cos^2 44^{\circ} + \cos^2 46^{\circ}) + \cos^2 45^{\circ}\\ \boxed{\text{Note that }\cos^2 n^{\circ} + \cos^2 (90 - n)^{\circ} = 1}\\ = 1 + 1 + 1 + 1 + \cdots + 1 + \dfrac{1}{2}\\ \boxed{\text{There are }\color{red}44\color{black}\text{ 1's}}\\ = 44 \dfrac{1}{2}\\ = \boxed{\dfrac{89}{2}}\)

\(\phantom{=}|(1 + 2i)^8|\\ = |1 + 2i|^8\\ = \left(\sqrt 5\right)^8\\ = 625\)

\(\displaystyle \sum_{k = 1}^9 \cos 3x_k\\ = \displaystyle \sum_{k = 1}^9 \left(4\cos^3 x_k-3\cos x_k\right)\\ = 4\displaystyle \sum_{k = 1}^9 \cos^3 x_k - 3\sum_{k = 1}^9 \cos x_k\\ =4\displaystyle \sum_{k = 1}^9 \cos^3 x_k\\ \le 4 \left(\displaystyle \sum_{k = 1}^9 \cos x_k\right)^3\\ = 0\\ \therefore \text{Maximum value is 0.}\)

We know that f(k) is a quadratic function in k.

Let f(k) = ak² + bk + c.

\(\phantom{=}\displaystyle \sum_{k = 1}^n f(k)\\ = a \displaystyle\sum_{k=1}^n k^2 + b\displaystyle\sum_{k=1}^n k + c\displaystyle\sum_{k=1}^n 1\\ = \dfrac{an(n + 1)(2n + 1)}{6} + \dfrac{bn(n + 1)}{2} + cn\\ = \dfrac{n}{6} \left(a(2n^2 + 3n + 1) + 3b(n + 1) + 6c\right)\\ = \dfrac{n(2an^2 + (3a + 3b)n + a + 3b + 6c)}{6}\\ \text{Compare with }n^3,\\ n^3 = \dfrac{n(2an^2 + (3a + 3b)n + a + 3b + 6c)}{6}\\ 6n^3 = 2an^3 + 3(a + b)n^2 + (a + 3b + 6c)n\\ \text{Comparing coefficients, }\\ 2a = 6 \implies a = 3\\ 3(3 + b) = 0 \implies b = -3\\ 3 + 3(-3) + 6c = 0 \implies c= 1\\ \therefore f(k) = 3k^2 -3k + 1\)

It will take 30 minutes to turn 1/3 of a round,

So it will take 6(30) = 180 minutes to turn 6(1/3) = 2 rounds.