Hey, If you guys can just answer this question correctly please. It will mean so much to me! I’ve been thinking here for 3 hours trying to find an answer to the question! I’ll do 30 points for now! Belinda is thinking about buying a car for $45,000. The table below shows the projected value of two different cars for 3 years.

Number of years:
1
2
3
Car 1 (value in dollars):
40,500
36,450
32,805
Car 2 (value in dollars):
42,000
39,000
36,000

Part A: What type of function, linear or exponential, can be used to describe the value of each of the cars after a fixed number of years? Explain your answer.

Part B: Write one function for each car to describe the value of the car f(x), in dollars, after x years.

Part C: Belinda wants to purchase a cat that would have the greatest value in 13 years. Will there be any significant difference in the value of either car after 13 years? Explain your answer, and show the value of each car after 13 years.

Thanks!

Part A:

The decline in the value of the first car is represented by an exponential function, since it does not have a constant rate of decrease as in the case of a linear function. The second car's depreciation in value may be modeled by a linear equation due to the constant difference between the yearly value.

Part B:

Let the function for car 1 be:

New value = original value * (depreciation factor)^x. For x = 1:

Depreciation factor = new value / original value

Depreciation factor = 40,500 / 45,000

Depreciation factor = 0.9

For car 1:

f(x) = 45,000(0.9)^x

For car 2:

Yearly decrease in value is $3,000, so:

f(x) = 45,000 - 3,000x

Part C:

We plug in 14 for x in both of our equations to find the value after 14 years of each car.

Car 1:

f(14) = 45,000(0.9)^14

f(14) = $10,294.56

Car 2:

f(14) = 45,000 - 3,000(14)

f(14) = $3,000

There is a significant difference between the values of the two cars after 14 years, greater than $7,000, so car 1 should be bought.

Step-by-step explanation:

Brainliest??

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answer:   2 multiplied by x =24

step-by-step explanation: