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A lil algebra help

  • Thread starter DB
  • Start date
  • #1
DB
501
0
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,.
 

Answers and Replies

  • #2
Doc Al
Mentor
44,871
1,119
If [tex]a = \frac{1}{b^3}[/tex]
Then [tex]b^3 = \frac{1}{a}[/tex]

Use this idea and keep going.
 
  • #3
DB
501
0
thnx 4 all the help doc,
i get
[tex]d_2=0.041^{\frac{1}{3}}[/tex]
 
  • #4
Doc Al
Mentor
44,871
1,119
DB said:
i get
[tex]d_2=0.041^{\frac{1}{3}}[/tex]
That doesn't seem right. Consider the following:
[tex][\frac{d_2}{0.1}]^3 = \frac{d_2^3}{0.001}[/tex]
 
  • #5
DB
501
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o i see wat i did, so [tex]d_2=[4.1*10^{-4}]^\frac{1}{3}[/tex]?
 
  • #6
Doc Al
Mentor
44,871
1,119
Much better.
 
  • #7
Ouabache
Science Advisor
Homework Helper
1,340
7
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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