mathematics?

Question

Perform the indicated operations and write the answer in standard form.

(6-4i)/(3-i)

Give your answer using the form Ai + B.

please answer and explain this

Answer 1

(6-4i)/(3-i)
=(6-4i)(3+i)/[(3-i)(3+i)]
=(22-6i)/10
=11/5 - 3i/5
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snowy,

[(3-i)(3+i)] = 9-i^2 = 10

Answer 2

First, you need to make the denominator a real number.

You do that by multiplying both the numerator and the denominator by the conjugate of the denominator (which would be 3 + i). In other words:
(6 - 4i) / (3 - i)
= (6 - 4i)(3 + i) / (3 - i)(3 + i)
= (18 + 6i - 12i -4i^2) / (9 - i^2)
= (18 - 6i + 4) / (9 + 1)
= (22 - 6i) / 10
= 11/5 - (3/5)i

Answer 3

I assume that "standard form" in this case means that you would not have an "i" in the denominator of your fraction. To simplify that, you would use the "difference of two squares".

(6-4i)/(3-i) * (3+i)/(3+i)

=(6-4i)(3+i)/(3-i)(3+i)

=[18 - 12i + 6i - 4(-1)]/[9-(-1)]

=(22-6i)/10

= (22/10) - (3/5)i

= 2.2 - 0.6i

Hope that helps.

Answer 4

To clear the denominator, we multiply numerator
and denominator by the conjugate of 3-i or 3 + i.
This gives us
(6-4i)(3+i)/10
= (22-6i)/10 = (11-3i)/5.

Answer 5

(6-4i)/(3-i) multiply numerator & denominator by 3+i
(6-4i)(3+i)/(9-i^2) use FOIL
(18+6i-12i-4i^2)/10
(18-6i+4)/10
2.2-.6i or
-.6i+2.2

Answer 6

76

Answer 7

multiply the numerator and denominator by 3+i

Answer 8

(6-4i)/(3-i)
= (6 - 4i)(3 + i)/[(3 - i)(3 + i)]
= (18 - 12i + 6i + 4)/[9 - 3i + 3i + 1], please note i^2 = -1
= (22 -6i)/10
= 2.2 - 0.6i
= -0.6i + 2.2

Answer 9

(6-4i)(3+i)/((3-i)(3+i)

=(18+6i-12i+4)/(9-1)

=(22-6i)/8
=11/4-3/4i