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Sorry I don't know anything about that!
Hope this helps somehow, Ayitzdaisy!
2,3,5,6 are all exterior angles because they are out of the triangle
The answer is 540.. I think
Bring u outside the bracket so u (32 -v + 8u)
Y=-x^2-2
dy/dx=-2x and d2y/dx2=-2
Since acceleration is a negative constant, when the velocity equals zero, it will be an absolute maximum for y(x)
dy/dx=0 when -2x=0, x=0
y(0)=-2
So the vertex is at the absolute maximum, which is the point (0,-2)
