I believe the answer would be 70 but make sure just in case.
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:
All you do is change signs write 2x minus 14 equals negative 10-14 equals four then you write 2x over two equals four over two then u write x equals negative two
20% of 160= 32
160-32= 128
The new amount is 128 bananas.
52/8= what comes closest to 52, well 6x8=48, 7x8=56, we use 48, 52-48=4
the quotient is 6
and the remainder is 4