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.