document.write( "Question 1210578: A standard 55 gallon drum is holding oil. A portion of the oil is removed. A stick with inch measurements is placed in the drum. The remaining level is 2 7/8 inches. Determine amount (gallons) remaining in drum.\r
\n" ); document.write( "\n" ); document.write( "Check my calculations, please.\r
\n" ); document.write( "\n" ); document.write( "Dimensions:\r
\n" ); document.write( "\n" ); document.write( "Diameter; 22.5\"\r
\n" ); document.write( "\n" ); document.write( "Height; 33.5\"\r
\n" ); document.write( "\n" ); document.write( "Drum is cylindrical, and standing vertical.\r
\n" ); document.write( "\n" ); document.write( "Volume formula; pi * r^2 * h.\r
\n" ); document.write( "\n" ); document.write( "3.1416 * 16.75 * 16.75 * 2 7/8 (2.875) = 2534.06 cu. in.\r
\n" ); document.write( "\n" ); document.write( "2534.06 / 231 = 10.969 = 11 gal.\r
\n" ); document.write( "\n" ); document.write( "Where am I correct ? Answer given is less.
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Algebra.Com's Answer #853820 by josgarithmetic(39736)\"\" \"About 
You can put this solution on YOUR website!
....., t the formula relies on SQUARING the radius. Notice, \"r%5E2\" in there.\r
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\n" ); document.write( "\n" ); document.write( "To put in the number values directly, full drum volume in cubic inches should be\r
\n" ); document.write( "\n" ); document.write( "\"%283.14159%29%2822.5%2F2%29%5E2%2A33.5\"\r
\n" ); document.write( "\n" ); document.write( "\"13319.86\" cubic inches, not rounding too much...\r
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\n" ); document.write( "\n" ); document.write( "The level being down to \"2%267%2F8\" inch,\r
\n" ); document.write( "\n" ); document.write( "\"13319.86%282.875%2F33.5%29\"\r
\n" ); document.write( "\n" ); document.write( "\"1143\" cubic inches.\r
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\n" ); document.write( "\n" ); document.write( "You can look for the conversion to gallons.\r
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\n" ); document.write( "\n" ); document.write( "about 4.95 gallons
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