Answer this question <br>PLEASE ANSWER FAST!!

I WILL GOVE YOU BRAINLIEST IF CORRECT YOU HAVE TO EXPLAIN WHY IT IS THO

Assoli18 [71]

Answer:

G

Step-by-step explanation:

If you do a common denominator of 30 for all of them

F would be 10/30

g would be 8/30

h would be 24/30

and j is obviously 14/30

3 0

3 years ago

Read 2 more answers

Help me and i’ll mark you brainliest plus you get 20 points !!!!

Anton [14]

Answer:

Angle bisector

Step-by-step explanation:

Find the measure of the angle COA. By angle addition postulate,

$m\angle COX=m\angle AOX+m\angle COA$

From the diagram,

$m\angle COX=80^{\circ}\\ \\m\angle AOX=40^{\circ},$

then

$80^{\circ}=m\angle COA+40^{\circ}\\ \\m\angle COA=80^{\circ}-40^{\circ}=40^{\circ}$

Find the measure of the angle BOA. By angle addition postulate,

$m\angle BOX=m\angle AOX+m\angle BOA$

From the diagram,

$m\angle BOX=60^{\circ}\\ \\m\angle AOX=40^{\circ},$

then

$60^{\circ}=m\angle BOA+40^{\circ}\\ \\m\angle BOA=60^{\circ}-40^{\circ}=20^{\circ}$

Find the measure of the angle COB. By angle addition postulate,

$m\angle COX=m\angle BOX+m\angle COB$

From the diagram,

$m\angle BOX=60^{\circ}\\ \\m\angle COX=80^{\circ},$

then

$80^{\circ}=m\angle COB+60^{\circ}\\ \\m\angle COB=80^{\circ}-60^{\circ}=20^{\circ}$

This means, the measures of angles COB and BOA are the same and are equal half the measure of angle COA, so angles COB and BOA are congruent. This means, the ray OB is the angle bisector of angle COA

6 0

3 years ago

Use Stokes' Theorem to evaluate C F · dr where C is oriented counterclockwise as viewed from above. F(x, y, z) = 2yi + xzj + (x

Lady_Fox [76]

By Stokes' theorem, the line integral of $\vec F$ over $C$ is given by the surface integral of the curl of $\vec F$ over $S$ , where $S$ is the region of intersection of the plane $z=y+6$ and the cylinder $x^2+y^2=1$ with $S$ having positive/upward orientation.

Parameterize $S$ by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath+(u\sin v+6)\,\vec k$

with $0\le u\le1$ and $0\le v\le2\pi$ .

Take the normal vector to $S$ to be

$\vec s_u\times\vec s_v=-u\,\vec\jmath+u\,\vec k$

The curl of $\vec F$ is

$\nabla\times\vec F(x,y,z)=(1-x)\,\vec\imath-\vec\jmath+(z-2)\,\vec k$

Then the line integral is equivalent to

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_0^{2\pi}\int_0^1\bigg((1-u\cos v)\,\vec\imath-\vec\jmath+(u\sin v+4)\,\vec k\bigg)\cdot\bigg(-u\,\vec\jmath+u\,\vec k\bigg)\,\mathrm du\,\mathrm dv$