- #1

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[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,.

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- Thread starter DB
- Start date

- #1

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[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,.

- #2

Doc Al

Mentor

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

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

Use this idea and keep going.

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

Use this idea and keep going.

- #3

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thnx 4 all the help doc,

i get

[tex]d_2=0.041^{\frac{1}{3}}[/tex]

i get

[tex]d_2=0.041^{\frac{1}{3}}[/tex]

- #4

Doc Al

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That doesn't seem right. Consider the following:DB said:i get

[tex]d_2=0.041^{\frac{1}{3}}[/tex]

[tex][\frac{d_2}{0.1}]^3 = \frac{d_2^3}{0.001}[/tex]

- #5

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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

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Much better.

- #7

Ouabache

Science Advisor

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Keep them as fractions and you should find you can easily take their cube root later.

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