Question

1. solve the compound inequality. $6 + k > -3$ and $-3k > 9$

a. $k \geq -8$
b. $k \leq 0$
c. $0 < k \leq 10$
d. $-9 < k < -3$

Explanation:

Step1: Solve \(6 + k > -3\)

Subtract 6 from both sides: \(k > -3 - 6\)
\(k > -9\)

Step2: Solve \(-3k > 9\)

Divide both sides by -3 (remember to reverse the inequality sign): \(k < \frac{9}{-3}\)
\(k < -3\)

Step3: Find the intersection

We need \(k > -9\) and \(k < -3\), so the solution is \(-9 < k < -3\)

Answer:

D. \(-9 < k < -3\)