Question

which of the numbers shown make the inequality $12 - c \leq 8$ true? drag each into the correct box. keyboard help solutions of $12 - c \leq 8$ 8 4 12 not solutions of $12 - c \leq 8$ 1 8 2 0 11 8 11 5 4 1 12 2 0

Explanation:

Step1: Solve the inequality

Start with the inequality \(12 - c \leq 8\). Subtract 12 from both sides: \(-c \leq 8 - 12\), which simplifies to \(-c \leq -4\). Multiply both sides by -1 (remember to reverse the inequality sign): \(c \geq 4\).

Step2: Check each number

• For \(c = 8\): \(12 - 8 = 4 \leq 8\), so it's a solution.
• For \(c = 4\): \(12 - 4 = 8 \leq 8\), so it's a solution.
• For \(c = 12\): \(12 - 12 = 0 \leq 8\), so it's a solution? Wait, no, wait. Wait, our solution is \(c \geq 4\). Wait, 12 is greater than 4, so \(12 - 12 = 0 \leq 8\), yes. Wait, but let's re - check the numbers. Wait, the left - hand numbers are 8, 11, 5, 4, 1, 12, 2, 0. Wait, no, the numbers to check are 8, 11, 5, 4, 1, 12, 2, 0 (from the left - hand side boxes). Wait, let's list all numbers: 8, 11, 5, 4, 1, 12, 2, 0.

Check each:

• \(c = 8\): \(12-8 = 4\leq8\) → solution.
• \(c = 11\): \(12 - 11=1\leq8\) → solution? Wait, no, wait our inequality solution was \(c\geq4\). Wait, I made a mistake in solving the inequality. Let's re - solve \(12 - c\leq8\). Subtract 12: \(-c\leq8 - 12=-4\). Multiply by - 1: \(c\geq4\) (since multiplying by a negative reverses the inequality). So \(c\) must be greater than or equal to 4.

So:

• \(c = 8\): \(8\geq4\) → \(12 - 8 = 4\leq8\) → solution.
• \(c = 11\): \(11\geq4\) → \(12 - 11 = 1\leq8\) → solution.
• \(c = 5\): \(5\geq4\) → \(12 - 5 = 7\leq8\) → solution.
• \(c = 4\): \(4\geq4\) → \(12 - 4 = 8\leq8\) → solution.
• \(c = 1\): \(1<4\) → \(12 - 1 = 11>8\) → not a solution.
• \(c = 12\): \(12\geq4\) → \(12 - 12 = 0\leq8\) → solution.
• \(c = 2\): \(2<4\) → \(12 - 2 = 10>8\) → not a solution.
• \(c = 0\): \(0<4\) → \(12 - 0 = 12>8\) → not a solution.

Wait, but in the "Solutions of \(12 - c\leq8\)" box, there are already 8, 4, 12. And in the "Not Solutions" box, there are 1, 8? Wait, no, the "Not Solutions" box has 1, 8, 2, 0, 11? Wait, no, the numbers in the "Not Solutions" box (right - hand lower) are 1, 8, 2, 0, 11? Wait, no, the original numbers to drag are 8, 11, 5, 4, 1, 12, 2, 0 (from the left - hand side).

So solutions ( \(c\geq4\) ): 8, 11, 5, 4, 12.

Non - solutions ( \(c < 4\) ): 1, 2, 0.

Wait, but in the given "Solutions of \(12 - c\leq8\)" box, there are 8, 4, 12. And "Not Solutions" has 1, 8, 2, 0, 11. Wait, maybe I misread the numbers. Let's re - examine the image.

Left - hand numbers (to drag): 8, 11, 5, 4, 1, 12, 2, 0.

"Solutions of \(12 - c\leq8\)" box (upper right) has 8, 4, 12 (already there? Maybe the initial box has some numbers, but we need to drag the correct ones.

So according to \(c\geq4\):

Numbers \(\geq4\): 8, 11, 5, 4, 12.

Numbers \(<4\): 1, 2, 0.

So the solutions are 8, 11, 5, 4, 12. The non - solutions are 1, 2, 0.

Answer:

Solutions of \(12 - c\leq8\): 4, 5, 8, 11, 12

Not Solutions of \(12 - c\leq8\): 0, 1, 2