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(6x^2y^5/30x^3y^3)^-2
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\[(\frac{ 6x^2y^5 }{ 30x^3y^3 })^{-2}\]
It is easier if you break it up \[(\frac{ 6 }{ 30 })(\frac{ x^2 }{ x^3 })(\frac{ y^5 }{ y^3 })^-2\]
\[\frac{ 6 }{ 30 }= \frac{1}{5}\] \[\frac{x^2}{x^3}=\frac{1}{x}\] \[\frac{y^5}{y^2}=y^3\]
You get \[(\frac{y^2}{5x})^-2\]
Sorry that's y^3 not 2.
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\[\left(\frac{\left(6 x^2 y^5\right)}{\left(30 x^3 y^3\right)}\right){}^{\wedge}-2 \]\[\left(\frac{y^2}{5 x}\right){}^{\wedge}-2 \]\[\frac{25 x^2}{y^4} \]
Yep.
Thanks again I understand. I am a online student and I need someone to break down each problem for me.
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