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3 years ago

14

You have 3/4 of a pie leftover after your party. Each guest takes home 1/6 of the leftover pie. What fraction of the original pi

e does each guest take home ?

1 answer:

Mazyrski [523]3 years ago

7 0

What are you looking for how many they took or how many in total is left

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Which graph represents the solution set for the system x + 2y 3, x + y 4, and 3x − 2y 4?

Lerok [7]

x + 2y + 3 = 0
Subtract x from both sides.
2y + 3 = -x
Subtract 3 from both sides.
2y = -x - 3
Divide by 2 on both sides
y = -(x+3)/2

x + y + 4 = 0
Subtract x and 4 from both sides
y = -x - 4

3x - 2y + 4 = 0
Subtract 3x and 4 from both sides.
-2y = -3x -4
Divide by -2 from both sides.
y = -(3x + 4) / 2

The answer is the graph that contains these slopes and lines on the graph, which was not provided.

4 0

3 years ago

Evaluate the surface integral. s x2 + y2 + z2 ds s is the part of the cylinder x2 + y2 = 4 that lies between the planes z = 0 an

Leya [2.2K]

Parameterize the lateral face $T_1$ of the cylinder by

$\mathbf r_1(u,v)=(x(u,v),y(u,v),z(u,v))=(2\cos u,2\sin u,v$

where $0\le u\le2\pi$ and $0\le v\le3$ , and parameterize the disks $T_2,T_3$ as

$\mathbf r_2(r,\theta)=(x(r,\theta),y(r,\theta),z(r,\theta))=(r\cos\theta,r\sin\theta,0)$
$\mathbf r_3(r,\theta)=(r\cos\theta,r\sin\theta,3)$

where $0\le r\le2$ and $0\le\theta\le2\pi$ .

The integral along the surface of the cylinder (with outward/positive orientation) is then

$\displaystyle\iint_S(x^2+y^2+z^2)\,\mathrm dS=\left\{\iint_{T_1}+\iint_{T_2}+\iint_{T_3}\right\}(x^2+y^2+z^2)\,\mathrm dS$
$=\displaystyle\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}((2\cos u)^2+(2\sin u)^2+v^2)\left\|{{\mathbf r}_1}_u\times{{\mathbf r}_2}_v\right\|\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+0^2)\left\|{{\mathbf r}_2}_r\times{{\mathbf r}_2}_\theta\right\|\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+3^2)\left\|{{\mathbf r}_3}_r\times{{\mathbf r}_3}_\theta\right\|\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle2\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r^3\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r(r^2+9)\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle4\pi\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv+2\pi\int_{r=0}^{r=2}r^3\,\mathrm dr+2\pi\int_{r=0}^{r=2}r(r^2+9)\,\mathrm dr$
$=136\pi$

7 0

3 years ago

A method of solving a system of equations in which one variable is replaced by an expression using the other variable as a repre

Tanya [424]

I believe it is referred to as The Substitution Method.

3 0

3 years ago

Read 2 more answers

Failinggggggggggg mathhhhhhhhhhhhhhhhhhhhhhhhh

azamat

2. (D)
3. (B) Hope it helps

3 0

3 years ago

Read 2 more answers

spayn [35]

D Oh no! Something went wrong while adding your answer
It's too short. Write at l<span>Remember that it's against our guidelines to copy content from other websites, books or any other material to which you don't own the rights to. We know you have the power to solve these problems on your own!</span>east 20 characters to explain it well

4 0

3 years ago