Question

22.
Makes s the subject
$\sqrt{p} \: is \: equals \: to \: \sqrt[r]{w \: - as ^{2}}$
​

Answer 1

Step-by-step explanation:

$\sqrt{p} = \sqrt[r]{w - {as}^{2} }$

Find raise each side of the expression to the power of r

That's

$( \sqrt{p} )^{r} = (\sqrt[r]{w - {as}^{2} } ) ^{r}$

we have

$( \sqrt{p} )^{r} = w - {as}^{2}$

Send w to the left of the equation

$( \sqrt{p} )^{r} - w = -{as}^{2}$

Divide both sides by - a

We have

${s}^{2} = -\frac{( \sqrt{p} )^{r} - w}{a}$

Find the square root of both sides

We have the final answer as

$s = \sqrt{ -\frac{( \sqrt{p} )^{r} - w }{a} }$

Hope this helps you