{VERSION 3 0 "APPLE_PPC_MAC" "3.0" }
{USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 
1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 
0 }{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 }
{CSTYLE "" -1 256 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 
"" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 
1 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 }
{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 
0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Output" 0 11 1 
{CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 3 0 -1 -1 -1 0 0 
0 0 0 0 -1 0 }}
{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 8 "Kate Hix" }}{PARA 0 "" 0 "
" {TEXT -1 14 "Love Wingfield" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 5 "Pg 82" }}{PARA 0 "" 0 "" {TEXT -1 4 "# 23
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 445 "Carb
on dating is used to determine the age of a fossil.  For example a hum
anoid skull was found in a cave in South Africa along the with the rem
ains of a campfire.  Archeologists believe the age of the skull was fo
und to be the same age as the campfire.  It is determined that only 2%
 of the original amount of carbon-14 remians in the burnt wood of the \+
campfire.  Estimate the age of the skull if the half-life of carbon-14
 is about 5600 years." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }{TEXT 259 17 "Question Summary:" }{TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 256 17 "Determin
e the age" }{TEXT -1 31 " of a fossil by carbon dating.\n" }{TEXT 257 
2 "2%" }{TEXT -1 79 " of the original carbon-14 remains in the fossil.
\nThe half-life of carbon-14 is" }{TEXT 258 12 " 5600 years." }{TEXT 
-1 1 "\n" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
46 "The equation we use to solve this problem is: " }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 37 "dsolve(\{diff(p(t),t)=(k*p(t))\},p(t));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"pG6#%\"tG*&%$_C1G\"\"\"-%$expG6#*
&%\"kGF*F'F*F*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 31 " Where k and _C
l are constants." }}{PARA 0 "" 0 "" {TEXT -1 15 "time = t = 5600" }}
{PARA 0 "" 0 "" {TEXT -1 64 "Carbon function = P(t) =_C1 / 2,  because
 carbon has a half-life" }}{PARA 0 "" 0 "" {TEXT -1 29 "        plug i
n P(5600)=_C1/2" }}{PARA 0 "" 0 "" {TEXT -1 23 "       and solve for k
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 41 "_k:=solve((_C1 / 2)=(_C1*exp(k*5600)),k);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%#_kG,$-%#lnG6#\"\"##!\"\"\"%+c" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 13 "Now plug \+
in k" }}{PARA 0 "" 0 "" {TEXT -1 63 "Change the left side of the equat
ion from (_Cl/2) to (.02*_C1) " }}{PARA 0 "" 0 "" {TEXT -1 55 "because
 the carbon constant is being multiplied by 2%. " }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 65 "age_of_fossil:=solve ( (.02*_C1)=(_C1*exp((-1
/5600)*ln(2)*t)),t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%.age_of_foss
ilG$\"+m%f0;$!\"&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
{MARK "0 3 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 }