Question

Which is the graph of the linear inequality y < 3x + 1? On a coordinate plane, a solid straight line has a positive slope and goes through (negative 1, negative 2) and (0, 1). Everything to the left of the line is shaded. On a coordinate plane, a solid straight line has a positive slope and goes through (negative 1, negative 2) and (0, 1). Everything to the right of the line is shaded. On a coordinate plane, a dashed straight line has a positive slope and goes through (negative 1, negative 2) and (0, 1). Everything to the left of the line is shaded. On a coordinate plane, a dashed straight line has a positive slope and goes through (negative 1, negative 2) and (0, 1). Everything to the right of the line is shaded.

Answer 1

For this case we have the following inequality: y < 3x + 1 < br/ >

What we must do is to evaluate a point of the Cartesian plane and verify if it is in the shaded region.

The shaded region represents the solution of the system of equations.

For the point (0, 0) we have:

0 < 3(0) + 1 < br / >

0 < 0 + 1 < br / >

0 < 1 < br / >

Therefore, the point (0, 0) is in the shaded region because it satisfies the inequality.

Then, the points that are on the line, are not part of the solution because the sign is of less strict.

Hope I helped ~~Laurel

Answer 2

Answer:

it is D!!!!!!!!!!! trust me, its right on edge:))))))

Step-by-step explanation: