Question

Simplify the following:

Answer 1

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!

Answer 2

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