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Mandarinka [93]
3 years ago
15

192 beads

Mathematics
1 answer:
nirvana33 [79]3 years ago
8 0

Answer:

Step-by-step explanation:

b + n = 192

n = 5b

b + 5b = 192

6b = 192

b = 192/6

b = 32 <==== 5 beads per bracelet

n = 5b

n = 5(32)

n = 160 <==== 160 beads per necklace

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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

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3 years ago
Help I don’t understand this ‍♀️
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Hello there!
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wlad13 [49]

The factor of the polynomial, the area of the pool and the simplified expression are all algebraic expressions

<h3>How to simplify the expressions?</h3>

<u>Factor of the polynomial</u>

The polynomial is given as:

2x^3 + 6x^2 + 6x + 18

Factor out 2

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Factorize

2x^3 + 6x^2 + 6x + 18 = 2(x^2(x + 3) + 3(x + 3))

Factor out x + 3

2x^3 + 6x^2 + 6x + 18 = 2(x^2 + 3)(x + 3)

Expand

2x^3 + 6x^2 + 6x + 18 = (2x^2 + 6)(x + 3)

Hence, 2x^2 + 6 is a factor of the polynomial 2x^3 + 6x^2 + 6x + 18

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<u>Simplify the expression</u>

The expression is given as:

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Answer:

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