document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=289724>Question 289724</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I am an adult who is reviewing my high school algebra II course from many years ago.  I am stuck on this problem in Chapter 2 (Solving Equations and Inequalities):  \"The hands of clock are together at noon.  When will they be together again for the first time?\"<BR>\n" );
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document.write( "Obviously, you can turn the hands of an actual clock and see that the answer is about 5 minutes after one, but how to make an equation out of that?  I'm not seeing how a position on the clock face can be a quantitive variable.  It must have something to do with the distance around the circle, and it must have something to do with the fact that the minute hand travels 12 times as fast as the hour hand.  But that's as far as I can get with it.  All the other travel and distance problems that I have done have been straight line paths, but in this problem, one hand laps the other.  The answer in the back of the book is five and 5/11ths minutes after one o'clock, but knowing that does not help me at all.  Even knowing the answer, I still don't see what the equation is.  \r<BR>\n" );
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document.write( "Can someone help me, please?  Really, I would like to have a hint even better than to be given the full equation.  If I can figure this one out, the next problem concerns three runners lapping each other on a track, so it is similar but slightly harder. If I can do this clock problem, I think I can do the track runner problem.  \r<BR>\n" );
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document.write( "Thank you.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #209791 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=209791\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Garamond\" size=\"+2\">\r<BR>\n" );
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document.write( "The equation you need for the clock is:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ m\ =\ \frac{60}{5.5}h\">\r<BR>\n" );
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document.write( "Where h is the hour number nearest the hour hand, noon being hour 0, and then m is the number of minutes where the hands are exactly superimposed.\r<BR>\n" );
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document.write( "For your situation, h = 1, so <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large m\ =\ \frac{60}{5.5}\ =\ 5\frac{5}{11}\ =\ 5.\overline{45}\">\r<BR>\n" );
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document.write( "The equation follows from the fact that the angle between the two hands of a clock is found by:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \cos^{-1}(\cos{5.5x})\">\r<BR>\n" );
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document.write( "where <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x\"> is the time of day expressed in minutes past noon, and you  have to set that expression equal to zero because when the hands are superimposed the angle between them is zero.\r<BR>\n" );
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document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi} + 1 = 0\"><BR>\n" );
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