A teacher decides to purchase a new car and considers two options. Option one is the new Zoomba for $60,000 with an expected depreciation of 2% per month. Option two is the new Starfish for $40,000 with an expected depreciation of $200 per month. The teacher plans to keep the car for three years and chooses the option that is expected to retain the most value. Which option did the teacher choose (exponential or linear), and what was the value of the chosen option after three years?

A Linear system is a system in which the degree of the variable in the equation is one. It may contain one, two, or more than two variables.

A teacher decides to purchase a new car and considers two options.

The teacher plans to keep the car for three years and chooses the option that is expected to retain the most value.

Option one is the new Zoomba for $60,000 with an expected depreciation of 2% per month.

The value of Zoomba after three years will be

[tex]\rm Zoomba\ value = 60000*(1-0.02)^{3*12}\\\\\\Zoomba\ value = 60000*(0.98)^{36}\\\\\\Zoomba\ value = 60000*0.4832\\\\\\Zoomba\ value = 28992.79[/tex]

The value of the Zoomba after three years is 28,992.79 dollars.

Option two is the new Starfish for $40,000 with an expected depreciation of $200 per month.

Let x be the number of months. Then the value of the car is the linear function. Then we have the equation,

[tex]\rm Starfish \ value = 40000 - 200x[/tex]

The value of the Starfish car after three years (36 months) will be

[tex]\rm Starfish \ value = 40000 - 200x\\\\\rm Starfish \ value = 40000 - 200*36\\\\\rm Starfish \ value = 40000 - 7200\\\\\rm Starfish \ value = 32800[/tex]

The value of the Starfish car after three years is 32,800 dollars.

It is clear that the depreciation of the Zoomba car is more than the Starfish car.

Thus, the teacher will choose a Starfish car.

More about the linear system link is given below.

https://brainly.com/question/20379472