anonymous
  • anonymous
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
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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mathstudent55
  • mathstudent55
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.
anonymous
  • anonymous
\[\sqrt{-9}\div(4-7i)-(6-6i)\]
anonymous
  • anonymous
@Nnesha @mathstudent55 @nincompoop @Jhannybean

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More answers

anonymous
  • anonymous
@DanJS
misty1212
  • misty1212
HI!!
mathstudent55
  • mathstudent55
\(\dfrac{\sqrt{-9}}{4 - 7i} - (6 - 6i) \)
misty1212
  • misty1212
\[\frac{\sqrt{-9}}{4+7i}\]??
mathstudent55
  • mathstudent55
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.
mathstudent55
  • mathstudent55
\(\dfrac{\sqrt{-9}}{4 - 7i} - (6 - 6i) \) \(=\dfrac{3i}{4 - 7i} \times \dfrac{4 + 7i}{4 + 7i} - (6 - 6i) \) Do you follow so far?
anonymous
  • anonymous
yes i am flowing
mathstudent55
  • mathstudent55
Now we actually do the multiplication.
mathstudent55
  • mathstudent55
\(=\dfrac{3i(4 + 7i)}{(4 - 7i)(4 + 7i)} - (6 - 6i) \) \(=\dfrac{12i + 21i^2}{16 - (7i)^2} - (6 - 6i) \) \(=\dfrac{12i - 21}{16 + 49} - (6 - 6i) \)
mathstudent55
  • mathstudent55
\(=\dfrac{12i - 21}{65} - (6 - 6i) \)
mathstudent55
  • mathstudent55
\(=\dfrac{12i - 21}{65} - \dfrac{65}{65}(6 - 6i) \) \(=\dfrac{12i - 21}{65} - \dfrac{390-390i}{65} \) \(=\dfrac{12i - 21- (390-390i)}{65} \) \(=\dfrac{12i - 21- 390+390i}{65} \) \(=\dfrac{-411 + 402i}{65} \) \(=-\dfrac{411}{65} + \dfrac{402}{65}i \)

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