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3 years ago

to convert feet per minute to inches per second which of the following conversion factors would be used select all that apply

Mathematics

1 answer:

Arisa [49]3 years ago

7 0

1 feet = 12 inches

1 min = 60 s

1 feet/min = 12 in/60 s = 1 in/5 s = 0.2 in/s

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I don't think this is correct plz check me

Readme [11.4K]

Answer:

The answer is the first one

Step-by-step explanation:

The cost of an adult ticket is $15

The price of an adult ticket is 1/2 the price of a student ticket plus $8

x is the cost of a student ticket, so

1/2(x) + $8 = $15

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3 years ago

Find the value of x in each case. PLEASE SHOW WORK!!!

myrzilka [38]

$m\angle BCA=180^o-5x$

$\angle CAB=\angle ACD\\\\m\angle CAB=33^o$

In the triangle ABC:

$m\angle ABC+m\angle BCA+m\angle CAB=180^o\\\\2x+180^o-5x+33^o=180^o\\\\-3x+213^o=180^o\ \ \ |-213^o\\\\-3x=-33^o\ \ \ |:(-3)\\\\x=11^o$

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3 years ago

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Darina [25.2K]

Answer:

I dont really do this kind of work but i think its A

Step-by-step explanation:

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3 years ago

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Andreyy89

Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

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

4 years ago