When originally purchased, a vehicle costing $24,840 had an estimated useful life of 8 years and an estimated salvage value of $2,600. after 4 years of straight-line depreciation, the asset's total estimated useful life was revised from 8 years to 6 years and there was no change in the estimated salvage value. the depreciation expense in year 5 equals: ?

Originally, the depreciation is the original price of $24,840 minus the salvage value of $2600 divided by the number of years of useful life which was 8 years. This means that (24840-2600)/8 = $2780 was being written off each year, so, at the end of year 4, (4*2780) = $11, 120 was written off. This leaves the value of the car at (24840 - 11120) = $13720 at the start of year 5. Since we are moving from 8 years to 6 years useful life, there are only 2 more years over which we can depreciate. The current value at the end of year 4 minus the salvage value, then divided by the 2 years useful life remaining gives us a depreciation expense of (13720 - 2600)/2 = $5560 in year 5

  b.   the average rate of change of f(x) is less than the average rate of change of g(x).

step-by-step explanation:

since the length of the interval is the same in each case, we only need to compare the change in y-value to see which is greater.

f(-2) = -3f(1) = 6 . . . estimated from f(x) = (x+2)^2-3, which matches points on the graph (-1,-2) and (0, 1)

so the change in y-value for function f is (6 )) = 9.

the table tells us

g(-2) = 1g(1) = 31

so the change in y-value for function g is (31 -1) = 30.

the change in y-value of f(x) on the interval is less than the change in y-value of g(x) on the same interval, so the average rate of change of f(x) is less than the average rate of change of g(x).