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Mathematics 51 Online
OpenStudy (anonymous):

is (6/7,square root of 13/7) on the unit circle?

OpenStudy (anonymous):

check to see if \[(\frac{6}{7})^2+(\frac{\sqrt{13}}{7})^2=1\]is it unlikely that you can memorize all the points on the unit circle, as there are an infinite number of them

OpenStudy (anonymous):

to tell if a point is on the unit circle, see if it satisfies the equation x^2+y^2 = 1

OpenStudy (anonymous):

there is no \(\pi\) in any coordinate on the unit circle, since \(\pi^2>1\)

OpenStudy (anonymous):

in fact lets check \[\left(\frac{6}{7}\right)^2+\left(\frac{\sqrt{13}}{7}\right)^2=\frac{36}{49}+\frac{13}{49}=\frac{36+13}{49}=\frac{49}{49}=1\] so it is for sure

OpenStudy (anonymous):

Thanks you guys so much!

OpenStudy (anonymous):

@satellite73 you are correct. pi is an angle measurement not a point. My mix up

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