Answer:
Option C [tex]-6<x\leq -3[/tex]
Step-by-step explanation:
we have
[tex]21\leq -3(x-4)<30[/tex]
The compound inequality can be divided into two inequality
[tex]21\leq -3(x-4)[/tex] -----> inequality A
[tex]-3(x-4)<30[/tex] ----> inequality B
Solve inequality A
[tex]21\leq -3x+12[/tex]
[tex]9\leq -3x[/tex]
Divide by -3 both sides
when you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol
[tex]-3\geq x[/tex]
Rewrite
[tex]x\leq -3[/tex]
The solution of the inequality A is the interval (-∞,-3]
Solve the inequality B
[tex]-3x+12<30[/tex]
[tex]-3x<18[/tex]
Divide by -3 both sides
when you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol
[tex]x>-6[/tex]
The solution of the inequality B is the interval [-6,∞)
The solution of the compound inequality is
[-6,∞) ∩ (-∞,-3]=(-6,-3]
[tex]-6<x\leq -3[/tex]
