Plz
which ordered pairs are solutions to the inequality 2x+3y≥−1 ?
select each correct answer.
(2, −1)
(−6, 0)
(−2, 1)
(0, 1)
(0, −1)

(2,-1)

(-2,1)

(0,1)

Step-by-step explanation:

Let's plug in the points given and see which satisfy it (make it true).

2x+3y>=-1

Test (2,-1)

2(2)+3(-1)>=-1

4    + (-3)>=-1

           1>=-1  This is true 1 is greater than -1 so (2,-1) works!

Test (-6,0)

2(-6)+3(0)>=-1

-12   +  0  >=-1

         -12 >=-1 This is not true. -12 is not greater than -1.

Test (-2,1)

2(-2)+3(1)>=-1

-4   +  3 >=-1

         -1 >=-1  This is true. -1 does equal -1.

Test (0,1)

2(0)+3(1)>=-1

 0  +  3 >=-1

         3>=-1  This is true.  3 is greater than -1.

Test (0,-1)

2(0)+3(-1)>=-1

0   + (-3)>=-1

         -3>=-1  This is false.  -3 is not greater than -1.

answer: a) (0,0), (1,1), (2,8), and (3,27).

step-by-step explanation:

0^3=0

1^3=1

2^3=8

3^3=27

in effect, the exponential equation for this would me y=3^x. each x-coordinate cubed is giving the y coordinate, so a is your answer : )