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TBNRfrags

  • one year ago

Help pl0x http://prntscr.com/7j31oq

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  1. UsukiDoll
    • one year ago
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    \[\frac{12x^4y^3z^2w}{2xy^2}\]

  2. UsukiDoll
    • one year ago
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    all we need to do is use division for the numbers only and use one of the exponent rules for the variables. \[\frac{x^a}{x^b} \rightarrow x^{a-b}\] where x = variable a and b = numbers so for the numbers only we have 12/2, so what's 12 divided by 2

  3. TBNRFrags
    • one year ago
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    6

  4. UsukiDoll
    • one year ago
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    yes.. so for the variables only we noticed that only x and y are present on the numerator and denominator, so we use the exponent law to figure this part out ... so if we let a = 4 and b = 1 from earlier \[\frac{x^4}{x^1} \rightarrow x^{4-1}\] we just have to subtract.. so what's 4-1 ?

  5. TBNRFrags
    • one year ago
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    3

  6. UsukiDoll
    • one year ago
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    alright so we have \[x^3\]

  7. UsukiDoll
    • one year ago
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    we do the same thing for the y variable because that exists on the numerator and denominator \[\frac{y^3}{y^2} \rightarrow y^{3-2}\] so all we need to do is subtract.. what is 3 -2 ?

  8. TBNRFrags
    • one year ago
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    1

  9. UsukiDoll
    • one year ago
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    yes so now we have just \[y^1 \] or you can write it as y only because it's to the first power... those 3 arithmetics we've just done are only necessary for this problem because z and w aren't present in the denominator, so we just leave them alone.

  10. UsukiDoll
    • one year ago
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    \[6x^3yz^2w\]

  11. UsukiDoll
    • one year ago
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    and that's our final answer.

  12. TBNRFrags
    • one year ago
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    tyvm<3

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spraguer (Moderator)
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