a new building that costs $1,000,000 has a useful life of 25 years and a scrap value of $600,000. using straight-line depreciation, find the equation for the value v in terms of t, where t is in years. (make sure you use t and not x in your answer.)
v(t) =
find the value after 1 year, after 2 years, and after 20 years.
value after 1 year $
value after 2 years $
value after 20 years $

You have two points, the starting point (0, 1, 000,000) and the end point (25, 600,000)

Find the slope, which id the difference in Y over the difference in x:

m = (600,000 - 1,000,000) / (25-0)

= -400,000 / 25

= -16,000

Now you have V -1,000,000 = -16,000(x-0)

Rewrite in terms of V(t):

V(t) = -16,000t + 1,000,000

Now replace t with the years and solve:

1 year = V(1) = -16000(1) + 1,000,000 = -16,000 + 1,000,000 = $984,000

2 years = V(2) = -16,000(2) + 1,000,000 = -32,000 + 1,000,000 = $968,000

20 years = V(20) = -16,000(20) + 1,000,000 = -320,000 + 1,000,000 = $680,000

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2