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castortr0y [4]
3 years ago
11

Use the table of values to find the function's values. Ja) Ifx 0, then f0) = If fe) = 27, then x= -3 33 -2 17 0 -15 2 27​

Mathematics
1 answer:
Paha777 [63]3 years ago
8 0

Answer:

-15 and 3

Step-by-step explanation:

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8:15 am is the answer because if u add up all of the timings it will become an hour and a half
And when u minus an hour and a half from 10:45 u get 8:15
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Jerry has a credit card debt of $15,600 that he would like to reduce by applying $8500 of his inheritance money to the balance
aev [14]

Answer:

Jerry would still be in debt by 7100

Step-by-step explanation:

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6 0
2 years ago
Please help me with this question
Alenkinab [10]

Answer:

A

Step-by-step explanation:

Remark

f and d are equal (and acute) because they are corresponding angles.

82 and f are supplementary, so we can find f

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f + 82 = 180

f = 180 - 82

f = 82

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5 0
2 years ago
Write the expression as a single natural logarithm.3In3+3Inc
k0ka [10]

Answer:

ln(27c^3)

Step-by-step explanation:

Given: 3ln(3)+3ln(c)

If there is a coefficient in front of a log or ln, that means it becomes an exponent.

ln(3^3)+ln(c^3)

Simplify the exponent.

ln(27)+ln(c^3)

When there is addition between two logarithms or natural logarithms, it means they multiply together.

ln(27c^3)

3 0
3 years ago
Read 2 more answers
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
Sophie [7]
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

And since we have

\nabla\cdot\mathbf f(x,y,z)=1

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
=\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}\int_{z=2}^{z=3-r^2}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta
=\displaystyle2\pi\int_{r=0}^{r=1}r(3-r^2-2)\,\mathrm dr
=\dfrac\pi2

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
\implies\mathbf s_u\times\mathbf s_v=u\,\mathbf k

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
=\displaystyle-4\pi\int_{u=0}^{u=1}u\,\mathrm du
=-2\pi

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
6 0
3 years ago
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