Chapter 1.1, Problem 93E

a.

To determine

Calculate the linear equation giving the value V of the equipment for each year t during its $10$ years of use.

Answer to Problem 93E

The linear equation of the V is $V=-2300t+25000$ .

Explanation of Solution

Given: It is given in the question that a school district purchase a high − volume printer,copier , and scanner for $\$25000$ .After $10$ years ,the equipment will have to be replaced .its value at that time is expected to be $\$2000$ .

Concept Used:

In this,use the slope formula ans the slope − intersept form and the formula used is $m=\frac{y_2-y_1}{x_2-x_1}$ and $y=mx+c$ .

Calculation:

In this,use the slope formula due to depreciation per year,

  $m=\frac{y_2-y_1}{x_2-x_1}=\frac{2000-25000}{10-0}=\frac{-23000}{10}=-2300$

So ,the depreciation per year is $-\$2300$ .

The starting price for the printer of this will be $\$25000$ and the c = $25000$ .

Now, just plug that and the starting value into the standard linear equation and will end up with

  $y=-2300(x)+25000$

Or It can also be in the form of $V=at+b$ , $\Rightarrow V=-2300t+25000$ .

Conclusion:

The equation is $V=-2300t+25000$ .

b.

To determine

Draw the graph of the linear equation representing the depreciation of the equipment,and use the value or trace feature to complete the table.Also,verify the answer algebraically by using the equation found in part (b).

Answer to Problem 93E

The table is :

t	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$
V	$25000$	$22700$	$20400$	$18100$	$15800$	$13500$	$11200$	$8900$	$6600$	$4300$	$2000$

Explanation of Solution

Given:

It is given in the question that a school district purchase a high − volume printer,copier , and scanner for $\$25000$ .After $10$ years ,the equipment will have to be replaced .its value at that time is expected to be $\$2000$ .

Concept Used:

In this , use the equation obtained in part (a) $V=-2300t+25000$ ,put the value of t Calculation:

Now,we full fill the table by taking the values from $0to10$ ,by putting the equation obtained in part (b) i.e, $V=-2300t+25000$

  $\begin{array}{l}Putt=0,V=-2300(0)+25000\Rightarrow 25000\\ Putt=1,V=-2300(1)+25000\Rightarrow 22700\\ Putt=2,V=-2300(2)+25000\Rightarrow 20400\\ Putt=3,V=-2300(3)+25000\Rightarrow 18100\\ Putt=4,V=-2300(4)+25000\Rightarrow 15800\\ Putt=5,V=-2300(5)+25000\Rightarrow 13500\\ Putt=6,V=-2300(6)+25000\Rightarrow 11200\\ Putt=7,V=-2300(7)+25000\Rightarrow 8900\\ Putt=8,V=-2300(8)+25000\Rightarrow 6600\\ Putt=9,V=-2300(9)+25000\Rightarrow 4300\\ Putt=10,V=-2300(10)+25000\Rightarrow 2000\end{array}$

The graph of the expression is :

  

Conclusion:

The table is :

t	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$
V	$25000$	$22700$	$20400$	$18100$	$15800$	$13500$	$11200$	$8900$	$6600$	$4300$	$2000$