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andrey2020 [161]
1 year ago
14

Solve this equation -4(X-26)=-200

Mathematics
2 answers:
Inessa05 [86]1 year ago
7 0
-4(x - 26) = -200
-4x = -174
x = 43.5
jek_recluse [69]1 year ago
5 0

Answer:

x = 76

Step-by-step explanation:

First, distribute -4:

-4x + 104 = -200

Isolate x:

-4x = -200 - 104 = -304

x = -304/-4 = 76

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What is the change in each termof the sequence? 5, -2, -9, -16, -23​
nordsb [41]

Answer:

-7

Step-by-step explanation:

To find the common difference, take the second term and subtract the first term

-2 - 5 = -7

Check by subtracting the second term from the third term

-9 - (-2) = -9+2 = -7

The common difference is -7

7 0
3 years ago
Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of
tresset_1 [31]

Because I've gone ahead with trying to parameterize S directly and learned the hard way that the resulting integral is large and annoying to work with, I'll propose a less direct approach.

Rather than compute the surface integral over S straight away, let's close off the hemisphere with the disk D of radius 9 centered at the origin and coincident with the plane y=0. Then by the divergence theorem, since the region S\cup D is closed, we have

\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iiint_R(\nabla\cdot\vec F)\,\mathrm dV

where R is the interior of S\cup D. \vec F has divergence

\nabla\cdot\vec F(x,y,z)=\dfrac{\partial(xz)}{\partial x}+\dfrac{\partial(x)}{\partial y}+\dfrac{\partial(y)}{\partial z}=z

so the flux over the closed region is

\displaystyle\iiint_Rz\,\mathrm dV=\int_0^\pi\int_0^\pi\int_0^9\rho^3\cos\varphi\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=0

The total flux over the closed surface is equal to the flux over its component surfaces, so we have

\displaystyle\iint_{S\cup D}\vec F\cdot\mathrm d\vec S=\iint_S\vec F\cdot\mathrm d\vec S+\iint_D\vec F\cdot\mathrm d\vec S=0

\implies\boxed{\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=-\iint_D\vec F\cdot\mathrm d\vec S}

Parameterize D by

\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec k

with 0\le u\le9 and 0\le v\le2\pi. Take the normal vector to D to be

\vec s_u\times\vec s_v=-u\,\vec\jmath

Then the flux of \vec F across S is

\displaystyle\iint_D\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^9\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^{2\pi}\int_0^9(u^2\cos v\sin v\,\vec\imath+u\cos v\,\vec\jmath)\cdot(-u\,\vec\jmath)\,\mathrm du\,\mathrm dv

=\displaystyle-\int_0^{2\pi}\int_0^9u^2\cos v\,\mathrm du\,\mathrm dv=0

\implies\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\boxed{0}

8 0
3 years ago
I just need help on this.<br>​
Arturiano [62]

Answer:

.42 is an answer that will work.

4 0
2 years ago
Read 2 more answers
Use differentials to estimate the amount of paint needed (in m3) to apply a coat of paint 0.04 cm thick to a hemispherical dome
pogonyaev

The amount of paint is the volume of paint needed.

The amount of paint needed is 0.065312 cubic meters

<h3>How to determine the amount of paint needed</h3>

The volume of a hemisphere is:

V = \frac 23\pi r^3

Differentiate the above equation

V' = 2\pi r^2 r'

The above equation represents the estimate of the amount of paint needed.

Where:

  • r represents the radius (r = 52/2 m)
  • r' represents the thickness (r' = 0.04 cm)

So, we have:

V' = 2\pi r^2 r'

V' = 2 * 3.14 * (52/2\ m)^2 * (0.04cm)

V' = 2 * 3.14 * (26\ m)^2 * (0.04cm)

Express cm as m

V' = 2 * 3.14 * (26\ m)^2 * (0.0004m)

V' = 0.065312 m^3

Hence, the amount of paint needed is 0.065312 cubic meters

Read more about volumes at:

brainly.com/question/10171109

3 0
2 years ago
2(n+5)= -2 I need the full explination
lakkis [162]

Answer:

n=-6

Step-by-step explanation:

2n + 10 = -2

2n=-12

n=-6

Hope this helps!

If not I'm sorry.

3 0
2 years ago
Read 2 more answers
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