document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=387320>Question 387320</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Use the exponential decay model A=Pe^(kt) to solve the following proublem.<BR>\n" );
document.write( "The half-life of plutonium-238 is days. The initial amount is 100 grams.<BR>\n" );
document.write( "a)find the value of K to four places.<BR>\n" );
document.write( "b) write the formula for the amount of plutonium-238 present at the end of T days.<BR>\n" );
document.write( "c)How much of the initial amount will remain after 30 days? Round the answer to the nearest tenth gram.\r<BR>\n" );
document.write( "\n" );
document.write( "I know this seems like alot but I swear it's all one proublem just has Three parts to it, been having trouble, thank you sooo much in advance!!! :) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #273808 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=273808\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Use the exponential decay model A=Pe^(kt) to solve the following proublem.<BR>\n" );
document.write( "The half-life of plutonium-238 is days. The initial amount is 100 grams.<BR>\n" );
document.write( "a)find the value of K to four places.<BR>\n" );
document.write( "(1/2)P = P*e^(k*238)<BR>\n" );
document.write( "e^(238k) = 1/2<BR>\n" );
document.write( "Take the natural log of both sides to get:<BR>\n" );
document.write( "238k = ln(1/2)<BR>\n" );
document.write( "k = -0.002912<BR>\n" );
document.write( "---------------------------<BR>\n" );
document.write( "b) write the formula for the amount of plutonium-238 present at the end of T days.<BR>\n" );
document.write( "A(t) = P*e^(-0.002912t)<BR>\n" );
document.write( "---------------------------<BR>\n" );
document.write( "c)How much of the initial amount will remain after 30 days? Round the answer to the nearest tenth gram.<BR>\n" );
document.write( "A(30) = 100*e^(-0.002912*30<BR>\n" );
document.write( "---<BR>\n" );
document.write( "A(30) = 100*0.916337<BR>\n" );
document.write( "A(30) = 91.6337 grams<BR>\n" );
document.write( "==========================<BR>\n" );
document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
document.write( "============= <BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );