Question

the number of grams of a certain radioactive substance present at time t is given by the formula $a = 500e^{-0.0004t}$, where t is the number of years. find the number of grams that are present at time t = 0 and at time t = 800.

the number of grams present at time t = 0 is
( round to the nearest hundredth as needed. )

the number of grams present at time t = 800 is
( round to the nearest hundredth as needed. )

Explanation:

Step1: Find grams at \( t = 0 \)

Substitute \( t = 0 \) into \( A = 500e^{-0.024t} \).
Since \( e^{0}=1 \), we have \( A = 500\times1 = 500.00 \).

Step2: Find grams at \( t = 800 \)

Substitute \( t = 800 \) into the formula:
\( A = 500e^{-0.024\times800} \).
First, calculate the exponent: \( -0.024\times800 = -19.2 \).
Then, \( e^{-19.2}\approx0 \) (since \( e^{-x}\) approaches 0 as \( x \) increases).
So, \( A\approx500\times0 = 0.00 \).

Answer:

The number of grams present at time \( t = 0 \) is \( \boldsymbol{500.00} \).
The number of grams present at time \( t = 800 \) is \( \boldsymbol{0.00} \).