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3 years ago

One-fourth of the sum of 2p and 11

Mathematics

1 answer:

scZoUnD [109]3 years ago

3 0

Answer:

What’s the answer

Step-by-step explanation:u have to tell me

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Please help, giving 15 points

GaryK [48]

Answer: 15, because it is 16-1

5 0

3 years ago

1. Which of the following values of n will make the below equation true? n × (−2+5 ) = −8+20

Salsk061 [2.6K]

Answer:

1a 2a

Step-by-step explanation:

n × (−2+5 ) = −8+20

n × 3 = 12

n = 4

2 + (n+6) = −1+6

n + 8 = 5

n = -3

7 0

2 years ago

Let f(x,y,z) = ztan-1(y2) i + z3ln(x2 + 1) j + z k. find the flux of f across the part of the paraboloid x2 + y2 + z = 3 that li

Sophie [7]

Consider the closed region $V$ bounded simultaneously by the paraboloid and plane, jointly denoted $S$ . By the divergence theorem,

$\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV$

And since we have

$\nabla\cdot\mathbf f(x,y,z)=1$

the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have

$\displaystyle\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\iiint_V\mathrm dV$
$=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=2}^{z=3-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle2\pi\int_{r=0}^{r=1}r(3-r^2-2)\,\mathrm dr$
$=\dfrac\pi2$

Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by $D$ , we have

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-\iint_D\mathbf f\cdot\mathrm dS$

Parameterize $D$ by

$\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+2\,\mathbf k$
$\implies\mathbf s_u\times\mathbf s_v=u\,\mathbf k$

which would give a unit normal vector of $\mathbf k$ . However, the divergence theorem requires that the closed surface $S$ be oriented with outward-pointing normal vectors, which means we should instead use $\mathbf s_v\times\mathbf s_u=-u\,\mathbf k$ .

Now,

$\displaystyle\iint_D\mathbf f\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(-u\,\mathbf k)\,\mathrm dv\,\mathrm du$
$=\displaystyle-4\pi\int_{u=0}^{u=1}u\,\mathrm du$
$=-2\pi$

So, the flux over the paraboloid alone is

$\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-(-2\pi)=\dfrac{5\pi}2$

6 0

3 years ago

What set of reflections would carry triangle ABC onto itslelf?

riadik2000 [5.3K]

<h3>Answer: Choice A</h3>

y axis, x axis, y axis, x axis

============================

Explanation:

Reflecting an object over the y axis twice will have it end up in its starting position. The same can be said for the x axis as well. It doesn't matter that the x axis reflections aren't grouped next to each other, nor the y. So in a sense, two x axis reflections undo each other, so do the y axis reflections, and we end up with the same image as shown in the diagram.

6 0

3 years ago

Please help with #20

abruzzese [7]

Answer:

The answer is 9 or 8 because they both are on the same fx

6 0

2 years ago