Question

A small business buys a computer for \(4000. After 4 years the value of the computer is expected to be \)200. For accounting purposes the business uses, linear depreciationto assess the value of the computer at a given time. This means that if Vis the value of the computer at time t, then a linear equation is used to relate Vand t.

(a)Find a linear equation that relates V andt

(b)Sketch a graph of this linear equation.

(c)What do the slope and V-intercept of the graph represent?

(d)Find the depreciated value of the computer 3 years from the date of purchase.

Short answer

(a) The linear equation is$V=50t+4000.$

(b) The answer is the graph of$V=50t+4000.$

(c) The slope of the line is 50 and the V-intercept is 4000.

(d) After 3 years, the value is $4150.

Step-by-step solution

Part a. Step 1. Concept of linear equation.

A linear equation is an algebraic equation in which each term has an exponent of one and the graphing of the equation results in a straight line.

Part a. Step 2. Two point form of linear equation.

If two points$A(x_1,y_1)$ and$B(x_2,y_2)$ are given, the equation of$AB$ is$\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right).$

Part a. Step 3. Forming the linear equation.

Let $V_1=\$4000,t_1=0$

$V_2=\$\left(4000+200\right),t_2=4$

$V_2=\$4200$

Thus, the linear equation is

$V-4000=\frac{4200-4000}{4-0}\left(t-0\right)$

$$\begin{gathered} V-4000=\frac{200}{4}\left(t\right) \\ V=50t+4000 \end{gathered}$$

Part b. Step 1. Concept of graph.

Graph is a mathematical representation of a network which describes the relationship between lines and points.

Part b. Step 2. Concept of graphing a linear equation.

The graph of linear equation in two variables is a line.

Part b. Step 3. Sketching the graph.

The graph of the linear equation$V=50t+4000$ is as follows,

Part c. Step 1. Concept of Slope.

Suppose a non-vertical line has a slope m and y-intercept b. This means that the line intersects the y-axis at the point $\left(0,b\right),$so the point slope form of the equation of the line with $x=0$and $y=b$becomes

$\begin{array}{c}(y-b)=m(x-0)\\ y=mx+b\end{array}$

Part c. Step 2. Concept of y-intercept.

The y-coordinate is a point where a line curve or a surface intersects the y-axis.

Part c. Step 3. Determining the slope and the intercept.

The given linear equation is

$V=50t+4000$

Then, the slope of the line is 50 and the V-intercept is 4000.

Part d. Step 1. Concept of linear equation.

A linear equation is an algebraic equation in which each term has an exponent of one and the graphing of the equation results in a straight line.

Part d. Step 2. Concept to determine two points.

If the value of x is given, then we have to evaluate the y by substituting the value in the equation and solve for y.

Part d. Step 3. Evaluating the depreciated value.

The given linear equation is

$V=50t+4000$

Substituting $t=3,$we get

$\begin{array}{l}V=50\left(3\right)+4000\\ V=150+4000\\ V=4150\end{array}$

Hence, after 3 years, the value of the computer is $4150.