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A lil algebra help

  1. Sep 14, 2005 #1

    DB

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    to solve for d_2:
    [tex]\frac{336}{81}=\frac{175}{100}*\frac{1}{[\frac{d_2}{0.1}]^3}[/tex]
    [tex]\sim 2.4=\frac{1}{[\frac{d_2}{0.1}]^3}[/tex]

    then i dont know wat to do :mad: i know its simple algebra, i just dont see it,.
     
  2. jcsd
  3. Sep 14, 2005 #2

    Doc Al

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    Staff: Mentor

    If [tex]a = \frac{1}{b^3}[/tex]
    Then [tex]b^3 = \frac{1}{a}[/tex]

    Use this idea and keep going.
     
  4. Sep 14, 2005 #3

    DB

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    thnx 4 all the help doc,
    i get
    [tex]d_2=0.041^{\frac{1}{3}}[/tex]
     
  5. Sep 14, 2005 #4

    Doc Al

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    Staff: Mentor

    That doesn't seem right. Consider the following:
    [tex][\frac{d_2}{0.1}]^3 = \frac{d_2^3}{0.001}[/tex]
     
  6. Sep 14, 2005 #5

    DB

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    o i see wat i did, so [tex]d_2=[4.1*10^{-4}]^\frac{1}{3}[/tex]?
     
  7. Sep 14, 2005 #6

    Doc Al

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    Staff: Mentor

    Much better.
     
  8. Sep 14, 2005 #7

    Ouabache

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    Science Advisor
    Homework Helper

    Even better if you don't convert your constants to a decimal so soon.
    Keep them as fractions and you should find you can easily take their cube root later. :wink:
     
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