Question

Yeah uh I have no idea how to do this.....

Answer 1

Answer:

$\frac{9b}{a^{2} }$$\sqrt{\frac{1}{a} }$ should be the answer of the problem that you posted

if you try to solve using $\sqrt{x}$    as as $x^{\frac{1}{2} }$

then $(a^{-5)} ^{\frac{1}{2} }$ =  $(a^{-5/2)}$ = $\frac{1}{a^{\frac{5}{2} } }$

$\frac{1}{a^{\frac{5}{2} } }$ * 9b = $\frac{9b}{a^{\frac{5}{2 } } }$

Step-by-step explanation:

$9 \sqrt{(a^-5 b^2)}$

$x^{-1} = \frac{1}{x}$

$a^{-5} = \frac{1}{a^{5} }$

$\sqrt{b^{2} } = b$

$a^{4}$ a's can be removed from the radicle (one will be left in because you have a^5)

$\frac{9b}{a^{2} }$$\sqrt{\frac{1}{a} }$

Answer 2

Answer:

9b /a^5/2

Step-by-step explanation:

9√(a^-5 b^2)

9 sqrt( a^ -5 b^2)

Rewriting sqrt as ^1/2

9 ( a^ -5 b^2)^1/2

9 ( a^ -5) ^1/2 ( b^2)^1/2

we know that  an exponent to an exponent means multiply

9 a^ (-5*1/2)  b^(2*1/2)

9 a^-5/2 b^1

9b a^ -5/2

We know that x^-y = 1/x^y

9b /a^5/2