Answer:
1.) 12000
2.) 6750
3.) 2.41
Step-by-step explanation:
Given the Equation :
f(t)=12,000(3/4)^t. ; where, t = time ; f(t) = worth of car at time, t
When, car was purchased, t = 0
t = 0
f(0) = 12000(3/4)^0
= 12000 * 1
= 12000
2.)
f(2) ; this mean the worth of the car after 2 years :
f(2)=12,000(3/4)^t.
12000(3/4)^2
12000 * 0.5625
= 6750
When car will be worth 6000
f(t) = 6000
f(t)=12,000(3/4)^t.
6000 = 12,000(3/4)^t
6000 / 12000 = (3/4)^t
1/2 = (3/4)^t
Take the log of both sides :
Log(0.5) = log(0.75)^t
log(0.5) = tlog(0.75)
- 0.301029 = - 0.124938t
t = - 0.301029 / - 0.124938
t = 2.4094
t = 2.41 years
90-80=10. 9-6=1 and 0-0=0 so it equals 10. Hopefully this helped!! =“)
Answer:
Step-by-step explanation:
One of the numbers multiplied to find the product called multiplicand.
The number being divided called Dividend
The number that divides another number called Divisor
ince the problem is only asking for 4 years, we can just calculated it out year by year. Recall the formula for compounding interest: A = P(1+r)n, where A is the total amount, P is the principle (amount you start with), r is the interest rate per period of time, and n is the number of periods (in this case, r is annual interest rate, so n is number of years). At the beginning (Year 0), Lou starts off with 10000: A = 10000 At the end of Year 1, Lou earned interest on that amount, plus he has deposited another 5000: A = 10000(1.08) + 5000 End of Year 2, Lou's interest from the year 0 amount has compounded, he has started earning interest on the amount deposited last year, and he deposits another 5000: A = 10000(1.08)2 + 5000(1.08) + 5000 End of Year 3, same idea. Lou has earned compounding interest on all existing deposits, and deposits another 5000: A = 10000(1.08)3 + 5000(1.08)2 + 5000(1.08) + 5000 End of Year 4, same idea: A = 10000(1.08)4 + 5000(1.08)3 + 5000(1.08)2 + 5000(1.08) + 5000 = 36135.45
