document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=706525>Question 706525</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 80 feet of lumbar is used to enclose a rectangulr field. The field is twice as long as it is wide. What are the dimensions of the field? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #435251 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley79><B>checkley79(3341)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley79><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=435251\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> L=2W<BR>\n" );
document.write( "P=2L+2W<BR>\n" );
document.write( "80=2(2W)+2W<BR>\n" );
document.write( "80=4W+2W<BR>\n" );
document.write( "80=6W<BR>\n" );
document.write( "W=80/6<BR>\n" );
document.write( "W=13.33 FT FOR THE WIDTH.<BR>\n" );
document.write( "L=2*13.33=26.66 FT. FOR THE LENGTH.<BR>\n" );
document.write( "PROOF:<BR>\n" );
document.write( "80=2*13.33+2*26.66<BR>\n" );
document.write( "80=26.66+53.32<BR>\n" );
document.write( "80=80 <BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );