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3 years ago

5

D'Naisa mixed her favorite shade of orange paint. She had 111 gallon of orange paint and added 222 quarts of yellow paint and 33

3 pints of red paint.

2 answers:

tatiyna3 years ago

6 0

I Hate it when Khan academy does this... The Answer is 7 1/2

belka [17]3 years ago

5 0

What’s the question though?

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Which of the following are geometric sequences?

Nana76 [90]

B. 10, 5, 2.5, 1.25, 0.625, 0.3125

5 0

3 years ago

Find the gcf of 14,35,84

Paha777 [63]

Answer:

7

Step-by-step explanation:

3 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

Find the centroid of the region bounded by the given curves y=8sin(4x), y=8cos(4x)

Luba_88 [7]

A cosine is just a sine shifted to the left by π/2. A cosine of 4x is shifted to the left by only π/8 because of the factor 4. Sketch them.

The region we're looking for is this sausage-shaped part between the cos and the sin.

The x intercepts are at π/8 for the cosine and π/4 for the sine. The midpoint between them is at (π/8 + π/4)/2 = 3/16π.

The region is point symmetric around the x axis, so the y coordinate of the centroid is 0.

So the centroid is at (3/16π, 0)

5 0

3 years ago

BRAINLIEST TO CORRECT ANSWER!!! NO links as answers, or you will be reported!! (question 4)

In-s [12.5K]

Answer:

Hello! Your answer would be, C) Divide by 2

Step-by-step explanation:

And plz don't be mean. :(

Hope I helped! Brainiest plz!♥ Have a nice afternoon. Hope you make a 100%! -Amelia

8 0

2 years ago