document.write( "Question 211405This question is from textbook College Algebra
\n" ); document.write( ": Can you please help me with this problem? Find the dimensiona of a rectangle whose area is 180cm^2 and whose perimeter is 54 cm. Please show me all the steps and reply as soon as possible this is due tues. before 10 am.
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Algebra.Com's Answer #159766 by rapaljer(4671)\"\" \"About 
You can put this solution on YOUR website!
Let w= width
\n" ); document.write( "L = length\r
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\n" ); document.write( "\n" ); document.write( "Perimeter: 2W + 2L = 54\r
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\n" ); document.write( "\n" ); document.write( "Divide this by 2: w+ L = 27\r
\n" ); document.write( "\n" ); document.write( "Solve for w: W = 27 - L\r
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\n" ); document.write( "\n" ); document.write( "Now, Area = L*W\r
\n" ); document.write( "\n" ); document.write( "L*W = 180\r
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\n" ); document.write( "\n" ); document.write( "Substitute W = 27-L to get an equation in one variable:
\n" ); document.write( "L * (27-L) = 180\r
\n" ); document.write( "\n" ); document.write( "27L - L^2 = 180\r
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\n" ); document.write( "\n" ); document.write( "This is a quadratic equation, so set it equal to zero. I recommend you move everything to the right side to avoid the negative L^2 coefficient:
\n" ); document.write( "0 = L^2 - 27 - 180\r
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\n" ); document.write( "\n" ); document.write( "Now,since this seems to be a take home problem that is DUE in Tuesday morning, YOU solve the equation!!\r
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\n" ); document.write( "\n" ); document.write( "R^2
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