Question

1. solve the compound inequality: $10 - 10x \geq 40$ and $8x + 9 > -71$

a. $x \geq -9$
b. $-4 < x \leq 4$
c. $-10 < x \leq -3$
d. no solution

Explanation:

Step1: Solve \(10 - 10x \geq 40\)

Subtract 10 from both sides: \( - 10x \geq 40 - 10\)
Simplify: \( - 10x \geq 30\)
Divide both sides by -10 (reverse inequality): \(x \leq - 3\)

Step2: Solve \(8x + 9 > - 71\)

Subtract 9 from both sides: \(8x > - 71 - 9\)
Simplify: \(8x > - 80\)
Divide by 8: \(x > - 10\)

Step3: Find intersection

From step1, \(x \leq - 3\); from step2, \(x > - 10\). So \( - 10 < x \leq - 3\)

Answer:

C. \( -10 < x \leq -3\)