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:

By the looks of this the answer is E
Derek weighs 155
155+220
This is a geometric sequence of the form:
a(n)=(x+3)(-2x)^(n-1) because each term is -2x times the previous term...so
a(8)=(x+3)(-128x^7)
a(8)=-128x^8-384x^7