Which set of ordered pairs could be generated by an exponential function?
(0, 0), (1, 1), (2, 8), (3, 27)
(0, 1), (1, 2), (2,5), (3, 10)
(0, 0), (1,3), (2, 6), (3,9)
(0,1), (1, 3), (2, 9), (3, 27)

The last one.  I found all the equations just for added features.

Step-by-step explanation:

The first set of points is of the equation y=x^3. You have to cube the x's to get to the y's.  This is a cubic function.

The second set of points is of the equation y=x^2+1. You have to square the x's then add one to it to get y.  This is a quadratic function.

What helped me here was looking at the first and then second difference of the y-coordinates. The first differences was 1 , 3 , 5 so the second differences was 2, 2 which means it is a quadratic form.

The third set of points is of the equation y=3x.  You have to multiply the x's by 3 get the y's.  This is a linear equation.

So it has to be the fourth set of points. So it should be of the form y=A*B^x.

(0,1) is on that equation so 1=A*B^0 which means 1=A(1) or so A=1.

So your equation is in the form y=B^x.

Plugging in the next point (1,3) will give be B.  3=B^1 so B=3.

So the equation is y=3^x. This does work.

(0,1) gives us 1=3^0 which is true.

(1,3) gives us 3=3^1 which is true.

(2,9) gives us 9=3^2 which is true.

(3,27) gives us 27=3^3 which is true.

The last one could have been in the form of y=A*B^x+C but we would have wounded up with a contradiction with earlier reasoning if it was.

you need to rotate this triangle about the origin 270 degrees clockwise to create the full square.

w< 4

step-by-step explanation: