document.write( "Question 408536: I think this is right for this problem. Can someone pls check?\r
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document.write( "50/4=12.5x2+7=32
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document.write( "32/2=16-7=9
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document.write( "W=32
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document.write( "L=9\r
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document.write( "The length of a rectangle is 7\" greater than its width. The perimeter of the rectangle is 50\". Find the length and width of the rectangle.
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Algebra.Com's Answer #287780 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! The length of a rectangle is 7\" greater than its width. The perimeter of the rectangle is 50\". Find the length and width of the rectangle.\r \n" ); document.write( "\n" ); document.write( "------ \n" ); document.write( "Perimeter = 2(length + width) \n" ); document.write( "--- \n" ); document.write( "Let width be \"x\". \n" ); document.write( "The length is \"x+7\". \n" ); document.write( "--- \n" ); document.write( "50 = 2(x+7 + x) \n" ); document.write( "25 = 2x+7 \n" ); document.write( "2x = 18 \n" ); document.write( "x = 9 (width) \n" ); document.write( "x+7 = 16 (length) \n" ); document.write( "====================== \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |