# If c (x) = StartFraction 5 Over x minus 2 EndFraction and d(x) = x + 3, what is the domain of (cd)(x)?Which

###### Question:

## Answers

The domain is [tex]\{x[/tex] ∈ [tex]R|x\neq 2\}[/tex]

Step-by-step explanation:

Given the fucntion [tex]c(x)[/tex]:

[tex]c(x)=\frac{5}{x-2}[/tex]

And the function [tex]d(x)[/tex]:

[tex]d(x) = x + 3[/tex]

You must multiply them in order to get [tex](cd)(x)[/tex]. Then, this is:

[tex](cd)(x)=(\frac{5}{x-2})(x + 3)\\\\(cd)(x)=\frac{(5)(x+3)}{x-2}\\\\(cd)(x)=\frac{5x+15}{x-2}[/tex]

You can notice that it is a Rational function.

Since the denominator cannot be zero, the domain of this function will be all the real numbers except those that makes the denominator zero.

Then, you must equate the denominator to zero and solve for "x":

[tex]x-2=0\\x=2[/tex]

Therefore, the domain is [tex]\{x[/tex] ∈ [tex]R|x\neq 2\}[/tex]

[tex]\frac{x+5}{x+3}[/tex]

Step-by-step explanation:

hope this helps!

the answer is b on edg2020

Step-by-step explanation:

The domain of the function (cd)(x) will be all real values of x except x = 2.

Step-by-step explanation:

The two functions are [tex]c(x) = \frac{5}{x - 2}[/tex] and d(x) = x + 3

So, (cd)(x) = [tex](\frac{5}{x - 2})(x + 3) = \frac{5(x + 3)}{x - 2}[/tex]

Then, for x = 2 the function (cd)(x) will be undefined as zero in the denominator will make the function (cd)(x) undefined.

Therefore, the domain of the function (cd)(x) will be all real values of x except x = 2. (Answer)