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- anonymous

##### 1 Attachment

- mathstudent55

|dw:1377144471353:dw|

- anonymous

\[14=.5(112-X)\]
Is that how i would do it?

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## More answers

- anonymous

Or \[14=.5(x-112)\]

- mathstudent55

The first one.

- anonymous

Okay...So 84?

- mathstudent55

|dw:1377145019880:dw|

- mathstudent55

Yes

- anonymous

is 84 X? i feel like im missing something..

- mathstudent55

Yes, x = 84. Now you still need y.

- anonymous

i wouldnt use the same theorem right?

- mathstudent55

No. This a different case. You need a different theorem for this one.

- mathstudent55

y is the measure of an angle. It is not a length. Perimeter or circumference is a length.

- anonymous

Okay...So then I'm clueless.

- mathstudent55

An angle formed by a tangent and a chord that intersect on the circle is half the measure of the intercepted arc.

- anonymous

Right!

- mathstudent55

|dw:1377145667642:dw|

- anonymous

But we need to find the arc...right?

- mathstudent55

What is the degree measure of arc CD?

- mathstudent55

What is the degree measure of an entire circle as an arc?

- anonymous

360

- mathstudent55

Right.
Which parts of the circle below do you know the measures of?
|dw:1377145802550:dw|

- anonymous

SO 360-112-84

- anonymous

164?

- mathstudent55

|dw:1377145885955:dw|

- anonymous

So 164, right?

- mathstudent55

Correct.
Now you need half of that for y.

- anonymous

82!

- mathstudent55

Bingo!

- anonymous

Thank you!

- mathstudent55

You deserve a medal.

- mathstudent55

wlcm

- anonymous

I just need someone to guide me threw the first couple of times then i start getting it lol

- mathstudent55

That's the whole idea of studying and having a teacher or a tutor.

- anonymous

Do you know anything about probability distribution?And Im studying online, so i miss out on that just a little bit!

- mathstudent55

No, sorry, probability is not my thing.

- anonymous

Alright, thanks for your help!

- mathstudent55

wlcm

- anonymous

Thanks to you and @Directrix I understand geometry!

- mathstudent55

That's great, thanks. Keep asking questions. That's how you learn. You can tag me any time. If I'm not on, I'll see it the next time I'm on, and if you didn't get any response yet, I'll try to help.

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