Question

Is (0,0) a solution to this system? y ≥ x^2 - 4 y < 2x - 1

Answer 1

The answer is C
If we replace the variables with 0 the equation looks like this
0 >/ 0^2-4 and there isn't a possible answer if you put it in the calculator
and 0<2(0)-1= 0< (-1) and in mathematics a negative number is a real number that is less than zero. I hope this helps please tell me if I'm wrong.