All categories

3 years ago

30 POINTS!!!!

Mathematics

2 answers:

sweet-ann [11.9K]3 years ago

5 0

Answer:

the third option ( 0,4)

Step-by-step explanation:

Simply use the equation

(x1 + x2 ) / 2 and (y1+y2) /2

label the coordinates that join the line GH as x1, y1 , x2 and y2 respectively.

x1 = -3 y1 = 3

x2 = 3 y2 = 5

(-3 + 3) /2 and (3+5) /2

0 and 4

(0,4)

hope this answers your question

AnnZ [28]3 years ago

3 0

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ G(\stackrel{x_1}{-3}~,~\stackrel{y_1}{3})\qquad H(\stackrel{x_2}{3}~,~\stackrel{y_2}{5}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{3-3}{2}~~,~~\cfrac{5+3}{2} \right)\implies \left(\cfrac{0}{2}~,~\cfrac{8}{2} \right)\implies (0~,~4)$

You might be interested in

HELP WILL GIVE BRAINLIEST!!!!

iris [78.8K]

Answer:

2 x 20^2

Step-by-step explanation:

I did it on paper and looked it up to make sure

5 0

3 years ago

Read 2 more answers

Khalid and Jesse took different overnight trains. The graph shows the relationship between the total distance Khalid traveled an

Usimov [2.4K]

Answer:

Jesse - 150 miles per hour.

Step-by-step explanation:

On the graph, it shows that the rise is 140 and the run is 1. 140/1 is 140, making the slope for Khalid is 140. The equation (y = 150x) shows that the slope is 150 (y = mx). Since 150 is greater than (>) 140, Jesse's train is more faster than Khalid since she is moving at 150mph rather than 140mph.

Also, is this for Pearson Realize?

4 0

4 years ago

Mathematics achievement test scores for 300 students had a mean and a variance equal to 550 and 3,600, respectively. If the dist

blondinia [14]

Answer:

number of scores between 490 and 610 is = 0.68 x 300 = 204

Step-by-step explanation:

The concept of Probability distribution is applied as it relates to the Z-scores.

Z = X - Mean /Standard deviation

The detailed steps and appropriate substitution is as shown in the attached file.

6 0

3 years ago

Use stoke's theorem to evaluate∬m(∇×f)⋅ds where m is the hemisphere x^2+y^2+z^2=9, x≥0, with the normal in the direction of the

ludmilkaskok [199]

By Stokes' theorem,

$\displaystyle\int_{\partial\mathcal M}\mathbf f\cdot\mathrm d\mathbf r=\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S$

where $\mathcal C$ is the circular boundary of the hemisphere $\mathcal M$ in the $y$ - $z$ plane. We can parameterize the boundary via the "standard" choice of polar coordinates, setting

$\mathbf r(t)=\langle 0,3\cos t,3\sin t\rangle$

where $0\le t\le2\pi$ . Then the line integral is

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=0}^{t=2\pi}\mathbf f(x(t),y(t),z(t))\cdot\dfrac{\mathrm d}{\mathrm dt}\langle x(t),y(t),z(t)\rangle\,\mathrm dt$
$=\displaystyle\int_0^{2\pi}\langle0,0,3\cos t\rangle\cdot\langle0,-3\sin t,3\cos t\rangle\,\mathrm dt=9\int_0^{2\pi}\cos^2t\,\mathrm dt=9\pi$

We can check this result by evaluating the equivalent surface integral. We have

$\nabla\times\mathbf f=\langle1,0,0\rangle$

and we can parameterize $\mathcal M$ by

$\mathbf s(u,v)=\langle3\cos v,3\cos u\sin v,3\sin u\sin v\rangle$

so that

$\mathrm d\mathbf S=(\mathbf s_v\times\mathbf s_u)\,\mathrm du\,\mathrm dv=\langle9\cos v\sin v,9\cos u\sin^2v,9\sin u\sin^2v\rangle\,\mathrm du\,\mathrm dv$

where $0\le v\le\dfrac\pi2$ and $0\le u\le2\pi$ . Then,

$\displaystyle\iint_{\mathcal M}\nabla\times\mathbf f\cdot\mathrm d\mathbf S=\int_{v=0}^{v=\pi/2}\int_{u=0}^{u=2\pi}9\cos v\sin v\,\mathrm du\,\mathrm dv=9\pi$

as expected.

7 0

3 years ago

Linear regressions please help me i’m going to fail my class

lyudmila [28]

The equation has the form Y= a + bX,

Where:

Y is the dependent variable (that's the variable that goes on the Y axis).

X is the independent variable (i.e. it is plotted on the X axis).

b is the slope of the line.

a is the y-intercept.

3 0

3 years ago