anonymous
  • anonymous
http://gyazo.com/b8b1ef0f12bc8d0f8667d816336da9b2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
Hint: The domain and range of \[\Large y = \sqrt{x}\] is [0,infinity)
anonymous
  • anonymous
huh lol sorry

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jim_thompson5910
  • jim_thompson5910
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
jim_thompson5910
  • jim_thompson5910
if you want negative outputs from a square root function, you have to tack on a negative sign on the outside
anonymous
  • anonymous
so false?
jim_thompson5910
  • jim_thompson5910
yeah there is no way to get -sin(theta) when it should be +sin(theta)
anonymous
  • anonymous
yayay thanks
anonymous
  • anonymous
it wasnt false ):
jim_thompson5910
  • jim_thompson5910
hmm let me think
anonymous
  • anonymous
it's ok i already submitted
jim_thompson5910
  • jim_thompson5910
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
anonymous
  • anonymous
it's ok lol dont worry :)

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