Solving for d_2: Simple Algebra Help

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Homework Help Overview

The discussion revolves around solving for the variable d_2 in a simple algebraic equation involving fractions and cube roots. Participants are exploring algebraic manipulation and the implications of their calculations.

Discussion Character

  • Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between variables and how to isolate d_2. There are attempts to manipulate the equation and questions about the correctness of intermediate steps.

Discussion Status

Some participants have provided guidance on algebraic transformations, while others are reflecting on their calculations and considering alternative approaches. There is an ongoing exploration of the problem without a clear consensus on the final form of d_2.

Contextual Notes

Participants are navigating through algebraic expressions and considering whether to maintain fractions or convert to decimals, which may affect their calculations.

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 don't know wat to do :mad: i know its simple algebra, i just don't see it,.
 
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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.
 
thnx 4 all the help doc,
i get
[tex]d_2=0.041^{\frac{1}{3}}[/tex]
 
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]
 
o i see wat i did, so [tex]d_2=[4.1*10^{-4}]^\frac{1}{3}[/tex]?
 
Much better.
 
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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