Check the picture below. So the parabola looks more or less like so.
let's recall that the vertex is half-way between the focus point and the directrix, at "p" units away from both.
Let's notice that the focus point is below the directrix, that means the parabola is vertical, namely the squared variable is the "x", and it also means that it's opening downwards as you see in the picture, namely that "p" is negative, in this case "p" is 1 unit, and thus is -1.
![\bf \textit{parabola vertex form with focus point distance} \\\\ \begin{array}{llll} 4p(x- h)=(y- k)^2 \\\\ \stackrel{\textit{we'll use this one}}{4p(y- k)=(x- h)^2} \end{array} \qquad \begin{array}{llll} vertex\ ( h, k)\\\\ p=\textit{distance from vertex to }\\ \qquad \textit{ focus or directrix} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} h=-2\\ k=5\\ p=-1 \end{cases}\implies 4(-1)(y-5)=[x-(-2)]^2\implies -4(y-5)=(x+2)^2 \\\\\\ y-5=-\cfrac{1}{4}(x+2)^2\implies y=-\cfrac{1}{4}(x+2)^2+5](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bparabola%20vertex%20form%20with%20focus%20point%20distance%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%204p%28x-%20h%29%3D%28y-%20k%29%5E2%20%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bwe%27ll%20use%20this%20one%7D%7D%7B4p%28y-%20k%29%3D%28x-%20h%29%5E2%7D%20%5Cend%7Barray%7D%20%5Cqquad%20%5Cbegin%7Barray%7D%7Bllll%7D%20vertex%5C%20%28%20h%2C%20k%29%5C%5C%5C%5C%20p%3D%5Ctextit%7Bdistance%20from%20vertex%20to%20%7D%5C%5C%20%5Cqquad%20%5Ctextit%7B%20focus%20or%20directrix%7D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20h%3D-2%5C%5C%20k%3D5%5C%5C%20p%3D-1%20%5Cend%7Bcases%7D%5Cimplies%204%28-1%29%28y-5%29%3D%5Bx-%28-2%29%5D%5E2%5Cimplies%20-4%28y-5%29%3D%28x%2B2%29%5E2%20%5C%5C%5C%5C%5C%5C%20y-5%3D-%5Ccfrac%7B1%7D%7B4%7D%28x%2B2%29%5E2%5Cimplies%20y%3D-%5Ccfrac%7B1%7D%7B4%7D%28x%2B2%29%5E2%2B5)
Answer:
2 km east and 10 km north
Step-by-step explanation:
I did the activity and got the right answer.
Answer:
umm idk let me get back to you
Step-by-step explanation:
Since eleven is odd, there must be a centra element, which in this case is the sixth. So, we can write 11 consecutive integers as

Since the arithmetic mean is the sum of the values, divided by how many they are, we have

So, the mean of 11 consecutive integers is the 6th integer.