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1 year ago

Suppose y varies directly with x where y=21 when x=7 find y when x = 15

Mathematics

1 answer:

Romashka [77]1 year ago

3 0

Golfkrieg ist die ganze Woche im Auto oder ist das 56357

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Which expression has the same value as the one below?

marta [7]

Answer:

answer is B 38-18

Step-by-step explanation:

38 + (-18)

38-18

5 0

3 years ago

Read 2 more answers

If y has moment-generating function m(t) = e 6(e t −1) , what is p(|y − µ| ≤ 2σ)?

Lostsunrise [7]

Since $\mathrm M_Y(t)=e^{6(e^t-1}$ , we know that $Y$ follows a Poisson distribution with parameter $\lambda=6$ .

Now assuming $\mu,\sigma$ denote the mean and standard deviation of $Y$ , respectively, then we know right away that $\mu=6$ and $\sigma=\sqrt6$ .

So,

$\mathbb P(|Y-\mu|\le2\sigma)=\mathbb P(6-2\sqrt6\le Y\le6+2\sqrt6)=\dfrac{66366}{175e^6}\approx0.940028$

6 0

3 years ago

A show uses 12 gallons of water in three minutes complete the table and the graph.

gregori [183]

Answer:

2 mins= 8 gallons 3 mins= 12 gallons 4 mins= 16 gallons 5 mins= 20 gallons 6 mins= 24 gallons

Step-by-step explanation:

12 divided by 3 equals 4 so just multiply 4 by each minute

5 0

2 years ago

Which of the following equations is the result after the first step in solving 3x + 6 = 12?

tensa zangetsu [6.8K]

3x=18 because you add the like terms 6 and 12

7 0

3 years ago

What's the flux of the vector field F(x,y,z) = (e^-y) i - (y) j + (x sinz) k across σ with outward orientation where σ is the po

emmasim [6.3K]

$\displaystyle\iint_\sigma\mathbf F\cdot\mathrm dS$
$\displaystyle\iint_\sigma\mathbf F\cdot\mathbf n\,\mathrm dS$
$\displaystyle\iint_\sigma\mathbf F\cdot\left(\frac{\mathbf r_u\times\mathbf r_v}{\|\mathbf r_u\times\mathbf r_v\|}\right)\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dA$
$\displaystyle\iint_\sigma\mathbf F\cdot(\mathbf r_u\times\mathbf r_v)\,\mathrm dA$

Since you want to find flux in the outward direction, you need to make sure that the normal vector points that way. You have

$\mathbf r_u=\dfrac\partial{\partial u}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=\mathbf k$
$\mathbf r_v=\dfrac\partial{\partial v}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=-2\sin v\,\mathbf i+\cos v\,\mathbf j$

The cross product is

$\mathbf r_u\times\mathbf r_v=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\0&0&1\\-2\sin v&\cos v&0\end{vmatrix}=-\cos v\,\mathbf i-2\sin v\,\mathbf j$

So, the flux is given by

$\displaystyle\iint_\sigma(e^{-\sin v}\,\mathbf i-\sin v\,\mathbf j+2\cos v\sin u\,\mathbf k)\cdot(\cos v\,\mathbf i+2\sin v\,\mathbf j)\,\mathrm dA$
$\displaystyle\int_0^5\int_0^{2\pi}(-e^{-\sin v}\cos v+2\sin^2v)\,\mathrm dv\,\mathrm du$
$\displaystyle-5\int_0^{2\pi}e^{-\sin v}\cos v\,\mathrm dv+10\int_0^{2\pi}\sin^2v\,\mathrm dv$
$\displaystyle5\int_0^0e^t\,\mathrm dt+5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv$

where $t=-\sin v$ in the first integral, and the half-angle identity is used in the second. The first integral vanishes, leaving you with

$\displaystyle5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv=5\left(v-\dfrac12\sin2v\right)\bigg|_{v=0}^{v=2\pi}=10\pi$

5 0

3 years ago