document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=218148>Question 218148</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The formula r = 2(sqrt of 5L)  can be used to approximate the speed r, in miles per hour, of a car that has left skid marks of length L, in feet. How far will a car skid going 50 miles per hour? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #164368 by </B><A HREF=/tutors/aboutme.mpl?userid=RAY100><B>RAY100(1637)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=RAY100><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=164368\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> r= 2*sqrt(5L)<BR>\n" );
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document.write( "for  r=50<BR>\n" );
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document.write( "50 =2*sqrt(5L)<BR>\n" );
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document.write( "25= sqrt(5L),,,,square  both  sides<BR>\n" );
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document.write( "625 = 5L<BR>\n" );
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document.write( "125  ft = L<BR>\n" );
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document.write( "check,,,r= 2*sqrt(5*125) = 2*sqrt625 = 2*25 =50,,,,,ok<BR>\n" );
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