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:
2
Step-by-step explanation:
Answer:
qrs
Step-by-step explanation:
Answer: 129
Step-by-step explanation: First find the area of the circle. Since we know the area of a circle is πr² (pi multiplied by the radius squared) we can find the area of the half circle. The radius is 5 (because radius is half of the diameter)
Area=3.14(5)²
=3.14(25)
=78.5
But since this is just a half circle, divide this by 2
That will get 39.25
Then add this to the area of the rectangle (L*W) which is 90
That will get about 129
You can use systems of equations for this one.
We are going to use 'q' as the number of quarters Rafael had,
and 'n' as the number of nickels Rafael had.
You can write the first equation like this:
3.50=0.05n+0.25q
This says that however many 5 cent nickels he had, and however many
25 cent quarters he had, all added up to value $3.50.
Our second equation is this:
q=n+8
This says that Rafael had 8 more nickels that he had quarters.
We can now use substitution to solve our system.
We can rewrite our first equation from:
3.50=0.05n+0.25q
to:
3.50=0.05n+0.25(n+8)
From here, simply solve using PEMDAS.
3.50=0.05n+0.25(n+8) --Distribute 0.25 to the n and the 8
3.50=0.05n+0.25n+2 --Subtract 2 from both sides
1.50=0.05n+0.25n --Combine like terms
1.50=0.30n --Divide both sides by 0.30
5=n --This is how many NICKELS Rafael has.
We now know how many nickels he has, but the question is asking us
how many quarters he has.
Simply substitute our now-known value of n into either of our previous
equations (3.50=0.05n+0.25q or q=n+8) and solve.
We now know that Rafael had 13 quarters.
To check, just substitute our known values for our variables and solve.
If both sides of our equations are equal, then you know that you have
yourself a correct answer.
Happy math-ing :)
