![x^{2} +10x+30](https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%2B10x%2B30)
I always start by moving the last term to the other side
![x^{2} +10x=-30](https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%2B10x%3D-30)
To complete the square you take term 'b' , divide it by 2 then square it
![( \frac{10}{2} )^2](https://tex.z-dn.net/?f=%28%20%5Cfrac%7B10%7D%7B2%7D%20%29%5E2)
= 25
We have to add 25 to what we have and what you do to one side you have to do to the other
![x^{2} +10x+25=-30+25](https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%2B10x%2B25%3D-30%2B25)
![x^{2} +10x+25=-5](https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%2B10x%2B25%3D-5)
Now we factor what is on the left side of the equation
![(x+5)^2=-5](https://tex.z-dn.net/?f=%28x%2B5%29%5E2%3D-5)
The last step is to move our right side of the equation back over
![(x+5)^2+5](https://tex.z-dn.net/?f=%28x%2B5%29%5E2%2B5)
- which is your answer :)))
Direct variation means that the line goes through the origin of the graph, so you know that the line must go through point (0,0). Using that along with the given point, you can find the slope of the line.
m =
![\frac{y_2 - y_1}{x_2 - x_1}](https://tex.z-dn.net/?f=%20%5Cfrac%7By_2%20-%20y_1%7D%7Bx_2%20-%20x_1%7D%20)
m =
![\frac{18 - 0}{3 - 0}](https://tex.z-dn.net/?f=%20%5Cfrac%7B18%20-%200%7D%7B3%20-%200%7D%20)
m =
![\frac{18}{3}](https://tex.z-dn.net/?f=%20%5Cfrac%7B18%7D%7B3%7D%20)
m = 6
Now that you know the slope of the line you are looking for is 6, you can plug that into a point-slope form equation and find the equation of your line.
y - y_1 = m (x - x_1) Plug in either set of coordinate. I chose (3, 18).
y - 18 = 6 (x - 3) Use he Distributive Property
y - 18 = 6x - 18 Add 18 to both sides
y = 6x
The equation of a direct variation line that includes the point (3, 18) is
y = 6x.
Answer: statement a). True b). Impossible to say c). False d). 5 students
34/3+12/3+2-6+9/6+4+7
Remember PEMDAS:
1. (34/3)+(12/3)+2-6+(9/6)+4+7
11.33+4+2-6+1.66+4+7
The rest can be added together into a calculator (or by hand if you are not supposed to use one)
The center of a circle given the equation in the form (x-h)^2 + (y-k)^2 = r^2 is (h,k). x-8 shows that h is 8, y+2 shows that k is -2, meaning the center is (8, -2)