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3 years ago

Which graph shows the line y=-3x+1? graph a graph b

Mathematics

1 answer:

SCORPION-xisa [38]3 years ago

8 0

Answer:
This is what your graph would look like...

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Can someone pleaseeee help and if you’re correct i’ll give brainliest

Ilya [14]

Answer:

πr²= 12.56

r²=12.56/3.14 = 4

r= 2

therefore , diameter = 2r = 2×2= 4 mm

5 0

3 years ago

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23. If 25% of a number is 75. find the number. A. 700 B. 600 C. 500 D, 400 E. 300

nadya68 [22]

Answer:

23.E

Step-by-step explanation:

please click the heart and rate excellent and brainleist to ☻❤☺️☺️❤☻

4 0

3 years ago

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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

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

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3 years ago

What is an equation of the line that passes through the points (6, 4) and (4,1)?

netineya [11]

Answer:

3/2 x-5

Step-by-step explanation:

7 0

3 years ago

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Jennifer scores a 77 on her algebra 2 test. The mean on the test was 73. The standard deviation on the test was 5. What was her

MA_775_DIABLO [31]

Answer:

0.80

Step-by-step explanation:

The z-score is found from the formula ...

Z = (X -μ)/σ

Z = (77 -73)/5 = 4/5 = 0.80

Jennifer's Z-score was 0.80.

7 0

4 years ago