Question

the amount of a radioactive isotope present at time t is given by ( a(t) = 600e^{-0.02896t} ) grams, where t is the time in years that the isotope decays. the initial amount present is 600 grams. complete parts (a) through (c)
(a) how many grams remain after 15 years?
g (type an integer or decimal rounded to two decimal places as needed)

Explanation:

Step1: Substitute \( t = 15 \) into the formula

We have the formula \( A(t)=600e^{- 0.02896t} \). Substitute \( t = 15 \) into it: \( A(15)=600e^{-0.02896\times15} \).
First, calculate the exponent: \( - 0.02896\times15=-0.4344 \).
So, \( A(15)=600e^{-0.4344} \).

Step2: Calculate the value of \( e^{-0.4344} \)

We know that \( e^{-x}=\frac{1}{e^{x}} \). Using a calculator, \( e^{-0.4344}\approx0.6479 \).

Step3: Calculate the value of \( A(15) \)

Multiply 600 by \( e^{-0.4344} \): \( A(15)=600\times0.6479 = 388.74 \) (rounded to two decimal places).

Answer:

\( 388.74 \)