Respuesta :

tramserran

|4r + 8| ≥ 32

4r + 8 ≥ 32         or         4r + 8 ≤ -32       inside of absolute value could be + or -

4r       ≥ 24         or          4r       ≤ -40       subtracted 8 from both sides  

r       ≥  6           or            r       ≤ -10         divided both sides by 4

Graph: ←---------------- -10        6 ------------------→

Interval Notation: (-∞, -10] U [6, ∞)

Okay here are the steps to solving this inequality:

First- Break down the problem into these 2 equations

               4r + 8 ≥ 32    →    - (4r + 8) ≥ 32


Second- Solve the 1st equation: 4r + 8 ≥ 32

                    r ≥ 6


Third- Solve the 2nd equation: - (4r + 8) ≥ 32

                    r ≤ -10


Lastly- Collect all solutions

          r ≥ 6   and   r ≤ -10

       

                         

                         

Okay here are the steps to solving this inequality:

First- Break down the problem into these 2 equations

               4r + 8 ≥ 32    →    - (4r + 8) ≥ 32


Second- Solve the 1st equation: 4r + 8 ≥ 32

                    r ≥ 6


Third- Solve the 2nd equation: - (4r + 8) ≥ 32

                    r ≤ -10


Lastly- Collect all solutions

          r ≥ 6   and   r ≤ -10

       

                         

                         

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