document.write( "Question 184476: Solve the following application problem (show your equation and then your solution). Round answers to nearest thousandths and properly label your answer as well.
\n" ); document.write( "The length of a rectagle is 5 cm. more than its width. If the area of the rectange is 90 cm^2, find the dimensions. \r
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Algebra.Com's Answer #138478 by checkley77(12844)\"\" \"About 
You can put this solution on YOUR website!
L=W+5
\n" ); document.write( "L*W=90
\n" ); document.write( "(W+5)W=90
\n" ); document.write( "W^2+5W-90=0
\n" ); document.write( "USING THE QUADRATIC EQUATION W = (-b +- sqrt( b^2-4*a*c ))/(2*a) }WE GET:
\n" ); document.write( "W=(-5+-SQRT(5^2-4*1*-90])/2*1
\n" ); document.write( "W=(-5+-SQRT[25+360])/2
\n" ); document.write( "W=(-5+-SQRT385)/2
\n" ); document.write( "W=(-5+-19.62)/2
\n" ); document.write( "W=(-5+19.62)/2
\n" ); document.write( "W=14.62/2
\n" ); document.write( "W=7.31 cm. IS THE WIDTH.
\n" ); document.write( "L=7.31+5=12.31 cm. IS THE LENGTH.
\n" ); document.write( "PROOF:
\n" ); document.write( "7.31*12.31=90
\n" ); document.write( "90~90\r
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