anonymous
  • anonymous
I just need someone to come behind me and check my work, and maybe help if needed. \[\left( \frac{ x ^{-6}y ^{3} }{ x ^{-3}y ^{3}} \right)^{1/3}\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\left( \frac{ x ^{-2}y }{ x ^{-1}y } \right)\]
anonymous
  • anonymous
keep going
anonymous
  • anonymous
\[x ^{-2 - (-1)} = x ^{-3}\]

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anonymous
  • anonymous
So now would it be \[\frac{ y }{ x ^{3}}\]
Australopithecus
  • Australopithecus
You have made a mistake, the best way to deal with these questions is to bring all the variables up from the denominator using the rule, \[x^{-1} = \frac{1}{x}\]
Australopithecus
  • Australopithecus
Then just add all the exponents on the same variables
anonymous
  • anonymous
So \[\frac{ 3 }{ x } ?\]
Australopithecus
  • Australopithecus
for example: \[\frac{xy^{2}}{z^{2}x^{-2}} = x^{1}z^{-2}y^{2}x^{2}\]
Australopithecus
  • Australopithecus
now it is easy to add variables, for x 1 + 2 = 3 so we have x^(3) so we have (x^1)z^-2(y^2) = ((x^1)(y^2))/z^-2
Australopithecus
  • Australopithecus
oops (x^3)z^-2(y^2) = ((x^3)(y^2))/z^-2
Australopithecus
  • Australopithecus
made a mistake
anonymous
  • anonymous
I'm sorry, I don't really get this principle. What does that last equation you posted mean?
Australopithecus
  • Australopithecus
wow I keep making mistakes (x^3)z^-2(y^2) = ((x^3)(y^2))/z^2
anonymous
  • anonymous
All of the parentheses are confusing me there.
Australopithecus
  • Australopithecus
I will give you another example then: \[\frac{c^2y^{-2}}{c^{-3}y^{3}} = c^2c^{3}y^{-2}y^3\] so for c variable, 2 + 3 = 5 for the y variable -2 + (3) = 1 so we have, \[c^{5}y^{1} = c^{5}y\]
Australopithecus
  • Australopithecus
*Hint* Any variable raised to the power 0 is equal to 1 \[x^0 = 1\]
anonymous
  • anonymous
Okay I get it now, thanks!
Australopithecus
  • Australopithecus
You just need to apply the two rules i gave you and these problems become super easy
Australopithecus
  • Australopithecus
\[x^{-1} = \frac{1}{x^{1}}\] and \[x^{1} = \frac{1}{x^{-1}}\] Remembering, \[x^1 = x\]

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