Which is the graph of the linear inequality y ≥ −x − 3? On a coordinate plane, a solid straight line has a negative slope and goes through (negative 3, 0) and (0, negative 3). Everything to the right of the line is shaded. On a coordinate plane, a solid straight line has a negative slope and goes through (negative 3, 0) and (0, negative 3). Everything to the left of the line is shaded. On a coordinate plane, a dashed straight line has a negative slope and goes through (negative 3, 0) and (0, negative 3). Everything to the right of the line is shaded. On a coordinate plane, a dashed straight line has a negative slope and goes through (negative 3, 0) and (0, negative 3). Everything to the left of the line is shaded.

"On a coordinate plane, a solid straight line has a negative slope and goes through (negative 3, 0) and (0, negative 3). Everything to the right of the line is shaded."

The given inequality is:

y ≥ −x − 3

So first we know that the line will pass through (-3, 0) and (0, -3) (you can check it by just evaluating the line).

Now, y can be equal to or larger than any point on that line, so the line must be a solid line (meaning that the points on the line are solutions) and y is larger than that, then all the points above the line are solutions, meaning that we must shade the region above (or at the right) of the line.

The graph can be seen below:

So the correct option is the first one:

"On a coordinate plane, a solid straight line has a negative slope and goes through (negative 3, 0) and (0, negative 3). Everything to the right of the line is shaded."

If you want to learn more, you can read:

https://brainly.com/question/17113339