Asked by Derrick
A construction company purchased some equipment costing $150,000. The value of the equipment depreciates at a rate of 8% per year.
Write a formula that models the value of the equipment each year.
Answers
Answered by
GPT-4o mini
The value of the equipment can be modeled using the formula for exponential decay, which is given by:
\[
V(t) = V_0 \times (1 - r)^t
\]
where:
- \( V(t) \) is the value of the equipment at time \( t \) (in years),
- \( V_0 \) is the initial value of the equipment,
- \( r \) is the rate of depreciation (expressed as a decimal),
- \( t \) is the time in years.
For your specific case:
- \( V_0 = 150,000 \)
- \( r = 0.08 \)
Substituting these values into the formula gives:
\[
V(t) = 150,000 \times (1 - 0.08)^t
\]
Simplifying further:
\[
V(t) = 150,000 \times (0.92)^t
\]
So the formula that models the value of the equipment each year is:
\[
V(t) = 150,000 \times (0.92)^t
\]
\[
V(t) = V_0 \times (1 - r)^t
\]
where:
- \( V(t) \) is the value of the equipment at time \( t \) (in years),
- \( V_0 \) is the initial value of the equipment,
- \( r \) is the rate of depreciation (expressed as a decimal),
- \( t \) is the time in years.
For your specific case:
- \( V_0 = 150,000 \)
- \( r = 0.08 \)
Substituting these values into the formula gives:
\[
V(t) = 150,000 \times (1 - 0.08)^t
\]
Simplifying further:
\[
V(t) = 150,000 \times (0.92)^t
\]
So the formula that models the value of the equipment each year is:
\[
V(t) = 150,000 \times (0.92)^t
\]
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