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anonymous

  • one year ago

If f(x) = 1 – x, which value is equivalent to |f(i)|? options are: A.0 B.1 C. √2 D. √-1

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  1. freckles
    • one year ago
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    \[|a+bi|=\sqrt{a^2+b^2}\]

  2. chrisdbest
    • one year ago
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    B should be the closest

  3. freckles
    • one year ago
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    it actually isn't B

  4. chrisdbest
    • one year ago
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    Bcuz i = sqrt -1, and the absolut value of a negative number is a positive number

  5. freckles
    • one year ago
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    \[f(i)=1-i \\ |f(i)|=|1-i|=?\]

  6. anonymous
    • one year ago
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    sorry guys but im still in the blue here ahahah

  7. chrisdbest
    • one year ago
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    Oh so its A!!!

  8. freckles
    • one year ago
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    \[|1-i|=\sqrt{1^2+1^2}=?\]

  9. chrisdbest
    • one year ago
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    Cuz 1 - 1 = 0!

  10. chrisdbest
    • one year ago
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    Bam!! Medal me please

  11. freckles
    • one year ago
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    or if you wanted to look at it as \[\sqrt{1^2+(-1)^2} \text{ instead of } \sqrt{1^2+1^2}\]

  12. anonymous
    • one year ago
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    ayyyy lmao

  13. freckles
    • one year ago
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    all I'm asking you to do is do the 1+1 underneath the radical

  14. freckles
    • one year ago
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    \[|a+bi|=\sqrt{a^2+b^2} \ \\ \text{ and you have } \\ |1+-1i|=\sqrt{1^2+(-1)^2}=\sqrt{1^2+1^2}=?\] can you finish this ?

  15. anonymous
    • one year ago
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    OOOOOOOO THANKS MAN LMAO I DONT KNOW HOW I DIDNT SEE THAT ahha its √2

  16. freckles
    • one year ago
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    yep :)

  17. chrisdbest
    • one year ago
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    Dang I didn't even see that answer choice. I'm Dyslexic, I thought it said \[\sqrt{21}\]

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