Answer:
The Answer Is (C.)
Step-by-step explanation:
Hope This Helps :)
Have A Great Day :)
The answer is c.(0,1.50) because $5.25 and $6.25 if u use a calculator u will see your answer
Answer:
Carbon-14 loses around 10% ( 0.1 in decimal form) of its mass, after one millennium.
Then if we start with a mass A of Carbon-14, after one millennium we will have a mass equal to:
A - A*0.1 = A*(0.9)
After another millennium, we will have a mass equal to:
A*(0.9) - A*(0.9)*0.1 = A*(0.9)^2
And so on, this is an exponential decay.
We already can see the pattern here, after x millenniums, the mass of carbon-14 will be:
M(x) = A*(0.9)^x
Now, in this problem we have 600 grams of carbon-14, then the equation for the mass will be:
y = M(x) = 600g*(0.9)^x
And the graph of this equation is shown below.
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