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Mathematics 13 Online
OpenStudy (haleyelizabeth2017):

Simplify. Write in radical form.

OpenStudy (andrewthecookie):

Haley hey :D

OpenStudy (haleyelizabeth2017):

\[\Large\left( \frac{ x^3y^{-2} }{ xy } \right)^{-\frac{ 1 }{ 5 }}\]

OpenStudy (anonymous):

-5 root this whole thing

OpenStudy (haleyelizabeth2017):

I know I need to do that, I'm just confused on what I would get.....

OpenStudy (anonymous):

Give me one second

OpenStudy (anonymous):

well first off the x^3 y^-2 / xy

OpenStudy (anonymous):

How would x and y effect the x and y on the top?

OpenStudy (haleyelizabeth2017):

some cancel

OpenStudy (haleyelizabeth2017):

\(\Large\left( \frac{ x^2y^{-2} }{ y } \right)^{-\frac{ 1 }{ 5 }}\)

OpenStudy (haleyelizabeth2017):

I'm not sure about the y...

OpenStudy (anonymous):

the y would make it y^-3 on the top

OpenStudy (haleyelizabeth2017):

okay....

OpenStudy (anonymous):

Go ahead and rationalize the denominator.

OpenStudy (haleyelizabeth2017):

so \((x^2y^{-3})^{-1/5}\)?

OpenStudy (anonymous):

I mean put y into the denominator

OpenStudy (haleyelizabeth2017):

I'm confused

OpenStudy (anonymous):

( x^2 / y^3 ) ^ -(1/5)

OpenStudy (anonymous):

right now

OpenStudy (haleyelizabeth2017):

\[\Large\left( \frac{ x^2 }{ y^3 } \right)^{-\frac{ 1 }{ 5 }}\] So this?

OpenStudy (anonymous):

Yup

jimthompson5910 (jim_thompson5910):

|dw:1418862971738:dw| so that's how he got y^(-3)

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