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  1. anonymous
    • one year ago
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    @jim_thompson5910

  2. jim_thompson5910
    • one year ago
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    Hint: The domain and range of \[\Large y = \sqrt{x}\] is [0,infinity)

  3. anonymous
    • one year ago
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    huh lol sorry

  4. jim_thompson5910
    • one year ago
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    basically you're not allowed to plug in negative x values into a square root function and the outputs of the square root function is never negative

  5. jim_thompson5910
    • one year ago
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    if you want negative outputs from a square root function, you have to tack on a negative sign on the outside

  6. anonymous
    • one year ago
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    so false?

  7. jim_thompson5910
    • one year ago
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    yeah there is no way to get -sin(theta) when it should be +sin(theta)

  8. anonymous
    • one year ago
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    yayay thanks

  9. anonymous
    • one year ago
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    it wasnt false ):

  10. jim_thompson5910
    • one year ago
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    hmm let me think

  11. anonymous
    • one year ago
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    it's ok i already submitted

  12. jim_thompson5910
    • one year ago
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    the only thing I can think of is to test angles in various quadrants ex: Test Q1. if theta = pi/4, then sqrt(1 - cos^2(theta) ) = -sin(theta) sqrt(1 - cos^2(pi/4) ) = -sin(pi/4) sqrt(1 - 1/2) = -sqrt(2)/2 sqrt(1/2) = -sqrt(2)/2 sqrt(2)/2 = -sqrt(2)/2 that last equation is false, so Q1 doesn't work. Check the other quadrants this way

  13. anonymous
    • one year ago
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    it's ok lol dont worry :)

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spraguer (Moderator)
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