document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=950262>Question 950262</A>: <I><SPAN STYLE=\"word-break: break-all;\"> if the length of a rectangle measure measure is twice as much as a width and the perimeter is 72 feet find the dimensions of the rectangle </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #580211 by </B><A HREF=/tutors/aboutme.mpl?userid=macston><B>macston(5194)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=macston><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/cgi-bin/show-face.mpl?user=macston><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=580211\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> W=width; L=length=2W; P=perimeter=2(L+W)=72 feet<BR>\n" );
document.write( "P=2(L+W) Substitute for L<BR>\n" );
document.write( "72 feet=2(2W+W) Divide each side by 2.<BR>\n" );
document.write( "36 feet=3W Divide each side by 3.<BR>\n" );
document.write( "12 feet=W ANSWER 1: The width is 12 feet.<BR>\n" );
document.write( "L=2w=2(12 ft)=24 ft ANSWER 2: The length is 24 feet.<BR>\n" );
document.write( "CHECK<BR>\n" );
document.write( "P=2(L+W)<BR>\n" );
document.write( "72 feet=2(24 feet+12 feet)<BR>\n" );
document.write( "72 feet=2(36 feet)<BR>\n" );
document.write( "72 feet=72 feet<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );