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Math Help - Dividing with Brackets And Indicies

  1. #1
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    Dividing with Brackets And Indicies

    My Question is this:

    (12pq-r)- / 144q^7
    -------------_________
    ----------------5p

    Note: Last part is 144q^7 OVER 5p, not all over 5p

    I have attempeted the first part and ended up with -1728p^4q^-9r.
    Am i on the right track here? If so where do i go from here? How do i do the 144q^7 over 5p, and then incorporate it back into the -1728... part?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mcaelli View Post
    My Question is this:

    (12pq-r)- / 144q^7
    -------------_________
    ----------------5p

    Note: Last part is 144q^7 OVER 5p, not all over 5p

    I have attempeted the first part and ended up with -1728p^4q^-9r.
    Am i on the right track here? If so where do i go from here? How do i do the 144q^7 over 5p, and then incorporate it back into the -1728... part?
    If I am reading this right:
    \frac{(12p^2q - 3r)^{-3}}{\frac{144q^7}{5p^2}}

    Get rid of the fraction in the denominator:
    = \frac{(12p^2q - 3r)^{-3}}{\frac{144q^7}{5p^2}} \cdot \frac{5p^2}{5p^2}

    = \frac{5p^2(12p^2q - 3r)^{-3}}{144q^7}

    Now take care of the negative exponent:
    = \frac{5p^2}{144q^7(12p^2q - 3r)^3}


    -Dan
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  3. #3
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    Ah the first bit with the brackets is : (144p^2 q^-3 r)^-3.
    Does this alter what you have done very much?
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  4. #4
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    Hello, mcaelli!

    Simplify: . \frac{(12p^2q^{-3}r)^{-3}}{\frac{144q^7}{5p^2}}

    We have: . \frac{12^{-3}(p^2)^{-3}(q^{-3})^{-3}r^{-3}}{1} \cdot \frac{5p^2}{144q^7}  \;=\;\frac{12^{-3}p^{-6}q^9r^{-3}}{1} \cdot\frac{5p^2}{144q^7}

    . . = \;\frac{q^9}{12^3p^6r^3}\cdot\frac{5p^2}{12^2q^7} \;=\;\frac{5p^2q^9}{12^5p^6q^7r^3} \;=\;\frac{5q^2}{12^5p^4r^3}

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