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jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1Hint: The domain and range of \[\Large y = \sqrt{x}\] is [0,infinity)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1basically 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
 one year ago
Best ResponseYou've already chosen the best response.1if you want negative outputs from a square root function, you have to tack on a negative sign on the outside

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1yeah there is no way to get sin(theta) when it should be +sin(theta)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1hmm let me think

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it's ok i already submitted

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.1the 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
 one year ago
Best ResponseYou've already chosen the best response.0it's ok lol dont worry :)
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