pupsy63 pupsy63
  • 08-12-2020
  • Mathematics
contestada


1.On factorising (3a^2bc + 9ab²c + 21abc^2), we get :

(a) 3abc (3a + 3b + 7c)
(b) 3abc (a + b + c)
(c) abc (a + 3b + 7c)
(d) 3abc (a + 3b + 7c)

Respuesta :

konnievas
konnievas konnievas
  • 08-12-2020
actually it’s none of these.

24a^2bc+9ab2c
48abc+9ab2c

answer: 3abc ^ (16+3b)
Answer Link
1774928
1774928 1774928
  • 10-02-2021

Answer:

Fast

Step-by-step explanation:

You told me "Please answer fast." ;-;

Answer Link

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