Do a gallon man 1G 4Q 2P 2C
Answer:
![A(t)=A_{0}e^{\frac{ln(\frac{1}{2})}{22}t}](https://tex.z-dn.net/?f=A%28t%29%3DA_%7B0%7De%5E%7B%5Cfrac%7Bln%28%5Cfrac%7B1%7D%7B2%7D%29%7D%7B22%7Dt%7D)
Step-by-step explanation:
This half life exponential decay equation goes by the formula:
![A(t)=A_{0}e^{kt}](https://tex.z-dn.net/?f=A%28t%29%3DA_%7B0%7De%5E%7Bkt%7D)
Where
![k=\frac{ln(\frac{1}{2})}{Half-Life}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7Bln%28%5Cfrac%7B1%7D%7B2%7D%29%7D%7BHalf-Life%7D)
Since half life is given as 22, we plug that into "Half-Life" in the formula for k and then plug in the formula for k into the exponential decay formula:
So,
![k=\frac{ln(\frac{1}{2})}{Half-Life}\\k=\frac{ln(\frac{1}{2})}{22}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7Bln%28%5Cfrac%7B1%7D%7B2%7D%29%7D%7BHalf-Life%7D%5C%5Ck%3D%5Cfrac%7Bln%28%5Cfrac%7B1%7D%7B2%7D%29%7D%7B22%7D)
Now
![A(t)=A_{0}e^{\frac{ln(\frac{1}{2})}{22}t}](https://tex.z-dn.net/?f=A%28t%29%3DA_%7B0%7De%5E%7B%5Cfrac%7Bln%28%5Cfrac%7B1%7D%7B2%7D%29%7D%7B22%7Dt%7D)
third choice is correct.
Answer:
y
Step-by-step explanation: