Number 28 is the only question i need , with steps

The domain is all real numbers where

Step-by-step explanation:

So g(x) must exist before plugging it into f(x).

Let's find where g(x) doesn't exist.

is a quadratic expression.

is the discriminant and will tell us if will have any solutions.  I'm trying to solve this equation because I want to figure out what to exclude from the domain.  Depending on what we may not have to go full quadratic formula on this problem.

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Since the discriminant is negative, then there are no real numbers that will make the denominator 0 here.  So we have no real domain restrictions on g.

Let's go ahead and plug g into f.

I replaced g(x) with (1/(x^2+6x+10)).

I replaced old input,x, in f with new input (1/(x^2+6x+10)).

Let's do some simplification now.

We do not like the mini-fraction inside the bigger fraction so we are going to multiply by any denominators contained within the mini-fractions.

I'm multiplying top and bottom by (x^2+6x+10).

Using distributive property:

We are going to distribute a little more:

Combine like terms on the bottom there (-2 and 90):

Now we can see if we have any domain restrictions here:

So again the bottom will never be zero because doesn't have any real solutions.  We know this because the discriminant is negative.

The domain is all real numbers where

the solution is the point

step-by-step explanation:

we have

> equation a

> equation b

adds equation a and equation b

substitute the value of x in equation b to find the value of y

the solution is the point