If f(x)=2-x^1/2 and g(x)=x^2-9 what is the domain of g(x) / f(x)

Let's consider what we are asked. The domain is defined as a set of points that satisfy the equation on the x-ordinate.
So, essentially, we need to find if and what the restriction on x is.

Let's now consider just f(x) because g(x) is completely irrelevant to the question.

f(x) = 2 - x^(1/2)
Since f(x) can never be 0 for a defined function, let's consider when f(x) = 0 to find a restriction on x.

2 - x^(1/2) = 0
2 = x^(1/2)
+-4 = x

But we can only take the positive 4 because inside a square root has to always be positive (unless you're dealing with complex numbers), so the only restriction is that x cannot be equal to 4.

Therefore, our domain is: x >= 0; x =/= 4

Given: A triangle QRS

Solution: 1.The side opposite ∠Q is RS.→True

2.The side opposite ∠Q is RQ.→False

3.The hypotenuse is QR.→ False,as not given which kind of triangle it is , and if it is a right angled triangle which angle is of 90°.

4.The side adjacent to ∠R is SQ.→False ,There are two adjacent sides adjacent to ∠R which are QR and RS.

5.The side adjacent to ∠Q is QS.→True, but there is also another side adjacent to ∠Q which is QR.