<span>Part
A:
a) What do the x-intercepts and maximum value of the graph represent?
The x-intercepts are the distances at which the ball is on the ground.
First, at x = 0, that is when the ball is kicked; second, at x = 30, when the ball falls (return) to the ground.
b) What are the intervals where the function is increasing and decreasing,
and what do they represent about the distance and height? (6 points)
The function is increasing in the interval (0, 15) and is decreasing in the interval (15,30)
The increasing interval (0,15) is the horizontal distance from the point the the ball was kicked until it reached its highest altitude, this is where the ball was going upward.
The decreasing interval (15,30) is the horizontal distance from the point where the ball reached its highest altitude until it landed on the ground, this is where the ball was falling down.
Part B: What is an approximate average rate of change of the graph from x
= 22 to x = 26, and what does this rate represent
On the graph you can read that at x = 22, f(x) ≈ 12, and at x = 26 f(x) ≈ 7.
So, an approximate rate of change from x = 22 to x = 26 is given by the equation below:
change on f(x) 7 - 12
average rate of change = --------------------- = ----------- = -5/4
change of x 26 - 22
That rate represents that the ball fell about 5 ft per 4 ft in that interval.
</span>
Answer:
C
Step-by-step explanation:
Answer:
b
Step-by-step explanation:
trust me my guy ik this is ez
Answer:
We want to find:
![\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n}](https://tex.z-dn.net/?f=%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B%5Csqrt%5Bn%5D%7Bn%21%7D%20%7D%7Bn%7D)
Here we can use Stirling's approximation, which says that for large values of n, we get:
Because here we are taking the limit when n tends to infinity, we can use this approximation.
Then we get.
![\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n} = \lim_{n \to \infty} \frac{\sqrt[n]{\sqrt{2*\pi*n} *(\frac{n}{e} )^n} }{n} = \lim_{n \to \infty} \frac{n}{e*n} *\sqrt[2*n]{2*\pi*n}](https://tex.z-dn.net/?f=%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B%5Csqrt%5Bn%5D%7Bn%21%7D%20%7D%7Bn%7D%20%3D%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B%5Csqrt%5Bn%5D%7B%5Csqrt%7B2%2A%5Cpi%2An%7D%20%2A%28%5Cfrac%7Bn%7D%7Be%7D%20%29%5En%7D%20%7D%7Bn%7D%20%3D%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7Bn%7D%7Be%2An%7D%20%2A%5Csqrt%5B2%2An%5D%7B2%2A%5Cpi%2An%7D)
Now we can just simplify this, so we get:
![\lim_{n \to \infty} \frac{1}{e} *\sqrt[2*n]{2*\pi*n} \\](https://tex.z-dn.net/?f=%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B1%7D%7Be%7D%20%2A%5Csqrt%5B2%2An%5D%7B2%2A%5Cpi%2An%7D%20%5C%5C)
And we can rewrite it as:
The important part here is the exponent, as n tends to infinite, the exponent tends to zero.
Thus:
300=x+1.9x
I got this by dividing the problem into two parts x[the speed of the first half] plus 1.9x [the speed of the second half increased by 1.9] both of those together should give you 300.
I went further and solved this...If you placed 103.5 into this problem for x you should get 300.15!
