Does the point (6, 0) satisfy the equation y = x2?

Replace x in the equation with the x value of the point (6) and solve. If it equals the y value (0) it is a solution if it noes not equal (0) it is not a solution.

Y = 6^2 = 36

36 is not 0 so (6,0) is not a solution

No, point (6, 0) is not on the equation.

Step-by-step explanation:

To do this question the easiest way, you would use your scientific/graphing calculator and type in your equation. But you can do this with your mind.

Since the equation y = x^2 does not have any number in it (such as m = slope) it does not start anywhere. You will put it in the origin which is (0, 0) from there, you can tell that the equation will not reach (6, 0), but only (1, 1).

check the picture below.

well, the angle at the vertex d is clearly not a right-angle, let's check for the one at the vertex f which does look like,   however if it's indeed, that means ef is perpendicular to df, in which case their slopes will be negative reciprocal of each other.

so the negative reciprocal of the slope of ef is 1, however the slope of df is not 1, is 1/2, so nope, they're not perpendicular lines.

with btw u said 100pts but it clearly says

i answerd and just got 10pts

no hate