Difference between revisions of "Proportion/Intermediate"

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==Problem==
 
==Problem==
<math>x</math> is directly proportional to the sum of the squares of <math>y</math> and <math>z</math> and inversely proportional to <math>y</math> and the square of <math>z</math>. If <math>x = 8</math> when <math>y = 1/2</math> and <math>z = \sqrt {3}/2</math>, find <math>y</math> when <math>x = 1</math> and <math>z = 6</math>.  
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<math>x</math> is directly proportional to the sum of the squares of <math>y</math> and <math>z</math> and inversely proportional to <math>y</math> and the square of <math>z</math>. If <math>x = 8</math> when <math>y = 1/2</math> and <math>z = \sqrt {3}/2</math>, find <math>y</math> when <math>x = 1</math> and <math>z = 6</math>. (Thanks to Bicameral of the AoPS forum for this one)
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==Solution==
 
==Solution==
 
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Revision as of 17:02, 9 October 2007

Problem

$x$ is directly proportional to the sum of the squares of $y$ and $z$ and inversely proportional to $y$ and the square of $z$. If $x = 8$ when $y = 1/2$ and $z = \sqrt {3}/2$, find $y$ when $x = 1$ and $z = 6$. (Thanks to Bicameral of the AoPS forum for this one)

Solution

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