Answer:
1. B. a=14, b=2
2. D. a=1, b=1.4
Step-by-step explanation:
The exponential growth function can be represented as
![y=a\cdot b^x,](https://tex.z-dn.net/?f=y%3Da%5Ccdot%20b%5Ex%2C)
where b is the growth factor.
1. When the function has equation
![g(x)=14\cdot 2^x,](https://tex.z-dn.net/?f=g%28x%29%3D14%5Ccdot%202%5Ex%2C)
then
![a=14,\\ \\b=2](https://tex.z-dn.net/?f=a%3D14%2C%5C%5C%20%5C%5Cb%3D2)
The initial amount is the value of the function at x=0:
![g(0)=14\cdot 2^0=14\cdot 1=14](https://tex.z-dn.net/?f=g%280%29%3D14%5Ccdot%202%5E0%3D14%5Ccdot%201%3D14)
The growth factor is b=2
2. When the function has equation
![f(t)=1.4^t,](https://tex.z-dn.net/?f=f%28t%29%3D1.4%5Et%2C)
then
![a=1,\\ \\b=1.4](https://tex.z-dn.net/?f=a%3D1%2C%5C%5C%20%5C%5Cb%3D1.4)
The initial amount is the value of the function at t=0:
![f(0)=1.4^0=1](https://tex.z-dn.net/?f=f%280%29%3D1.4%5E0%3D1)
The growth factor is b=1.4
Answer:
im not exactly sure what you mean by that, could you please rephrase and expand on the pregunta
Step-by-step explanation:
Explanation:
There are 12 months in a year, if we chose 25 random dates we'll have two dates per month
2x12=24
and one extra day that will fit in another month.
The principle used for this kind of problem is the<em> pigeonhole principle</em>, supposing we have a number k of pigeonholes and n pigeons to be placed in them, If the number of pigeons is bigger than the pigeonholes, there will be at least one pigeonhole with more than one pigeon.
This happens with our problem, 25 days is a larger number than 12 months two times (24), this means we'll choose at least one month three times.
I hope you find this information useful! good luck!
Answer:
Line segment: AD (without arrow)
Ray: BA (with one sided arrow)
Rays with common endpoint: AC and AD (with one sided arrow)
Intersection: AE (double sided arrow)
Collinear points: B, A, C
PLEASE MARK AS BRAINLIEST
Answer:
(fg)(x) = (4x^2 + x^4)(x^2 + 4)^(1/2)
Step-by-step explanation:
Mathematically;
(fg)(x) = f(x) * g(x)
so we have;
4x^2 +x^4 * √(x^2 + 4)
But √(x^2 + 4) = (x^2 + 4)^(1/2)
So we have;
(fg)(x) = (4x^2 + x^4)(x^2 + 4)^(1/2)