# How do you simplify (9r^4)^0.5?

$3 {r}^{2}$
$0.5 = \frac{1}{2}$, and $\frac{1}{2}$ as an exponent is equivalent to a square root:
${a}^{\frac{1}{2}} = \sqrt{a}$ (when $a \ge 0$)
So, ${\left(9 {r}^{4}\right)}^{0.5} = \sqrt{9 {r}^{4}} = \sqrt{9} \cdot \sqrt{{r}^{4}} = 3 {r}^{2}$