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Ksju [112]
3 years ago
9

Need help please to find X

Mathematics
1 answer:
bagirrra123 [75]3 years ago
4 0

Answer:

\boxed{x = 86 \degree}

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Which graph represents a function
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Answer:

2

i think

Step-by-step explanation:

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3 years ago
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Use Stokes' Theorem to evaluate C F · dr F(x, y, z) = xyi + yzj + zxk, C is the boundary of the part of the paraboloid z = 1 − x
Serggg [28]

I assume C has counterclockwise orientation when viewed from above.

By Stokes' theorem,

\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S

so we first compute the curl:

\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k

\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k

Then parameterize S by

\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k

where the z-component is obtained from

1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v

with 0\le u\le\dfrac\pi2 and 0\le v\le\dfrac\pi2.

Take the normal vector to S to be

\vec r_v\times\vec r_u=2\cos u\cos v\sin^2v\,\vec\imath+\sin u\sin v\sin(2v)\,\vec\jmath+\cos v\sin v\,\vec k

Then the line integral is equal in value to the surface integral,

\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S

=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv

=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}

6 0
3 years ago
ASAP! BRAINLIEST!
nasty-shy [4]

Answer:

The average rate of change from 3 to 6 storms = 0.04

Step-by-step explanation:

Let number of Storms = N

And Predicted Gas Price = P

it is required to find the average rate of change from 3 to 6 storms.

At N = 3 ⇒ P = $2.44

At N = 6 ⇒ P = $2.56

So, the average rate of change = \frac{2.56-2.44}{6-3}=\frac{0.12}{3}=0.04

5 0
3 years ago
PYTHAGORAS THEROM AND TRIGONOMETRY RATIO​
marysya [2.9K]

Answer:

1) ΔACD is a right triangle at C

=> sin 32° = AC/15

⇔ AC = sin 32°.15 ≈ 7.9 (cm)

2) ΔABC is a right triangle at C, using Pythagoras theorem, we have:

AB² = AC² + BC²

⇔ AB² = 7.9² + 9.7² = 156.5

⇒ AB = 12.5 (cm)

3)  ΔABC is a right triangle at C

=> sin ∠BAC = BC/AB

⇔ sin ∠BAC = 9.7/12.5 = 0.776

⇒ ∠BAC ≈ 50.9°

4) ΔACD is a right triangle at C

=> cos 32° = CD/15

⇔ CD = cos32°.15

⇒ CD ≈ 12.72 (cm)

Step-by-step explanation:

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Which statement is true about setting up a proportion to solve for the missing measure?
LiRa [457]

Answer: B)

Corresponding parts must be in the same position.

Step-by-step explanation:

7 0
3 years ago
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