7
algebra

A car depreciates linearly. Value new: $20,000. Value after 5 years: $10,000. What is the value after 8 years?

A

$4,000

B

$6,000

C

$8,000

D

$2,000

Correct Answer: A

Choice A is the correct answer. Find rate and project.

  1. Change: 10,000−20,000=−10,00010,000 - 20,000 = -10,000 in 5 years.
  2. Rate: −10,000/5=−2,000-10,000 / 5 = -2,000 per year.
  3. Model: V=−2000t+20,000V = -2000t + 20,000.
  4. Calculate: t=8t=8. V=−2000(8)+20,000=−16,000+20,000=4,000V = -2000(8) + 20,000 = -16,000 + 20,000 = 4,000.

💡 Strategic Tip: Or step by step: Year 5 is 10k. Year 6 is 8k. Year 7 is 6k. Year 8 is 4k.

Choice B is incorrect.Choice C is incorrect.Choice D is incorrect.

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Linear Functions Set 97