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

.
1. a) equation of the line :
y - intercept = -7
so, it will pass through point (0, -7)
and if we plug the value of x as 5, we get
so, it will pass through point (5, -5) too
now, just plot the points (0 , -7) and (5 , -5) and join them.
2. b) equation of line is :
here, y - intercept = 5
so the line passes through point (0 , 5)
now, Plugging the value of x = 1 we get :
so, the given line passes through point (1 , 2)
plotting the points, we can get our required line.
Answer:
f
−
1(x)=x−2 if you are talking about the inverse
please comment what is the purpose of this excersise so i can help you further
Step-by-step explanation:
A(R+T)=W
(A×R)+(A×T)=W
AR+AT=W
-AR from both sides
AT=W-AR
divide both sides by A
T=(W-AR)/A
X= 28
explanation: pythagorean theorem