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Digiron [165]
3 years ago
5

Please help! asap! thank you so much! I have photos attached!

Mathematics
1 answer:
schepotkina [342]3 years ago
4 0

Answer:


Step-by-step explanation:

Given problems are absolute value problems, So we need to plug the values of given parameters and get the final result.

We have given here,

a =-2 , b = 3 ,  c = -4 and d = -6

Now we know that An absolute function always gives a positive value.

Let's apply this strategy in the given problems.

1. ║a+b║

Plug a= -2 and b = 3

We get, ║-2+3║=║1║= 1

2. 5║c+b║

Plug c= -4 and b=3

i.e.   5║-4 + 3║= 5║-1║=5×1 = 5

3. a+b║c║

Plug values a= -2 , b=3 and c=-4

i.e -2 +3║-4║ = -2 + 3×4 = -2 + 12 = 10

4. ║a+c║÷(-d)

i.e ║-2 + (-4)║÷(-6) = ║-6║÷(-6) = 6÷(-6) = -1

5. 3║a+d║+b

i.e 3║-2+(-6)║+3 = 3║-8║+3 = 3×8 +3 = 27

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Answer:

x=7 is the line parallel to the given line x=8 . Parallel lines are the lines which are always same distance apart and never intersect. The given line is x=8 with intersects the x axis at x=8 and it is parallel to y axis with slope infinite or not defined.

Step-by-step explanation:

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3 years ago
Let f(x,y,z) = ztan-1(y2) i + z3ln(x2 + 1) j + z k. find the flux of f across the part of the paraboloid x2 + y2 + z = 3 that li
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Consider the closed region V bounded simultaneously by the paraboloid and plane, jointly denoted S. By the divergence theorem,

\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm dS=\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV

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the volume integral will be much easier to compute. Converting to cylindrical coordinates, we have

\displaystyle\iiint_V\nabla\cdot\mathbf f(x,y,z)\,\mathrm dV=\iiint_V\mathrm dV
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Then the integral over the paraboloid would be the difference of the integral over the total surface and the integral over the disk. Denoting the disk by D, we have

\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-\iint_D\mathbf f\cdot\mathrm dS

Parameterize D by

\mathbf s(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+2\,\mathbf k
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which would give a unit normal vector of \mathbf k. However, the divergence theorem requires that the closed surface S be oriented with outward-pointing normal vectors, which means we should instead use \mathbf s_v\times\mathbf s_u=-u\,\mathbf k.

Now,

\displaystyle\iint_D\mathbf f\cdot\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=2\pi}\mathbf f(x(u,v),y(u,v),z(u,v))\cdot(-u\,\mathbf k)\,\mathrm dv\,\mathrm du
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So, the flux over the paraboloid alone is

\displaystyle\iint_{S-D}\mathbf f\cdot\mathrm dS=\frac\pi2-(-2\pi)=\dfrac{5\pi}2
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divide them:

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