Answer:
Step-by-step explanation:
Given expression is [tex](\frac{2n}{6n+4})(\frac{3n+2}{3n-2})[/tex]
First simplify the given expression as below ;
[tex](\frac{2n}{6n+4})(\frac{3n+2}{3n-2})[/tex]
[tex]=(\frac{2n}{2(3n+2)})(\frac{3n+2}{3n-2})[/tex] ( here by taking the common term 2 outside the factor )
[tex]=(\frac{n}{1})(\frac{1}{3n-2})[/tex] ( here by simplifying the factors )
[tex]=\frac{n}{3n-2}[/tex]
Therefore [tex](\frac{2n}{6n+4})(\frac{3n+2}{3n-2})=\frac{n}{3n-2}[/tex]
Therefore the result of the given simplified expression is [tex]\frac{n}{3n-2}[/tex]
The denominator of the resulting function is 3n-2
Given the function expressed as:
[tex](\frac{2n}{6n+4} )(\frac{3n+2}{3n-2} )[/tex]
This expression can be written as:
[tex]g(n)=(\frac{2n}{6n+4} )(\frac{3n+2}{3n-2} )\\g(n)=(\frac{2n}{2(3n+2)} )(\frac{3n+2}{3n-2} )\\g(n)=\frac{2n}{n}\times \frac{3n+2}{(3n+2)(3n-2)} \\[/tex]
Cancel out the common function at the numerator and denominator.
[tex]g(n)=2\times \frac{1}{(3n-2)} \\\g(n)=\frac{2}{3n-2}[/tex]
From the resulting function, the denominator of the resulting function is 3n-2
Learn more on product of functions here: https://brainly.com/question/10404935
