document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=179348>Question 179348</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please help.<BR>\n" );
document.write( "In 1920, the record for a certain race was 45.4 sec. In 1960, it was 44.6 sec.<BR>\n" );
document.write( "Let R(t) = the record in the race and t= the number of years since 1920.\r<BR>\n" );
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document.write( "a.)Find a linear finction that fits the data.<BR>\n" );
document.write( "b.) Use the functionin (a) to predict the record in 2003 and in 2006.<BR>\n" );
document.write( "c.) Find the year when the record will be 43.4 sec. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #134269 by </B><A HREF=/tutors/aboutme.mpl?userid=Mathtut><B>Mathtut(3670)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Mathtut><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=134269\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> from 1920-1960 there is 40 years<BR>\n" );
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document.write( "the difference between 45.4-44.6=.8<BR>\n" );
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document.write( "that difference took place over a 40 year span so we have to divide the difference by 40 to get a yearly average of the change<BR>\n" );
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document.write( ".8/40=.02.  this is how much on average per year that the race time is falling<BR>\n" );
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document.write( "equation: R(t)=-.02t+45.4<BR>\n" );
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document.write( "b)2003 would be 83 years:R(83)=-.02(83)+45.4=-1.66+45.4=43.74sec<BR>\n" );
document.write( "  2006 would be 86 years:R(86)=-.02(86)+45.4=-1.72+45.4=43.68sec<BR>\n" );
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document.write( "c)43.4=-.02t+45.4<BR>\n" );
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document.write( "-.02t=-2<BR>\n" );
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document.write( "t=100 years......so in the year 2020 this formula predicts a time of 43.4sec </SPAN>\n" );
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