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Mathematics 19 Online
OpenStudy (anonymous):

Can someone please help me?

OpenStudy (anonymous):

OpenStudy (anonymous):

sorry I don't know how to do this yet I would but sry...

OpenStudy (anonymous):

it's okay

OpenStudy (anonymous):

@ganeshie8 can you help her?

OpenStudy (anonymous):

@agent0smith

OpenStudy (anonymous):

did you try bumping the question?

OpenStudy (anonymous):

yeah twice people just look at it for a while and leave

OpenStudy (anonymous):

same thing happens to me all the time

OpenStudy (anonymous):

what you can always do is reask the question

OpenStudy (anonymous):

Hmm yeah I guess but that just seems kind of pointless, yesterday when I was asking questions it took a lil over an hour for anyone to understand how to do it.

OpenStudy (solomonzelman):

\[a^2+b^2=c^2\]

OpenStudy (anonymous):

when I did that I got 59.4

OpenStudy (agent0smith):

It makes a right triangle, since AB is tangent to the circle. Use pythag as @SolomonZelman said.

OpenStudy (solomonzelman):

the missing side, is a leg, not a hypotenuse.

OpenStudy (anonymous):

ooo

OpenStudy (solomonzelman):

\[b^2=c^2-a^2~~~~that gives,~~~b=±\sqrt{c^2-a^2}\] in this case just + not -, b/c distance is always positive.

OpenStudy (anonymous):

thank you.

OpenStudy (solomonzelman):

\[b=\sqrt{(11.7)^2-(6)^2}\] Okay, just tell us what you get as the final ans. to make sure...

OpenStudy (anonymous):

10.0

OpenStudy (solomonzelman):

yes, good. I would approximate it to 10.04 (2 decimal places) but it's very good either way :)

OpenStudy (anonymous):

you got it @secretcat :)

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