Question

help again thanks! Anyone out there that can help?

Answer 1

\[\left(\begin{matrix}2x^3y ^{-3} \over\ x^{-2}y^4\end{matrix}\right)^{-5}\]

Answer 2

\[
\left(\begin{matrix}2x^3y ^{-3} \over\ x^{-2}y^4\end{matrix}\right)^{-5}\\
(2x^5y ^{-7})^{-5}\\
2^{-5}x^{-25}y^{35}\\
\dfrac{y^{35}}{32x^{25}}
\]

Answer 3

you mean 2 not 32 and flip them right

Answer 4

nm

Answer 5

your right

Answer 6

4 rules help you simplify this. Use these rules in this order:

1) With a negative exponent, take the reciprocal of the base, then make the exponent positive.
2) With an exponent around parenthesis, as long as the parenthesis are multiplied things and not added things, you can distribute the exponent to all of the factors inside and apply the exponent individually.
3) When you have two things with the same base and you divide them, you can just keep the base, but subtract the exponents like this:
\[\frac{a^b}{a^c} = a^{b-c}\]
4) Finally, if something has a negative exponent, you can move it from the numerator to the denominator or vice versa and change it to positive.

Answer 7

thank that was very helpful im going to write that down! :D

Answer 8

Excellent =D I'm glad.