MCQ
$\cos 405^{\circ}=?$
- A$\frac{-1}{\sqrt{2}}$
- B$-\sqrt{2}$
- C$\sqrt{2}$
- D$\frac{1}{\sqrt{2}}$
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The centroid of the triangle with vertices (2, 6), (-5, 6) and (9, 3) is: