All categories

4 years ago

I'm lost I need help

Mathematics

1 answer:

dangina [55]4 years ago

3 0

So set both equations equal to each other then solve out of x and plug it back into the x equation to find the angles so start with x+5=2x+10 then subtract x from 2x and you will have 1x then subtract 10 from 5 leaving you with negative 5 then x equals negative 5 and then plug it back into the equation so for example 2 times negative 5 equals negative 10 plus positive 10 equals 0 so therefore x equals 0

Read more about circle equation at:

brainly.com/question/1559324

#SPJ1

3 0

2 years ago

Draw 2 squares. The first one should have an area of 12 square units.The second one should have an area that is DOUBLE that of t

statuscvo [17]

Step-by-step explanation:

Given the information:

1st square: 12 square units
2nd square: DOUBLE that of the first square = 2*12 = 24 square units

From that, we can determine the length of the side in each square:

1/ $x^{2} = 12$

<=> x = 2 $\sqrt{3}$

2/ $a^{2} = 24$

<=> a = 2 $\sqrt{6}$

Please have a look at the attached photo.

Hope it will find you well.

6 0

3 years ago

Evaluate the surface integral:S

rjkz [21]

Assuming $S$ does not include the plane $z=0$ , we can parameterize the region in spherical coordinates using

$\mathbf r(u,v)=\left\langle3\cos u\sin v,3\sin u\sin v,3\cos v\right\rangle$

where $0\le u\le2\pi$ and $0\le v\le\dfrac\pi/2$ . We then have

$x^2+y^2=9\cos^2u\sin^2v+9\sin^2u\sin^2v=9\sin^2v$
$(x^2+y^2)=9\sin^2v(3\cos v)=27\sin^2v\cos v$

Then the surface integral is equivalent to

$\displaystyle\iint_S(x^2+y^2)z\,\mathrm dS=27\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^2v\cos v\left\|\frac{\partial\mathbf r(u,v)}{\partial u}\times \frac{\partial\mathbf r(u,v)}{\partial u}\right\|\,\mathrm dv\,\mathrm du$

We have

$\dfrac{\partial\mathbf r(u,v)}{\partial u}=\langle-3\sin u\sin v,3\cos u\sin v,0\rangle$
$\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle3\cos u\cos v,3\sin u\cos v,-3\sin v\rangle$
$\implies\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle-9\cos u\sin^2v,-9\sin u\sin^2v,-9\cos v\sin v\rangle$
$\implies\left\|\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}\|=9\sin v$

So the surface integral is equivalent to

$\displaystyle243\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv\,\mathrm du$
$=\displaystyle486\pi\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv$
$=\displaystyle486\pi\int_{w=0}^{w=1}w^3\,\mathrm dw$

where $w=\sin v\implies\mathrm dw=\cos v\,\mathrm dv$ .

$=\dfrac{243}2\pi w^4\bigg|_{w=0}^{w=1}$
$=\dfrac{243}2\pi$

4 0

3 years ago

For a school play, the teacher asked the class to set up chairs in 20 rows with 25 chairs in each row. After setting up all the

Sholpan [36]

They set up 495 chairs.

I hope this helps! :)

7 0

3 years ago

Read 2 more answers

Sketch the smallest positive such angle. As well as find the six trig functions.

Licemer1 [7]

X-246-c4779952vhjjj23467890865

5 0

3 years ago