Does the point (3,-1) lie on the circle (x+1)^2 + (y-1)^2=16

No

Step-by-step explanation:

Simply substitute the values of (3,-1) into the equation and see if it is valid.

(x+1)² + (y-1)²   (for x =3 and y = -1)

= (3+1)² + [ -1 -1 ]²

= 4² + ( -2 )²

=16 + 4 = 20 (this is not equal to 16 as given in the equation.

Hence the point does NOT lie on the circle.

No.

Step-by-step explanation:

The circle of that equation lies on the point (3,1) as its furthest point to the right (x-axis) and therefore could never also lie on the point (3,-1).

Your equation is in center radius form, which is as follows:

Noting that your radius is 4 (since your radius squared is 16).

To graph your circle, simply go to your origin (h, k) which in this case is (-1, 1) and then count out in any direction (up, down, left, or right -- no diagonals) 4 units (since your radius is 4). This will give you the four outermost edges of your circle. Simply fill in the gaps from there, and you'll have sketched your circle.

1e+27

step-by-step explanation:

lockers 1, 4, 9, 16, 25, 40, and 49 are open

i ran through it by the other answer is wrong