Question

Which choice is equivalent to the quotient shown here for acceptable values of x

Answer 1

Answer:

D

Step-by-step explanation:

Using the rule of radicals

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

given

$\frac{\sqrt{30(x-1)} }{\sqrt{5(x-1)^2} }$

= $\sqrt{\frac{30(x-1)}{5(x-1)^2} }$

[ cancel 30 and 5 by 5 and (x - 1) / (x - 1)² by (x - 1) ]

= $\sqrt{\frac{6}{x-1} }$ → D

Answer 2

Answer:

D. $\sqrt\frac{6}{x-1}$

Step-by-step explanation:

Given:

$\frac{\sqrt{30(x-1)} }{\sqrt{5(x-1)^2} }$

We can write $\sqrt{xy} = \sqrt{x} \sqrt{y}$

Using this property we can rewrite the given expression as

= $\frac{\sqrt{30}\sqrt{x-1} }{\sqrt{5} \sqrt{x-1}\sqrt{x-1} }$

Now we can simplify √30/√5 = √6 and we can cancel out √(x - 1) both in the numerator and in the denominator, so we get

= $\sqrt\frac{6}{x-1}$

Therefore, the answer is D. $\sqrt\frac{6}{x-1}$