MaxWong

Let x be the third length of the triangle.

By triangle inequality, 

$x < 17 + 11 \\ x < 28$

The greatest integer solution of this inequality is $x = 27$.

Therefore the longest possible third side is 27

x + y + z = 1.

$\dfrac{x + y + z}3 = \dfrac13$.

By AM-GM inequality, 

$\dfrac{x + y +z}3 \geq \sqrt[3]{xyz}\\ \sqrt[3]{xyz} \leq\dfrac13\\ xyz \leq \dfrac1{27}$

Maximum value = $\dfrac1{27}$

$(a + b)^{-1}(a^{-1} + b^{-1}) \equiv (a + b)^{-1}(a + b)(ab)^{-1}\equiv(ab)^{-1}\equiv2\pmod{n}$

$\quad \dfrac1{\sqrt2 +\sqrt5+\sqrt7}\\ =\dfrac{\sqrt2-\sqrt5+\sqrt7}{(\sqrt2+\sqrt7)^2-\sqrt5^2}\\ =\dfrac{\sqrt2-\sqrt5+\sqrt7}{4+2\sqrt{14}}\\ =\dfrac{(\sqrt2-\sqrt5+\sqrt7)(2-\sqrt{14})}{(4+2\sqrt{14})(2-\sqrt{14})}\\ =\dfrac{\sqrt{70}-2\sqrt5-5\sqrt2}{2(2^2 - \sqrt{14}^2)}\\ = \dfrac{2\sqrt5 + 5\sqrt 2 - \sqrt{70}}{20}$

$\lceil 2x \rceil$ may not be equal to $2 \lceil x \rceil$. But indeed, this equation has no solutions.

(a) Because $z\bar{z} = |z|^2$, $z\bar{z} = 1^2$. This means $\overline z= \dfrac1z$.

Similarly, with the exact same steps, $\overline w= \dfrac1w$.

(b) We write z and w in polar form. Let $z = \cos \theta + i\sin \theta$ and $w = \cos \phi + i\sin\phi$.

$\dfrac{z + w}{zw + 1} = \dfrac{(\cos \theta + \cos \phi) + i(\sin \theta + \sin \phi)}{(\cos \theta + i\sin \theta)(\cos \phi + i\sin \phi) + 1} = \dfrac{(\cos \theta + \cos \phi) + i(\sin \theta + \sin \phi)}{(\cos \theta \cos \phi - \sin \theta \sin \phi + 1) + i(\sin \theta \cos \phi + \cos \theta \sin \phi)}$

Simplifying, $\dfrac{z + w}{zw + 1} = \dfrac{((\cos \theta + \cos \phi) + i(\sin \theta + \sin \phi))(\cos(\theta + \phi) + 1 - i\sin(\theta + \phi))}{(\cos(\theta + \phi) + 1)^2+\sin^2 (\theta + \phi)}$

This means $\operatorname{Im}\left(\dfrac{z + w}{zw + 1}\right) = \dfrac{(\cos(\theta + \phi) + 1)(\sin \theta + \sin \phi) - (\cos \theta + \cos \phi)\sin(\theta + \phi)}{(\cos(\theta + \phi) + 1)^2 + \sin^2(\theta + \phi)}$

Notice that 

$(\cos(\theta + \phi) + 1)(\sin \theta + \sin \phi) \\= \sin \theta + \sin \phi + \sin \theta \cos \theta \cos \phi -\sin^2 \theta \sin \phi + \sin \phi \cos \theta \cos \phi - \sin \theta \sin^2 \phi\\=\sin \theta \cos^2 \phi + \sin \phi \cos^2 \theta+\sin \theta \cos \theta \cos \phi + \sin \phi \cos \theta \cos \phi$

Also, 

$(\cos \theta + \cos \phi)\sin(\theta + \phi)\\ = \cos \theta \sin \theta \cos \phi + \cos \theta \sin \phi \cos \theta+\cos \phi \sin \theta \cos \phi + \cos \phi \sin \phi \cos \theta\\ = \cos \theta \sin \theta \cos \phi + \cos^2 \theta \sin \phi+\cos^2 \phi \sin \theta + \cos \phi \sin \phi \cos \theta$

This means $(\cos(\theta + \phi) + 1)(\sin \theta + \sin \phi) = (\cos \theta + \cos \phi)\sin(\theta + \phi)$.

Then it immediately follows that $\operatorname{Im}\left(\dfrac{z + w}{zw + 1}\right) = 0$, which means $\dfrac{z + w}{zw + 1}$ is a real number.

We first focus on the first and last part of the equality.

(y - 2)/3 = 24/(y - 1)

(y - 2)(y - 1) = 72

y^2 - 3y + 2 - 72 = 0

y^2 - 3y - 70 = 0

(y - 10)(y + 7) = 0

y = 10 or y = -7

Case 1: y = 10

Then (10 - 2)/3 = (1 - 5x)/(10 - 4).

i.e., (1 - 5x)/6 = 8/3

(1 - 5x) = 16

x = -3

So (x, y) = (-3, 10) is a solution.

Case 2: y = -7

Then (-7 - 2)/3 = (1 - 5x)/(-7 - 4)

i.e., 1 - 5x = 33

x = -32/5

So (x, y) = (-32/5, -7) is another solution.

Perpendicular: Product of slopes = -1.

Let m be the slope of the line perpendicular to y = -2x + 5.

Slope of y = -2x + 5 is -2.

This means $-2\cdot m = -1$, i.e., $m = \dfrac12$.

The required line passes through (5, 5) and the slope is 1/2. We use the point-slope form of the equation of a straight line.

The equation of the required line is 

$y - 5 = \dfrac12 \left(x - 5\right)\\ 2y - 10 = x - 5\\ \boxed{x - 2y + 5 = 0}$

The velocity of the wave, the frequency, and the wavelength is related by the formula: $v = f\lambda$.

We substitute v = (speed of light) = c, f = $7.5\cdot 10^{14}$.

$3\cdot 10^8 = 7.5\cdot10^{14} \cdot \lambda\\ \lambda = 4 \cdot 10^{-7}\text{ m}$

Then, we use Einstein's equation $E = hf$ to find the energy of the photon.

$E = 6.6262\cdot 10^{-34}\cdot 7.5\cdot 10^{14}\\ E = 4.96965 \cdot 10^{-19} J$

What do I do with this function? 

Find the derivative? Find the domain? Please specify.