You would do three square in length and 31 in width
I assume there are some plus signs that aren't rendering for some reason, so that the plane should be

.
You're minimizing

subject to the constraint

. Note that

and

attain their extrema at the same values of

, so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.
The Lagrangian is

Take your partial derivatives and set them equal to 0:

Adding the first three equations together yields

and plugging this into the first three equations, you find a critical point at

.
The squared distance is then

, which means the shortest distance must be

.
Answer:
The information we need is that the angle between the sides (that is, angle TVU for triangle TVU and angle WVX for triangle WVX) need to be equal.
These angles are vertically opposite angles, because they are formed by two lines crossing, so these angles are equal.
Step-by-step explanation:
The SAS Congruence Theorem says that if two triangles have 2 equal sides and the angle between these sides are also equal, the triangles are congruent.
In this question, we know that the sides UV and VW are congruent, as V is the midpoint of UW. We also know that TV = VX, so now we have two equal sides for each triangle.
The information we need is that the angle between the sides (that is, angle TVU for triangle TVU and angle WVX for triangle WVX) need to be equal.
These angles are vertically opposite angles, because they are formed by two lines crossing, so these angles are equal.
Now we can conclude that the triangles are congruent.
Answer:
wait whats the question-
Step-by-step explanation:
86 years old because you subtract 2017-1931