1.On factorising (3a^2bc + 9ab²c + 21abc^2), we get :
(a) 3abc (3a + 3b + 7c)
(b) 3abc (a + b...
Mathematics, 08.12.2020 18:30, justinc10
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)
Answers: 2
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