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3 years ago

At a tournament, the bowling league is providing 160 sandwiches for its members.

Mathematics

1 answer:

kiruha [24]3 years ago

6 0

Roast Beef: 32
Vegetarian:48
Hope this helps :)

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

Romashka-Z-Leto [24]

By the divergence theorem, the surface integral over $S_2$ is

$\displaystyle\iint_{S_2}\mathbf F\cdot\mathrm dS=\iiint_R\nabla\cdot\mathbf F\,\mathrm dR$

where $R$ denotes the space bounded by $S_2$ . Assuming the vector field is given to be

$\mathbf F(x,y,z)=z^2x\,\mathbf i+(y^3+\tan z)\,\mathbf j+(x^2z+y^2)\,\mathbf k$

then

$\nabla\cdot\mathbf F=\dfrac\partial{\partial x}[z^2x]+\dfrac\partial{\partial y}[y^3+\tan z]+\dfrac\partial{\partial z}[x^2z+y^2]=z^2+3y^2+x^2$

Converting to spherical coordinates, we take

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}$

so that the triple integral becomes

$\displaystyle\int_{\varphi=0}^{\varphi=\pi/2}\int_{\theta=0}^{\theta=2\pi}\int_{\rho=0}^{\rho=3}(\rho^2\cos^2\varphi+3\rho^2\sin^2\theta\sin^2\varphi+\rho^2\cos^2\theta\sin^2\varphi)\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$
$=\displaystyle\int_0^{\pi/2}\int_0^{2\pi}\int_0^3\rho^4(\cos^2\varphi+2\sin^2\theta\sin^2\varphi+\sin^2\varphi)\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$
$=162\pi$

Now the integral over $S$ alone will be the difference of the integral over $S_2$ and the integral over $S_1$ , i.e.

$\displaystyle\iint_{S_2}=\iint_{S_1\cup S}=\iint_{S_1}+\iint_S\implies\iint_S=\iint_{S_2}-\iint_{S_1}$

We can parameterize the points in $S_1$ by

$\mathbf s(r,\theta)=\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=0\end{cases}$

so that the integral over $S_1$ is

$\displaystyle\iint_{S_1}\mathbf F\cdot\mathrm dS=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=3}\mathbf F(x(r,\theta),y(r,\theta),z(r,\theta))\cdot\left(\dfrac{\partial\mathbf s}{\partial r}\times\dfrac{\partial\mathbf s}{\partial\theta}\right)\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle\int_0^{2\pi}\int_0^3(r^3\sin^3\theta\,\mathbf j+r^2\sin^2\theta\,\mathbf k)\cdot(r\,\mathbf k)\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle\int_0^{2\pi}\int_0^3r^3\sin^2\theta\,\mathrm dr\,\mathrm d\theta$
$=\dfrac{81\pi}4$

So, the integral over $S$ alone evaluates to

$\displaystyle\iint_S=\iint_{S_2}-\iint_{S_1}=162\pi-\dfrac{81\pi}4=\dfrac{567\pi}4$

3 0

3 years ago

Using ƒ(x) = 2x + 7 and g(x) = x − 3, find ƒ(g(-2)).

Eva8 [605]

Answer:

We conclude that:

f(g(-2)) = -3

Step-by-step explanation:

Given

f(x) = 2x + 7

g(x) = x-3

To determine

f(g(-2)) = ?

In order to determine f(g(-2)), first we need to determine g(-2)

so substituting x = -2 in g(x) = x-3

g(-2) = -2 - 3

= -5

Thus,

f(g(-2)) = f(-5)

now substitute x = -5 in the function f(x) = 2x+7

f(x) = 2x+7

f(-5) = 2(-5) + 7

f(-5) = -10 + 7

f(-5) = -3

f(g(-2)) = f(-5) = -3

Therefore, we conclude that:

f(g(-2)) = -3

4 0

3 years ago

Find the sum of the first 12 terms of the sequence. Show all work for full credit.

xeze [42]

Answer:

The sum of 12 terms of AP = -318

Step-by-step explanation:

Points to remember

Sum of n terms of an AP

Sₙ = n/2[ 2a + (n - 1)d]

Where n - number of terms

a - first term

d - common difference

<u>To find the sum of 12 terms</u>

The given AP is 1, -4, -9, -14, . . .

a = 1, d= -5 and n = 12

Sₙ = n/2[ 2a + (n - 1)d]

= 12/2[2*1 + (12 - 1)*(-5)]

= 6[2 + (11*(-5))]

= 6[2 - 55]

= 6 * (-53)

= -318

The sum of 12 terms of AP = -318

6 0

3 years ago

Evaluate the following expression. 7^0

gregori [183]

Hey there!

The correct answer would be 1.
Any number value that is raised to the power of 0 would equal 1.

Hope this helps and have a nice day! :)

8 0

3 years ago

Learning Task 2. Which of the following represents a quadratic function? 1. f(x) = 8x + 5 2. f(x) = x2 - 2x + 7 х -5 4 3 -2 . 3.

Aloiza [94]

A quadratic function is a function which had the largest power of the equation 2.

7 0

1 year ago