Answer:
V(t) = 25000 * (0.815)^t
The depreciation from year 3 to year 4 was $2503.71
Step-by-step explanation:
We can model V(t) as an exponencial function:
V(t) = Vo * (1+r)^t
Where Vo is the inicial value of the car, r is the depreciation rate and t is the amount of years.
We have that Vo = 25000, r = -18.5% = -0.185, so:
V(t) = 25000 * (1-0.185)^t
V(t) = 25000 * (0.815)^t
In year 3, we have:
V(3) = 25000 * (0.815)^3 = 13533.58
In year 4, we have:
V(4) = 25000 * (0.815)^4 = 11029.87
The depreciation from year 3 to year 4 was:
V(3) - V(4) = 13533.58 - 11029.87 = $2503.71
Answer:
4 in 2,417
4 in 2,147
Step-by-step explanation:
the place value of 4 in 2,417 is HUNDREDS
while the place value of 4 in 2,147 is TENS
Therefore the place value of 4 in 2,417 is ten times the value of 4 in 2,147
In a parallelogram diagonals bisect each other,
=>AT=TX=>4y-2=14=>4y=14+2=>4y =16=>y=16/4=4
And ZT=TY=>2x+12=6x-12
=>12+12=6x-2x
=>24=4x
=>24/4=X
=>6=X
Hope this helps u...!!!
Answer:
I= -20p^2 + 840p
Step-by-step explanation:
When the ticket price is $2 there are 800 passengers daily, but every $0.1 increase in ticket price the number of passengers will be decreased by 2.
You can put information into these equations of:
passenger- = (800-2x)
ticket price= p = $2 + 0.1x
Income is calculated by multiplying the number of the passenger with the ticket price. The answer will be expressed in terms of the ticket price, so we need to remove x from the passenger equation.
p= $2 +0.1x
p-$2 = 0.1x
x= 10p- $20
If p= ticket price, the function for the number of passengers it will be:
passenger = (800-2x)
passenger = 800- 2(10p- $20)
passenger =800- 20p+40
passenger =840- 20p
The function of I will be:
I= passenger x ticket price
I= 840- 20p * p
I= -20p^2 + 840p
