-6(a-3b-9c)
9
2ab - 6ac + 3b - 9c = 2a(b - 3c) + 3(b - 3c) = (2a + 3)(b - 3c).
Well I got this: 2c+7=9+6-7c+2c 2c+7=15-9c 11c+7=15 (add 9c) 11c=8 ( Subtract 7) 11---11 ( Divide 11) C=0.72
9C = 234
-6(a-3b-9c)
It is: 6-11+5-8c+9c = c
9c is.
(12c5 - 18c3 + 9c) / (-3c) = -4c4 + 6c2 - 3
3a + 9 =3X(a + 3)
Yes; the factored form would be (9c+4)(9c+4) or just (9c+4)2 Since the two factors are the same, the beginning trinomial 81c2+72c+16 is a perfect square trinomial
9
2ab - 6ac + 3b - 9c = 2a(b - 3c) + 3(b - 3c) = (2a + 3)(b - 3c).
2ab - 6ac + 3b - 9c = b(2a + 3) - 3c(2a + 3) = (b - 3c)(2a + 3)
Well I got this: 2c+7=9+6-7c+2c 2c+7=15-9c 11c+7=15 (add 9c) 11c=8 ( Subtract 7) 11---11 ( Divide 11) C=0.72
9C = 234
18 degrees, C