Question

**NEEd Help??!!

1. Hank bought a $24,000 car when he graduated from college. His car will lose half of its value in 6.58 years. At what rate does Hank’s car depreciate?

A) 30%
B) 20%
C) 10%
D) 5%

2. A new car worth $20,000 loses 20% of its value every year. Is the value of the car represented by a linear or exponential function?
A) linear
B) exponential
C) both linear and exponential
D) neither linear or exponential

Answer 1

12000=24000(1-r)^6.58
Solve for r
R=0.09998*100=9.9998%
Round your answer
R=10%

To solve the second question we use the formula of
A=p(1-r)^ t

Answer 2

Answer: 1.C) 10% (approx)

2. B) Exponential

Step-by-step explanation:

1) Since, The initial price of the car, P = $24,000

And, the total time, t = 6.8 years.

Final price of the car after 6.8 years, A = 12,000

Let the rate of depreciation is r %

Thus,
`A= P (1-(r)/(100))^(6.58)`

⇒
`12,000= 24,000 (1-(r)/(100))^(6.58)`

⇒
`0.5 = (1+(r)/(100))^(6.58)`

⇒
`(0.5)^(1/6.58) = 1-(r)/(100)`

⇒
`(0.5)^(1/6.58) = 1-(r)/(100)`

⇒
`0.90001709913 = 1-(r)/(100)`

⇒
`r=100-90.001709913=9.99829008677\approx 10\%`

Thus, Option C is correct.

2) Here, The price of car initially, P = $20,000

Rate of decay, r = 20%

Thus, the function that shows the price of car,

`f(x)= 20,000 (1-(20)/(100))^x`

`f(x)= 20,000 (1-(1)/(5))^x`

`f(x)= 20,000 (1.2)^x`

Since, the function is decreasing by the constant percent not the constant rate.

Therefore,By the property of exponential function,

f(x) is an exponential function.

Thus, Option B is correct.