Question

2:35 AM Sun Aug 11 100% dsbearingantarioca Assignment 1 Complete the following assignment and submit your work to the dropbox. 1. When a new item is purchased, its value typically decreases over time. We call this depreciation. Vehicles, computers, and machinery are all examples of items that depreciate in value. The value of a sports car is given by the equation V = 35000(0.85)" + 1000, where V is the value in dollars, and n is the age of the car, in years. a. Complete a table of values showing the value of the car over the first eight years. b. Graph the data from part a. c. Use either your graph or the equation to answer the following questions. (Be sure to justify your answers.) i. How long will it take for the value of the car to reach $5000? ii. How much value does the car lose over the first ten years? ili. Does the value of the car ever reach $0? Explain. d. Do you think all cars depreciate at the same rate? What factors influence the rate of depreciation? 2. The population of Humpback whales was once thought to be as high as 125 000, although this number was drastically reduced due to whaling. New whaling laws have allowed the population to increase again. Since 1979, the population has experienced exponential growth that can be modelled by the equation P = 7500(1.031)", where P represents the population, and n represents the years since 1979. (i.e. for 1979, n = 0) a. Use the equation to estimate the population in the years 1979, 1985, and 1995. b. According to the mathematical model, in what year will the population reach 20,000 whales? C. Give reasons why this model may not accurately predict the future population.

Answer 1

os V= 35000/0,85)*71000 age (n) I value of the car (Le) 30.750 26 28705 22 494.375 19 270.2167 16529.6859 14200.233 12220.198 10537. 168 20750 1060

C. 5000 - 35000055512, 4ooo (0,05) n=13-345yr after 20 ye Vo = 36000 Vo = 784060 36000-7890.60 2810944 never nonimam velas gaas 4 $ 1300 because when we hereas value on then (0.85) ► o

P= 75001-038) al in 1979 P = 7500 In 1985 ; n = 6 P = 90071686 In p= 1995 ; n=16 P= 12 223.6219 6) 20000 = 7500(1:03) 200+ (1031) & = (1-031) n = 32.12 (approx)