Answer:
Answer:.1.77
Step-by-step explanation:
e^x = 5.9
Take the natural logarithm of both sides
x = ln (59/10)
x = 1.77
(found this)
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:
2.8 units
Step-by-step explanation:
Think of this distance as the hypotenuse of a right triangle that has a vertical leg and a horizontal one as well.
Going from P to Q, the change in x is 2 and the change in y is also 2.
Thus, by the Pythagorean Theorem, this desired distance is:
d = √(2^2 + 2^2) = 2√2 units, or 2.8 units