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Mathematics 24 Online
OpenStudy (anonymous):

m^2=13 I put no solution for this one. Am I correct in doing this?

OpenStudy (anonymous):

No. I thought you had this before except with c instead of m

OpenStudy (phi):

take the square root of both sides you only get into trouble if you take the square root of a negative number. 13 is not negative, so you are ok.

OpenStudy (anonymous):

No I never had a problem like this yet

OpenStudy (anonymous):

√m √13*1?

OpenStudy (anonymous):

√m √3.606

OpenStudy (phi):

\[ m^2=13 \] \[ \sqrt{m^2} = ±\sqrt{13} \] \[ m = ±\sqrt{13} \]

OpenStudy (anonymous):

ahhhhh damn it!

OpenStudy (anonymous):

Thank you again

OpenStudy (phi):

or if you like \[ \sqrt{ m \cdot m } = ± \sqrt{13} \] \[ m = ± \sqrt{13} \]

OpenStudy (anonymous):

Thank you phi

OpenStudy (anonymous):

So it m=√13, and m=-√13 right

Directrix (directrix):

@treydwg -----> Yes, the answer you posted is correct. >> So it m=√13, and m=-√13 right

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