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3 years ago

6

In a survey 13 out of 20 teacher respond yes to a proposal for a new after school club in the same survey 37 out of 50!students

respond yes

1 answer:

boyakko [2]3 years ago

8 0

Answer:Yes

Step-by-step explanation:13 out of 20

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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,

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

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

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

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3 years ago

WILL GIVE BRAINLIEST<br><br>(3x – 14) (2x + 10) <br><br>Find the value of x.

Brut [27]

Answer:

(3x - 14) = (2x+ 10) { Because the angles are vertically opposite to each other}

3x-2x= 14+10

x=24

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3 years ago

I don’t understand how to factor trinomials

gayaneshka [121]

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3 years ago

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You go to a store to pick up milk and cereal. You see that the milk costs $4

PolarNik [594]

Use the estimate but that should be more than the exact price

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3 years ago

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I NEED HELP WILL GIVE BRAINLIEST!! PLEASE HELP ME ASAP!!!!

GenaCL600 [577]

Answer:

A. The length of the second leg is 8.5 inches

B. The length of the three-dimensional diagonal is 9.9 inches

Step-by-step explanation:

Let us revise the relation between the hypotenuse and the two legs of a right triangle

(hypotenuse)² = (vertical leg)² + (horizontal leg)²

∵ The length of the rectangular box = 8 inches

∵ The width of the rectangular box = 3 inches

∵ The height of the rectangular box = 5 inches

∵ Length and width are perpendicular to each other

∴ The Δ whose legs are 3 and 8 is a right triangle

In the right Δ whose legs are 3 and 8

∵ (hypotenuse)² = (3)² + (8)²

∴ (hypotenuse)² = 9 + 64

∴ (hypotenuse)² = 73

- Take √ for both sides

∴ hypotenuse = $\sqrt{73}$ = 8.544003745

- Round it to the nearest tenth of one inch

∴ hypotenuse = 8.5 inches

A.

The 3-dimensional diagonal is the hypotenuse of a right triangle whose legs are the vertical edge and the hypotenuse of the right triangle whose legs are 3 and 8

∵ The hypotenuse of the right triangle whose legs are 3 and

8 is 8.5 inches

∴ The length of the second leg is 8.5 inches

B.

In the right triangle whose hypotenuse is the 3-dimensional diagonal and legs are the vertical edge , the hypotenuse of the right triangle whose legs are 3 and 8

∵ (3-dimensional diagonal)² = (5)² + (73)²

∴ (3-dimensional diagonal)² = 25 + 73

∴ (3-dimensional diagonal)² = 98

- Take √ for both sides

∴ 3-dimensional diagonal = $\sqrt{98}$ = 9.899494937

- Round it to the nearest tenth of an inch

∴ 3-dimensional diagonal = 9.9 inches

∴ The length of the three-dimensional diagonal is 9.9 inches

<em>V.I.N: you can find the length of the three-dimensional diagonal by using this rule → </em> $d=\sqrt{l^{2}+w^{2}+h^{2}}$ <em> </em>

3 0

3 years ago