$$\\\boxed{a^3+b^3=(a+b)(a^2-ab+b^2)}\\\\ x^3+\left(\frac{1}{8}\right)^3 =\left(x+\frac{1}{8}\right)\left(x^2-\frac{x}{8}+\frac{1}{64}\right)$$
I'll give you more points too saseflower because you always give such good answers:D
Anon's answer is correct. :)
Here is another answer :) they will give the same answer.
$$\\1.06^x=2\\\\ log(1.06^x)=log2\\\\ xlog1.06=log2\\\\ x=\frac{log2}{log1.06}\\\\$$
the web2 calc is really good for many things but its graphing features are not great.
I suggest that you use
https://www.desmos.com/calculator
Desmos is a beautiful graphing calculator. :)
$$\\12<3y<6 \\ 4
this has no solutions.
if 4<y then y>4
Y cannot be greater than 4 and less than 2 at the same time. :)
Use Desmos
The spanner has the degrees/radian tucked into it.
Maybe this will help. There are a lot of good resources on the net.
http://www.mathsisfun.com/combinatorics/combinations-permutations.html
Hi SoulReaper. It is easier to multiply both sides by 10 first to get rid of all the fractions.
But the answer will be the same. :))
I'm all dolled up, packed my essential.
I feel soooo good
Who's on the menu tonight.
Where's that Zee-man
I am so excited (and hungry).