No, it isn't a right triangle.
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:
45 cm
Step-by-step explanation:

Answer:
A
Step-by-step explanation:
Since we're dealing with perpendicular lines, we take the recipricol of the m value of the original equation.
So the slope we're dealing with is 1/2.
And the equation we have so far is
y = 1/2x + b
Now we solve for b by plugging in the point (4,2).
2 = 1/2(4) + b
2 = 2 + b
b = 0
So the final equation is
y = 1/2x
Answer:
2x bigger
Step-by-step explanation:
because if you split the rectange in half, it would be 2x?