document.write( "Question 254246: If you build a circular pool with a volume of 1000 cubic feet, what is the approximate radius of the pool? \n" ); document.write( "
Algebra.Com's Answer #186566 by Theo(13342)\"\" \"About 
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depends on the depth of the pool.\r
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\n" ); document.write( "\n" ); document.write( "If the pool is circular, the formula for volume is \"pi%2Ar%5E2%2Ah\"\r
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\n" ); document.write( "\n" ); document.write( "you get \"v+=+pi%2Ar%5E2%2Ah\"\r
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\n" ); document.write( "\n" ); document.write( "v = 1000 so you get \"1000+=+pi%2Ar%5E2%2Ah\"\r
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\n" ); document.write( "\n" ); document.write( "you can solve for r or you can solve for h.\r
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\n" ); document.write( "\n" ); document.write( "if you solve for r, you get \"r%5E2\" = \"1000+%2F+%28pi%2Ah%29\"\r
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\n" ); document.write( "\n" ); document.write( "take square root of both sides of equation and you get:\r
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\n" ); document.write( "\n" ); document.write( "r = +/- \"sqrt%281000%2F%28pi%2Ah%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "if you solve for h, you get \"h+=+1000+%2F+%28pi%2Ar%5E2%29\"\r
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\n" ); document.write( "\n" ); document.write( "assuming the depth of the pool is 6 feet all around, the radius would be:\r
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\n" ); document.write( "\n" ); document.write( "r = +/- \"sqrt%281000%2F%28pi%2A6%29%29+=+7.283656204+\" feet.\r
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\n" ); document.write( "\n" ); document.write( "graph of equation for the radius is shown below:\r
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\n" ); document.write( "\n" ); document.write( "y value represents the radius of the pool.
\n" ); document.write( "x value represents the depth of the pool.\r
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\n" ); document.write( "\n" ); document.write( "you can see that when the depth of the pool is 1 foot, the radius of the pool is:\r
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\n" ); document.write( "\n" ); document.write( "\"sqrt%281000%2F%28pi%29%29+=+17.84124116\"\r
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\n" ); document.write( "\n" ); document.write( "\"graph%28600%2C600%2C-15%2C15%2C-20%2C20%2Csqrt%281000%2F%28pi%2Ax%29%29%29\"\r
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