document.write( "Question 83801: The formula for the surface area of a sphere is
\n" ); document.write( "\"A=4%2Api%2Ar%5E2\", where A is surface area and r is the
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\n" ); document.write( "\n" ); document.write( "a) use the formula \"A=4%2Api%2Ar%5E2\" to express r in terms of A and
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\n" ); document.write( "\n" ); document.write( "b) suppose the surface area of a child's ball is
\n" ); document.write( " about 113 square inches. Find the radius of the
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Algebra.Com's Answer #60269 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
The formula for the surface area of a sphere is
\n" ); document.write( "\"A=4%2Api%2Ar%5E2\", where A is surface area and r is the
\n" ); document.write( "radius of the sphere\r
\n" ); document.write( "
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\n" ); document.write( "\n" ); document.write( "a) use the formula \"A=4%2Api%2Ar%5E2\" to express r in terms of A and
\n" ); document.write( " rationalize the denominator. Explain your answer\r
\n" ); document.write( "\n" ); document.write( " Since \"A=4%2Api%2Ar%5E2\" we can solve for r by these steps\r
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\n" ); document.write( "\n" ); document.write( "\"A=4%2Api%2Ar%5E2\" Start with the given equation\r
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\n" ); document.write( "\n" ); document.write( "\"A%2F%284%2Api%29=cross%28%284%2Api%29%2F%284%2Api%29%29r%5E2\" Divide both sides by \"4%2Api\"\r
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\n" ); document.write( "\n" ); document.write( "\"sqrt%28A%2F%284%2Api%29%29=r\" Take the square root of both sides\r
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\n" ); document.write( "\n" ); document.write( "\"sqrt%28A%29%2Fsqrt%28%284%2Api%29%29=r\" Break up the square root using the identity \"sqrt%28x%2Fy%29=sqrt%28x%29%2Fsqrt%28y%29\"}\r
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\n" ); document.write( "\n" ); document.write( "\"sqrt%28A%29%2F%28sqrt%284%29%2Asqrt%28pi%29%29=r\" Break up the square root using the identity \"sqrt%28x%2Ay%29=sqrt%28x%29%2Asqrt%28y%29\"}\r
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\n" ); document.write( "\n" ); document.write( "\"sqrt%28A%29%2F%282%2Asqrt%28pi%29%29=r\" Take the square root of 4 to get 2\r
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\n" ); document.write( "\n" ); document.write( "\"%28sqrt%28pi%29%2Fsqrt%28pi%29%29%28sqrt%28A%29%2F%282%2Asqrt%28pi%29%29%29=r\" Now rationalize the denominator by multiplying the numerator and denominator by \"sqrt%28pi%29\". This is equivalent to \"%28sqrt%28pi%29%29%5E2=pi\" which in other words eliminates the square root in the denominator\r
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\n" ); document.write( "\n" ); document.write( "\"%28sqrt%28pi%29sqrt%28A%29%2F%282%2Api%29%29=r\" Multiply and reduce\r
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\n" ); document.write( "\n" ); document.write( "\"sqrt%28pi%2AA%29%2F%282%2Api%29=r\" Combine the square roots in the numerator\r
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\n" ); document.write( "\n" ); document.write( "So now we have an equation in terms of A which is\r
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\n" ); document.write( "\n" ); document.write( "\"r=sqrt%28pi%2AA%29%2F%282%2Api%29\"\r
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\n" ); document.write( "\n" ); document.write( "b) suppose the surface area of a child's ball is
\n" ); document.write( " about 113 square inches. Find the radius of the
\n" ); document.write( " ball and explain your steps.\r
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\n" ); document.write( "\n" ); document.write( "Since we have an equation in terms of A, we can plug in the given surface area (113 sq in) in for A and solve for r\r
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\n" ); document.write( "\n" ); document.write( "\"r=sqrt%28pi%2A113%29%2F%282%2Api%29\" Plug in A=113\r
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\n" ); document.write( "\n" ); document.write( "Now lets approximate \"pi\" to 3.14. From here on, all of our values will be approximations. So even though I don't show it (I'm unable to), keep this in mind.\r
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\n" ); document.write( "\n" ); document.write( "\"r=sqrt%283.14%2A113%29%2F%282%2A3.14%29\" Replace \"pi\" with 3.14\r
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\n" ); document.write( "\n" ); document.write( "\"r=sqrt%28354.82%29%2F%286.28%29\" Multiply\r
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\n" ); document.write( "\n" ); document.write( "\"r=18.837%2F%286.28%29\" Take the square root\r
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\n" ); document.write( "\n" ); document.write( "\"r=2.999\" Divide\r
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\n" ); document.write( "\n" ); document.write( "So the for a ball with a surface area of 113 sq in., the radius is about 3 inches
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