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I was supposed to solve for z and im not sure if im doing this right.
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|dw:1418788358009:dw|
\[a^2 + b^2 = c^2?\]
yeah I did all of that and I got stuck at 64 + 196=c^2
add and take the square root
64+196=260 160=c^2 Now we take the square root on both sides \[\sqrt{260}=\sqrt{c^2}\] \[\sqrt{260}=c\]
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Thanks
well we can simplify sqrt 260 to \[\sqrt{260}=\sqrt{4*65}=2 \sqrt{65}\]
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