9
advanced-math

A car purchased for $20,000 depreciates at a rate of 15% per year. Which equation models its value V(t)V(t)?

A

V(t)=20000(0.15)tV(t) = 20000(0.15)^t

B

V(t)=20000(1.15)tV(t) = 20000(1.15)^t

C

V(t)=20000(0.85)tV(t) = 20000(0.85)^t

D

V(t)=2000015tV(t) = 20000 - 15t

Correct Answer: C

Choice C is the correct answer. Depreciation means decay.

  1. Identify Rate: r=15%=0.15r = 15\% = 0.15.
  2. Calculate Factor: b=1r=10.15=0.85b = 1 - r = 1 - 0.15 = 0.85.
  3. Form Equation: V(t)=20000(0.85)tV(t) = 20000(0.85)^t.

💡 Strategic Tip: For depreciation, the base must be less than 1.

Choice A is incorrect because 0.15 would mean losing 85% per year. Choice B is incorrect because 1.15 means the car gains value. Choice D is incorrect because it represents linear depreciation.