H(t) = -16t² + 16 t + 480
Factor out -16
-16(t² - t - 30)
t² - t - 30
(t + 5)(t - 6) ⇒ t(t-6) + 5(t-6) ⇒ t² -6t + 5t - 30 ⇒ t² - t - 30
t + 5 = 0 t - 6 = 0
t = -5 t = 6
The value of t is either -5 or 6 however time can't be measured in negative form; then, it took Rose 6 seconds to hit the water.
<u>The answer is 808/99</u>
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:
Answer:
The hens are the independent variable
Step-by-step explanation:
An independent variable is something that isn't dependent on anything else. Whose laying the eggs? The hens right? So the number of eggs laid is dependent on the number of hens laying them. Basically the hens are the independent variable because the number of eggs is relying on the number of hens laying them.
