SOLUTION: The depreciated value VV of a machine is a linear function of time t in years. A machine that is purchased for $100000 today will be worth approximately $67681 in 33 years. Write t

Algebra ->  Linear-equations -> SOLUTION: The depreciated value VV of a machine is a linear function of time t in years. A machine that is purchased for $100000 today will be worth approximately $67681 in 33 years. Write t      Log On


   

Question 1087074: The depreciated value VV of a machine is a linear function of time t in years. A machine that is purchased for $100000 today will be worth approximately $67681 in 33 years. Write the linear function V(t) that represents the value of the machine at a given time t.
My work---> 100,1000---->67,681 3yrs
67,681v+3=100,000 this is what I got but I'm not sure.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I think you meant to say "The depreciated value V" instead of "The depreciated value VV". I changed the VV to V.

The value today is V = 100000 dollars when the time is t = 0, which is the starting time value.
The value in 33 years, when t = 33, is V = 67681 dollars

So we have two ordered pairs (0,100000) and (33,67681). I'm going to treat t as x, and treat V as y. So each ordered pair goes from the form (t,V) to (x,y). Using the (x,y) form we can use the slope formula

First point = (x1,y1) = (0,100000)
Second point = (x2,y2) = (33,67681)
m+=+%28y%5B2%5D+-+y%5B1%5D%29%2F%28x%5B2%5D+-+x%5B1%5D%29
m+=+%2867681+-+100000%29%2F%2833+-+0%29
m+=+%28-32319%29%2F%2833%29
m+=+-10773%2F11
m+=+-979.3636 Use a calculator here. The decimal form is approximate (the '36' portion repeats forever)

So the slope is roughly m+=+-979.3636

The y intercept is b+=+100000 because this is the starting value (at t = 0)

So we go from y+=+mx%2Bb to y+=+-979.3636x%2B100000

The last thing to do is replace x with t and replace y with V(t) to get V%28t%29+=+-979.3636t%2B100000

Therefore the value function is V%28t%29+=+-979.3636t%2B100000

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Let's check the function. Plug t = 0 into the function to get
V%28t%29+=+-979.3636t%2B100000
V%280%29+=+-979.3636%2A0%2B100000
V%280%29+=+0%2B100000
V%280%29+=+100000
So that matches with the fact the initial value is $100,000

Now plug in t = 33
V%28t%29+=+-979.3636t%2B100000
V%2833%29+=+-979.3636%2A%2833%29%2B100000
V%2833%29+=+-32318.9988%2B100000
V%2833%29+=+67681.0012
V%2833%29+=+67681.00
Which matches with the value after 33 years. So the answer is confirmed

Side Note: if you use the fraction form for the slope then you won't run into rounding errors. The fact that we're rounding to 2 decimal places means that we don't have to worry about precision too much as long as the slope is expressed to 3 decimal places or more. You're probably wondering what the slope means? If so, then the slope is simply the rate of value decay or drop. In this case, the slope -979.3636 means the value V(t) is decreasing by $979.36 each year. The actual drop is a bit more than that but you can only round to the nearest hundredth for money problems like this.