5 is d and idk question 5 sorry
F=ir^t
139=134r^10
139/134=r^10
r=(139/134)^(1/10) then:
f=134(139/134)^(t/10) so in 2014, t=24 so
f=134(139/134)^(2.4)
f≈146 million (to nearest million)
Some will say that you have to use the exponential function, but it really gives you the same answer...even for continuous compounding :)...
A=Pe^(kt)
139=134e^(10k)
139/134=e^(10k)
ln(139/134)=10k
k=ln(139/134)/10 so
A=134e^(t*ln(139/134)/10) when t=24
A=134e^(2.4*ln(139/134))
A≈146 million (to nearest million)
The only real reason or advantage to using A=Pe^(kt) is when you start getting into differential equations...
The first two is worth 2,000 and the second is worth 200.
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
