Question

Which Choice is equivalent to the quotient shown here when x is greater or equal to 0

Answer 1

$\sqrt{ \frac{27x}{48} } = \sqrt{ \frac{ {3}^{3}x}{ {2}^{4} \times 3 } } = \sqrt{ \frac{ {3}^{2}x }{ {2}^{4} } } = \frac{3 \sqrt{x} }{4}$

Answer 2

Answer:

C

Step-by-step explanation:

using the rules of radicals

• $\frac{\sqrt{a} }{\sqrt{b} }$ ⇔ $\sqrt{\frac{a}{b} }$

• $\sqrt{a}$ × $\sqrt{b}$ ⇔ $\sqrt{ab}$

given $\sqrt{27x}$ ÷ $\sqrt{48}$

simplifying the radicals

$\sqrt{27x}$ = $\sqrt{9(3)x}$ = 3$\sqrt{3x}$

$\sqrt{48}$ = $\sqrt{16(3)}$ = 4$\sqrt{3}$, hence

$\frac{3\sqrt{3x} }{4\sqrt{3} }$

= $\frac{3\sqrt{x} }{4}$ → C