document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=578806>Question 578806</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If $P are invested at an interest rate r compounded annually, at the end of 2 years the amount will be A=P(1+r). At what interest rate will $1000 increase to $1440 in 2 years  ?  A=$1440 and P=$1000.<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 #370779 by </B><A HREF=/tutors/aboutme.mpl?userid=dfrazzetto><B>dfrazzetto(283)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=dfrazzetto><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=dfrazzetto><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=370779\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> According to the stated formula (which does not look how annually compounded interest works FYI):\r<BR>\n" );
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document.write( "A=P(1+r)<BR>\n" );
document.write( "1440 = 1000(1+r)<BR>\n" );
document.write( "1440 = 1000 + 1000r\r<BR>\n" );
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document.write( "1000r = 440<BR>\n" );
document.write( "r= 440/1000=.44 = 44% </SPAN>\n" );
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