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2 years ago

4. Find the value of x. A) 6 √ 2 B) 12 √ 3 C) 24 √ 3 D) 12

Mathematics

2 answers:

Schach [20]2 years ago

7 0

A^2+b^2=c^2
24^2=12^2+x^2
sqrt(576-144)=x
sqrt(432)=x
12sqrt(3)=x
the answer is c

gregori [183]2 years ago

4 0

24\3...........................

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2 years ago

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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)

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

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

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

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###

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.

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3 years ago

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Step-by-step explanation:

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2 years ago

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scoray [572]

Answer:

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Step-by-step explanation:

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2 years ago

A circle with radius 3 has a sector with a central angle of 60°.

victus00 [196]

Answer:

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Step-by-step explanation:

The radius of the circle is 3 and the angle of the sector is 60°.

The area of a sector is given as:

$A = \frac{\alpha }{360} * \pi r^2$

where α = central angle of sector

r = radius of circle

Hence, the area of the sector is:

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The area of the sector is 4.71

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3 years ago