Let's solve the given problem step by step.
We start with the decay model which describes the amount of carbon-14 present after [tex]\( t \)[/tex] years:
[tex]\[ A = 16e^{-0.000121t} \][/tex]
Here,
- [tex]\( A \)[/tex] is the amount of carbon-14 present after [tex]\( t \)[/tex] years.
- [tex]\( 16 \)[/tex] grams is the initial amount of carbon-14.
- [tex]\( e \)[/tex] is the base of natural logarithms.
- [tex]\( -0.000121 \)[/tex] is the decay constant.
- [tex]\( t \)[/tex] is the time in years, which is given as 8490 years.
Now, let's substitute [tex]\( t = 8490 \)[/tex] into our model to determine the amount of carbon-14 remaining.
[tex]\[ A = 16e^{-0.000121 \times 8490} \][/tex]
We calculate the exponent first:
[tex]\[ -0.000121 \times 8490 \][/tex]
This product evaluates approximately to:
[tex]\[ -1.02729 \][/tex]
Then we calculate [tex]\( e^{-1.02729} \)[/tex]. Evaluating this exponential term, we get:
[tex]\[ e^{-1.02729} \approx 0.358225762 \][/tex]
Now, multiply this by the initial amount of carbon-14:
[tex]\[ A = 16 \times 0.358225762 \][/tex]
Doing the multiplication:
[tex]\[ A \approx 5.727612185 \][/tex]
Thus, the amount of carbon-14 present after 8490 years is approximately:
[tex]\[ 5.727612185 \, \text{grams} \][/tex]
Rounding this to the nearest whole number, we get:
[tex]\[ A \approx 6 \, \text{grams} \][/tex]
So, after 8490 years, the amount of carbon-14 present will be approximately 6 grams.
