document.write( "Question 272992: I'm stuck on this question.
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document.write( "A rect. has a perimeter of 68 cm, and it's area is 273 sq.cm. Find the dimensions of this rect. (For purposes of this problem only, the width is shorter than the length.) \n" );
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Algebra.Com's Answer #199690 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! Write the equation for each statement \n" ); document.write( ": \n" ); document.write( "A rect. has a perimeter of 68 cm, \n" ); document.write( "2L + 2W = 68 \n" ); document.write( "Simplify, divide by 2 \n" ); document.write( "L + W = 34 \n" ); document.write( "or \n" ); document.write( "L = (34-W); use this form for substitution \n" ); document.write( ": \n" ); document.write( "it's area is 273 sq.cm. \n" ); document.write( "L * W = 273 \n" ); document.write( "Substitute (34-W) for L \n" ); document.write( "(34-W)*W = 273 \n" ); document.write( "-W^2 + 34W = 273 \n" ); document.write( "Arrange as a quadratic equation \n" ); document.write( "-W^2 + 34W - 273 = 0 \n" ); document.write( "multiply by -1, it's easier to factor \n" ); document.write( "W^2 - 34W + 273 \n" ); document.write( "Factors to \n" ); document.write( "(W - 13)(W - 21) = 0 \n" ); document.write( "Two solutions \n" ); document.write( "W = 13 \n" ); document.write( "and \n" ); document.write( "W = 21 \n" ); document.write( ": \n" ); document.write( "W = 13 is the width and L = 21 is the length \n" ); document.write( "; \n" ); document.write( ": \n" ); document.write( "You can check this by finding the perimeter and area with these values.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |