document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1105763>Question 1105763</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Time(y)      0    1    2    3    4    5<BR>\n" );
document.write( "Balance($) 5000 5600 6272 7025 7868 8812\r<BR>\n" );
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document.write( "Write an equation to represent this relation, is it linear or not explain. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #720654 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=720654\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> it is not linear.  If it were, the change from 0 to 1 (+600) would be the same for every single year, and it increases to (+672) for the second year and (+944) for the fifth year.<BR>\n" );
document.write( "This is compound interest and the initial increase of 600 in one year is 600/5000 or 0.12 increase.<BR>\n" );
document.write( "y=5000(1.12)^n, where n is between 0 and 5<BR>\n" );
document.write( "when n=0, y=5000<BR>\n" );
document.write( "exponential equation. </SPAN>\n" );
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