<span> Circumference of
the tire:</span>
<span> PI x diameter =
3.14 x 26 = 81.6814 Round to 81.68</span>
Speed of the car in inches per minute:
46*5280*12/60 = 48576 inches per minute
48576/81.68 = 594.711
<span> Round to 594.71
revolutions per minute</span>
No, because 2/5 in decimal form is 0.4 and 6/10 in decimal form is 0.6, so it's obviously not equal.
Hope I helped! <3
Answer:
21
Step-by-step explanation:
prim factors: 2, 3, 5, and 11
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:
![n! = \sqrt{2*\pi*n} *(\frac{n}{e} )^n](https://tex.z-dn.net/?f=n%21%20%3D%20%5Csqrt%7B2%2A%5Cpi%2An%7D%20%2A%28%5Cfrac%7Bn%7D%7Be%7D%20%29%5En)
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:
![\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n}](https://tex.z-dn.net/?f=%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B1%7D%7Be%7D%20%2A%282%2A%5Cpi%2An%29%5E%7B1%2F2n%7D)
The important part here is the exponent, as n tends to infinite, the exponent tends to zero.
Thus:
![\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n} = \frac{1}{e}*1 = \frac{1}{e}](https://tex.z-dn.net/?f=%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20%5Cfrac%7B1%7D%7Be%7D%20%2A%282%2A%5Cpi%2An%29%5E%7B1%2F2n%7D%20%3D%20%5Cfrac%7B1%7D%7Be%7D%2A1%20%3D%20%5Cfrac%7B1%7D%7Be%7D)
It comes from integrating by parts twice. Let
![I = \displaystyle \int e^n \sin(\pi n) \, dn](https://tex.z-dn.net/?f=I%20%3D%20%5Cdisplaystyle%20%5Cint%20e%5En%20%5Csin%28%5Cpi%20n%29%20%5C%2C%20dn)
Recall the IBP formula,
![\displaystyle \int u \, dv = uv - \int v \, du](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20u%20%5C%2C%20dv%20%3D%20uv%20-%20%5Cint%20v%20%5C%2C%20du)
Let
![u = \sin(\pi n) \implies du = \pi \cos(\pi n) \, dn](https://tex.z-dn.net/?f=u%20%3D%20%5Csin%28%5Cpi%20n%29%20%5Cimplies%20du%20%3D%20%5Cpi%20%5Ccos%28%5Cpi%20n%29%20%5C%2C%20dn)
![dv = e^n \, dn \implies v = e^n](https://tex.z-dn.net/?f=dv%20%3D%20e%5En%20%5C%2C%20dn%20%5Cimplies%20v%20%3D%20e%5En)
Then
![\displaystyle I = e^n \sin(\pi n) - \pi \int e^n \cos(\pi n) \, dn](https://tex.z-dn.net/?f=%5Cdisplaystyle%20I%20%3D%20e%5En%20%5Csin%28%5Cpi%20n%29%20-%20%5Cpi%20%5Cint%20e%5En%20%5Ccos%28%5Cpi%20n%29%20%5C%2C%20dn)
Apply IBP once more, with
![u = \cos(\pi n) \implies du = -\pi \sin(\pi n) \, dn](https://tex.z-dn.net/?f=u%20%3D%20%5Ccos%28%5Cpi%20n%29%20%5Cimplies%20du%20%3D%20-%5Cpi%20%5Csin%28%5Cpi%20n%29%20%5C%2C%20dn)
![dv = e^n \, dn \implies v = e^n](https://tex.z-dn.net/?f=dv%20%3D%20e%5En%20%5C%2C%20dn%20%5Cimplies%20v%20%3D%20e%5En)
Notice that the ∫ v du term contains the original integral, so that
![\displaystyle I = e^n \sin(\pi n) - \pi \left(e^n \cos(\pi n) + \pi \int e^n \sin(\pi n) \, dn\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20I%20%3D%20e%5En%20%5Csin%28%5Cpi%20n%29%20-%20%5Cpi%20%5Cleft%28e%5En%20%5Ccos%28%5Cpi%20n%29%20%2B%20%5Cpi%20%5Cint%20e%5En%20%5Csin%28%5Cpi%20n%29%20%5C%2C%20dn%5Cright%29)
![\displaystyle I = \left(\sin(\pi n) - \pi \cos(\pi n)\right) e^n - \pi^2 I](https://tex.z-dn.net/?f=%5Cdisplaystyle%20I%20%3D%20%5Cleft%28%5Csin%28%5Cpi%20n%29%20-%20%5Cpi%20%5Ccos%28%5Cpi%20n%29%5Cright%29%20e%5En%20-%20%5Cpi%5E2%20I)
![\displaystyle (1 + \pi^2) I = \left(\sin(\pi n) - \pi \cos(\pi n)\right) e^n](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%281%20%2B%20%5Cpi%5E2%29%20I%20%3D%20%5Cleft%28%5Csin%28%5Cpi%20n%29%20-%20%5Cpi%20%5Ccos%28%5Cpi%20n%29%5Cright%29%20e%5En)
![\implies \displaystyle I = \frac{\left(\sin(\pi n) - \pi \cos(\pi n)\right) e^n}{1+\pi^2} + C](https://tex.z-dn.net/?f=%5Cimplies%20%5Cdisplaystyle%20I%20%3D%20%5Cfrac%7B%5Cleft%28%5Csin%28%5Cpi%20n%29%20-%20%5Cpi%20%5Ccos%28%5Cpi%20n%29%5Cright%29%20e%5En%7D%7B1%2B%5Cpi%5E2%7D%20%2B%20C)