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Mathematics 20 Online
OpenStudy (anonymous):

What is the value of x? Round to the nearest tenth. _____cm http://static.k12.com/calms_media/media/1582000_1582500/1582242/1/d312962d0705dffa9b47eba97a44e3dcfceee918/MS_IMC-150209-150419.jpg

OpenStudy (anonymous):

do it by pethagoras theoram u will get the answer

OpenStudy (anonymous):

Just use the Pythagorean theorem. Which is \[a^2+b^2=c^2\]

OpenStudy (anonymous):

so it's 70?

OpenStudy (anonymous):

How did you get 70?

OpenStudy (anonymous):

oh dang it I think I did that wrong...

OpenStudy (anonymous):

Okay so first I do \[10^{2} + 12^{2} =244\]

OpenStudy (anonymous):

It would be \[10^2+12^2=c^2\]\[244=c^2\]\[\sqrt{244}=c\]

OpenStudy (anonymous):

okay. I got 15.62

OpenStudy (anonymous):

That would be it, just round it to 15.6.

OpenStudy (anonymous):

Okay Thank You So Much! :)

OpenStudy (anonymous):

No problem :)

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