Answer:
a) P' = P
where t is step of 6 months
b) 7.7 years
c)1064.67 rabbits/year
Step-by-step explanation:
The differential equation describing the population growth is
Where t is the range of 6 months, or half of a year.
P(t) would have the form of
where 
is the initial population
After 6 month (t = 1), the population is doubled to 48
Therefore
where t is step of 6 months
b. We can solve for t to get how long it takes to get to a population of 1,000,000:
So it would take 15.35 * 0.5 = 7.7 years to reach 1000000
c. 
We need to resolve for k if t is in the range of 1 year. In half of a year (t = 0.5), the population is 48
Therefore, 
At the mid of the 3rd year, where t = 2.5, we can calculate P'
rabbits/year
Answer:
so you graph at (0,2), and go up one, over two for every other point you graph.
Step-by-step explanation:
rise over run.
basically, the m is the slope, and b is the y intercept (where it intersects the y axis).
so when looking at how to graph the line based on 1/2 or any other number, I use rise/run, which is basically telling us tha we need to go up one and over 2 (1/2)
if that makes any sense, let me know if you have any further questions. you got this, youre doing your best!!
slope is the start of your endless pain.
Answer:
20 + 4y
Step-by-step explanation:
The 4 multiplies the 5 and the y, i.e., it distributes over the terms of the expression in ( ).
-15x -8y=-27 Thas the answer
Answer:
You did not post the options, but i will try to answer this in a general way.
Because we have two solutions, i know that we are talking about quadratic equations, of the form of:
0 = a*x^2 + b*x + c.
There are two easy ways to see if the solutions of this equation are real or not.
1) look at the graph, if the graph touches the x-axis, then we have real solutions (if the graph does not touch the x-axis, we have complex solutions).
2) look at the determinant.
The determinant of a quadratic equation is:
D = b^2 - 4*a*c.
if D > 0, we have two real solutions.
if D = 0, we have one real solution (or two real solutions that are equal)
if D < 0, we have two complex solutions.
