document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=2633>Question 2633</A>: <I><SPAN STYLE=\"word-break: break-all;\"> half-life of silicon 32 is 710 years....40 grams present..<BR>\n" );
document.write( "how much in 200 years? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #1130 by </B><A HREF=/tutors/aboutme.mpl?userid=khwang><B>khwang(438)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=khwang><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=1130\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  Let Q = Qo e^(-kt),<BR>\n" );
document.write( " when t =0, Q = 40, so Qo = 40,<BR>\n" );
document.write( " So, Q = 40 e^(-kt)<BR>\n" );
document.write( " when t=710, Q = 1/2 Q0 = 20,<BR>\n" );
document.write( " solve 20 = 40 e^(-710k), or 1/2 = e^(-710k)<BR>\n" );
document.write( " we get k = ln2/710 = 0.000976264 = 9.76 *10^(-4)\r<BR>\n" );
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document.write( " hence,Q = 40 e^(- 9.76 *10^(-4)t)<BR>\n" );
document.write( " when t = 200 yrs, Q = 40 e^(- 9.76 *10^(-4)*200)<BR>\n" );
document.write( "                     = 32.91 grams\r<BR>\n" );
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document.write( " Kenny<BR>\n" );
document.write( " \r<BR>\n" );
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document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );