Question

Which choice is equivalent to the product below when x > 0?

Answer 1

The product below is equivalent when x > 0 is 1/9

       

Resolution - Explanation

Square root is a real number x multiplied by itself - which results in a perfect value, where it is possible to calculate the real proof (which is the square root).

             

Given the expression, $\large \sf \sqrt{\dfrac{1}{x^{2} } } \cdot \sqrt{\dfrac{x^{2} }{81} }$, first step: we will calculate the root of the numerator and denominator of this fraction:

$\\\large \sf \sqrt{\dfrac{1}{x^{2} } } \cdot \sqrt{\dfrac{x^{2} }{81} }$

$\large \sf \dfrac{\sqrt{1} }{\sqrt{x^{2} } } \rightarrow \dfrac{1}{x}$

$\large \sf \dfrac{\sqrt{x^{2} } }{\sqrt{81 } } \rightarrow \dfrac{x}{9}$

Step two: rearranging the expression and canceling the common factors x, we will have,:

$\\\large \sf \dfrac{1}{\not x} \cdot \dfrac{\not x}{9}$

$\pink{\boxed{\large \sf \dfrac{1}{9} }}$

Therefore, the final answer to this multiplication will be 1/9.