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Math Help - Divide the expression

  1. #1
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    Divide the expression

    Can someone help me divide the following expression: 112a^0b^3c^-4/ 48a^3b^0c^3 ? Thanks much!
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  2. #2
    Super Member bigwave's Avatar
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    misunderstood first initial post

    Quote Originally Posted by jay1 View Post
    Can someone help me divide the following : 112a^0b^3c^-4/ 48a^3b^0c^3 ? Thanks much!
    <br />
\frac{112a^0b^3c^{-4}}{48a^3b^0c^3} <br />
\Rightarrow \frac{112}{48}\ \frac{a^0}{a^3}\ \frac{b^3}{b^0}\ \frac{c^{-4}}{c^3}<br />
\Rightarrow \frac{7}{3}\frac{1}{a^3}\frac{b^3}{1}\frac{1}{c^7}<br />
\Rightarrow \frac{7b^3}{3a^3c^7}<br />
    Last edited by bigwave; January 18th 2010 at 08:46 PM. Reason: to show solved expression
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  3. #3
    Senior Member Stroodle's Avatar
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    I think that it may be:

    \frac{112a^0b^3c^{-4}}{48a^3b^0c^3}

    If this is correct, then remember that any value to the power of zero equals one, and \frac{c^{-4}}{c^3}=\frac{1}{c^7}
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by jay1 View Post
    Can someone help me divide the following expression: 112a^0b^3c^-4/ 48a^3b^0c^3 ? Thanks much!
    \frac{112a^0b^3c^-4}{48a^3b^0c^3}.

    So, what we should do here is subtract all powers of like bases in the denominator from those that are in the numerator...

    So....

    \frac{112a^0b^3c^{-4}}{48a^3b^0c^3}=\frac{112a^{0-3}b^{3-0}c^{-4-3}}{48}=\frac{112a^{-3}b^3c^{-7}}{48}

    So, note that the negative exponenets go into the denominator and become positive leaving us with

    \frac{112b^3}{48a^3c^7}

    Now, reduce the numerical coefficients

    \frac{7b^3}{3a^3c^7}
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  5. #5
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    That makes sense now Von Nemo..Thanks! Now, could someone help me verify my answer on performing the indicated operations and simplfying: 5a+3/a^2 +13a +12 - 2a-3/a^2 + 13a + 12 I have: -3(a+2)/(a+12)(a+1) ?
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  6. #6
    Senior Member Stroodle's Avatar
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    You should really use some parentheses.

    Do you mean: \frac{5a+3}{a^2}+13a+12-\frac{2a-3}{a^2}+13a+12 ??
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  7. #7
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    Quote Originally Posted by Stroodle View Post
    You should really use some parentheses.

    Do you mean: 5a+\frac{3}{a^2}+13a+12-2a-\frac{3}{a^2}+13a+12 ??
    I'm sorry.. It's (5a +3 /a^2 +13a +12) - (2a - 3/a^2 +13a +12) My answer was 3(a+2)/(a+12)(a+1) ?? Thanks!
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  8. #8
    Senior Member Stroodle's Avatar
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    If you mean to simplify \frac{5a+3}{a^2+13a+12}-\frac{2a-3}{a^2+13a+12}

    then \frac{3(a+2)}{(a+12)(a+1)} is correct.
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