document.write( "Question 841849: I have been searching this site for hours for a solution to this:
\n" ); document.write( "The rectangular cross-sectional area of a metal brace is to have an area of 2250
\n" ); document.write( "square millimeters. The length is to be one and one half times the width. Compute the length and width dimensions to the nearest millimeter. I am in
\n" ); document.write( "desperate need of an answer! Thank you. \n" ); document.write( "

Algebra.Com's Answer #507281 by LinnW(1048) $\"\"$ $\"About$

You can put this solution on YOUR website!
Set L = length , W = width
\n" ); document.write( "L = 1.5*W
\n" ); document.write( "area = L*W
\n" ); document.write( "2250 = L*W
\n" ); document.write( "substitute 1.5W for L
\n" ); document.write( "2250 = (1.5W)(W)
\n" ); document.write( " $\"2250+=+1.5W%5E2\"$
\n" ); document.write( "divide each side by 1.5
\n" ); document.write( " $\"1500+=+W%5E2\"$
\n" ); document.write( " $\"W+=+sqrt%281500%29\"$ , approximately 38.73
\n" ); document.write( " $\"L=+1.5sqrt%281500%29\"$ , approximately 58.02 \n" ); document.write( "

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