can someone please explain this to me thoroughly because i keep getting confused on how o simplify this expression

- AngelaB97

- jamiebookeater

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- AngelaB97

\[(x^2/y^-2)^-2 (2y^-3/x^2)^3\]

- AngelaB97

|dw:1441669920633:dw|

- AngelaB97

it is supposed to be 8/x^10y^13 bu i keep getting 8/x^10y^5

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## More answers

- Jhannybean

\[\large \left(\frac{x^2}{y^{-2}}\right)^{-2}\left(\frac{2y^{-3}}{x^2}\right)^3\]Change the variables inside the parenthesis so they are all positive. \[\large ( x^2y^2)^{-2}\left(\frac{2}{x^2y^3}\right)^3\]Now that we've got all the variables INSIDE the parenthesis with positive powers, now we take care of the outer powers. Take the first one and put it over 1. \[\frac{1}{(x^2y^2)^2}\cdot \left(\frac{2}{x^2y^3}\right)^3\]

- Jhannybean

Now let's distribute the powers through all variables inside the parenthesis. \[\frac{1}{(x^2)^2(y^2)^2} \cdot \frac{2^3}{(x^2)^3(y^3)^3}\]

- AngelaB97

so since the exponent outside of the parentheses is negative aren't you supposed to flip it so it becomes y^2/x^2 ?

- AngelaB97

r it that where i am making the mistake?

- Jhannybean

- AngelaB97

- Jhannybean

Hmm... not exactly. We take care of each step individually.

- AngelaB97

see, its (x^2/y^-2) isn't it supposed to be (y^2/x^2)^2

- AngelaB97

- Jhannybean

So we can write out \(\left(\dfrac{x^2}{y^{-2}}\right)^{-2}\) as \(\dfrac{1}{\left( \dfrac{x^2}{\dfrac{1}{y^2}}\right)^2}\)

- Jhannybean

Do you see whats happening here?

- Jhannybean

I don't think it's proper to say \(\left(\dfrac{x^2}{y^{-2}}\right)^{-2} \iff \left(\dfrac{y^2}{x^2}\right)^2\)

- AngelaB97

oh i understand now

- AngelaB97

can you also help me with another problem please?

- Jhannybean

Are you finished with this one?

- AngelaB97

yes i am

- Jhannybean

Alright :)

- AngelaB97

okay so it's (2y^-1z/z^2)^-1 (y/3z^2)^2

- Jhannybean

This is a new problem?

- AngelaB97

yes

- Jhannybean

\[\left(\frac{2y^{-1}z}{z^2}\right)^{-1}\cdot \left(\frac{y}{3z^2}\right)^2\]

- Jhannybean

So like our previous problem, what will you want to do first?

- AngelaB97

flip the problem?

- Jhannybean

No, what do we do to all the negative variables within the parenthesis?

- AngelaB97

we witch them dont we?

- Jhannybean

By switch you mean... make them positive? :)

- AngelaB97

yes

- Jhannybean

You are correct. Therefore we will have \[\left(\frac{2z}{yz^2}\right)^{-1}\cdot \left(\frac{y}{3z^2}\right)^2\]

- AngelaB97

yes so then it becomes (yz^2/2z)^1 ?

- AngelaB97

multiplied by (y/3z^2)^2

- AngelaB97

so first you make the numbers inside the parentheses positive, then you flip it whenever there's a negative exponent over the parentheses?

- Jhannybean

\(\color{#0cbb34}{\text{Originally Posted by}}\) @AngelaB97
so first you make the numbers inside the parentheses positive, `then you flip it whenever there's a negative exponent over the parentheses?`
\(\color{#0cbb34}{\text{End of Quote}}\)
Im not understanding what you mean by that highlighted portion.

- AngelaB97

as in whenever you have a negative exponent both outside and inside the parentheses, you have to deal with the numbers inside the parentheses?

- Jhannybean

Yes, you deal with the numbers inside the ( ) first, making them + , then we take care of the outside ( ) by putting over 1. [i.e 1/ ( ) ]

- Jhannybean

Does that make sense?

- AngelaB97

yes exactly

- Jhannybean

Cool! \(\checkmark\) Glad you're following along.

- Jhannybean

Now we have: \[\left(\frac{2z}{yz^2}\right)^{-1}\cdot \left(\frac{y}{3z^2}\right)^2\] We're going to put the first one over 1 to make it +. \[\frac{1}{\left(\dfrac{2z}{yz^2}\right)} \cdot \left(\frac{y}{3z^2}\right)^2\]

- Jhannybean

SOrry, bear with me, Im lagging a lot right now.

- AngelaB97

don't worry you're helping me alot

- Jhannybean

This becomes: \[\frac{yz^2}{2z} \cdot \left(\frac{y}{3z^2}\right)^2\]

- Jhannybean

Now we can reduce our fraction. Remember, \(\dfrac{x^m}{x^n} = x^{m-n} \therefore \dfrac{z^2}{z} = z^{2-1} = z\) Now we can rewrite this as: \[\frac{yz}{2} \cdot \left(\frac{y}{3z^2}\right)^2\]

- AngelaB97

then yz/2 * y^2/9z^4

- Jhannybean

exactly.

- Jhannybean

We get \[=\frac{y^3z}{18z^4}\] \[\frac{z}{z^4} =z^{1-4} =z^{-3}\] \[=\frac{y^3}{18z^{-3}} \iff \boxed{\frac{y^3z^3}{18}}\]

- Jhannybean

Whoop whoop wait a minute.

- Jhannybean

Typo.

- Jhannybean

we got \(z^{-3}\) therefore it would be: \[\frac{y^3z^{-3}}{18} \iff \boxed{\frac{y^3}{18z^3}} \]

- AngelaB97

gotcha, thanks sosososo much!!! :) <3

- AngelaB97

you helped me out so much today

- Jhannybean

No problem :) I've got to head off now. SO good luck on the rest!

- Jhannybean

Im glad you participated instead of blatantly asking for the answer.

- AngelaB97

thank you!!

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