How do you simplify ${(9 r^{4})}^{0.5}$ $(9r^4)^0.5$ ?

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Answer 1 · 2017-01-15T17:31:35

$3 r^{2}$ $3r^2$

$0.5 = \frac{1}{2}$ $0.5 = 1/2$ , and $\frac{1}{2}$ $1/2$ as an exponent is equivalent to a square root:

$a^{\frac{1}{2}} = \sqrt{a}$ $a^(1/2) = sqrta$ (when $a \geq 0$ $a >= 0$ )

So, ${(9 r^{4})}^{0.5} = \sqrt{9 r^{4}} = \sqrt{9} \cdot \sqrt{r^{4}} = 3 r^{2}$ $(9r^4)^(0.5) = sqrt(9r^4) = sqrt9 * sqrt(r^4) = 3r^2$

How do you simplify (9r^4)^0.5(9r4)0.5(9r^4)^0.5?