7i(x+3i) +5
2007-01-04 13:01:51 · 8 answers · asked by Siny T 1 in Science & Mathematics ➔ Mathematics
7i(x+3i) +5
= 7ix + 21i^2 + 5
Do the i's signify it is a complex number, or is i just another variable?
If so, i^2 = -1
= 7ix - 21 + 5
= 7ix - 16
2007-01-04 13:04:10 · answer #1 · answered by Tom :: Athier than Thou 6 · 1⤊ 0⤋
First: use distribution > multiply 7i with both terms in the parenthesis >
7i(x) + 7i(3i) + 5
Sec: solve according to order of operations >
7ix + 21i^2 + 5
Third: i^2 is -1 > replace i^2 with -1 in the expressions >
7ix + 21(-1) + 5
7ix - 21 + 5
7ix - 16
2007-01-04 21:05:59 · answer #2 · answered by ♪♥Annie♥♪ 6 · 0⤊ 0⤋
Dear
you just apply the regular rules of algebra and keep in mind
that i * i = -1
so,
7i (x + 3i) + 5 = 7xi + 21*(i^2) + 5 = 7xi + 21*(-1) + 5 = 7xi - 21 + 5 = 7xi - 16
Note:
a good question is halfe the way to a solution.
2007-01-04 21:17:43 · answer #3 · answered by Mohamed K 2 · 0⤊ 0⤋
7i (x + 3i) + 5 = 7xi + 21*(i^2) + 5 = 7xi + 21*(-1) + 5 = 7xi - 21 + 5 = 7xi - 16
Good luck! =)
2007-01-04 21:04:00 · answer #4 · answered by Jess 2 · 2⤊ 0⤋
7ix - 21 + 5
= 7xi - 16
2007-01-04 21:08:17 · answer #5 · answered by whatever_9123 1 · 0⤊ 0⤋
7xi-21+5
=7xi-16
2007-01-04 21:04:08 · answer #6 · answered by raj 7 · 1⤊ 0⤋
7xi-16
2007-01-04 21:05:21 · answer #7 · answered by Anonymous · 1⤊ 0⤋
You guys are good!
2007-01-04 21:04:57 · answer #8 · answered by debra62 2 · 1⤊ 0⤋