document.write( "Question 171016: The length of a rectangle is 1ft less than twice its width. The area is 55ft^2 find the perimeter. \n" ); document.write( "
Algebra.Com's Answer #126264 by checkley77(12844)![]() ![]() ![]() You can put this solution on YOUR website! L=2W-1 \n" ); document.write( "L*W=55 \n" ); document.write( "(2W-1)W=55 \n" ); document.write( "2W^2-W-55=0 \n" ); document.write( "(2W-11)(W+5)=0 \n" ); document.write( "2W=11 \n" ); document.write( "W=11/2 \n" ); document.write( "W=5.5 FT. ANS. \n" ); document.write( "L=2*5.5-1 \n" ); document.write( "L=11-1 \n" ); document.write( "L=10 FT. ANS. \n" ); document.write( "PROOF: \n" ); document.write( "10*5.5=55 \n" ); document.write( "55=55 \n" ); document.write( "THUS THE PERIMETERIS: \n" ); document.write( "2L+2W=PERIMETER. \n" ); document.write( "2*10+2*5.5=20+11=31 FT. IS THE PERIMETER. \n" ); document.write( " \n" ); document.write( " |