Question

the function $f(t) = 610(0.9975)^{\frac{t}{10}}$ represents the change in a quantity over $t$ years. what does the constant 0.9975 reveal about the rate of change of the quantity?
answer attempt 2 out of 2
the function is decaying exponentially at a rate of 0.25 % every year.

Explanation:

Step1: Recognize decay form

The function is $f(t) = 610(0.9975)^{\frac{t}{10}}$, matching exponential decay $A(t)=A_0(1-r)^k$, where $1-r=0.9975$.

Step2: Find 10-year decay rate

Solve for $r$: $r = 1 - 0.9975 = 0.0025 = 0.25\%$. This is the decay rate over 10 years.

Step3: Confirm time interval

The exponent $\frac{t}{10}$ means the base corresponds to a 10-year period.

Answer:

The function is decaying exponentially at a rate of 0.25% every 10 years.