document.write( "Question 352012: The perimeter of a rectangle is 82 yards.The width is 15 yards less than the length.What are the dimensions of the rectangle? \n" ); document.write( "
Algebra.Com's Answer #251548 by mananth(16949) You can put this solution on YOUR website! let length = l \n" ); document.write( "width = l-15 \n" ); document.write( "... \n" ); document.write( "Perimeter = 2*(L+W) \n" ); document.write( "2*(L+L-15)=82 \n" ); document.write( "2*(2l-15)=82 \n" ); document.write( "4l-30=82 \n" ); document.write( "4l=82+30 \n" ); document.write( "4l=112 \n" ); document.write( "l= 28 yards the length \n" ); document.write( "width = l-15 = 13 yards \n" ); document.write( ".... \n" ); document.write( "C H E C K \n" ); document.write( "13+13+28+28=82 \n" ); document.write( "... \n" ); document.write( "m.ananth@hotmail.ca \n" ); document.write( " |