Hello there,
I hope you and your family are staying safe and healthy during this winter season.
![x^2 + y^2 -6x+14y-1=0](https://tex.z-dn.net/?f=x%5E2%20%2B%20y%5E2%20-6x%2B14y-1%3D0)
We need to use the Quadratic Formula*
, ![\frac{-b-\sqrt{b^2} -4ac }{2a}](https://tex.z-dn.net/?f=%5Cfrac%7B-b-%5Csqrt%7Bb%5E2%7D%20-4ac%20%7D%7B2a%7D)
Thus, given the problem:
![a = 1, b=-6, c=y^2+14y-1](https://tex.z-dn.net/?f=a%20%3D%201%2C%20b%3D-6%2C%20c%3Dy%5E2%2B14y-1)
So now we just need to plug them in the Quadratic Formula*
, ![x=\frac{6-\sqrt{(-6)^2-4(y^2+14y-1)} }{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B6-%5Csqrt%7B%28-6%29%5E2-4%28y%5E2%2B14y-1%29%7D%20%7D%7B2%7D)
As you can see, it is a mess right now. Therefore, we need to simplify it
, ![x = \frac{6-2\sqrt{10-y^2-14y} }{2}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B6-2%5Csqrt%7B10-y%5E2-14y%7D%20%7D%7B2%7D)
Now that's get us to the final solution:
, ![x=3-\sqrt{10-y^2-14y}](https://tex.z-dn.net/?f=x%3D3-%5Csqrt%7B10-y%5E2-14y%7D)
It is my pleasure to help students like you! If you have additional questions, please let me know.
Take care!
~Garebear
Same line so the correct answer is b
BCD should have to equal 24 degrees because angles ACB and BCD look Complimentary and a complimentary is 2 angles that equal 90 degrees when combined since ACB and BCD look like a 90 degree angle I would say that BCD is 24 degrees