document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1140193>Question 1140193</A>: <I><SPAN STYLE=\"word-break: break-all;\">  Question Help <BR>\n" );
document.write( "The​ half-life of radium is 1690 years. If 80 grams are present​ now, how much will be present in 420 years?<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #760688 by </B><A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><B>ankor@dixie-net.com(22740)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><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=ankor@dixie-net.com><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=760688\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The​ half-life of radium is 1690 years.<BR>\n" );
document.write( " If 80 grams are present​ now, how much will be present in 420 years?<BR>\n" );
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document.write( "The radioactive decay formula<BR>\n" );
document.write( "A = Ao*2^(-t/h), where<BR>\n" );
document.write( "A = amt remaining after t time<BR>\n" );
document.write( "Ao - initial amt<BR>\n" );
document.write( "t = time of decay<BR>\n" );
document.write( "h = half-life of substance<BR>\n" );
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document.write( "A = 80*2(-420/1690)<BR>\n" );
document.write( "use calc<BR>\n" );
document.write( "A = 80 * .84176<BR>\n" );
document.write( "A = 67.32 grams after 420 yrs </SPAN>\n" );
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