In which of the given sets of numbers are all the numbers solutions of 18+2x≥36

Question

?(1 point)
Responses

9, 10, 11, 12
9, 10, 11, 12

9, 8, 7, 6
9, 8, 7, 6

0, 3, 9, 12
0, 3, 9, 12

8, 7, 6

Answer 1

To solve the inequality \( 18 + 2x \geq 36 \):

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

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

This means that any number \( x \) must be greater than or equal to 9 to satisfy the inequality.

Now, let's evaluate each set of numbers:

1. Set: 9, 10, 11, 12

   • All numbers (9, 10, 11, 12) are greater than or equal to 9.
   • This set is valid.

2. Set: 9, 8, 7, 6

   • 9 is valid (\( 9 \geq 9 \)), but 8, 7, and 6 are not valid.
   • This set is not valid.

3. Set: 0, 3, 9, 12

   • 0 and 3 are not valid, but 9 and 12 are valid.
   • This set is not valid.

4. Set: 8, 7, 6

   • None of these numbers (8, 7, 6) are valid since they are all less than 9.
   • This set is not valid.

Thus, the only set in which all numbers are solutions to the inequality \( 18 + 2x \geq 36 \) is: 9, 10, 11, 12.