1. Sep 14, 2005 #1

   DB

   

   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 i know its simple algebra, i just dont see it,.

    

   DB, Sep 14, 2005
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3. Sep 14, 2005 #2

   Doc Al

   

   Staff: Mentor

   If [tex]a = \frac{1}{b^3}[/tex]
   Then [tex]b^3 = \frac{1}{a}[/tex]
   
   Use this idea and keep going.

    

   Doc Al, Sep 14, 2005
4. Sep 14, 2005 #3

   DB

   

   thnx 4 all the help doc,
   i get
   [tex]d_2=0.041^{\frac{1}{3}}[/tex]

    

   DB, Sep 14, 2005
5. Sep 14, 2005 #4

   Doc Al

   

   Staff: Mentor

   That doesn't seem right. Consider the following:
   [tex][\frac{d_2}{0.1}]^3 = \frac{d_2^3}{0.001}[/tex]

    

   Doc Al, Sep 14, 2005
6. Sep 14, 2005 #5

   DB

   

   o i see wat i did, so [tex]d_2=[4.1*10^{-4}]^\frac{1}{3}[/tex]?

    

   DB, Sep 14, 2005
7. Sep 14, 2005 #6

   Doc Al

   

   Staff: Mentor

   Much better.

    

   Doc Al, Sep 14, 2005
8. Sep 14, 2005 #7

   Ouabache

   

   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.

    

   Ouabache, Sep 14, 2005

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