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3 years ago

14

Linda had 45 fliers to post around town. She posted 1/3 of them. This week she posted 2/3 of the remaining fliers. How many flie

rs has she not posted?

1 answer:

garik1379 [7]3 years ago

4 0

She has not posted 10 fliers.

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What is the measure to

antoniya [11.8K]

The answer is 90 it’s a right angle :)

3 0

3 years ago

The monthly fees for the local pool are $8 per month

Liono4ka [1.6K]

Given:

Monthly fees for the local pool are $8 per month and $2 per visit.

Hector pays $34 in pool fees total for the month.

To find:

The number of times he visit the pool.

Solution:

We have,

Monthly fee of pool = $8

Additional fee = $2 per visit

Let Hector visit x times.

Additional fee for x times = $2x

Total fee = Monthly fee + Additional fee

$34=8+2x$

$34-8=2x$

$26=2x$

Divide both sides by 2.

$\dfrac{26}{2}=x$

$13=x$

Therefore, Hector visit the pool 13 times.

3 0

3 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 is the perimeter of the figure? <br> 229 4/9 ft<br> 229 2/3 ft<br> 230 2/3 ft <br> 231 ft

mote1985 [20]

The answer would be 230 2/3, because to get the perimeter you have to add all the sides together, 60 5/6 + 59 1/3 + 56 1/6 + 54 1/3 = 230 2/3

6 0

3 years ago

Find the slope of the line that passes through (-3,-2) and (-3,2)

fiasKO [112]

Answer:

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Step-by-step explanation:

The x value does not change, which means it is a vertical line

Vertical lines have an undefined slope

we can check using

m = (y2-y1)/(x2-x1)

=(2- -2)/( -3 - -3)

= (2+2)/( -3+3)

4/0

undefined

7 0

3 years ago