document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=278412>Question 278412</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Using application of rational expressions?\r<BR>\n" );
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document.write( "I can run 5 miles in 1 hour. I used to run 5 miles in 1 hour 30 minutes. How shorter was the time to run in 5 miles in 1 hour? \r<BR>\n" );
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document.write( "5 miles = 1 hour<BR>\n" );
document.write( "5 miles = 1 hour 30 minutes<BR>\n" );
document.write( "average miles/time = 5+5(60+90)<BR>\n" );
document.write( "= 10/90<BR>\n" );
document.write( "=1/9 miles/minute<BR>\n" );
document.write( "=1/9 * 60<BR>\n" );
document.write( "= 60/90 miles per hour\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 #202589 by </B><A HREF=/tutors/aboutme.mpl?userid=JBarnum><B>JBarnum(2146)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=JBarnum><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=JBarnum><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=202589\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> ok the way i see it is that 30 mins is half of an hour, so if he runs now at 5mi/hr then half of 5miles =1/2 hr which is 2.5 miles. so if he used to run 5miles in 1.5hrs and now runs 5miles in 1 hr then hes runing 2.5mi/hr faster. </SPAN>\n" );
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