document.write( "Question 81536This question is from textbook Coll. Algebra
\n" ); document.write( ": Hello:\r
\n" ); document.write( "\n" ); document.write( "Here is my problem that I need assistance with. Thank you.\r
\n" ); document.write( "\n" ); document.write( "The period T (in seconds) of a simple pendulum is a function of its length L (in feet), given by T(L) = 2 pi square root of L/G where G = 32.2 feet per second. Per second is the acceleration of gravity. Express length L as a function of the period T. \n" ); document.write( "

Algebra.Com's Answer #58378 by tutor_paul(519) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"T=2%28pi%29sqrt%28L%2F32.2%29\"$
\n" ); document.write( "The problem is asking you to solve for L in terms of T.
\n" ); document.write( "First get the square root term alone on one side of the equation:
\n" ); document.write( " $\"T%2F%282pi%29=sqrt%28L%2F32.2%29\"$
\n" ); document.write( "Square both sides to get rid of that square root:
\n" ); document.write( " $\"T%5E2%2F%284pi%5E2%29=L%2F32.2\"$
\n" ); document.write( "Solve for L:
\n" ); document.write( " $\"highlight%28L%28t%29=%28%2832.2%29T%5E2%29%2F4pi%5E2%29\"$
\n" ); document.write( "Good Luck,
\n" ); document.write( "tutor_paul@yahoo.com \n" ); document.write( "

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