document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=981454>Question 981454</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Exponential growth word problem: I believe I answered a, b, c, and d correctly. Don't know how to answer e. Had to put all questions since they are related.\r<BR>\n" );
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document.write( "a. A penny is 1.55 mm thick. About how many pennies can you stack on top of each other in order to get one meter? Round to nearest hundredth.(There are 100mm in a meter). My answer: 64.52\r<BR>\n" );
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document.write( "b. Imagine that you stack pennies on top of each other every day. Each day, you double the amount of pennies in the stack. This means that on day 0, there is 1 penny, on day 1, there is 2 pennies, on day 2 there is 4 pennies, etc. What exponential function is being modeled here? My answer: Exponential growth y=c(1+r)^t\r<BR>\n" );
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document.write( "c. If this pattern continues, in how many days will the stack of pennies be about 1 meter tall? My answer: 6 days.\r<BR>\n" );
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document.write( "d. The distance from the earth to the moon is 405,696,000 meters. About how many stacked pennies would it take to reach the moon? Round to the nearest hundredth. My answer: 6,287,910.73\r<BR>\n" );
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document.write( "e. After about how many days will the stack of pennies be tall enough to reach the moon?  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #602413 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=602413\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  There are 100 cm in a meter, 1000mm.  There are 645.16 pennies in a meter.\r<BR>\n" );
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document.write( "c(1+r)^t<BR>\n" );
document.write( "c=1.55<BR>\n" );
document.write( "1.55*2^t<BR>\n" );
document.write( "let t=6 days<BR>\n" );
document.write( "1.55*64=99.2 mm (this is 0.1 meter), want 1000 mm. <BR>\n" );
document.write( "1000=1.55*2^t<BR>\n" );
document.write( "divide by 1.55 and take ln of both sides.  I will round but don't round until end.<BR>\n" );
document.write( "645.16=2^t<BR>\n" );
document.write( "6.47=t ln 2; ln2=0.693<BR>\n" );
document.write( "9.333 days=t<BR>\n" );
document.write( "2^9=512<BR>\n" );
document.write( "512*1.55mm=793.6mm<BR>\n" );
document.write( "1024*1.55 mm=1587.2 mm.  Between the 9th and 10th days.<BR>\n" );
document.write( "405696000meters*645.16 pennies/meter=2.617*10^11 pennies\r<BR>\n" );
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document.write( "2.617X10^11=1.55*2^t<BR>\n" );
document.write( "divide by 1.55<BR>\n" );
document.write( "1.6886 X10 ^11=2^t<BR>\n" );
document.write( "ln of both sides<BR>\n" );
document.write( "25.8524=t ln 2<BR>\n" );
document.write( "divide by 0.693 both sides<BR>\n" );
document.write( "t=37.30 days\r<BR>\n" );
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document.write( "The two areas are 1000 mm per meter and thinking of rolls of 50 pennies, that are only a few inches long.  That is where I get the 645 pennies.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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