Question

Simplify 10 times the square root of quantity 3 x end quantity plus 4 times the square root of quantity 3 x end quantity plus 5 times the square root of quantity 3 x end quantity. 9 times the square root of quantity 3 x end quantity 9 times the square root of quantity 6 x end quantity 19 times the square root of quantity 3 x end quantity 19 times the square root of quantity 9 x cubed

Answer 1

$\\ \rm\dashrightarrow10\sqrt{3x}+4\sqrt{3x}+5\sqrt{3x}+9\sqrt{3x}+9\sqrt{6x}+19\sqrt{3x}+19\sqrt{9x^3}$

$\\ \rm\dashrightarrow (10+4+5+9+19)\sqrt{3x}+9\sqrt{6x}+19\sqrt{3^2x^2x}$

$\\ \rm\dashrightarrow 47\sqrt{3x}+9\sqrt{2x}\sqrt{3x}+57x\sqrt{x}$

$\\ \rm\dashrightarrow (47+9\sqrt{2x})\sqrt{3x}+57x\sqrt{x}$

Answer 2

Answer:

$19\sqrt{3x}$

Step-by-step explanation:

Given expression:

$10 \sqrt{3x}+4\sqrt{3x}+5\sqrt{3x}$

Given answer options:

• $9\sqrt{3x}$

• $9\sqrt{6x}$

• $19\sqrt{3x}$

• $19\sqrt{9x^3}$

To simplify the given expression, factor out $\sqrt{3x}$:

$\implies 10 \sqrt{3x}+4\sqrt{3x}+5\sqrt{3x}$

$\implies \sqrt{3x}(10+4+5)$

Add the numbers in the parentheses:

$\implies \sqrt{3x}(19)$

Rearrange to standard form:

$\implies 19\sqrt{3x}$

Therefore, the answer is the third option.