MathProblemSolver101

$\begin{align*} x^8 - y^8 &= (x^4)^2 - (y^4)^2 \\ &= (x^4 - y^4)(x^4 + y^4) \\ &= ((x^2)^2 - (y^2)^2)(x^4 + y^4) \\ &= (x^2 - y^2)(x^2 + y^2)(x^4 + y^4) \\ &= (x-y)(x+y)(x^2 + y^2)(x^4 + y^4)  \end{align*}$

Can you take it from here?

Hint: Manipulate $x^2 + y^2$ into $(x+y)^2 - 2xy$ and do the same thing with $x^4 + y^4.$ You can substitute your equation for $x^2 + y^2$ and do some further manipulation and you are done. If you have degrees $>1,$ you can square root it.

$\frac{2x+3}{4} = \frac{x+3}{2}$

$2x + 3 = 2x + 6$

$0 = -3$
$\boxed{\varnothing}$

Let the two boys and Jill be one: There are $\binom{3}{2} = 3$ possibilities to permute them.

Then there are $5-3 + 1 = 3$ people. There is $\binom{5}{3} = 10.$

the answer is $10\cdot 3 = \boxed{30}$

$x = \frac{k-14}{5}$

$\binom{8}{3}=56$

X-30 = 0.2x - 14

0.8 x  = 16

x = 20

20+40= 60

(8/4)*12 = 2 *12= 24

X^2 + 2x + 3 = x^2-x-20

3x = -23

x= -23/3

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$\tan(\theta) = \frac{58}{35}$

$\theta = \tan^{-1} (\frac{58}{35})$

$\boxed{\theta = 59}$

Let ? be $x$

$\tan(x) = \frac{13}{20}$

$x = \tan^{-1}(\frac{13}{20})$ is about $33^{\circ}$