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4. The value of a car decreases linearly with time. The car was bought new for $62,000 and had a value of $41,000 after6 years.A. [4 pts] Create the equation that models this problem. Let V(t) be the value of the car at time t years after it wasbought.B. [3 pts] Find its value 11 years after it was bought. Use the equation you created in Part A above

Respuesta :

Answer:

A. Equation: V(t) = -3500t + 62 000

B. $23,500

Explanation:

We are told that the price of the car can be modelled as a linear equation. This means, we can write

[tex]V(t)=mx+b[/tex]

where m is the slope of the line and b is the y-intercept.

Now we know that the points (0, 62 000) and (6, 41 000) lie on the line. Therefore, the slope of the line is

[tex]m=\frac{41,000-62,000}{6-0}=-3500[/tex]

Therefore, the equation thus far we have is

[tex]V(t)=-3500t+b[/tex]

Now what is the y-intercept b? the y-intercept is found by putting t = 0 into the equation. Luckily for us though, we know that the point (0, 62,000) lies on the line. This tells us that b = 62,000. Therefore, the equation of the line is

[tex]\boxed{V\left(t\right)=-3500t+62000.}[/tex]

Part B.

Now that we have the equation that models the price of the car, we can find the price after 11 years by putting t = 11 into the above equation. This gives

[tex]V(11)=-3500(11)+62000[/tex]

The right hand simplifies to give

[tex]V(11)=$ 23,500 $[/tex]

which is our answer!

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