Answer:
See below for answers and explanations
Step-by-step explanation:
4! = 4*3*2*1 = 24
0! = 1
2! + 3! = 2*1 + 3*2*1 = 2 + 6 = 8
3! * 4! = 3*2*1 * 4*3*2*1 = 6 * 24 = 144
10! / 7! = 10*9*8*7*6*5*4*3*2*1 / 7*6*5*4*3*2*1 = 10*9*8 / 1 = 720
Answer:
y=2x
Step-by-step explanation:
slope = rise/run = 2/1 = 2
<span>(3.5, 3) is the circumcenter of triangle ABC.
The circumcenter of a triangle is the intersection of the perpendicular bisectors of each side. All three of these perpendicular bisectors will intersect at the same point. So you have a nice self check to make sure your math is correct. Now let's calculate the equation for these bisectors.
Line segment AB:
Slope
(4-2)/(1-1) = 2/0 = infinity.
This line segment is perfectly vertical. So the bisector will be perfectly horizontal, and will pass through ((1+1)/2, (4+2)/2) = (2/2, 6/2) = (1,3).
So the equation for this perpendicular bisector is y = 3.
Line segment BC
(2-2)/(6-1) = 0/5 = 0
This line segment is perfectly horizontal. So the bisector will be perfectly vertical, and will pass through ((1+6)/2,(2+2)/2) = (7/2, 4/2) = (3.5, 2)
So the equation for this perpendicular bisector is x=3.5
So those two bisectors will intersect at point (3.5,3) which is the circumcenter of triangle ABC.
Now let's do a cross check to make sure that's correct.
Line segment AC
Slope = (4-2)/(1-6) = 2/-5 = -2/5
The perpendicular will have slope 5/2 = 2.5. So the equation is of the form
y = 2.5*x + b
And will pass through the point
((1+6)/2, (4+2)/2) = (7/2, 6/2) = (3.5, 3)
Plug in those coordinates and calculate b.
y = 2.5x + b
3 = 2.5*3.5 + b
3 = 8.75 + b
-5.75 = b
So the equation for the 3rd bisector is
y = 2.5x - 5.75
Now let's check if the intersection with this line against the other 2 works.
Determining intersection between bisector of AC and AB
y = 2.5x - 5.75
y = 3
3 = 2.5x - 5.75
8.75 = 2.5x
3.5 = x
And we get the correct value. Now to check AC and BC
y = 2.5x - 5.75
x = 3.5
y = 2.5*3.5 - 5.75
y = 8.75 - 5.75
y = 3
And we still get the correct intersection.</span>
and 
. The squared distance between any point in this region 
and the point (1, 1) is
![=\dfrac{\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}[r^2-2r(\cos\theta-\sin\theta)+2]r\,\mathrm dr\,\mathrm d\theta}{\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}r\,\mathrm dr\,\mathrm d\theta}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B%5Cdisplaystyle%5Cint_%7B%5Ctheta%3D0%7D%5E%7B%5Ctheta%3D2%5Cpi%7D%5Cint_%7Br%3D0%7D%5E%7Br%3D1%7D%5Br%5E2-2r%28%5Ccos%5Ctheta-%5Csin%5Ctheta%29%2B2%5Dr%5C%2C%5Cmathrm%20dr%5C%2C%5Cmathrm%20d%5Ctheta%7D%7B%5Cdisplaystyle%5Cint_%7B%5Ctheta%3D0%7D%5E%7B%5Ctheta%3D2%5Cpi%7D%5Cint_%7Br%3D0%7D%5E%7Br%3D1%7Dr%5C%2C%5Cmathrm%20dr%5C%2C%5Cmathrm%20d%5Ctheta%7D)
