MaxWong

This looks hard but I will give it a try.

$\text{Let }x=\log\left(e^{(1+2i)}\right)\\ 10^x= e^{1+2i}\\ 10^x=e\times (e^i)^2$

$e^{i\pi}+1=0\\ (e^i)^\pi + 1=0\\ (e^i)^\pi=-1\\ e^i=(-1)^{\frac{1}{\pi}}$

$10^x=e\times (-1)^{\frac{2}{\pi}}\\ 10^x = e\times (1)^{\frac{1}{\pi}}\\ 10^x = e\\ x = \log e$

$\begin{array}{rll}4-\frac{x}{2}&=&\frac{x}{3}+6\\24-3x&=&2x+36\\5x&=&-12\\x&=&-\dfrac{12}{5}\end{array}$

g(x) = x+5

f(x) = 1/x

It is likely that f(x+5) = 1/x+5

Therefore the composite function is: $(f\circ g )(x)= \dfrac{1}{x+5}$

Pythagoras Theorem forms a relationship between the 3 sides of the triangles

$a^2+b^2=c^2$

Where c is the length of hypotenuse, a and b is the length of the other 2 sides.

$60 \div 157\\\approx 0.382$

Sorry for the messy pic I made it all by myself.

Good job anonymous 12-year-old. Me, a 13-year-old, couldn't have thought of that complex proof :O

Nope. 

cofunction of cot 37.8° would be the cofunction of tan 52.2°

Thanks Alan :D

5+2-(-13)

= 7 + 13

= 20

Guest#1 made a little mistake.