The 15th term of the arithmetic sequence 
Option B is correct
The nth term of an arithmetic sequence is given as:
The first value, a = 22
Since 
The common difference, d = 3
The 15th term of the arithmetic sequence = 64
Learn more here: brainly.com/question/24072079
Answer: For y it's y=4x−12 and for x it's x=12+y/4
Step-by-step explanation:
SOLVING FOR X
8x-2y=24
2(4x−y)=24
4x−y=24/2
4x−y=12
4x=12+y
x=12+y/4
SOLVING FOR Y
8x-2y=24
2(4x−y)=24
4x−y=24/2
4x−y=12
−y=12−4x
y=−12+4x
y=4x−12
Hope this helps, have a BLESSED AND WONDERFUL DAY! As well as a great Black History Month and Valentine's day! :-)
- Cutiepatutie ☺❀❤
The function that best defines this arithmetic sequence is:
<h3>What is an arithmetic sequence?</h3>
- In an <em>arithmetic sequence,</em> the <u>difference between consecutive terms is always the same</u>, called common difference d.
The recursive function that defines the sequence is:
- In which
is the first term.
In this problem, the sequence is: {9, 5, 1, -3}
- Each term is the previous term subtracted by 4, hence
.
- The first term is 9, hence
.
Hence, the function is:
You can learn more about arithmetic sequences at brainly.com/question/6561461
Step-by-step explanation:
(a) dP/dt = kP (1 − P/L)
L is the carrying capacity (20 billion = 20,000 million).
Since P₀ is small compared to L, we can approximate the initial rate as:
(dP/dt)₀ ≈ kP₀
Using the maximum birth rate and death rate, the initial growth rate is 40 mil/year − 20 mil/year = 20 mil/year.
20 = k (6,100)
k = 1/305
dP/dt = 1/305 P (1 − (P/20,000))
(b) P(t) = 20,000 / (1 + Ce^(-t/305))
6,100 = 20,000 / (1 + C)
C = 2.279
P(t) = 20,000 / (1 + 2.279e^(-t/305))
P(10) = 20,000 / (1 + 2.279e^(-10/305))
P(10) = 6240 million
P(10) = 6.24 billion
This is less than the actual population of 6.9 billion.
(c) P(100) = 20,000 / (1 + 2.279e^(-100/305))
P(100) = 7570 million = 7.57 billion
P(600) = 20,000 / (1 + 2.279e^(-600/305))
P(600) = 15170 million = 15.17 billion
