3 years ago

I NEED HELP PLZ DUE IN 5 min

Mathematics

1 answer:

OverLord2011 [107]3 years ago

7 0

Answer:

2y^2 + 18x

Step-by-step explanation:

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6x + -2 =16 solve for x

nasty-shy [4]

Answer:

Step-by-step explanation:

+2 =18

18/6 =3

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3 years ago

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Four friends go to the movies and pay $11 per ticket. They also each buy a snack

Setler79 [48]

Answer:

Each snack combo costed $8

Step-by-step explanation:

4 friends* 11 per ticket = $44 for four tickets

76 - 44 = $32 spent on all the snack combos

$32 ÷ 4 friends = $8 spent on each snack combo

So, each snack combo costed $8

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3 years ago

) Set up a double integral for calculating the flux of F=3xi+yj+zk through the part of the surface z=−5x−2y+2 above the triangle

Fynjy0 [20]

The surface (call it $S$ ) is a triangle with vertices at the points

$x=0,y=0\implies z=2\implies(0,0,2)$

$x=0,y=2\implies z=-2\implies(0,2,-2)$

$x=2,y=0\implies z=-8\implies(2,0,-8)$

Parameterize $S$ by

$\vec s(u,v)=(1-v)(2,0,-8)+v\bigg((1-u)(0,2,-2)+u(0,0,2)\bigg)=(2-2v,2v-2uv,-8+6v+4uv)$

with $0\le u\le1$ and $0\le v\le1$ . Take the normal vector to $S$ to be

$\vec s_v\times\vec s_u=(20v,8v,4v)$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^1\int_0^1(6-6v,2v-2uv,-8+6v+4uv)\cdot(20v,8v,4v)\,\mathrm du\,\mathrm dv$

$=\displaystyle8\int_0^1\int_0^1(11v-10v^2)\,\mathrm du\,\mathrm dv=\boxed{\frac{52}3}$

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4 years ago

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Afina-wow [57]

It would be the first option

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3 years ago

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