document.write( "Question 364453: The width of a rectangle is 5 times its length. If the width of the rectangle is doubled, then doubled again, the new perimeter is which percent of the original perimeter?
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Algebra.Com's Answer #259850 by mananth(16946) You can put this solution on YOUR website! The width of a rectangle is 5 times its length. If the width of the rectangle is doubled, then doubled again, the new perimeter is which percent of the original perimeter? \n" ); document.write( ".. \n" ); document.write( "let length be x \n" ); document.write( "width = 5x \n" ); document.write( "perimeter = 2*(L+W) \n" ); document.write( "=2*6x \n" ); document.write( "=12x \n" ); document.write( ".. \n" ); document.write( "2*2*5x= 20x \n" ); document.write( "length = x \n" ); document.write( "... \n" ); document.write( "perimeter = 2*(20x+x) \n" ); document.write( "=42x \n" ); document.write( "Ratio = 42x/12x \n" ); document.write( "2.5 *100 = 250 % \n" ); document.write( "... \n" ); document.write( "m.ananth@hotmail.ca \n" ); document.write( " |