the product of (3-2i) and (7+6i) is?
回答 (6)
2009-01-04 9:44 am
✔ 最佳答案
(3-2i)(7+6i)=21+4i-12i^2
=21+4i-12(-1) i^2=-1
=21+4i+12
=33+4i answer @(^_^)@
2009-01-04 6:00 pm
33 + 4i
2009-01-04 5:49 pm
(3 - 2i)(7 + 6i)
= 3*7 - 2i*7 + 3*6i - 2i*6i
= 21 - 14i + 18i - 12i^2
= 21 - 12(-1) + 4i
= 21 + 12 + 4i
= 33 + 4i
= 3*7 - 2i*7 + 3*6i - 2i*6i
= 21 - 14i + 18i - 12i^2
= 21 - 12(-1) + 4i
= 21 + 12 + 4i
= 33 + 4i
2009-01-04 5:35 pm
(3-2i)(7+6i) = (3)(7) + (3)(6i) - (2i)(7) - (2i)(6i)
................. = 21 + 18i - 14i - 12i²
................. = 21 + 18i - 14i + 12
................. = 33 + 4i
................. = 21 + 18i - 14i - 12i²
................. = 21 + 18i - 14i + 12
................. = 33 + 4i
2009-01-04 5:29 pm
32 + 4i
2009-01-04 5:28 pm
(3 - 2i)(7 + 6i)
FOIL it out.
21 + 18i - 14i - 12i^2
Use the fact that i^2 = -1.
21 + 18i - 14i - 12(-1)
21 + 4i + 12
32 + 4i
FOIL it out.
21 + 18i - 14i - 12i^2
Use the fact that i^2 = -1.
21 + 18i - 14i - 12(-1)
21 + 4i + 12
32 + 4i
收錄日期: 2021-05-01 11:44:31
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