Question

- -2u^2v^4.uv^-2 OVER (u^-3)^3

Answer 1

Is it this:
\[\frac{-2u^2v^4\times(uv)^{-2}}{(u^{-3})^3}\]
?

Answer 2

yes

Answer 3

And what you need? :-)

Answer 4

I need to factorr I believe

Answer 5

sorry I need to simplify

Answer 6

hey keith can u plz help me on mine

Answer 7

I keep getting 2u^12v^2 but the two should be negative

Answer 8

Haha alright! well you need to know two laws:
\(x^{-a}=\frac{1}{x^a}\)
And
\((x^a)^b=x^{ab}\)
So we can proceed:
\[\frac{-2u^2v^4\times(uv)^{-2}}{(u^{-3})^3}=\frac{-2u^2v^4}{(u^{-9})(uv)^2}=\frac{-(u^9)2u^2v^4}{(uv)^2}=\frac{-2u^{11}v^4}{u^2v^2}=-2u^9v^2\]

Answer 9

Im sorry but that's very confusing

Answer 10

Haha its okay my bad. So you start with The original equation.
Then the second step I remembered that \((u^{-3})^3=u^{-9}\) and that \((uv)^{-2}=\frac{1}{(uv)^2}\) Okay?

Answer 11

okay

Answer 12

And then the second step I realized \(\frac{1}{u^{-9}}=u^9\)

Answer 13

Cool?

Answer 14

Uhhh okay

Answer 15

yo I just need help with the last step, cause it goes to - (-2u^12v^2) so wouldnt negaitve and negative make positve 2?

Answer 16

And then the third step I realized that \((u^9)(u^2)=u^{11}\) and that \((uv)^2=u^2v^2\)
And the last step I divided everything out

Answer 17

achaa alright alright thanks ! makes a bit more sense now

Answer 18

Lol great im glad!

Answer 19

thank you :)