alexwing285
contestada

What is the quotient?
[tex] \frac{n+3}{2n-6} * \frac{n+3}{3n-9} [/tex]

A. 2/3
B. 3/2
C. [tex] \frac{(n+3)^2}{6(n-3)^2} [/tex]
D. [tex] \frac{6(n-3)^2}{(n+3)^3} [/tex]

Respuesta :

AlphasLuna
To  get the answer to this problem you first factor out the common term 2

(n+3)/(2(n-3) ) * (n+3)/(3n-9)

then factor out the common term 3

(n+3)/(2 (n-3) ) * (n+3)/(3 (n-3) )

then simplify

( (n+3)(n+3)/(2 (n-3)*3(n-3) )

Use product rule : x^a x^b= x^a+b

( (n+3)^2/(2 (n-3)*3(n-3) )

simplify 2 (n-3) * 3(n-3) to 6(n-3)(n-3)

( (n+3)^2/(6 (n-3)(n-3) )

use product rule again : x^a x^b=x^a+b

so you get the answer : ( (n+3)^2)/(6 (n-3)^2 )

so the orrect answer to this problem is answer choice (C)

Answer is c on e2020