its multiplying complex numbers
2007-01-16 14:29:59 · 23 answers · asked by armando p 1 in Science & Mathematics ➔ Mathematics
6i
2007-01-16 14:33:00 · answer #1 · answered by lollypopgirl411 2 · 2⤊ 1⤋
3i + 3i^2 + 3 + 3i =
3i^2 + 6i + 3
2007-01-16 14:46:17 · answer #2 · answered by Malvi 2 · 0⤊ 1⤋
FOIL it out.
=3i+3i^2+3+3i
=3i +3(-1)+3+3i
= 3i -3+3+3i
=6i
2007-01-16 14:33:18 · answer #3 · answered by Wolfshadow 3 · 2⤊ 1⤋
(3i+3)(1+i)
= 3i + 3i^2 + 3 + 3i
since i^2 = -1
we have: 6i + 3 + 3(-1)
= 6i + 3 -3
= 6i
2007-01-16 14:38:41 · answer #4 · answered by bochevik 2 · 1⤊ 0⤋
(3i+3)(1+i) =3i+3i^2+3+3i
=3i - 3+3+3i =6i.Answer
2007-01-16 14:35:47 · answer #5 · answered by Anonymous · 1⤊ 1⤋
(3i+3)(1+i) =
3i + 3i² + 3 + 3i =
3i² + 6i + 3 = 3(-1) + 6i + 3 = -3 + 6i + 3 = 6i
Answer is 6i.
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2007-01-16 14:37:50 · answer #6 · answered by aeiou 7 · 1⤊ 1⤋
(3i+3)(1+i)
multiply all the components with each other...
= (3i*1)+ (3i*i)+(3*1)+(3*i)
= 3i+3i^+3+3i
=3i^+6i+3
if you want to simplify it further...
=3(i^+2i+3)
2007-01-16 14:38:45 · answer #7 · answered by Ms. M 3 · 0⤊ 1⤋
3i + 3i squared plu 3 plus 3i
6i + 3i squared + 3
2007-01-16 14:34:07 · answer #8 · answered by Anonymous · 0⤊ 3⤋
I think its 3i+3
2007-01-16 14:33:03 · answer #9 · answered by Kevin P 1 · 0⤊ 3⤋
(3i+3) (1+i) = 3i+3i.i+3+3i
= i(9i+3)
= 9i.i+3i
= 12i.i
2007-01-16 14:36:13 · answer #10 · answered by love 3 · 0⤊ 2⤋
the answer is 6i
3i^2+6i+3
i^2= -1 so -3+6i+3= 6i
2007-01-16 14:34:59 · answer #11 · answered by ~_~ 2 · 1⤊ 1⤋