document.write( "Question 612919: The length of a rectangle is 2ft longer than its width. Find the dimensions of the rectangle such that the perimeter of the rectangle is 845ft \n" ); document.write( "
Algebra.Com's Answer #385792 by jim_thompson5910(35256)![]() ![]() ![]() You can put this solution on YOUR website! P = 2L + 2W\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "845 = 2(W+2) + 2W\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "845 = 2W+4 + 2W\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "845 = 4W+4\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "845-4 = 4W\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "841 = 4W\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "841/4 = W\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "W = 841/4\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Now that you know W, use it to find L\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Remember that L = W+2 \n" ); document.write( " |