When s is the open hemisphere x 2 + y 2 + z 2 = 1, z ≤ 0 , oriented by the inward normal pointing to the origin, then the bounda
ry orientation on ∂s is clockwise. true or false?
1 answer:
The figure below shows a diagram of this problem. First of all we graph the hemisphere. This one has a radius equal to 1. Given that z ≤ 0 a sphere will be valid only in the negative z-axis, that is, we will get a half of a sphere that is the hemisphere shown in the figure. We know that this hemisphere is oriented by the inward normal pointing to the origin, then we have a Differential Surface Vector called N, using the Right-hand rule <span>the boundary orientation is </span>counterclockwise.
Therefore, the answer above False
Therefore, the answer above False
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Answer:
Step-by-step explanation:
<h3>Solving linear equation with one variable:</h3>1) -4 + 3x = 4x - 8
Add 4 to both sides
-4 + 3x + 4 = 4x - 8 + 4
3x = 4x - 4
Subtract 4x from both sides,
3x - 4x = -4x + 4x - 4
-x = -4
2) -5x - 8 = 2
Add 8 to both sides
-5x - 8 + 8 = 2 + 8
-5x = 10
Divide both sides by (-5)
3) 12r - 14 = 5(2-r)
12r - 14 = 5*2 - 5*r
12r - 14 = 10 - 5r
Add 14 to both sides
12r - 14 + 14 = 10 - 5r + 14
12r = 24 - 5r
Add 5r to both sides
12r + 5r = 24
17r = 24
Divide both sides by 17
r = 24/17
4) 3x - 8 = -(17 + 2x)
3x - 8 = -17 - 2x
Add 8 to both sides
3x - 8 + 8 = -17 - 2x + 8
3x = -9 - 2x
Add 2x to both sides
3x + 2x = -9
5x = -9
Divide both sides by 5