document.write( "Question 987650: Hi! **P=pi (3.14)\r
\n" ); document.write( "\n" ); document.write( "A=2Pr^2 + 2Prh \r
\n" ); document.write( "\n" ); document.write( "Solve for r>0.\r
\n" ); document.write( "\n" ); document.write( "Thank you soooo much in advance. This was in a Calc BC summer packet. Due Tuesday!
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Algebra.Com's Answer #608381 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
\"A+=+2%2Api%2Ar%5E2+%2B+2%2Api%2Ar%2Ah\"\r
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\n" ); document.write( "\n" ); document.write( "\"A-A+=+2%2Api%2Ar%5E2+%2B+2%2Api%2Ar%2Ah-A\"\r
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\n" ); document.write( "\n" ); document.write( "\"0+=+2%2Api%2Ar%5E2+%2B+2%2Api%2Ar%2Ah-A\"\r
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\n" ); document.write( "\n" ); document.write( "\"2%2Api%2Ar%5E2+%2B+2%2Api%2Ar%2Ah-A+=+0\"\r
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\n" ); document.write( "\n" ); document.write( "\"2%2Api%2Ar%5E2+%2B+2%2Api%2Ah%2Ar-A+=+0\"\r
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\n" ); document.write( "\n" ); document.write( "The last equation is in the form \"ax%5E2%2Bbx%2Bc=0\" where \"a+=+2pi\", \"b+=+2pi%2Ah\" and \"c+=+-A\"\r
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\n" ); document.write( "\n" ); document.write( "Use the quadratic formula to solve for r\r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%28-2pi%2Ah%2B-sqrt%28%282pi%2Ah%29%5E2-4%2A2pi%2A%28-A%29%29%29%2F%282%282pi%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%28-2pi%2Ah%2B-sqrt%284pi%5E2%2Ah%5E2%2B8Api%29%29%2F%284pi%29\"\r
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\n" ); document.write( "\n" ); document.write( "Since \"r+%3E+0\", this means we only focus on the \"plus\" portion of the \"plus/minus\". So we end up with\r
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\n" ); document.write( "\n" ); document.write( "\"r+=+%28-2pi%2Ah%2Bsqrt%284pi%5E2%2Ah%5E2%2B8Api%29%29%2F%284pi%29\"
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