Question
The amount of a radioactive isotope present at time t is given by Upper A left parenthesis t right parenthesis equals 400 e Superscript negative 0.02833 t grams, where t is the time in years that the isotope decays. The initial amount present is 400 grams. Complete parts (a) through (c).
Question content area bottom
Part 1
(a) How many grams remain after 15 years?
enter your response here g (Type an integer or decimal rounded to two decimal places as needed.)
Question content area bottom
Part 1
(a) How many grams remain after 15 years?
enter your response here g (Type an integer or decimal rounded to two decimal places as needed.)
Answers
GPT-4o mini
To find the amount of the radioactive isotope remaining after 15 years, we can use the formula provided:
\[
A(t) = 400 e^{-0.02833 t}
\]
Substituting \( t = 15 \) into the equation:
\[
A(15) = 400 e^{-0.02833 \times 15}
\]
First, calculate the exponent:
\[
-0.02833 \times 15 = -0.42495
\]
Now substitute this back into the equation:
\[
A(15) = 400 e^{-0.42495}
\]
Next, we can calculate \( e^{-0.42495} \):
\[
e^{-0.42495} \approx 0.6533 \quad (\text{using a scientific calculator})
\]
Now, we can continue:
\[
A(15) = 400 \times 0.6533 \approx 261.32
\]
So, the amount of the radioactive isotope remaining after 15 years is approximately:
\[
\boxed{261.32} \text{ grams}
\]
\[
A(t) = 400 e^{-0.02833 t}
\]
Substituting \( t = 15 \) into the equation:
\[
A(15) = 400 e^{-0.02833 \times 15}
\]
First, calculate the exponent:
\[
-0.02833 \times 15 = -0.42495
\]
Now substitute this back into the equation:
\[
A(15) = 400 e^{-0.42495}
\]
Next, we can calculate \( e^{-0.42495} \):
\[
e^{-0.42495} \approx 0.6533 \quad (\text{using a scientific calculator})
\]
Now, we can continue:
\[
A(15) = 400 \times 0.6533 \approx 261.32
\]
So, the amount of the radioactive isotope remaining after 15 years is approximately:
\[
\boxed{261.32} \text{ grams}
\]