## anonymous one year ago plz help!

1. SanjanaP

What's up!

2. Hayleymeyer

alrighttty <3

3. anonymous

$\left(\begin{matrix}5x ^{-2}y ^{5} \\ 4x ^{6}y ^{3}\end{matrix}\right)$

4. anonymous

to the $^{-2}$

5. anonymous

Division? Matrix?

6. anonymous

division

7. anonymous

So $\huge( \large \frac{5x^{-2}y^{5}}{4x^{6}y^{3}} \huge )^{\large -2}~\large ?$

8. anonymous

yes

9. anonymous

Well first let's apply the exponent outside of the parentheses. We're gonna use two properties to do this. The first is $\large (\frac{x}{y})^{2}=(\frac{x^{2}}{y^{2}})$ and the second is $\large (a^{x})^{x}=(a)^{x \times x}$ Do you know how these will be applied?

10. anonymous

no

11. anonymous

It means that the -2 outside the parentheses will be multiplied by the exponents of both the numerator and denominators.

12. anonymous

oh

13. anonymous

thank you

14. anonymous

Do you think you will be able to take it from there?

15. anonymous

yes