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Math Help - Could you please double-check my work

  1. #1
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    Could you please double-check my work

    I'm working on a couple of problems here and I wanted to make sure my steps to my answers (in bold) are correct. The numbers in red are exponents.

    1. Problem: 8 a-3
    ..................4a2

    ................= 2
    ..................a5


    2. Problem:

    (3x-2 y2)2
    -12 x3 y-3

    (3)2 (x-2)2 (y2)2
    -12 x3 y-3

    ...3x4 y4
    -12 x3 y-3

    3x7
    4y7


    Thanks again.






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  2. #2
    Super Member Aryth's Avatar
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    Quote Originally Posted by slykksta View Post
    I'm working on a couple of problems here and I wanted to make sure my steps to my answers (in bold) are correct. The numbers in red are exponents.

    1. Problem: 8 a-3
    ..................4a2

    ................= 2
    ..................a5


    2. Problem:

    (3x-2 y2)2
    -12 x3 y-3

    (3)2 (x-2)2 (y2)2
    -12 x3 y-3

    ...3x4 y4
    -12 x3 y-3

    3x7
    4y7


    Thanks again.
    1. The work is fine.

    2. There is one error that messed up your answer.


    (3)2 (x-2)2 (y2)2
    -12 x3 y-3

    In this, you simplified it to:


    3x4 y4
    -12 x3 y-3

    Which is wrong, when you separate each component of the first part I copied, you end up getting the following on the numerator:

    (3)^2 = 9

    (x^-2)^2 = (x^-2)*(x^-2) = (x^(-2 + -2)) = (x^-4)

    (y^2)^2 = (y^2)*(y^2) = y^4

    The numerator is as such:

    9x^-4*y^4

    The whole fraction is:

    9x^-4*y^4
    -----------
    -12x^3*y^-3

    Now we have to use the law of inverses, which states that:

    x^-1 = 1/x^1

    So:

    x^-4 = 1/x^4

    And:

    1/y^-3 = y^3

    Now we have to rearrange the fraction:

    9y^4*y^3
    ----------
    -12x^3*x^4

    Therefore, we have:

    3y^7
    -----
    -4x^7


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  3. #3
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    Hello, slykksta!

    1)\;\;\frac{8a^{-3}}{4a^2}\;=\;\frac{2}{a^5} . . . . Right!

    2)\;\;\frac{(3x^{-2}y^2)^2} {-12x^3y^{-3}}

    We have: . \frac{3^2(x^{-2})^2(y^2)^2} {-12x^3y^{-3}} \;=\;\frac{9x^{-4}y^4}{-12x^3y^{-3}} \;=\;-\frac{3y^7}{4x^7}

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