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puteri [66]
3 years ago
8

Mr. Keller bought a car for $20,000. A study shows that a car will depreciate (go DOWN in value) by 15% each year. How much is M

r. Keller’s car worth after 5 years? A) Formula used: B)Substitute values: C) Final answer D) Does this final answer make sense compared to the original cost of the car? Why? E) DESCRIBE (using a complete sentence or two) how you can solve this problem graphically. Please answer all
Mathematics
2 answers:
Mariulka [41]3 years ago
7 0

AnswerA: the formula would be f(x) = P(R) ^T or f(x) = Principle(rate)^time

B: f(x) = 20,000(0.85)^5  

C: = 8,874.10625

D: Yes, the final answer makes sense compared to the origional cost of the car in relation to the formula. As well, time decreases the value of a car, so for the cost to be so low only makes sense due to the cars decrease in value or an extended and elongated amount of time.

E: You can solve this equation graphically by plotting th point at 20,000 and then taking 85% of 20,000 and plotting it each time until you get to the fifth year.

Step-by-step explanation:A: the formula would be f(x) = P(R) ^T or f(x) = Principle(rate)^time

B: f(x) = 20,000(0.85)^5  

C: = 8,874.10625

D: Yes, the final answer makes sense compared to the origional cost of the car in relation to the formula. As well, time decreases the value of a car, so for the cost to be so low only makes sense due to the cars decrease in value or an extended and elongated amount of time.

E: You can solve this equation graphically by plotting th point at 20,000 and then taking 85% of 20,000 and plotting it each time until you get to the fifth year.

Schach [20]3 years ago
5 0
A: the formula would be f(x) = P(R) ^T or f(x) = Principle(rate)^time

B: f(x) = 20,000(0.85)^5  

C: = 8,874.10625

D: Yes, the final answer makes sense compared to the origional cost of the car in relation to the formula. As well, time decreases the value of a car, so for the cost to be so low only makes sense due to the cars decrease in value or an extended and elongated amount of time. 

E: You can solve this equation graphically by plotting th point at 20,000 and then taking 85% of 20,000 and plotting it each time until you get to the fifth year.

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