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3 years ago

JOURNAL 5.3.4. PLZZZ HELP, I NEED 1-9

Mathematics

1 answer:

aniked [119]3 years ago

7 0

Answer:

Step-by-step explanation:

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Quick please no links or anything just the straight up answer answer both please

Serjik [45]

Answer: 42 degrees

Explanation: KFD=84 KFL is half of KFD(84)
84 divided by 2 = 42

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3 years ago

How to solve<br> ([120-2+(46-15)+2]-(9+4)

NNADVOKAT [17]

Answer:138

Step-by-step explanation:

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3 years ago

Can I please have some HELP!!! (Brainllest)

Solnce55 [7]

Answer:

its the last one I think.

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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

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

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

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

8 0

4 years ago

Write the standard form of the line that has a slope of - and y-intercept of -2. Include your work in your final answer. Type yo

ehidna [41]

You need to go from slope-intercept ( $y=mx+b$ ) to standard form ( $ax + by = c$ ).

slope-intercept form of your provided values:
$y=-x-2$
now, solve for x and y
$x+y=-2$

Your answer is $x+y=-2$ .

6 0

3 years ago

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