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anonymous

  • 5 years ago

(2x ^{-3} yz ^{-6}) (2x)^{-5} (I'm little lost on how to get y/16^8 z^6) ?.? only at this part (-32x^-5)/2x^-3yz^-6

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  1. anonymous
    • 5 years ago
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    |dw:1327351420225:dw|

  2. anonymous
    • 5 years ago
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    Thanks Pross

  3. anonymous
    • 5 years ago
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    :)

  4. anonymous
    • 5 years ago
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    (2^{-1}x ^{-2}y ^{-1})^{-2}(2x ^{-3}y)^{-2}(16x ^{-3}y ^{3})^{0}/ (2x ^{-3}y ^{-5)^{2}}

  5. anonymous
    • 5 years ago
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    first multiply each exponent in the first large parentheses times the -2. Then in the second large parentheses the zero exponent will make the whole term equal to 1.

  6. anonymous
    • 5 years ago
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    that should get you started.

  7. anonymous
    • 5 years ago
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    I starting to understand a little, now I'm little confuse on this part (4x^4y^2) (2x^6 y ^-2) (xy)/ 4x^-6 y^-10. ?.?

  8. anonymous
    • 5 years ago
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    |dw:1327353996471:dw| I am not sure that I have all of your parentheses correctly written out...take a look and see...

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