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2 years ago

10

7.385 rounded to the nearest tenth

2 answers:

Lana71 [14]2 years ago

3 0

Answer:

10 lol..................

Leni [432]2 years ago

3 0

Answer:

It should be 7.4

Step-by-step explanation:

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) 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

3 years ago

A function f() is graphed on the coordinate plane.

Mkey [24]

Answer:

i dont know but some one will help freeee points

Step-by-step explanation:

7 0

2 years ago

Write the equation of the line in slope-intercept form passing through the points (-4,2) and (1,12)

alexira [117]

Answer:

y = 2x + 10

Step-by-step explanation:

Find the slope:

12 - 2 / 1 - (-4)

10 / 5 = 2

Write in point-slope form:

y - 12 = 2(x - 1)

rearrange:

y - 12 = 2x - 2

y = 2x + 10

6 0

3 years ago

In physics, if a moving object has a starting position at so, an initial velocity of vo, and a constant acceleration a, the

emmasim [6.3K]

Answer:

$a=\frac{2S -2v_ot-2s_o}{t^2}$

Step-by-step explanation:

We have the equation of the position of the object

$S = \frac{1}{2}at ^2 + v_ot+s_o$

We need to solve the equation for the variable a

$S = \frac{1}{2}at ^2 + v_ot+s_o$

Subtract $s_0$ and $v_0t$ on both sides of the equality

$S -v_ot-s_o = \frac{1}{2}at ^2 + v_ot+s_o - v_ot- s_o$

$S -v_ot-s_o = \frac{1}{2}at ^2$

multiply by 2 on both sides of equality

$2S -2v_ot-2s_o = 2*\frac{1}{2}at ^2$

$2S -2v_ot-2s_o =at ^2$

Divide between $t ^ 2$ on both sides of the equation

$\frac{2S -2v_ot-2s_o}{t^2} =a\frac{t^2}{t^2}$

Finally

$a=\frac{2S -2v_ot-2s_o}{t^2}$

5 0

3 years ago

2+4=what is the answer to 2 + 4

Nitella [24]

I don’t know if this is a joke or not but the answer is 6

6 0

3 years ago