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2 years ago

Please help me name the pairs of vertical angles

Mathematics

2 answers:

vladimir2022 [97]2 years ago

7 0

Answer:

The answer is q

Step-by-step explana

babymother [125]2 years ago

4 0

There are 2 pairs of vertical angles and they are:

angle NOP and angle ROS

angle POR and angle NOS

To see this, it might help to use a different color to mark the segments NR and SP. A giant X forms. The angles NOP and ROS are opposite one another, so they are vertical angles.

Similarly, angles POR and NOS are opposite one another so they are the second pair of vertical angles.

Vertical angles are congruent, meaning that angle NOP = angle ROS. Also, angle POR = angle NOS

No other vertical angle pairs are present because the segments MO and QO do not extend past point O.

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2 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=5%28sin%28t%29%20%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%2Bycos%28t%29%29%29%3Dcos%28t%29%20%20%28sin%28

Nataliya [291]

Assuming you mean

$5\sin t\dfrac{\mathrm dy}{\mathrm dt}+5y\cos t=\cos t\sin^2t$

This ODE is linear in $y$ , and you can already contract the left hand side as the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[5y\sin t\right]=\cos t\sin^2t$

Integrating both sides with respect to $t$ yields

$5y\sin t=\displaystyle\int\cos t\sin^2t\,\mathrm dt$
$5y\sin t=\dfrac13\sin^3t+C$
$y=\dfrac1{15}\sin^2t+C\csc t$

Given that $y\left(\dfrac\pi2\right)=9$ , we have

$9=\dfrac1{15}\sin^2\dfrac\pi2+C\csc\dfrac\pi2$
$9=\dfrac1{15}+C$
$C=\dfrac{134}{15}$

so that the particular solution over the interval is

$y=\dfrac1{15}\sin^2t+\dfrac{134}{15}\csc t$

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