The time rate of change of a rabbit population P is proportional to the square root of PP. At time t=0 (months) the population numbers 100 rabbits and is increasing at the rate of 10 rabbits per month. Let P′=kP12 describe the growth of the rabbit population, where k is a positive constant to be found. Find the formulas for k and for the rabbit population P(t) after t months.

k = 1

P(t) = (t + 20)²

Step-by-step explanation:

P' = k√P = k P⁰•⁵

To solve for k,

P' = 10 rabbits/month

P = 100 rabbits

10 = k √100

10 = 10k

k = 1

To solve for P(t)

P' = dP/dt

(dP/dt) = kP⁰•⁵

dP/P⁰•⁵ = k dt

P⁻⁰•⁵ dP = k dt

∫ P⁻⁰•⁵ dP = ∫ k dt

2P⁰•⁵ = kt + c

At t = 0, P = 100

2(100)⁰•⁵ = 0 + c

2 × 10 = c

c = 20

2P⁰•⁵ = kt + c

2P⁰•⁵ = kt + 20

Recall, k = 1

2P⁰•⁵ = t + 20

P⁰•⁵ = (t + 20)

P(t) = (t + 20)²

i dont know if you cant find a way its best to actually add all of them start from 105 tho

step-by-step explanation:

answer: 1722

step-by-step explanation:

given: the mass of saturn=

the mass of mercury=

the number of times the mass of saturn is greater than the mass of mercury

hence, the mass of saturn is 1722 times greater than the mass of mercury.