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:
$0.558
Step-by-step explanation:
The expected value is the sum of the value of each outcome times the chance that it happens. In this case, there are two outcomes:
- Win $31 million
- Win $0
Then our expected value can be calculated as:
Answer:
dissect the trapezium into a square and triangle. since the triangle is a right angled one the base is 17-7=10cm
the height is found using Pythagorean theorem to be 24cm i.e 26²=x²+10²,x=√(676-100),x=√576,x=24cn
area of trapezium is (1/2)(a+b)h
a is the shorter side and b the longer side
(1/2)*(7+17)24
=288cm²
Formula: A = pq/2
p = d1
q = d2
Substitute with the given values and solve.
A = 12*14/2
A = 168/2
A = 84
Therefore, the ansewr is 84in²
Best of Luck!
Answer:
Step-by-step explanation:
Let's see how well I can explain this. 
is the same as a 30 degree angle which is in quadrant 1. If you picture the unit circle, right in the center of it is the origin. If you draw a straight line from 30 degrees and through the center (the origin), you will automatically "connect" with the reference angle of 30 (this is true for ALL angles on the unit circle). This puts us in quadrant 3. In quadrant 3, x is negative and so is y. So the terminal point of the reference angle for 30 degrees has the same exact values, but both of them are negative (again, because both x and y are negative in quadrant 3). I can't see your choices but the one you want looks like this:
