Question

Rationalize the denominator of square root of negative 9 over open parentheses 4 minus 7 i close parentheses minus open parentheses 6 minus 6 i close parentheses.

quantity of negative 3 minus 6 i over 5
quantity of 39 minus 6 i over negative 9
quantity of negative 39 minus 6 i over 17
quantity of 3 minus 6 i over 3

Answer 1

Please write the problem using numbers and mathematical symbols instead of words.
Use the equation editor, a copy or a scan of the original problem, or the draw tool.

Answer 2

\[\sqrt{-9}\div(4-7i)-(6-6i)\]

Answer 3

@Nnesha @mathstudent55
@nincompoop @Jhannybean

Answer 4

@DanJS

Answer 5

HI!!

Answer 6

\(\dfrac{\sqrt{-9}}{4 - 7i} - (6 - 6i) \)

Answer 7

\[\frac{\sqrt{-9}}{4+7i}\]??

Answer 8

To rationalize a denominator, multiply by a fraction which is the conjugate over the conjugate.
Also, let's take care of the square root of the negative number.

Answer 9

\(\dfrac{\sqrt{-9}}{4 - 7i} - (6 - 6i) \)

\(=\dfrac{3i}{4 - 7i} \times \dfrac{4 + 7i}{4 + 7i} - (6 - 6i) \)

Do you follow so far?

Answer 10

yes i am flowing

Answer 11

Now we actually do the multiplication.

Answer 12

\(=\dfrac{3i(4 + 7i)}{(4 - 7i)(4 + 7i)} - (6 - 6i) \)

\(=\dfrac{12i + 21i^2}{16 - (7i)^2} - (6 - 6i) \)

\(=\dfrac{12i - 21}{16 + 49} - (6 - 6i) \)

Answer 13

\(=\dfrac{12i - 21}{65} - (6 - 6i) \)

Answer 14

\(=\dfrac{12i - 21}{65} - \dfrac{65}{65}(6 - 6i) \)

\(=\dfrac{12i - 21}{65} - \dfrac{390-390i}{65} \)

\(=\dfrac{12i - 21- (390-390i)}{65} \)

\(=\dfrac{12i - 21- 390+390i}{65} \)

\(=\dfrac{-411 + 402i}{65} \)

\(=-\dfrac{411}{65} + \dfrac{402}{65}i \)