what is the domain and range of: square root of sinx?

Answer 1

since x can be anything you want, then the domain of x is -infinity to +infinity, now the answer to sinx, which is the range, is always going to be between -1 and +1, inclusive.
[-1, 1], use the brackets to indicate a closed interval that includes the end points of -1 & +1, paranthesis like the previous answer are incorrect.
now since the square root can only be used for positive numbers then that limits the values of x to ones that only produce a positive answer, this is 0 to 180 degrees and their muultiples, in radians, that is 0 to pi.
So your final answer becomes:

Domain = [0,180] * K degrees where K is any odd integer OR [0,pi]*K radians where K is any odd integer.

Range = [0,1]

Answer 2

sinx=+/_1

sqr domain sinx=0 to 1

Answer 3

y=sqrt(sin x)
for y to be defined, sin x should be positive. Therefore
pi<=x<=0
sin x, then lies in [0,1]
So value of y lies in the range [0,1]

Domain : [0,pi]
Range : [0,1]

Answer 4

D: (-infinity, +infinity)
R: (-1, 1)

Answer 5

d=(inf,inf)
r=[0,1]