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
Answer:
[tex]\frac{9b}{a^{2} }[/tex][tex]\sqrt{\frac{1}{a} }[/tex] should be the answer of the problem that you posted
if you try to solve using [tex]\sqrt{x}[/tex] as as [tex]x^{\frac{1}{2} }[/tex]
then [tex](a^{-5)} ^{\frac{1}{2} } \\[/tex] = [tex](a^{-5/2)}[/tex] = [tex]\frac{1}{a^{\frac{5}{2} } }[/tex]
[tex]\frac{1}{a^{\frac{5}{2} } }[/tex] * 9b = [tex]\frac{9b}{a^{\frac{5}{2 } } }[/tex]
Step-by-step explanation:
[tex]9 \sqrt{(a^-5 b^2)}[/tex]
[tex]x^{-1} = \frac{1}{x}[/tex]
[tex]a^{-5} = \frac{1}{a^{5} }[/tex]
[tex]\sqrt{b^{2} } = b[/tex]
[tex]a^{4}[/tex] a's can be removed from the radicle (one will be left in because you have a^5)
[tex]\frac{9b}{a^{2} }[/tex][tex]\sqrt{\frac{1}{a} }[/tex]
