What is the value of $(a+b)\sqrt{cd}$ if $a = 2$ , $b = 4$, $c = 3$, and $d = 12$?

Answer 1

First, substitute the given values into the expression $(a+b)\sqrt{cd}$:

$(2+4)\sqrt{3\cdot12}$
$= 6\sqrt{36}$
$= 6\cdot6$
$= 36$

Therefore, the value of $(a+b)\sqrt{cd}$ when $a = 2$, $b = 4$, $c = 3$, and $d = 12$ is $\boxed{36}$.