Solve the following inequality using both the graphical and algebraic approach: 0.5 x + 3 greater-than-or-equal-to 2 x minus 1.5 Graph A On a coordinate plane, a line goes through (0, 2) and (4, 4). Another line goes through (0, 0) and (2, 6). The lines intersect at (1, 1.5). Graph B On a coordinate plane, a line goes through (0, 3) and (4, 4). Another line goes through (1, 0) and (4, 6). The lines intersect at (3, 4). a. x greater-than-or-equal-to 3 Graph A b. x less-than-or-equal-to 3 Graph A c. x greater-than-or-equal-to 3 Graph B d. x less-than-or-equal-to 3 Graph B

d. x les-than-or-equal-to 3 Graph B

Graph B expressed as follows; The graph on the coordinate plane, a line goes through (0, 3) and (4, 5). Another line goes through (1, 0.5), and (4, 6.5). The lines intersect at (3, 4.5)

Step-by-step explanation:

The given inequality is expressed as follows;

0.5·x + 3 ≥ 2·x - 1.5

Let y₁ = 0.5·x + 3, and y₂ = 2·x - 1.5, we get;

For x = -1, 0, 1, 2, 3, 4, 5, 6

y₁ = 2.5, 3, 3.5, 4, 4.5, 5, 5.5, 6

y₂ = -3.5, -1.5, 0.5, 2.5, 4.5, 6.5, 8.5, 10.5

From the given data, the lines intersect at (3, 4.5)

The graph on the coordinate plane, a line goes through (0, 3) and (4, 5). Another line goes through (1, 0.5), and (4, 6.5). The lines intersect at (3, 4.5)

Please find attached the required inequality created with MS Excel

Therefore, we have;

3 + 1.5 ≥ 2·x - 0.5·x

4.5 ≥ 1.5·x

∴ 3 ≥ x

x ≤ 3

Therefore, with (typographical) correction, the best option is x les-than-or-equal-to 3 Graph B

inequality:

620-37.4p ≥ 450

the minimum amount of packages needed to be removed from the elevator is 5.

step-by-step explanation: