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2 years ago

Hey help me with this question plzzzz

Mathematics

2 answers:

Wittaler [7]2 years ago

6 0

<h3>Two Answers: choice C, choice D</h3>

Look at where we don't have repeating x values. This happens with function C and function D. All the x values are unique for each choice mentioned.

In choices A, B, and E, the value x = -3 repeats itself. So we don't have a function for either of these. A function is only possible if any input (x) leads to exactly one output (y).

zavuch27 [327]2 years ago

5 0

The answer is c and d

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Use the Divergence Theorem to evaluate S F · dS, where F(x, y, z) = z2xi + y3 3 + sin z j + (x2z + y2)k and S is the top half of

GenaCL600 [577]

Close off the hemisphere $S$ by attaching to it the disk $D$ of radius 3 centered at the origin in the plane $z=0$ . By the divergence theorem, we have

$\displaystyle\iint_{S\cup D}\vec F(x,y,z)\cdot\mathrm d\vec S=\iiint_R\mathrm{div}\vec F(x,y,z)\,\mathrm dV$

where $R$ is the interior of the joined surfaces $S\cup D$ .

Compute the divergence of $\vec F$ :

$\mathrm{div}\vec F(x,y,z)=\dfrac{\partial(xz^2)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial k}=z^2+y^2+x^2$

Compute the integral of the divergence over $R$ . Easily done by converting to cylindrical or spherical coordinates. I'll do the latter:

$\begin{cases}x(\rho,\theta,\varphi)=\rho\cos\theta\sin\varphi\\y(\rho,\theta,\varphi)=\rho\sin\theta\sin\varphi\\z(\rho,\theta,\varphi)=\rho\cos\varphi\end{cases}\implies\begin{cases}x^2+y^2+z^2=\rho^2\\\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi\end{cases}$

So the volume integral is

$\displaystyle\iiint_Rx^2+y^2+z^2\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^3\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{486\pi}5$

From this we need to subtract the contribution of

$\displaystyle\iint_D\vec F(x,y,z)\cdot\mathrm d\vec S$

that is, the integral of $\vec F$ over the disk, oriented downward. Since $z=0$ in $D$ , we have

$\vec F(x,y,0)=\dfrac{y^3}3\,\vec\jmath+y^2\,\vec k$

Parameterize $D$ by

$\vec r(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath$

where $0\le u\le 3$ and $0\le v\le2\pi$ . Take the normal vector to be

$\dfrac{\partial\vec r}{\partial v}\times\dfrac{\partial\vec r}{\partial u}=-u\,\vec k$

Then taking the dot product of $\vec F$ with the normal vector gives

$\vec F(x(u,v),y(u,v),0)\cdot(-u\,\vec k)=-y(u,v)^2u=-u^3\sin^2v$

So the contribution of integrating $\vec F$ over $D$ is

$\displaystyle\int_0^{2\pi}\int_0^3-u^3\sin^2v\,\mathrm du\,\mathrm dv=-\frac{81\pi}4$

and the value of the integral we want is

(integral of divergence of <em>F</em>) - (integral over <em>D</em>) = integral over <em>S</em>

==> 486π/5 - (-81π/4) = 2349π/20

5 0

3 years ago

Is the answer option c?

Masja [62]

Answer:

Discrete

Step-by-step explanation:

A set can simply be defined as a list of values. There are many different types of sets, some are intervals, others are enumerated numbers. Often, a set composed of values that are listed is referred to as a discrete set. Discrete sets can be a list of information, or "isolated points". Knowing this information, the most logical answer to fill in the blank is "discrete".

"If a set is made up of isolated points and can be written as a list, it is called a <u>discrete</u> set."

7 0

2 years ago

In the formula C= prn, p stands for

umka21 [38]

Answer:

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

I NEED HELP IN THIS QUESTION PLEASE

aliina [53]

Looks like you got it right :) Each package contains 8 chocolates, so to find 8 containers, just multiply by 8 to get 64, which would be H.

5 0

2 years ago

Find the composition of

Rus_ich [418]

Answer:

[(x + 6), (y + 1)]

Step-by-step explanation:

Vertices of the quadrilateral ABCD are,

A → (-5, 2)

B → (-3, 4)

C → (-2, 4)

D → (-1, 2)

By reflecting the given quadrilateral ABCD across x-axis to form the image quadrilateral A'B'C'D',

Rule for the reflection of a point across x-axis is,

(x, y) → (x , -y)

Coordinates of the image point A' will be,

A(-5, 2) → A'(-5, -2)

From the picture attached, point E is obtained by translation of point A'.

Rule for the translation of a point by h units right and k units up,

A'(x+h, y+k) → E(x', y')

By this rule,

A'(-5 + h, -2 + k) → E(1, -1)

By comparing coordinates of A' and E,

-5 + h = 1

h = 6

-2 + k = -1

k = 1

That means

Rule for the translation will be,

[(x + 6), (y + 1)]

8 0

2 years ago