document.write( "Question 618708: the perimeter of a rectangleis 600 yards. What are the dimensions of the rectangle if the length is 50 yards more than the width? \n" ); document.write( "

Algebra.Com's Answer #389045 by ewatrrr(24785) $\"\"$ $\"About$

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document.write( "Hi,
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document.write( "the perimeter of a rectangleis 600 yards. What are the dimensions of the rectangle if the length is 50 yards more than the width ?
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document.write( "P = 2L + 2W = 2(x+50) + 2x = 600
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document.write( "          2(x+50) + 2x = 600
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document.write( "              4x = 500
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document.write( "               x = 125yd, the width.  the length is 175yd     (125yd + 50yd) \n" );
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