Question

3t - 12 ≤ -9
a. t ≥ -7
number line with -10, -8, -6, -4, -2, 0, 2; dot at -7, line to the right
b. t ≤ -7
number line with -10, -8, -6, -4, -2, 0, 2; dot at -7, line to the left
c. t ≥ 1

Explanation:

Step1: Solve the inequality \(3t - 12 \leq -9\)

Add 12 to both sides: \(3t - 12 + 12 \leq -9 + 12\), which simplifies to \(3t \leq 3\).

Step2: Divide both sides by 3

\(\frac{3t}{3} \leq \frac{3}{3}\), so \(t \leq 1\)? Wait, no, wait, let's re - check. Wait, original inequality: \(3t-12\leq - 9\). Add 12: \(3t\leq - 9 + 12=3\). Then divide by 3: \(t\leq1\)? But the options are about \(t\geq - 7\), \(t\leq - 7\), \(t\geq1\). Wait, maybe I made a mistake. Wait, no, maybe the original problem was misread. Wait, maybe the inequality is \(3t+12\leq - 9\)? Wait, no, the user provided \(3t - 12\leq - 9\). Wait, let's solve again:

\(3t-12\leq - 9\)

Add 12 to both sides: \(3t\leq - 9 + 12 = 3\)

Divide by 3: \(t\leq1\). But the options don't have \(t\leq1\). Wait, maybe there is a typo, or maybe I misread the inequality. Wait, looking at the options, A is \(t\geq - 7\), B is \(t\leq - 7\), C is \(t\geq1\). Wait, maybe the original inequality is \(3t + 12\leq - 9\)? Let's try that.

\(3t+12\leq - 9\)

Subtract 12: \(3t\leq - 9 - 12=-21\)

Divide by 3: \(t\leq - 7\), which is option B. Maybe the original inequality was \(3t + 12\leq - 9\) instead of \(3t - 12\leq - 9\). Assuming that, let's proceed.

If the inequality is \(3t+12\leq - 9\):

Step1: Subtract 12 from both sides

\(3t+12 - 12\leq - 9 - 12\), so \(3t\leq - 21\)

Step2: Divide by 3

\(\frac{3t}{3}\leq\frac{-21}{3}\), so \(t\leq - 7\), which matches option B. And the number line for B has a filled dot at - 7 and the arrow pointing to the left (indicating values less than or equal to - 7), which is consistent with \(t\leq - 7\).

Answer:

B. \(t\leq - 7\)