Question

Which choice is equivalent to the quotient below when. The problem is in the photo below.please

Answer 1

Answer:

C. $\sqrt{3-x}$

Step-by-step explanation:

The given quotient is

$\sqrt{9-x^2}\div \sqrt{3+x}$, when $-3<\:x\le3$

We can rewrite this as

$\frac{\sqrt{9-x^2} }{\sqrt{3+x} }$

We factor the numerator using difference of two squares

$\Rightarrow \frac{\sqrt{3^2-x^2}}{\sqrt{3+x}}$

$\Rightarrow \frac{\sqrt{(3-x)(3+x)}}{\sqrt{(3+x)}}$

$\Rightarrow \frac{\sqrt{(3-x)} \sqrt{(3+x)}}{\sqrt{(3+x)}}$

We cancel the common factors to get;

$\sqrt{3-x}$