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simplify the following:
a) ((w^-5)/(w^-9))^1/2
b) (m^6)^-2/3
c) (3x^-4y^5)/((2x^3y^-7))^-2

Sagot :

brookegallagher
a. [tex] (\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} [/tex]
b.[tex](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} } [/tex]
Hope this helps!
Panoyin
Problem A
[tex](\frac{w^{-5}}{w^{-9}})^{{\frac{1}{2}} [/tex] = [tex](\frac{w^{9}}{w^{5}})^{{\frac{1}{2}} [/tex] = [tex](w^{4})^{{\frac{1}{2}} [/tex] = [tex]\sqrt[2]{w^{4}} [/tex] = [tex]w^{2}[/tex]

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

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


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