document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=843414>Question 843414</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If I have 80 feet of fencing and I need to have 500 square feet of area. How would I do it? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #508088 by </B><A HREF=/tutors/aboutme.mpl?userid=JulietG><B>JulietG(1812)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=JulietG><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=JulietG><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=508088\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Your perimeter is 2Length + 2Width.  You know you have 80 feet of fencing, so 2L + 2W = 80.  Divide each side by 2 to get L + W = 40<BR>\n" );
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document.write( "Width * Length = area.  W * L = 500<BR>\n" );
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document.write( "What you need is 2 numbers that equal 40 when added and 500 when multiplied.<BR>\n" );
document.write( "What are the factors of 500?<BR>\n" );
document.write( "1,500<BR>\n" );
document.write( "2,250<BR>\n" );
document.write( "4,125<BR>\n" );
document.write( "5,100<BR>\n" );
document.write( "10,50<BR>\n" );
document.write( "20,25\r<BR>\n" );
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document.write( "None of these work.  Are you sure the 80 and 500 numbers are correct/ </SPAN>\n" );
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