document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=319158>Question 319158</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A dish of lasagna baked at 375°F is taken out of the oven into a kitchen that is 74°F. After 6 minutes, the temperature of the lasagna is 319.5°F. What will its temperature be 20 minutes after it was taken out of the oven? Round your answer to the nearest degree. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #228602 by </B><A HREF=/tutors/aboutme.mpl?userid=nyc_function><B>nyc_function(2741)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=nyc_function><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=nyc_function><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=228602\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> According to Newton's Law of Cooling, T - Ts = (To - Ts) e^(-kt), where Ts is the temperature of the surrounding medium, To is the initial temperature, T is the temperature at time t.\r<BR>\n" );
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document.write( "319.5 - 74 = (375 - 74) e^(-k • 6)<BR>\n" );
document.write( "245.5 = 301 e^(-6k)<BR>\n" );
document.write( "0.791935... = e^(-6k)<BR>\n" );
document.write( "-0.23327... = -6k<BR>\n" );
document.write( "0.038879225... = k\r<BR>\n" );
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document.write( "T - 74 = 301e^(-0.03887... • 30) = 93.7595488...<BR>\n" );
document.write( "T = 167.7595488... &#8776; 168°<BR>\n" );
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