Question

1. A resident in a particular city pays a water bill of $4.75 per month plus $1.56 per 1000 gallons of water used (up to 4000 gallons). If this city has 443,000 customers that use less than 4000 gallons a month, find its monthly revenue function R(x), where the total number of gallons x is measured in thousands.

2.

Amira's car is worth 14,360 now and expected to be worth 10,964 in 33 years. Assuming linear depreciation, find a linear depreciation function for this car and estimate its value in 55 years. What is the linear depreciation function for this car?

Answer #1

1.

Given, city pays a water bill of $4.75 per month = fixed cost

variable cost = $1.56 per 1000 gallons

implies, monthlyrevenue R(x) = $4.75+ $1.56 x, where x is measured in thousands.

2.

Amira's initial car worth = 14,360

after 33 years, final car worth= 10,964

difference in car worth in 33 years is, 14,360- 10, 964 =3396

linear depreciation per year is = 3396/33 = 102.91

Linear depreciation for x years is = 102.91x

So, the function for the car worth after x years is = 14,360 - 102.91x

for x= 55 years, the linear depreciation function is = 14,360 - 102.91(55)

Depreciation after 55 years is = 5660. Total car worth is 14360-5660 = 8700.

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