[tex]\sqrt{a}=b\iff b^2=a\ for\ a\geq0\ \wedge\ b\geq0\\---------------------\\\\Method\ \#1:\\\\\sqrt{9\cdot4}=\sqrt{36}=6\ because\ 6^2=36\\\\d)\ \sqrt9\cdot\sqrt4=3\cdot2=6\\\\\sqrt9=3\ because\ 3^2=9;\ \sqrt4=2\ because\ 2^2=4\\\\Answer:\ \boxed{d)\ \sqrt9\cdot\sqrt4}\\\\Method\ \#2:\\\\\text{use}\ \sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b}\\\\\sqrt{9\cdot4}=\sqrt9\cdot\sqrt4\\\\\text{Therefore your answer is}\ \boxed{d)}[/tex]
