Question: f(x) = 5,470(0.64)^{x/12}
The function f gives the value, in dollars, of a certain piece of equipment after x months of use. If the value of the equipment decreases each year by p% of its value the preceding year, what is the value of p?
A. 4
B. 5
C. 36
D. 64
1. **State the problem:** We have an exponential decay function $$f(x) = 5470(0.64)^{\frac{x}{12}}$$ where $$x$$ is in months. We want to find the yearly percentage decrease $$p$$ in the equipment's value.
2. **Understand the formula:** The function shows the value after $$x$$ months. The base $$0.64$$ raised to $$\frac{x}{12}$$ means the value is multiplied by $$0.64$$ every 12 months (1 year).
3. **Interpret the yearly decay:** After 1 year ($$x=12$$), the value is:
$$
f(12) = 5470 \times 0.64^{\frac{12}{12}} = 5470 \times 0.64
$$
This means the equipment retains $$64\%$$ of its value each year.
4. **Calculate the percentage decrease $$p$$:** The decrease $$p$$ is the percentage lost each year, so:
$$
p = 100\% - 64\% = 36\%
$$
5. **Answer:** The value of $$p$$ is $$36\%$$, which corresponds to choice C.
**Final answer:** $$p = 36\%$$