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    miamiamia's Avatar
    miamiamia Posts: 1, Reputation: 1
    New Member
     
    #1

    Feb 28, 2011, 04:24 PM
    how do you answer 28a^5/7a^3
    freshangel15's Avatar
    freshangel15 Posts: 6, Reputation: 3
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    #2

    Feb 28, 2011, 04:59 PM
    You divide the coefficients and subtract the powers. You should get 4a^2
    jcaron2's Avatar
    jcaron2 Posts: 986, Reputation: 204
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    #3

    Feb 28, 2011, 08:11 PM
    Comment on freshangel15's post
    Exactly.
    RPVega's Avatar
    RPVega Posts: 29, Reputation: 2
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    #4

    Mar 10, 2011, 12:02 PM
    I'm assuming you mean (28*a^5)/(7*a^3)? If that's the case, here's the
    answer: 4*(a^2).

    Explanation:

    First, divide 28 by 7, to get 4 in the numerator. Next, divide a^5 by a^3.
    To do this, subtract 5 from 3 to get 2. Remember, 1/(a^3)= a^(-3). Then you
    have (a^5)*(a^-3) in the numerator. To evaluate (a^5)*(a^-3), simply ADD the
    exponents of the base,a, which is a^(5-3)= a^2. Since 28/7=4, the answer is
    4*(a^2). I know it's confusing... This problem involves a mastery of the
    properties of exponents. Good luck!

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