Answer:
Step-by-step explanation:
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Answer:did you get the answer to this yet ?
Step-by-step explanation:
Answer:
a) r=-0.00012097
b) ![\large\ \boxed{Q(t)=Q_0e^{-0.00012097*t}}](https://tex.z-dn.net/?f=%5Clarge%5C%20%5Cboxed%7BQ%28t%29%3DQ_0e%5E%7B-0.00012097%2At%7D%7D)
c) 13,304.65 years
Step-by-step explanation:
a)
Q(t) satisfies the differential equation Q'(t) = rQ(t), hence
![\large \frac{Q'(t)}{Q(t)}=r](https://tex.z-dn.net/?f=%5Clarge%20%5Cfrac%7BQ%27%28t%29%7D%7BQ%28t%29%7D%3Dr)
Integrating on both sides
![\int \frac{Q'(t)}{Q(t)}dt=\int rdt\rightarrow log(Q(t))=rt+C](https://tex.z-dn.net/?f=%5Cint%20%5Cfrac%7BQ%27%28t%29%7D%7BQ%28t%29%7Ddt%3D%5Cint%20rdt%5Crightarrow%20log%28Q%28t%29%29%3Drt%2BC)
where C is a constant. Taking the exponential on both sides
![\large e^{log(Q(t))}=e^{rt+C}=e^Ce^{rt}\rightarrow Q(t)=C_0e^{rt}](https://tex.z-dn.net/?f=%5Clarge%20e%5E%7Blog%28Q%28t%29%29%7D%3De%5E%7Brt%2BC%7D%3De%5ECe%5E%7Brt%7D%5Crightarrow%20Q%28t%29%3DC_0e%5E%7Brt%7D)
In this case,
is the initial value Q(0), the amount of the initial value of carbon-14 and we have
![\large Q(t)=Q_0e^{rt}](https://tex.z-dn.net/?f=%5Clarge%20Q%28t%29%3DQ_0e%5E%7Brt%7D)
As we know that the half-life of carbon-14 is approximately 5,730 years we have
![\large Q(5,730)=\frac{Q_0}{2}\rightarrow Q_0e^{5,730r}=\frac{Q_0}{2}\rightarrow e^{5,730r}=\frac{1}{2}](https://tex.z-dn.net/?f=%5Clarge%20Q%285%2C730%29%3D%5Cfrac%7BQ_0%7D%7B2%7D%5Crightarrow%20Q_0e%5E%7B5%2C730r%7D%3D%5Cfrac%7BQ_0%7D%7B2%7D%5Crightarrow%20e%5E%7B5%2C730r%7D%3D%5Cfrac%7B1%7D%7B2%7D)
Taking logarithm on both sides
![\large log(e^{5,730r})=log(\frac{1}{2})\rightarrow 5,730r=-0.693147\rightarrow r=\frac{-0.693147}{5,730}](https://tex.z-dn.net/?f=%5Clarge%20log%28e%5E%7B5%2C730r%7D%29%3Dlog%28%5Cfrac%7B1%7D%7B2%7D%29%5Crightarrow%205%2C730r%3D-0.693147%5Crightarrow%20r%3D%5Cfrac%7B-0.693147%7D%7B5%2C730%7D)
and
![\large \boxed{r=-0.00012097}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7Br%3D-0.00012097%7D)
b)
![\large \boxed{Q(t)=Q_0e^{-0.00012097*t}}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7BQ%28t%29%3DQ_0e%5E%7B-0.00012097%2At%7D%7D)
c)
We want to find a value of t for which
Q(t) = 20% of
= ![\large\ 0.2Q_0](https://tex.z-dn.net/?f=%5Clarge%5C%200.2Q_0)
![\large Q(t)=0.2Q_0\rightarrow Q_0e^{-0.00012097*t}=0.2Q_0\rightarrow e^{-0.00012097*t}=0.2\rightarrow\\-0.00012097*t=log(0.2)\rightarrow t=\frac{log(0.2)}{-0.00012097}](https://tex.z-dn.net/?f=%5Clarge%20Q%28t%29%3D0.2Q_0%5Crightarrow%20Q_0e%5E%7B-0.00012097%2At%7D%3D0.2Q_0%5Crightarrow%20e%5E%7B-0.00012097%2At%7D%3D0.2%5Crightarrow%5C%5C-0.00012097%2At%3Dlog%280.2%29%5Crightarrow%20t%3D%5Cfrac%7Blog%280.2%29%7D%7B-0.00012097%7D)
and
![\large \boxed{t\approx 13,304.65\;years}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7Bt%5Capprox%2013%2C304.65%5C%3Byears%7D)
We need to remove those parenthesis! :)
-2[g(7)]
-2g * 7 (What I did was (2 x 7 = 14)
-14g