document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=276446>Question 276446</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose that $7000 is invested at an interest rate of 5.8% per year, compounded continuously. What is the balance after 9 years?\r<BR>\n" );
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document.write( "I need a detailed explanation please! 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 #201463 by </B><A HREF=/tutors/aboutme.mpl?userid=jirehcwe><B>jirehcwe(1)</B></A><img src=\"/images/stars/iconNewId_16x16.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jirehcwe><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=201463\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> balance=7000(1.058)^n<BR>\n" );
document.write( "n being the number of years.<BR>\n" );
document.write( "1.058 is given by 100%+5.8% and power n is through compounding interest.<BR>\n" );
document.write( "balance=7000(1.058)^9<BR>\n" );
document.write( "=7000(1.661)<BR>\n" );
document.write( "=11627.03 </SPAN>\n" );
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