Answer:
C. [tex]\sqrt{3-x}[/tex]
Step-by-step explanation:
The given quotient is
[tex]\sqrt{9-x^2}\div \sqrt{3+x}[/tex], when [tex]-3<\:x\le3[/tex]
We can rewrite this as
[tex]\frac{\sqrt{9-x^2} }{\sqrt{3+x} }[/tex]
We factor the numerator using difference of two squares
[tex]\Rightarrow \frac{\sqrt{3^2-x^2}}{\sqrt{3+x}}[/tex]
[tex]\Rightarrow \frac{\sqrt{(3-x)(3+x)}}{\sqrt{(3+x)}}[/tex]
[tex]\Rightarrow \frac{\sqrt{(3-x)} \sqrt{(3+x)}}{\sqrt{(3+x)}}[/tex]
We cancel the common factors to get;
[tex]\sqrt{3-x}[/tex]
