Question

Perfrom the indicated operation

Answer 1

Answer:

$\dfrac{c}{3(c+2)}$

Step-by-step explanation:

$\dfrac{c^2+4c}{c^2-4} \div \dfrac{3c+12}{c-2}=\dfrac{c^2+4c}{c^2-4} \times \dfrac{c-2}{3c+12}=\dfrac{(c^2+4c)(c-2)}{(c^2-4)(3c+12)}$

Factor $c^2-4$:    $c^2-4=(c-2)(c+2)$

$\implies\dfrac{(c^2+4c)(c-2)}{(c+2)(c-2)(3c+12)}$

Cancel the common factor  $(c-2)$:

$\implies\dfrac{(c^2+4c)}{(c+2)(3c+12)}$

Factor  $c^2+4c$:  $c^2+4c=c(c+4)$

Factor  $3c+12$:  $3c+12=3(c+4)$

$\implies\dfrac{c(c+4)}{3(c+2)(c+4)}$

Cancel the common factor  $(c+4)$:

$\implies\dfrac{c}{3(c+2)}$

Answer 2

Answer: (6,0)

Step-by-step explanation: