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Tamiku [17]
2 years ago
11

Multiply:

rac{8n^{2} }{m^{2}-16 }" alt="\frac{8n^{2} }{m^{2}-16 }" align="absmiddle" class="latex-formula">×\frac{m^{2}-4m }{6n}
Mathematics
1 answer:
Mkey [24]2 years ago
3 0

9514 1404 393

Answer:

  4mn/(3m+12)

Step-by-step explanation:

It is often helpful to factor expressions so that common factors can cancel.

  \dfrac{8n^2}{m^2-16}\times\dfrac{m^2-4m}{6n}=\dfrac{8mn^2(m-4)}{6n(m-4)(m+4)}=\dfrac{4mn}{3(m+4)}\\\\=\boxed{\dfrac{4mn}{3m+12}}

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I need help again :/
liq [111]

Answer:

They have a 0 in the ones place

Step-by-step explanation:

Because you start with 10, you will always have a 0 at the end.

30, 90, 270, etc.

7 0
3 years ago
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How do I write 5/3x+y=8 in standard form
aivan3 [116]

Answer:

5x+3y+24

Step-by-step explanation:

5 0
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Need help please. will give brainliest. what is 3x+y
yaroslaw [1]

Answer:

3x+y

Step-by-step explanation:

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8 0
2 years ago
I got zero, but that’s wrong.
meriva

Answer:

Expected value is 4,000

Step-by-step explanation:

To find expected value ⇒ multiply the value by it's probability

40% × ( -25,000 ) = - 10,000

Breaking means neither add nor subtract a given amount

20% × 0 = 0

35% × 40,000 = 14,000

∴ Expected value = -10,000 + 14,000 = 4,000

LOSS:                   -10,000

BREAK EVEN:       0

WIN:                       14,000

Expected value is 4,000

<em>hope this helps....</em>

3 0
2 years ago
Calculate s f(x, y, z) ds for the given surface and function. g(r, θ) = (r cos θ, r sin θ, θ), 0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π; f(x, y, z)
Triss [41]

g(r,\theta)=(r\cos\theta,r\sin\theta,\theta)\implies\begin{cases}g_r=(\cos\theta,\sin\theta,0)\\g_\theta=(-r\sin\theta,r\cos\theta,1)\end{cases}

The surface element is

\mathrm dS=\|g_r\times g_\theta\|\,\mathrm dr\,\mathrm d\theta=\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta

and the integral is

\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\int_0^{2\pi}\int_0^4((r\cos\theta)^2+(r\sin\theta)^2)\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta

=\displaystyle2\pi\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac\pi4(132\sqrt{17}-\sinh^{-1}4)

###

To compute the last integral, you can integrate by parts:

u=r\implies\mathrm du=\mathrm dr

\mathrm dv=r\sqrt{1+r^2}\,\mathrm dr\implies v=\dfrac13(1+r^2)^{3/2}

\displaystyle\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac r3(1+r^2)^{3/2}\bigg|_0^4-\frac13\int_0^4(1+r^2)^{3/2}\,\mathrm dr

For this integral, consider a substitution of

r=\sinh s\implies\mathrm dr=\cosh s\,\mathrm ds

\displaystyle\int_0^4(1+r^2)^{3/2}\,\mathrm dr=\int_0^{\sinh^{-1}4}(1+\sinh^2s)^{3/2}\cosh s\,\mathrm ds

\displaystyle=\int_0^{\sinh^{-1}4}\cosh^4s\,\mathrm ds

=\displaystyle\frac18\int_0^{\sinh^{-1}4}(3+4\cosh2s+\cosh4s)\,\mathrm ds

and the result above follows.

4 0
3 years ago
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