Answer:
Suppose a population of rodents satisfies the differential equation dP 2 kP dt = . Initially there are P (0 2 ) = rodents, and their number is increasing at the rate of 1 dP dt = rodent per month when there are P = 10 rodents.
How long will it take for this population to grow to a hundred rodents? To a thousand rodents?
Step-by-step explanation:
Use the initial condition when dp/dt = 1, p = 10 to get k;
Seperate the differential equation and solve for the constant C.
You have 100 rodents when:
You have 1000 rodents when:
Answer:
225 km^2
Step-by-step explanation:
Area of a trapezoid: ((Base 1 + Base 2) x h) / 2
(11 + 19) x 15 / 2
= 225 km^2
We can figure this out using the explicit formula.
n represents the term we are looking for.
f(1) represents the first term in the sequence, which in this case, is 7.
d represents the common difference, which in this case, is +3.
f(n) = 7 + 3(n - 1)
f(n) = 7 + 3n - 3
f(n) = 4 + 3n
Now, we can input 214 for n and solve.
f(214) = 4 + 3(214)
f(214) = 4 + 642
f(214) = 646
The 214th term in this sequence is 646.
Answer:
-x+11
Step-by-step explanation:
am not sure if this is correct but
2x-3x=-x
-7+18=11
-x+11
