SOLUTION: A company predicts that the value in dollars V and the time that a peice of equipment has been in use t are related by linear equation. If the equipment is valued at %1500 after 2
Question 437335: A company predicts that the value in dollars V and the time that a peice of equipment has been in use t are related by linear equation. If the equipment is valued at %1500 after 2 years and $300 after 10 years, find the linear model relating t and V.
Use the model to find the equipment model after 5 years.
Use the model to determine when the equipment will have a value of $100. Found 2 solutions by htmentor, ewatrrr:Answer by htmentor(1343) (Show Source):
You can put this solution on YOUR website! The linear function will have the general form
V(t) = at + b
where a and b are constants
Given: V(2) = 1500, V(10) = 300
So we have the following two equations:
1500 = 2a + b (1)
300 = 10a + b (2)
This gives us two equations in two unknowns, a and b.
By subtracting (2) from (1) we can eliminate b:
1200 = -8a -> a = -150
Now use one of the two equations above to solve for b:
Using (1), we get
1500 = -300 + b -> b = 1800
So, our equation is:
V(t) = -150t + 1800
So the value after 5 years is:
V(5) = -150(5) + 1800 = $1050
The time when the equipment will have a value of $100 is given by:
100 = -150t + 1800
Solve for t:
t = -1700/-150 = 11.33 years
The graph is shown below:
Hi
(2, 1500) |Using the point-slope formula,
(10, 300 ) m = 1200/-8 = -150 (depreciating $150 per year)
Using the standard slope-intercept form for an equation of a line y = mx + b
where m = -150 is the slope and b the y-intercept.
V(t) = -150t + $1800 ($1500+$300 = $1800 brand new: t = 0)
Use the model to find the equipment model after 5 years.
V(5) = -150*5 + $1800 = $1050
Use the model to determine when the equipment will have a value of $100.
100 = -150t + 1800
t = 1700/150 = 11 1/3 years
