A $1,600,000 building is depreciated linearly over 40 years. The scrap value will be $0.
1. Find the annual rate of depreciation.
a. -$1,600,000
b. -$40,000
c. $40,000
d. $1,600,000
e. $1,560,000
2. Find an expression for the building value in the t th year of use (0 < t < 40).

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- anonymous

- chestercat

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- anonymous

\[1,6000,000\div 40=?\]

- anonymous

these questions all have a certain sameness to them
is this math by boredom ?

- anonymous

I solved that and got 40,000. Then I used V= c-r which is 1600000 - 40000t but all of my possible answers start with 40,000 or -40,000 so I don't understand ....

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## More answers

- anonymous

oh i guess you have to put a minus sign in front

- anonymous

is one of them
\[y=-40,000 t+1,600,000\]?

- anonymous

Not boredom I am studying linear models and then once I get this down I move onto solving linear programming problems. I am just having a hard time getting this down ...

- anonymous

oh ok

- anonymous

yes that is one of my answers but I don't understand why if its v= c - r why this one is switched to v= r-c

- anonymous

or even if its v(t) = m(t) + b. that makes sense but in the examples my professor gives its all c - r(t)

- jim_thompson5910

#1
change in value = (new value) - (old value)
change in value = (0) - (1,600,000)
change in value = -1,600,000
change in years = (new year) - (old year)
change in years = (40) - (0)
change in years = 40
annual rate of depreciation = (change in value)/(change in years)
annual rate of depreciation = (-1,600,000)/(40)
annual rate of depreciation = -40,000
the item is losing $40,000 in value every year
-------------------------------------------------------
# 2
from #1, the slope is -40,000
the y-intercept is 1.6 million because this is the starting value
m = -40,000
b = 1.6 million
so the y=mx+b equation would be y = -40,000x + 1,600,000
x = time in years (0 < x < 40)
y = value of the item

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