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: 3x + 6 -4x = 1
Simplify to -x + 6 = 1 because 3x - 4x = -1x
Answer:
b
Step-by-step explanation:
∠ BAD = 90° and
∠ DAC + ∠ CAB = ∠ BAD, that is
x + 2x - 30 = 90
3x - 30 = 90 ( add 30 to both sides )
3x = 120 ( divide both sides by 3 )
x = 40 → b
Answer:
27KG
Step-by-step explanation:
Mouse- M , Cat - C, Dog - D
M + C = 10 ----- (1)
M + D = 20 -------(2)
C + D = 24 --------(3)
(1) - (2) => C - D = -10 ---- (4)
(3) + (4) => 2C = 14
=> C = 7
Substitute C in (1) => M + 7 = 10
=> M = 3
Substitute M in (2) => 3 + D = 20
=> D = 17
Therefore C + M + D = 7 + 3 + 17 = 27KG
Answer:
31:150
Step-by-step explanation:
First of all we convert the hours into minutes:
5×60=300
which gets us 42:300, which we can simplify to:
31:150
and is 31 is prime and 150 can't be divided into it without a remainder, it is in it's simplest form and it is the answer.