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3 years ago

How do I performe the operation and simplify?

Mathematics

1 answer:

klasskru [66]3 years ago

4 0

Remember that : aⁿ.aˣ = aⁿ⁺ˣ ( Example 5³.5⁶ = 5³⁺⁶ = 5⁹)

Apply the same logic for:
5⁴/₃ x 5⁸/₃

Add up the exponents: 4/3 + 8/3 = 12/3 = 4, then
5⁴/₃ x 5⁸/₃ = 5⁴

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5/6k+2/3=4/3 What is K? I don't understand how to do these

Drupady [299]

First, subtract 2/3 from both sides and you get $\frac{5}{6} k = \frac{2}{3}$
Then multiply both sides by 6/5 and you get $k = \frac{4}{5}$

8 0

3 years ago

Miranda owes max $120.00. She pays back 30% of what she owes. How much does she have left to pay back?

Mamont248 [21]

Answer:

I’m pretty sure the answer is $84

Step-by-step explanation:

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3 years ago

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Need help with all of these!!! ASAP!! 15, 16, 17, 18!!!

vodomira [7]

Answer:

15. A

16. B

17. B

18. C

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

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

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

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4 years ago

$14.00 for 7 minutes Unit rate

babunello [35]

Answer:

$2/min

Step-by-step explanation:

14 ÷ 7 = 2

Therefore, $2 per minute.

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3 years ago

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