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1 year ago

Which sets of ordered pairs represent a function? Select all that apply.

Mathematics

1 answer:

antoniya [11.8K]1 year ago

7 0

Answer:

B, C, and E

Step-by-step explanation:

In order for a function to be a function, an ordered pair cannot be on the same x coordinate

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PLEASE HELP!!!!!!!!!!!!!!!!!!!!!

Shalnov [3]

Answer:

D i think sorry if wrong

Step-by-step explanation:

8 0

2 years ago

Help with numbers 1/6 please

Maslowich

Answer:

1. A=56.25

Step-by-step explanation:

sry but i dont know the answer to six

7 0

2 years ago

Evaluate csc(-315°).

Maru [420]

Answer:

$\frac{2}{\sqrt{2} } = \sqrt{2}$

Step-by-step explanation:

Notice from the attached image that -315 degrees is the same as 45 degrees.

Recall as well that the csc function is defined as the reciprocal of the sine function: $csc(x)=\frac{1}{sin(x)}$

Then, you need to evaluate $\frac{1}{sin(45)} = 1/\frac{\sqrt{2} }{2} = \frac{2}{\sqrt{2} } =\sqrt{2}$

8 0

2 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

2 years ago

Consider the following regression line: T e s t s c o r e ^ = 698.9 − 2.28 S T R . You are told that the t-statistic on the slop

Amiraneli [1.4K]

Answer:

0.52

Step-by-step explanation:

Given the regression line :

Test score^ = 698.9 - 2.28STR

T statistic on slope Coefficient = 4.38

The standard error on slope Coefficient :

2.28 / 4.38

= 0.5205479

= 0.52

6 0

2 years ago