depreciates by 20% of its value in the value in the preceding year. Let Vn be the value of the bulldozer in the nth year. (let n=1 be the year the bulldozer is purchased.)What is the formula for Vn? and in what year will the value of the bulldozer be less than $100,000?pls show your solution so that i can understand it easily...pls...thank you
If it depreciates 20% each year, from year to year, you're going to multiply the previous year's price by (1 - .2) = .8. So, after n years, it will cost V(n) = 160,000*(.8)^n 100,000 = 160,000(.8)^n .625 = .8^n. log both sides. log (.625) = n*log(.8). n = log(.625)/log(.8). n = 2.106. So, a little after two years, the bulldozer will already be worth less than $100,000. Hope this helps!
V(n+1) = 0.8 V(n) year V(n+1) would be the current year and V(n) would be the last year if you work through it we find that in the fourth year the value is less than 100000
HOW MANY YEARS CAN YOU DEPRECIATE A DOZER
Vn=.8n + 160,000 In just over 2 years the bulldozer will be worth 100,000. So if you are going by years the answer would be. In three years the value of the bulldozer will be less than 100000. Another way to look at the problem would be so say that the dozer retains 80percent of its value from the previous year. 100-20=80.
