Verified Answer :

brookegallagher
A\((\frac{w^{-5}}{w^{-9} })^{ \frac{1}{2} } \\ ( \frac{w^{9} }{w^{5} })^{ \frac{1}{2} } \\ (w^{9-5})^{ \frac{1}{2} } \\ (w^{4})^{ \frac{1}{2} } \\ \sqrt{w^{4} } \\ w^{2}\)
b.\((m^{6})^{ \frac{-2}{3} } \\ (\frac{1}{m^{6} })^{\frac{2}{3} } \\ \frac{ \sqrt[3]{1^{2} } }{ \sqrt[3]{m^{6} }^{2} } \\ \frac{ \sqrt[3]{1} }{m^{2*2} } \\ \frac{1}{m^{4} }\)
Hope this helps!
Panoyin

Problem A
\((\frac{w^{-5}}{w^{-9}})^{{\frac{1}{2}}\) = \((\frac{w^{9}}{w^{5}})^{{\frac{1}{2}}\) = \((w^{4})^{{\frac{1}{2}}\) = \(\sqrt[2]{w^{4}}\) = \(w^{2}\)

Problem B
\((m^{6})^{-\frac{2}{3}}\) = \(\frac{1}{(m^{6})^{\frac{2}{3}}}\) = \(\frac{1}{\sqrt[3]{(m^{6})^{2}}}\) = \(\frac{1}{\sqrt[3]{m^{12}}}\) = \(\frac{1}{m^{4}}\)

Problem C
\((\frac{3x^{-4}y^{5}}{2x^{3}y^{-7}})^{-2}\) = \((\frac{3y^{12}}{2x^{7}})^{-2}\) = \(\frac{(3y^{12})^{-2}}{(2x^{7})^{-2}}\) = \(\frac{(2x^{7})^{2}}{(3y^{12})^{2-}}\) = \(\frac{4x^{14}}{9y^{24}}\)


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