Let's think about the information in the problem. The problem tells us a few key points:
- The number of rabbits grows exponentially
- We start with 20 rabbits (
,
) - After 6 months (
), we have 100 rabbits (
)
Since we know we are going to be working with an exponential model, we can start with a base exponential model:

is the principal, or starting amount
is the growth/decay rate (in this case, growth)
is the number of months
is the number of rabbits
Based on the information in the problem, we can create two equations:


The first equation tells us that
, or that we start with 20 rabbits. Thus, we can change the second equation to:


Now, we don't know
, but we want to, so let's solve for it.

![r = \sqrt[6]{5}](https://tex.z-dn.net/?f=r%20%3D%20%5Csqrt%5B6%5D%7B5%7D)
Now, the problem is asking us how many rabbits we are going to have after one year (
), so let's find that:
![a = 20 \cdot (\sqrt[6]{5})^{12}](https://tex.z-dn.net/?f=a%20%3D%2020%20%5Ccdot%20%28%5Csqrt%5B6%5D%7B5%7D%29%5E%7B12%7D)



After one year, we will have 500 rabbits.
Answer:
A: y = (x − 4)^2 − 4
Step-by-step explanation:
vertex=(4.-4)
A: y = (x − 4)^2 − 4
y=x^2-8x+16-4
y=x^2-8x+12 (a=1,b=-8,c=12)
the y intercept is (0,12)
vertex ( h, k)
h=-b/2a ⇒ h=-(-8)/2=4
plug the value of h in the equation y=x^2-8x+12
k=4²-8(4)+12
k=16-32+12
k=-4
v(4,-4)
Answer:
Which of the following sampling techniques is the most likely to produce a random sample, representative sample of all students at a high school?
A. Choosing every 10th name on the student roster
B. Choosing every 10th student arriving in a 9th grade homeroom
<u>C. Choosing the first 100 students who arrive at school
</u>
D. Asking students to call a phone number to answer survey questions
Step-by-step explanation:
Answer:
i wish i could help-
Step-by-step explanation:
For point (1, -2): -2 = -1 - 1 = -2. Therefore, point (1, -2) lies on the graph of the equation.
The graph of the equation is the set of points that are solutions to the equation.
A coordinate pair on the graph of the equation is a solution to the equation.
For the point (0, -1): -1 = -(0) - 1 = 0 - 1 = -1.
Therefore, The point ( 0, −1 ) lies on the graph of the equation.