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anonymous

  • one year ago

Simplify the expression

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  1. anonymous
    • one year ago
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    \[(x ^{10}y^-5)^{1/5}\div(x^2y^3)^{1/3}\]

  2. anonymous
    • one year ago
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    My answer is \[x^{4/3}\div2y\]

  3. Nnesha
    • one year ago
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    mhm

  4. Nnesha
    • one year ago
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    \[\huge\rm (xy)^m = x^m \times y^m\]remember this exponent rule x and y both raising to the m power

  5. anonymous
    • one year ago
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    Which is what I did

  6. anonymous
    • one year ago
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    The top became (x^2y^-1). I put the y on the bottom because it's negative. The bottom was (x^2/3y).

  7. anonymous
    • one year ago
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    have you got the right answer?

  8. anonymous
    • one year ago
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    is -5 is the power of y?

  9. Nnesha
    • one year ago
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    sorry my internet gt disconnected yesterday.

  10. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @ducksonquack The top became (x^2y^-1). I put the y on the bottom because it's negative. The bottom was (x^2/3y). \(\color{blue}{\text{End of Quote}}\) how did you get 3 coefficient at the bottom ?

  11. Nnesha
    • one year ago
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    \[\huge\rm \frac{ x^2 y^{-1} }{ x^{2 \times \frac{ 1 }{ 3 }} y^{\cancel{3} \times \frac{ 1 }{ \cancel{3}} } }\]

  12. Nnesha
    • one year ago
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    ahhh i see it's x^{2/3}y please put the parentheses to show that (2/3) is an exponent yes that's right

  13. Nnesha
    • one year ago
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    \[\huge\rm \frac{ x^2y^{-1} }{ x^{\frac{ 2 }{ 3 }} y^1 }\] when you divide same bases you should `subtract` their exponents

  14. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @ducksonquack My answer is \[x^{4/3}\div2y\] \(\color{blue}{\text{End of Quote}}\) your x exponent is right x^{4/3} is correct but it's not 2y i assume that's a typo

  15. anonymous
    • one year ago
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    Can you explain it to me then?

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