in the fom a+bi, where a and b are real numbers.
please help. pre cal problem.
2007-09-20 05:58:30 · 9 answers · asked by icygurl555 1 in Science & Mathematics ➔ Mathematics
First: use the Foil Method...
(1)(8)+(1)(-7i)+(9i)(8)+(9i)(-7i)
8 + (-7i) + 72i + (-63i^2)
8 - 7i + 72i - 63i^2
*Rule - (i^2) becomes > (-1)
8 - 7i + 72i - (63)(-1)
8 - 7i + 72i - (-63)
8 - 7i + 72i + 63
Sec: combine "like" terms....
8 + 63 - 7i + 72i = 71 + 65i
2007-09-20 08:47:00 · answer #1 · answered by ♪♥Annie♥♪ 6 · 1⤊ 0⤋
(1+9i)(8-7i)
=8+63+72i-7i
=71+65i. ANS.
2007-09-20 13:04:20 · answer #2 · answered by Anonymous · 0⤊ 0⤋
There are two basic facts that you need to know to solve this problem. The first is that i x i = -1, which is just the definition of an imaginary number. The second is the FOIL rule for distributive multiplication (see link below). Using the FOIL rule you get
(1x8) + (1x-7i) + (9ix8) + (9ix-7i) Then simplify each term:
8 + (-7i) + (72i) + (-63i^2)
8 + 65i + 63
71 + 65i
And that's it!
2007-09-20 13:10:26 · answer #3 · answered by endo_jo 4 · 0⤊ 1⤋
Use FOIL
(1+9i)(8-7i)
(1)(8) + (1)(-7i) + (9i)(8) + (9i)(-7i)
8 -7i +72i -63i^2
remember that i = sqrt(-1) therfore i^2 = -1
8 + 65i -63(-1)
8 + 63 + 65i
71 + 65i
2007-09-20 13:06:32 · answer #4 · answered by T 5 · 0⤊ 0⤋
multiply it like binomials
1*8+9i*8-1*7i-9i*7i
8+72i-7i+63 (i*i=-1 so changes the sign)
71+65i
2007-09-20 13:06:00 · answer #5 · answered by chasrmck 6 · 1⤊ 0⤋
= 8 - 7 i + 72 i - 63 i ²
= 8 + 65 i + 63
= 71 + 65 i
2007-09-20 14:18:24 · answer #6 · answered by Como 7 · 1⤊ 0⤋
expanded form is 8+72i-7i-63i^2 =8+65i-(63)(-1)
=8+65i+63=71+65i
2007-09-20 13:04:48 · answer #7 · answered by Kenneth H 3 · 1⤊ 0⤋
8 - 7i + 72 i + 63
71 + 65 i
2007-09-20 13:04:59 · answer #8 · answered by CPUcate 6 · 1⤊ 0⤋
(1+9i)(8-7i)
=1(8-7i)+9i(8-7i)
=8-7i+72i-63i^2
=8+65i+63
=71+65i
2007-09-20 13:09:10 · answer #9 · answered by invert_the_feelings 1 · 0⤊ 0⤋