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3 years ago

A1+a2+a3=180 solve for a1

2 answers:

8 0

A1 = 180 - (a2 + a3) ;
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or:
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A1 = 180 - a2 - a3.
_______________________.
Since there are no values given, we solve for "A1" by isolating "A1" and solving in terms of the remaining values.

Leviafan [203]3 years ago

6 0

Answer: a1=180-a2-a3....

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Two number cubes each have sides that are labeled 1 to 6. Isis rolls the 2 number cubes. What is the probability that the sum of

AURORKA [14]

Answer:

1/12 or 0.0833...

Step-by-step explanation:

There are 36 different possible outcomes for rolling the number cubes. To get a sum of 4 you would need to roll a 1 and 3 (2 possibilities, each cube could roll a 1 or a 3) or a 2 and a 2 (1 possibility, each has to roll the same number). Since there is no 0 or negative numbers, these are your only possibilities. 3 out of 36 possibilities is also equivalent to a 1/12 chance.

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2 years ago

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 F) - (integral over D) = integral over S

==> 486π/5 - (-81π/4) = 2349π/20

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3 years ago

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The answer is 90 miles per hour.

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Answer:

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Step-by-step explanation:

First I just drew out two squares, one was the first un-cut square and the second is the cut one. Since we know that the area of the smaller square is 144, and Area is just side^2 , we know that all the sides of the smaller square are 12. Now, all we have to do is add 8 to 12 to get the original lawn. The original lawn was a 20 x 20 ft lawn.

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2 years ago

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Answer:

Step-by-step explanation:

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