document.write( "Question 150914: If the diameter of a sphere is doubled, its surface becomes
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\n" ); document.write( "B.four times the orginal surface
\n" ); document.write( "C.eight times the orginal surface
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Algebra.Com's Answer #110848 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
Surface area of a sphere: \"S=4pi%2Ar%5E2\"\r
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\n" ); document.write( "\n" ); document.write( "Remember, the diameter is \"d=2r\". So if the diameter is doubled, then the new radius is \"2r\" units (instead of \"r\" units)\r
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\n" ); document.write( "\n" ); document.write( "\"S=4pi%2Ar%5E2\" Start with the given equation.\r
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\n" ); document.write( "\n" ); document.write( "\"S=4pi%2A%282r%29%5E2\" Replace the original \"r\" with \"2r\".\r
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\n" ); document.write( "\n" ); document.write( "\"S=4pi%2A4r%5E2\" Square 2r to get \"4r%5E2\"\r
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\n" ); document.write( "\n" ); document.write( "So the old equation is \"S%5Bold%5D=4pi%2Ar%5E2\" and the new one is \"S%5Bnew%5D=4pi%2A4r%5E2\" \r
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\n" ); document.write( "\n" ); document.write( "The ratio of the new surface area to the old surface area is then:\r
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\n" ); document.write( "\n" ); document.write( "\"S%5Bnew%5D%2FS%5Bold%5D=%284pi%2A4r%5E2%29%2F%284pi%2Ar%5E2%29\"\r
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\n" ); document.write( "\n" ); document.write( " Highlight the common terms\r
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\n" ); document.write( "\n" ); document.write( " Cancel out the common terms\r
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\n" ); document.write( "\n" ); document.write( "\"S%5Bnew%5D%2FS%5Bold%5D=4\" Simplify\r
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\n" ); document.write( "\n" ); document.write( "So the new surface area is simply 4 times the original surface area. So the answer is B
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