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Math Help - Evaluate

  1. #1
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    Evaluate

    evaluate (\sqrt[3]{0.216})(243^{-\frac{2}{5}})

    I am not too sure if I am on the right track...

    Attempt:

    (\sqrt[3]{0.216})(243^{-\frac{2}{5}})

    = \frac{\sqrt[3]{0.216}}{243^{\frac{2}{5}}}

    = \sqrt[3]{\frac{0.216}{243^{\frac{6}{5}}}}
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  2. #2
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    evaluate

    Quote Originally Posted by Punch View Post
    evaluate (\sqrt[3]{0.216})(243^{-\frac{2}{5}})

    I am not too sure if I am on the right track...

    Attempt:

    (\sqrt[3]{0.216})(243^{-\frac{2}{5}})

    = \frac{\sqrt[3]{0.216}}{243^{\frac{2}{5}}}

    = \sqrt[3]{\frac{0.216}{243^{\frac{6}{5}}}}

    first step is fine. Answer using calculator or logs. The numerator is .6 and the denominator is 9.


    bjh
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  3. #3
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    Quote Originally Posted by Punch View Post
    evaluate (\sqrt[3]{0.216})(243^{-\frac{2}{5}})

    I am not too sure if I am on the right track...

    Attempt:

    (\sqrt[3]{0.216})(243^{-\frac{2}{5}})

    = \frac{\sqrt[3]{0.216}}{243^{\frac{2}{5}}}

    = \sqrt[3]{\frac{0.216}{243^{\frac{6}{5}}}}
    Hi Punch,

    243=3^5

    243^{\frac{1}{5}}=3

    243^{\frac{6}{5}}=3^6

    0.216=\frac{216}{1000}=\frac{6^3}{10^3}=\frac{2^33  ^3}{2^35^3}=\frac{3^3}{5^3}


    \sqrt[3]{\frac{0.216}{243^{\frac{6}{5}}}}=\sqrt[3]{\frac{3^3}{5^33^6}}=\sqrt[3]{\frac{1}{5^33^3}}=\frac{1}{5(3)}=\frac{1}{15}
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  4. #4
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    is it necessary to simplify it so much?
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  5. #5
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    evaluate

    Quote Originally Posted by Punch View Post
    is it necessary to simplify it so much?

    Yes if you were asked to solve without logs or calculator


    bjh
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