$\displaystyle d=\sqrt{\frac{3h}{2}}$

I got

$\displaystyle h=\frac{2d^2}{3}$

is this correct?

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- Jun 1st 2010, 05:29 AMMukilabMake h the subject of the formula
$\displaystyle d=\sqrt{\frac{3h}{2}}$

I got

$\displaystyle h=\frac{2d^2}{3}$

is this correct? - Jun 1st 2010, 05:33 AMmathaddict