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Inessa [10]
3 years ago
5

A science test, which is worth 100 points, consists of 24 questions. Each question is worth either 3 points or 5 points. If xis

Mathematics
1 answer:
Luba_88 [7]3 years ago
7 0

x+y=24

3x + 5y = 100

Rewrite the first equation: x = 24-y

Replace x in the second equation and solve:

3(24-y) +5y = 100

72-3y +5y = 100

72 +2y = 100

2y = 28

y = 14

Y is the number of 5 point questions, so there are 14 5 point questions.

The answer would be:

The test contains 10 three-point questions and 14 five-point questions.

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Well it was a lucky guess but the answer is 75

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Use the Divergence Theorem to evaluate S F · dS, where F(x, y, z) = z2xi + y3 3 + sin z j + (x2z + y2)k and S is the top half of
GenaCL600 [577]

Close off the hemisphere S by attaching to it the disk D of radius 3 centered at the origin in the plane z=0. By the divergence theorem, we have

\displaystyle\iint_{S\cup D}\vec F(x,y,z)\cdot\mathrm d\vec S=\iiint_R\mathrm{div}\vec F(x,y,z)\,\mathrm dV

where R is the interior of the joined surfaces S\cup D.

Compute the divergence of \vec F:

\mathrm{div}\vec F(x,y,z)=\dfrac{\partial(xz^2)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial k}=z^2+y^2+x^2

Compute the integral of the divergence over R. Easily done by converting to cylindrical or spherical coordinates. I'll do the latter:

\begin{cases}x(\rho,\theta,\varphi)=\rho\cos\theta\sin\varphi\\y(\rho,\theta,\varphi)=\rho\sin\theta\sin\varphi\\z(\rho,\theta,\varphi)=\rho\cos\varphi\end{cases}\implies\begin{cases}x^2+y^2+z^2=\rho^2\\\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi\end{cases}

So the volume integral is

\displaystyle\iiint_Rx^2+y^2+z^2\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^3\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{486\pi}5

From this we need to subtract the contribution of

\displaystyle\iint_D\vec F(x,y,z)\cdot\mathrm d\vec S

that is, the integral of \vec F over the disk, oriented downward. Since z=0 in D, we have

\vec F(x,y,0)=\dfrac{y^3}3\,\vec\jmath+y^2\,\vec k

Parameterize D by

\vec r(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath

where 0\le u\le 3 and 0\le v\le2\pi. Take the normal vector to be

\dfrac{\partial\vec r}{\partial v}\times\dfrac{\partial\vec r}{\partial u}=-u\,\vec k

Then taking the dot product of \vec F with the normal vector gives

\vec F(x(u,v),y(u,v),0)\cdot(-u\,\vec k)=-y(u,v)^2u=-u^3\sin^2v

So the contribution of integrating \vec F over D is

\displaystyle\int_0^{2\pi}\int_0^3-u^3\sin^2v\,\mathrm du\,\mathrm dv=-\frac{81\pi}4

and the value of the integral we want is

(integral of divergence of <em>F</em>) - (integral over <em>D</em>) = integral over <em>S</em>

==>  486π/5 - (-81π/4) = 2349π/20

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A 39-foot piece of string (hypotenuse) is stretched from the top of a 18-foot flagpole to the ground. How far is the base of the
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Step-by-step explanation:

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The distance from the base of the pole to the end of the string is about 34.6 feet.

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