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kenny6666 [7]
3 years ago
9

F = xi+yj+2zk and σ is the portion of the cone z=x2+y2−−−−−−√ between the planes z=4 and z=5 oriented with normal vectors pointi

ng upwards. Find the flux of the flow field F across σ
Mathematics
1 answer:
trapecia [35]3 years ago
4 0
There is no question for me to answer
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Find the product. (2n + 2)(2n – 2)
vekshin1
<span>(2n + 2)(2n – 2)
= (2n)^2 -2^2
= 4n^2 - 4

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3 0
3 years ago
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If trapezoid ABCD was reflected over the y-axis, reflected over the x-axis, and rotated 180°, where would point A' lie?
Dmitrij [34]
A is the point (-4, 1)

i) reflection over the y-axis projects a point (a, b) to (-a, b):

(-4, 1) is projected to (4, 1), 

ii) reflection over the x axis projects a point (a, b) to (a, -b):

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8 0
3 years ago
20 POINTS!!!
choli [55]

Hello from MrBillDoesMath!

Answer:

x = 0, 7

Discussion:

f(g(x)) =

f ( x^2-7x) =

1/ ( x^2 - 7x)


The points NOT in the domain are those where the denominator,  x^2 - 7x = 0.

x^2 - 7x   = 0       =>       factor x from each term

x(x-7) = 0            =>        one or both terms must each 0

x = 0 or x =7


Thank you,

MrB


6 0
3 years ago
Which relationship describes angles 1 and 2?
umka2103 [35]
I belive its B,and D,
B because, the lines are supplementary(because supplementary means it equal 180 degrees in length, which is those two angles.)
D, because they are put in such an angle in which making them adjecant.
Hope I helped:-) Bye bye.
8 0
3 years ago
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Please I need 1,2, and 4
Virty [35]

Answer:what grade are you in?

Step-by-step explanation:

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3 years ago
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