Depreciation is the decrease or loss in value of an item due to age, wear, or market conditions. We usually consider depreciation on expensive items like cars. Businesses use depreciation as a loss when calculating their income and taxes. One company buys a new bulldozer for $148950. The company depreciates the bulldozer linearly over its useful life of 24 years. Its salvage value at the end of 24 years is $14550. A) Express the value of the bulldozer, V, as a function of how many years old it is, t. Make sure

V = -5600t + 148,950

Step-by-step explanation:

In this question, we are expected to give an equation or relation that relates the value of a bulldozer V to the time it took for it to have depreciated to that particular value.

This question is seeking to establish a linear model to relate the depreciated value V to the time taken t.

Now, since it is a linear model, let’s picture a plot. A graphical plot that has time t on the x axis and Value V on the y axis.

At year 0, when the bulldozer was newly bought, it has a value of $148,950.

Writing this in form of coordinates, we can say that the point here on the Cartesian plane would be (0, 148,950)

Now, after 24 years, depreciated value is $14,550. This means that the point we could have here is (24, 14,550)

Now, we can calculate the slope of the line to be modeled using these two points;

Gradient of linear line= (14,550-148,950)/(24 - 0) = -5,600

Mathematically, the equation of a linear model is y = mx + c

Where m is the gradient of the line and c is the y intercept of the line. Here, we can identify the gradient as -5,600 , y-intercept as value of bulldozer at when t is 0 years, y is our value V and x is our time t.

Putting the equation together, we have;

V = -5600t + 148,950

We should understand that our slope is negative because, the value of the bulldozer is expected to decrease as the years toll by

part a: 1.08

part b: 27.59