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3 years ago

And can you please show your answer thx

Mathematics

1 answer:

MrMuchimi3 years ago

5 0

Answer:

y=4x is the answer

Step-by-step explanation:

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lavatory tests show that the lives of light bulbs are normally distributed with a mean of 750 hours and a stoandard devotion of

maria [59]

Answer:h

Step-by-step explanation:6

5 0

3 years ago

Starting at noon, the temperature dropped steadily at a rate of 0.08 degrees Celsius every hour.

dangina [55]

Answer:

a. 55 hours

b. 1.9 degrees C

Step-by-step explanation:

This means for each hour the temperature drops 0.08. This is the expression 0.08x where x is number of hours.

a.) If the temperature has dropped 4.4, then 4.4 = 0.08x. Solving for x will give the number of hours it takes to drop 4.4.

4.4 = 0.08x

4.4 / 0.08 = x

55 = x

b.) If the temperature after 55 hours and dropping 4.4 is -2.5, then the temperature at the start would be -2.5 + 4.4 = 1.9. The temperature at noon was 1.9 degrees celsius.

6 0

3 years ago

Factorise fully 9x squared - 6xy

prisoha [69]

3x(3x-2y)

you have to find a common factor between the two numbers. in this case it's three, so you just divide each number by three.

4 0

3 years ago

Please tell me if I was right.

Gennadij [26K]

Im pretty sure you are i have done this question before and im pretty sure thats what i put and i got it right. plus just go with your gut

8 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