The distance between a point

on the given plane and the point (0, 2, 4) is

but since

and

share critical points, we can instead consider the problem of optimizing

subject to

.
The Lagrangian is

with partial derivatives (set equal to 0)




Solve for

:


which gives the critical point

We can confirm that this is a minimum by checking the Hessian matrix of

:


is positive definite (we see its determinant and the determinants of its leading principal minors are positive), which indicates that there is a minimum at this critical point.
At this point, we get a distance from (0, 2, 4) of
Answer:
no she dont
Step-by-step explanation:
she gotta have 125 cus theres 5 20's in 100 and theres only 4 5's in 20
Answer:
9005.90 m
Step-by-step explanation:
The radius of the collider is half the diameter or 2150 m. The arc length S subtended by the angle is given by
S = r×(theta) where theta is the angle in radians. We need to convert the degree angle into radian angle using the following:
240° × (2pi/360°) = 4.19 radians
Therefore, the distance traveled by the atoms is so equal to the arc length S so
S = (2150 m)×(4.19 rad) = 9005.90 m
Vertical is the to the question