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Also: Find the indicated sums and differences of the complex numbers:
(4-3i)-(8-2i)
2006-07-28 05:31:22 · 4 answers · asked by Brandon ツ 3 in Science & Mathematics ➔ Mathematics
For the first part we could multiply both sides by sqrt (6) and end up with sqrt (24)/6.
The second is -4-i.
2006-07-28 05:44:31 · answer #1 · answered by Nelson_DeVon 7 · 4⤊ 2⤋
What cases ?3 provides a rational variety? ?3 , that's what, because of the fact ?3?3 = 3 via definition of sq. root. So multiply the two numerator and denominator via ?3. the consequence is 7?3/3 or (7/3)?3 or (?147)/3. the 2d subject could properly be completed the comparable way, yet in line with risk this is extra proper to divide 30 via 5 and get 6. answer ?6.
2016-11-03 05:02:02 · answer #2 · answered by zubrzycki 4 · 0⤊ 0⤋
sqrt(4) = 2
sqrt(4)/sqrt(6) = 2/sqrt(6)
Now multiply the top and bottom by sqrt(6).
2/sqrt(6) = 2sqrt(6)/6
Now reduce 2/6 to 1/3
Answer: sqrt(6)/3
(4-3i)-(8-2i)
First distribute the negative sign
= 4 - 3i - 8 + 2i
Now combine like terms
= -4 - i
2006-07-28 15:22:39 · answer #3 · answered by MsMath 7 · 0⤊ 0⤋
sqrt(4)/sqrt(6) multiply numerator and the denomintor by sqrt(6)
we get sqrt4 x sqrt6/sqrt6xsqrt6
= 2xsqrt6/6
= sqrt6/3
2006-07-28 15:35:39 · answer #4 · answered by Subhash G 2 · 0⤊ 0⤋