Answer:
greater then
Step-by-step explanation:
For
![f](https://tex.z-dn.net/?f=f)
to be continuous at
![x=5](https://tex.z-dn.net/?f=x%3D5)
, we need to have
![\displaystyle\lim_{x\to5}f(x)=f(5)](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bx%5Cto5%7Df%28x%29%3Df%285%29)
Note that
![x\to5](https://tex.z-dn.net/?f=x%5Cto5)
means that
![x\neq5](https://tex.z-dn.net/?f=x%5Cneq5)
, but that
![x](https://tex.z-dn.net/?f=x)
is *approaching* 5. We're told that for
![x\neq5](https://tex.z-dn.net/?f=x%5Cneq5)
, we have
![f(x)=\dfrac{x^2-5x}{x^2-25}](https://tex.z-dn.net/?f=f%28x%29%3D%5Cdfrac%7Bx%5E2-5x%7D%7Bx%5E2-25%7D)
We can write
![\dfrac{x^2-5x}{x^2-25}=\dfrac{x(x-5)}{(x+5)(x-5)}=\dfrac x{x+5}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%5E2-5x%7D%7Bx%5E2-25%7D%3D%5Cdfrac%7Bx%28x-5%29%7D%7B%28x%2B5%29%28x-5%29%7D%3D%5Cdfrac%20x%7Bx%2B5%7D)
and the limit would be
![\displaystyle\lim_{x\to5}\frac x{x+5}=\dfrac5{5+5}=\dfrac5{10}=\dfrac12\neq1=f(5)](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bx%5Cto5%7D%5Cfrac%20x%7Bx%2B5%7D%3D%5Cdfrac5%7B5%2B5%7D%3D%5Cdfrac5%7B10%7D%3D%5Cdfrac12%5Cneq1%3Df%285%29)
and so
![f](https://tex.z-dn.net/?f=f)
is discontinuous.
Answer:
![\alpha=3](https://tex.z-dn.net/?f=%5Calpha%3D3)
Step-by-step explanation:
<u>Equation of a Circle</u>
A circle of radius r and centered on the point (h,k) can be expressed by the equation
![(x-h)^2+(y-k)^2=r^2](https://tex.z-dn.net/?f=%28x-h%29%5E2%2B%28y-k%29%5E2%3Dr%5E2)
We are given the equation of a circle as
![3x^2+3y^2-6\alpha x+12y-3\alpha=0](https://tex.z-dn.net/?f=3x%5E2%2B3y%5E2-6%5Calpha%20x%2B12y-3%5Calpha%3D0)
Note we have corrected it by adding the square to the y. Simplify by 3
![x^2+y^2-2\alpha x+4y-\alpha=0](https://tex.z-dn.net/?f=x%5E2%2By%5E2-2%5Calpha%20x%2B4y-%5Calpha%3D0)
Complete squares and rearrange:
![x^2-2\alpha x+y^2+4y=\alpha](https://tex.z-dn.net/?f=x%5E2-2%5Calpha%20x%2By%5E2%2B4y%3D%5Calpha)
![x^2-2\alpha x+\alpha^2+y^2+4y+4=\alpha+\alpha^2+4](https://tex.z-dn.net/?f=x%5E2-2%5Calpha%20x%2B%5Calpha%5E2%2By%5E2%2B4y%2B4%3D%5Calpha%2B%5Calpha%5E2%2B4)
![(x-\alpha)^2+(y+2)^2=r^2](https://tex.z-dn.net/?f=%28x-%5Calpha%29%5E2%2B%28y%2B2%29%5E2%3Dr%5E2)
We can see that, if r=4, then
![\alpha+\alpha^2+4=16](https://tex.z-dn.net/?f=%5Calpha%2B%5Calpha%5E2%2B4%3D16)
Or, equivalently
![\alpha^2+\alpha-12=0](https://tex.z-dn.net/?f=%5Calpha%5E2%2B%5Calpha-12%3D0)
There are two solutions for
:
![\alpha=-4,\ \alpha=3](https://tex.z-dn.net/?f=%5Calpha%3D-4%2C%5C%20%5Calpha%3D3)
Keeping the positive solution, as required:
![\boxed{\alpha=3}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Calpha%3D3%7D)
Answer:
$.999 for 1 pound
Step-by-step explanation:
If you divided both sides by 10. So 10/10= 1. 9.99/10= 0.999
Answer:
C
Step-by-step explanation: