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never [62]
3 years ago
10

Points A(8,4)&B(-2,4) lie on a line .AB is parallel to which axis

Mathematics
1 answer:
IRINA_888 [86]3 years ago
8 0
If the line is parallel they will have the same slope
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Simplify the following expression:<br><br> 48⋅47
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Answer:

2256

Step-by-step explanation:

you simply multiply the two together

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3 years ago
I need help with 15,19,17<br><br> Negative exponents
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2 years ago
Simplify<br> (3m²n4)3<br> O 9m5 n?<br> O 9m6n12<br> O 27mn12<br> O 27mn?
Lana71 [14]

Answer:

(3m2n4)^3

Step-by-step explanation:

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3 years ago
Consider the following region R and the vector field F. a. Compute the​ two-dimensional curl of the vector field. b. Evaluate bo
Shalnov [3]

Looks like we're given

\vec F(x,y)=\langle-x,-y\rangle

which in three dimensions could be expressed as

\vec F(x,y)=\langle-x,-y,0\rangle

and this has curl

\mathrm{curl}\vec F=\langle0_y-(-y)_z,-(0_x-(-x)_z),(-y)_x-(-x)_y\rangle=\langle0,0,0\rangle

which confirms the two-dimensional curl is 0.

It also looks like the region R is the disk x^2+y^2\le5. Green's theorem says the integral of \vec F along the boundary of R is equal to the integral of the two-dimensional curl of \vec F over the interior of R:

\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\iint_R\mathrm{curl}\vec F\,\mathrm dA

which we know to be 0, since the curl itself is 0. To verify this, we can parameterize the boundary of R by

\vec r(t)=\langle\sqrt5\cos t,\sqrt5\sin t\rangle\implies\vec r'(t)=\langle-\sqrt5\sin t,\sqrt5\cos t\rangle

\implies\mathrm d\vec r=\vec r'(t)\,\mathrm dt=\sqrt5\langle-\sin t,\cos t\rangle\,\mathrm dt

with 0\le t\le2\pi. Then

\displaystyle\int_{\partial R}\vec F\cdot\mathrm d\vec r=\int_0^{2\pi}\langle-\sqrt5\cos t,-\sqrt5\sin t\rangle\cdot\langle-\sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm dt

=\displaystyle5\int_0^{2\pi}(\sin t\cos t-\sin t\cos t)\,\mathrm dt=0

7 0
3 years ago
Which statement of the function is true ????
Lera25 [3.4K]
B is the answer when the parabola is facing down it is negative
3 0
3 years ago
Read 2 more answers
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