In which of the given sets of numbers are all the numbers solutions of 18+2x≥36 ?(1 point) Responses 0, 3, 9, 12 0, 3, 9, 12 9, 8, 7, 6 9, 8, 7, 6 9, 10, 11, 12 9, 10, 11, 12 8, 7, 6

Answer 1

First, let's solve the inequality \(18 + 2x \geq 36\).

1. Subtract 18 from both sides: \[ 2x \geq 36 - 18 \] \[ 2x \geq 18 \]

2. Divide both sides by 2: \[ x \geq 9 \]

Now we know that the solutions to the inequality \(18 + 2x \geq 36\) are all numbers \(x\) such that \(x \geq 9\).

Next, we can evaluate each of the given sets of numbers to see which ones contain only values greater than or equal to 9:

1. Set 1: \(0, 3, 9, 12\)

   • Contains: 0 (not a solution), 3 (not a solution), 9 (solution), 12 (solution)
   • Not all numbers are solutions.

2. Set 2: \(9, 8, 7, 6\)

   • Contains: 9 (solution), 8 (not a solution), 7 (not a solution), 6 (not a solution)
   • Not all numbers are solutions.

3. Set 3: \(9, 10, 11, 12\)

   • Contains: 9 (solution), 10 (solution), 11 (solution), 12 (solution)
   • All numbers are solutions.

4. Set 4: \(8, 7, 6\)

   • Contains: 8 (not a solution), 7 (not a solution), 6 (not a solution)
   • Not all numbers are solutions.

The only set where all numbers are solutions of the inequality \(18 + 2x \geq 36\) is Set 3: \(9, 10, 11, 12\).