Please I need this ASAP
2007-06-27 14:36:13 · 5 answers · asked by pedro 1 in Science & Mathematics ➔ Mathematics
i/(3+i) = [i(4+i)]/[(3+i)(4+i)]=(4i-1)/(11+7i)
(3+i)/(4+i) = [(3+i)(3+i)]/[(4+i)(3+i)]=(6i+8)/(11+7i)
i/(3+i) - (3+i)/(4+i) = (4i-1)/(11+7i) - (6i+8)/(11+7i) = -(2i-9)/(7i+11)
2007-06-27 14:42:51 · answer #1 · answered by ? 5 · 0⤊ 0⤋
i / (3+i) - (3+i) / (4+i)
= i (3-i) / [(3-i)(3+i)] - (3+i)(4-i) / [(4+i)(4-i)]
= (1+3i) / 10 - (13+i) / 17
= (17 + 51i - 130 - 10i) / 170
= (41/170) i - 113/170
2007-06-27 14:41:49 · answer #2 · answered by Scarlet Manuka 7 · 0⤊ 0⤋
(3i - i^2)/10 - (12+i -i^2)/17
=(3i +1)/10 - (13+i)/17
=(51i +17-130 - 10 i)/ 170
=(41i-113)/170
2007-06-27 15:04:04 · answer #3 · answered by Joe 1 · 0⤊ 0⤋
i/0/(4 + i)
flip it
i(4 + i)
_____
0
answer: 0
2007-06-27 14:41:16 · answer #4 · answered by Anonymous · 0⤊ 1⤋
What is it we're supposed to do?
2007-06-27 14:41:01 · answer #5 · answered by Anonymous · 0⤊ 2⤋