document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=145116>Question 145116</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 1.\r<BR>\n" );
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document.write( "i am supposed to simplify this.\r<BR>\n" );
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document.write( "1/3 the square root of 24 minus 2 the square root of 2/3.\r<BR>\n" );
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document.write( "i dont know what to do.\r<BR>\n" );
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document.write( "2.\r<BR>\n" );
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document.write( "i am supposed to solve this.<BR>\n" );
document.write( "the lenth of a rectangle is 2x width. the area is 64 cm squared. i need to find the dimensions and round to the nearest hundredth. I dont know what to do .\r<BR>\n" );
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document.write( "Please help\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #105825 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=105825\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%281%2F3%29sqrt%2824%29-2sqrt%282%2F3%29 ALIGN=MIDDLE  ALT=\"%281%2F3%29sqrt%2824%29-2sqrt%282%2F3%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%281%2F3%29%5E2=1%2F9 ALIGN=MIDDLE  ALT=\"%281%2F3%29%5E2=1%2F9\"   DISABLED_style= \"max-width:90%; height:auto;\" >, so <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%281%2F3%29sqrt%2824%29=sqrt%2824%2F9%29=sqrt%28%288%2A3%29%2F%283%2A3%29%29=sqrt%288%2F3%29 ALIGN=MIDDLE  ALT=\"%281%2F3%29sqrt%2824%29=sqrt%2824%2F9%29=sqrt%28%288%2A3%29%2F%283%2A3%29%29=sqrt%288%2F3%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >, so:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%281%2F3%29sqrt%2824%29-2sqrt%282%2F3%29=green%28sqrt%288%2F3%29-2sqrt%282%2F3%29%29=sqrt%284%2A%282%2F3%29%29-2sqrt%282%2F3%29=green%28sqrt%284%29%2Asqrt%282%2F3%29-2sqrt%282%2F3%29%29=2sqrt%282%2F3%29-2sqrt%282%2F3%29=green%280%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "2.  Area of a rectangle: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=LW ALIGN=MIDDLE  ALT=\"A=LW\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  If length is twice the width and the area is 64, then <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=L=2W ALIGN=MIDDLE  ALT=\"L=2W\"   DISABLED_style= \"max-width:90%; height:auto;\" > and the area is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A=%282W%29W=2W%5E2=64 ALIGN=MIDDLE  ALT=\"A=%282W%29W=2W%5E2=64\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Check your work by multiplying the value you determined for W by the value you determined for L.  You have solved the problem correctly if your result is 64 and L is twice the value of W.\r<BR>\n" );
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document.write( "Divide both sides by 2 and take the square root of both sides giving you the value of W.  Multiply that by 2 to get L. </SPAN>\n" );
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