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VLD [36.1K]
3 years ago
12

Expressions are equivalent???

Mathematics
1 answer:
Semenov [28]3 years ago
4 0

Answer:

only the first option is equivalent

Step-by-step explanation:

You might be interested in
An office building has the same number and size of offices on each floor. If there are 21 offices on the first 3 floors, then wh
zhenek [66]

Answer:

70 offices

Step-by-step explanation:

21 ÷ 3 to get amount per floor, 7 per floor, 7 × 10 = 70, 70 offices

6 0
2 years ago
Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of
tresset_1 [31]

Because I've gone ahead with trying to parameterize S directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over S straight away, let's close off the hemisphere with the disk D of radius 9 centered at the origin and coincident with the plane y=0. Then by the divergence theorem, since the region S\cup D is closed, we have

\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV

where R is the interior of S\cup D. \vec F has divergence

\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z

so the flux over the closed region is

\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0

\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}

Parameterize D by

\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec k

with 0\le u\le9 and 0\le v\le2\pi. Take the normal vector to D to be

\vec s_u\times\vec s_v=-u\,\vec\jmath

Then the flux of \vec F across S is

\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^{2\pi}\int_0^9(u^2\cos v\sin v\,\vec\imath+u\cos v\,\vec\jmath)\cdot(-u\,\vec\jmath)\,\mathrm du\,\mathrm dv

=\displaystyle-\int_0^{2\pi}\int_0^9u^2\cos v\,\mathrm du\,\mathrm dv=0

\implies\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\boxed{0}

8 0
3 years ago
Solve x2 + 14x = -40 by completing the square.
zalisa [80]

Answer: C x=-4 and -10

Step-by-step explanation:

6 0
2 years ago
Please help me i don’t know if I’m right
Scilla [17]

Answer:

   your choice is correct

Step-by-step explanation:

The magnitude of the difference between the target amount and the actual amount will be no greater than the allowed error.

  |x -34| ≤ 0.25

5 0
3 years ago
Please help ill give brainiest answer
trapecia [35]

Answer:

1: AAS, RQC 2: ASA, SRP

Step-by-step explanation:

(1) We are shown that two angles and one side are congruent in the order of AAS. Make sure you write the letters in terms of the corresponding angles. The angles and sides are congruent because the problem labels it for us. For example, B,A,C=R,Q,S. Answers: AAS, RQC.

(2) We are shown that two angles and one side are congruent in the order of ASA. Make sure you write the letters in terms of the corresponding angles again. For example, P,Q,R=P,S,R. The angles are congruent because the problem labels it for us. Side PR is congruent to side PR by reflexive property. Answers: ASA, SRP.

I hope this helped :) Good luck

4 0
3 years ago
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