document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=770300>Question 770300</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The temperature inside Daphne's refrigerator was 10 degrees at 1:00PM. If the temperature then started decreasing at a rate of one half of a degree per minute, what was the temperature at 2:20PM? Also what would be the temperature 6 hours after 2:20PM? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #469442 by </B><A HREF=/tutors/aboutme.mpl?userid=Alan3354><B>Alan3354(69443)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Alan3354><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=Alan3354><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=469442\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The temperature inside Daphne's refrigerator was 10 degrees at 1:00PM. If the temperature then started decreasing at a rate of one half of a degree per minute, what was the temperature at 2:20PM? Also what would be the temperature 6 hours after 2:20PM?<BR>\n" );
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document.write( "From 1:00 to 2:20 is 80 minutes<BR>\n" );
document.write( "80*0.5 = 40 deg decrease<BR>\n" );
document.write( "10 - 40 = -30 degs<BR>\n" );
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document.write( "6 more hours = 360 minutes<BR>\n" );
document.write( "360*0.5 = 180 deg decrease<BR>\n" );
document.write( "-30 - 180 = -210 degrees<BR>\n" );
document.write( "That's a cold refrigerator.<BR>\n" );
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