MaxWong

Definitions: 

\(\text{Arithmetic mean} = \dfrac{a + b}2\\ \text{Geometric mean} = \sqrt{ab}\\ \text{Harmonic mean} = \dfrac2{\dfrac1a + \dfrac1b}\)

We get, by using the definitions,

\(\begin{cases}a + b = 20\\\dfrac1a + \dfrac1b = \dfrac1{180}\end{cases}\)

From equation (2), 

\(\dfrac{a + b}{ab} = \dfrac1{180}\\ \dfrac{20}{ab} = \dfrac1{180}\\ ab = 3600\\ \text{Geometric mean} = \sqrt{ab} = \sqrt{3600} = \boxed{60}\)

Let d be the common difference.

\(\quad\dfrac{D^2 - A^2}{C^2 - B^2}\\ = \dfrac{(A + 3d)^2 - A^2}{(A + 2d)^2 - (A + d)^2}\\ = \dfrac{((A + 3d) - A)((A + 3d) + A)}{((A + 2d) - (A + d))((A + 2d) + (A + d))}\\ = \dfrac{3\color{red}d (2A + 3d)}{\color{red}d(2A+ 3d)}\\ = 3\)

\(a^2 - b^2 = (a - b)(a + b)\).

By Law of indices, \(9x^2 = 3^2 x^2 = (3x)^2\), and \(64 = 8^2\).

We use the formula and substitute a = 3x, b = 8.

\(9x^2 - 64 = (3x)^2 - 8^2 = (3x - 8)(3x +8)\)

Because AB and BC are horizontal and vertical lines,

AB = 9 - (-6) = 15 units

BC = 6 - (-2) = 8 units

By Pythagorean theorem,

AC = \(\sqrt{8^2 + 15^2}\) = 17 units

The perimeter is 8 + 15 + 17 = 40 units 

\(z^2 (iz + 1) + i(iz + 1) = 0\\ (z^2 + i)(iz + 1) = 0\\ z^2 = -i \text{ or } z = \dfrac{-1}{i} = i\\ \)

For z² = -i, let \(z = re^{i\theta}\).

\(r^2 e^{2i\theta} = 1\cdot e^{3i\pi/2}\\ r = 1\\ \theta = \dfrac{3\pi}4 \text{ or } \theta = \dfrac{7\pi}4\)

Therefore the other two solutions are \(z = e^{3i\pi/4}\) and \(z = e^{7i\pi/4}\), which is \(-\dfrac{\sqrt 2}2 + \dfrac{\sqrt 2}{2} i\) and \(\dfrac{\sqrt 2}2 - \dfrac{\sqrt 2}{2} i\).

Graph: https://www.desmos.com/calculator/bpbp0kmds9

\(f(x) = 1 - \dfrac 2{x + 1}\\ f(f(x)) = 1 - \dfrac2{1 - \dfrac2{x + 1} + 1} = 1 - \dfrac1{\dfrac x{x + 1}} = -\dfrac1x\\ f(f(f(x))) = -\dfrac1{\dfrac{x - 1}{x + 1}} = \dfrac{1 + x}{1 - x}\)

Now substitute x = 5.

Steps: 

1) Find the equation of the straight line first

2) Substitute y = 0 into the equation, solve for x.

3) The answer is the x-intercept.

\(rT = \displaystyle\sum_{k=0}^\infty kr^{k + 1} = \sum_{k = 1}^\infty (k - 1)r^k = \sum_{k = 1}^\infty kr^k -\sum_{k = 1}^\infty r^k = T - S + 1\)

\(T = \dfrac{1 - S}{r - 1}\)

\(T = \dfrac{1 - \dfrac{1}{1 - r}}{r - 1} = \dfrac{r}{(1 - r)^2}\)

\(\dfrac{7 - y}{-1 - (-6)} = \dfrac{-2 - y}{3 - (-6)} \text{ (why?)}\\ \dfrac{7 - y}5 = \dfrac{y + 2}{-9}\\ 9y - 63 = 5y + 10\\ 4y = 73\\ y = \dfrac{73}4 \)