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Math Help - really easy cancel I can't do!

  1. #1
    Member i_zz_y_ill's Avatar
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    really easy cancel I can't do!

    Im not sure how to times through by 3/3 on the bottom

    (2/3) / (1+(2/3)^2)^0.5

    trying to get 2/(9+4x^2)^0.5 If somebody could show me how that'd be much appreciated
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  2. #2
    GAMMA Mathematics
    colby2152's Avatar
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    Quote Originally Posted by i_zz_y_ill View Post
    Im not sure how to times through by 3/3 on the bottom

    (2/3) / (1+(2/3)^2)^0.5

    trying to get 2/(9+4x^2)^0.5 If somebody could show me how that'd be much appreciated
    What is \frac{3}{3} squared? It is \frac{9}{9}.

    Multiply that out through the "top" and "bottom" of the fraction.
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  3. #3
    Bar0n janvdl's Avatar
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    Quote Originally Posted by i_zz_y_ill View Post
    Im not sure how to times through by 3/3 on the bottom

    (2/3) / (1+(2/3)^2)^0.5

    trying to get 2/(9+4x^2)^0.5 If somebody could show me how that'd be much appreciated
    I'm sure you mean the following:

    \frac{ \frac{2}{3} }{ \sqrt{1 + \left( \frac{2x}{3} \right) ^{2}} }

    Observe the \frac{2}{3} in the numerator. We can move that 3 to the denominator of the entire function.

    Giving us:

    \frac{ 2 }{ 3 \sqrt{1 + \left( \frac{2x}{3} \right) ^{2}} }


    But we also know that:
    3 = \sqrt{9}

    Thus:
    \frac{ 2 }{ \sqrt{9} \sqrt{1 + \left( \frac{2x}{3} \right) ^{2}} }

    Simplify the denominator.

    \frac{ 2 }{ \sqrt{9(1 + \left( \frac{2x}{3} \right) ^{2})} }

    \frac{ 2 }{ \sqrt{9(1 + \left( \frac{4x^2}{9} \right) )} }

    \frac{ 2 }{ \sqrt{9 + 4x^2 } }
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