Answer:
x=11, y=13
Step-by-step explanation:
3x-2y=7
3x-2(x+2)=7 . - x + = - -2*+2=-4
3x-2x-4=7
1x-4=7
x=7+4
x=11
y=x+2
y=11+2
y=13
<span>i think not to sure
6 sqrt 2</span>
Answer:
Hope this helps
Step-by-step explanation:
Answer:
![\frac{\sqrt{\pi}}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B%5Cpi%7D%7D%7B4%7D)
Step-by-step explanation:
You are going to integrate the following function:
(1)
furthermore, you know that:
![\int_0^{\infty}e^{-x^2}=\frac{\sqrt{\pi}}{2}](https://tex.z-dn.net/?f=%5Cint_0%5E%7B%5Cinfty%7De%5E%7B-x%5E2%7D%3D%5Cfrac%7B%5Csqrt%7B%5Cpi%7D%7D%7B2%7D)
lets call to this integral, the integral Io.
for a general form of I you have In:
![I_n=\int_0^{\infty}x^ne^{-ax^2}dx](https://tex.z-dn.net/?f=I_n%3D%5Cint_0%5E%7B%5Cinfty%7Dx%5Ene%5E%7B-ax%5E2%7Ddx)
furthermore you use the fact that:
![I_n=-\frac{\partial I_{n-2}}{\partial a}](https://tex.z-dn.net/?f=I_n%3D-%5Cfrac%7B%5Cpartial%20I_%7Bn-2%7D%7D%7B%5Cpartial%20a%7D)
by using this last expression in an iterative way you obtain the following:
(2)
with n=2s a even number
for s=1 you have n=2, that is, the function g(x). By using the equation (2) (with a = 1) you finally obtain:
![\int_0^{\infty}x^2e^{-x^2}dx=\frac{(2(1)-1)!}{2^{1+1}(1^1)}\sqrt{\pi}=\frac{\sqrt{\pi}}{4}](https://tex.z-dn.net/?f=%5Cint_0%5E%7B%5Cinfty%7Dx%5E2e%5E%7B-x%5E2%7Ddx%3D%5Cfrac%7B%282%281%29-1%29%21%7D%7B2%5E%7B1%2B1%7D%281%5E1%29%7D%5Csqrt%7B%5Cpi%7D%3D%5Cfrac%7B%5Csqrt%7B%5Cpi%7D%7D%7B4%7D)