Question

media geometrica a numerelor a=16 radicaldin 3 si b=radical din 48

Answer 1

$m_g=\sqrt{ab}$

$b=\sqrt{48} \Rightarrow b = \sqrt{16\cdot3}=4\sqrt3$

[tex]ab=16\sqrt3\cdot4\sqrt3=64\cdot3 \\\;\\ m_g=\sqrt{ab} = \sqrt{64\cdot3}=8\sqrt3.[/tex]