anonymous
  • anonymous
- -2u^2v^4.uv^-2 OVER (u^-3)^3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Is it this: \[\frac{-2u^2v^4\times(uv)^{-2}}{(u^{-3})^3}\] ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
And what you need? :-)

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anonymous
  • anonymous
I need to factorr I believe
anonymous
  • anonymous
sorry I need to simplify
Candy13106
  • Candy13106
hey keith can u plz help me on mine
anonymous
  • anonymous
I keep getting 2u^12v^2 but the two should be negative
anonymous
  • anonymous
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: \[\frac{-2u^2v^4\times(uv)^{-2}}{(u^{-3})^3}=\frac{-2u^2v^4}{(u^{-9})(uv)^2}=\frac{-(u^9)2u^2v^4}{(uv)^2}=\frac{-2u^{11}v^4}{u^2v^2}=-2u^9v^2\]
anonymous
  • anonymous
Im sorry but that's very confusing
anonymous
  • anonymous
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?
anonymous
  • anonymous
okay
anonymous
  • anonymous
And then the second step I realized \(\frac{1}{u^{-9}}=u^9\)
anonymous
  • anonymous
Cool?
anonymous
  • anonymous
Uhhh okay
anonymous
  • anonymous
yo I just need help with the last step, cause it goes to - (-2u^12v^2) so wouldnt negaitve and negative make positve 2?
anonymous
  • anonymous
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
anonymous
  • anonymous
achaa alright alright thanks ! makes a bit more sense now
anonymous
  • anonymous
Lol great im glad!
anonymous
  • anonymous
thank you :)

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