Final answer:
The inequality \(\frac{3}{m} \le -\frac{2}{11}\) is true for an infinite number of negative integer values of m, starting from -16 and going to negative infinity.
Step-by-step explanation:
To determine for how many integers m the inequality \(\frac{3}{m} \le -\frac{2}{11}\) is true, we must consider the sign and value of the fraction on the right side of the inequality.
Since the fraction is negative, m must also be negative to have a positive divisor. Now, solving the inequality step by step:
- Multiply both sides by m, which is negative, remembering to reverse the inequality: m\(\frac{3}{m}\) \ge m\(-\frac{2}{11}\), simplifying to 3 \ge -\frac{2m}{11}.
- Multiplying both sides by -11 (which is negative, so the inequality sign flips again): -33 \le 2m.
- Divide by 2 to isolate m: -16.5 \le m.
- Remember, m must be an integer, so we take the next integer greater than -16.5, which is -16.
Therefore, we are looking for integers less than or equal to -16. These are -17, -18, -19, and so on.
There are infinitely many such integers. Hence, the inequality is true for an infinite number of integer values of m.