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anonymous
 one year ago
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).
anonymous
 one year ago
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
 one year ago
Best ResponseYou've already chosen the best response.0\[1,6000,000\div 40=?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0these questions all have a certain sameness to them is this math by boredom ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I solved that and got 40,000. Then I used V= cr which is 1600000  40000t but all of my possible answers start with 40,000 or 40,000 so I don't understand ....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh i guess you have to put a minus sign in front

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is one of them \[y=40,000 t+1,600,000\]?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Not 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
 one year ago
Best ResponseYou've already chosen the best response.0yes that is one of my answers but I don't understand why if its v= c  r why this one is switched to v= rc

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or 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
 one year ago
Best ResponseYou've already chosen the best response.1#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 yintercept 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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