Answer:
Explanation:
I hope you understand this
Consider the closed region

bounded simultaneously by the paraboloid and plane, jointly denoted

. By the divergence theorem,

And since we have

the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have




Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by

, we have

Parameterize
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by


which would give a unit normal vector of

. However, the divergence theorem requires that the closed surface

be oriented with outward-pointing normal vectors, which means we should instead use

.
Now,



So, the flux over the paraboloid alone is
First to get the equation you knew to understand one thing about perpendicular lines. The slope of the line is the opposite reciprocal of the perpendicular lines or the new slope is m = 10.
Then you use the formula
y = mx + b
you plug in your values from the point and the new slope.
(1,5) with new slope m
5= 10(1)+b
5-10=b
-5 = b
then make your new equation
y = 10x -5
that's your line that goes through point (1,5) and is perpendicular to the line given
Step-by-step explanation:
given :
2x - 3y = 11
-6x + 8y = 34
find : the solutions of the system by using Cramers Rule.
solutions:
in the matrix 2x2 form =>
[ 2 -3] [x] [11]
=
[-6 8] [ y] [34]
D =
| 2 -3 |
|-6 8 |
= 8×2 - (-3) (-6)
= 16-18 = -2
Dx = | 11 -3 |
| 34 8 |
= 11×8 - (-3) (34)
= 88 + 102
= 190
Dy = | 2 11 |
|-6 34 |
= 2×34 - (-6) (11)
= 68 + 66
= 134
x = Dx/D = 190/-2 = -95
y = Dy/D = 134/-2 = -67
the solutions = {-95, -67}