MaxWong

$\text{I am going to go through the 2 possibilities of what this question means.(5^x-3 and 5^(x-3))}$

$5^x-3=10\\ 5^x=13\\ x = log_513\\\text{Which is about 1.59}$

$5^{x-3}=10\\ x= log_510 + 3\\\text{Which is about 4.43}$

Oops I was just about the reveal the answer for the last one :P

$f(x+1)\\=4(x+1)^2+2(x+1)-2\\=4x^2+8x+4+2x+2-2\\=4x^2+10x+4$

$f(-3)\\=4(-3)^2+2(-3)-2\\=28$

I am sorry to say that your trick question is too easy for me and the answer is '? = 5' :P

Guest #5, I don't quite understand your answer. But Alan's answer did help me a lot. Thanks Alan. And Guest#5 can you give a more detailed explanation?

$\text{Why is }\sqrt{i}=\frac{1+i}{\sqrt2}? \text{I still don't understand how to obtain this answer?}$

Just discovered how to escape from Math mode of LaTeX while the above method won't work for me(The $ one)

\text{}

$\text{2^5}$\text{2^5}

$\text{The above one have completely escaped from math mode, so is this one}$

\text{The above one have completely escaped from math mode, so is this one}

LOL

$\left(z-\frac{2}{\sqrt{z}}\right)^9$

$\text{The general term is}$

$(9Cr)(z)^{(9-r)}\left(\frac{-2}{\sqrt{z}}\right)^r$

$=(9Cr)\left(\frac{z^9}{z^r}\right)\left(\frac{(-2)^r}{(\sqrt{z})^r}\right)$

$=(9Cr)\left(\frac{z^9*(-2)^r}{z^r*(\sqrt{z})^r}\right)$

$=(9Cr)\left(\frac{z^{18/2}*(-2)^r}{z^{(2r/2)}*(z)^{(r/2)}}\right)$

$=(9Cr)\left(\frac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)$

$=(9Cr)(z^{((18-3r)/2)}*(-2)^r)$

$\text{The constant term will be when}$

$(18-3r)/2=0$

$18-3r=0$

$r=6$

Just wanna try. This code is really crazy! Typing it twice takes me 30minutes. (I accidentally deleted when I have already finished)