Solve for z. 6 ≤ z + 18 < 10 Write your answer as a compound inequality with integers. Choices: (A)24 ≤ z < -8 (B)-12 ≤ z < -8 (C)-12 ≤ z < 28 (D)24 ≤ z < 28

Analyze Compound Inequality: Analyze the given compound inequality. The inequality 6 ≤ z + 18 < 10 involves two inequalities combined: one is ＂6 ≤ z + 18＂ and the other is ＂z + 18 < 10＂. We need to isolate z in both inequalities by performing the same operation on all parts of the compound inequality.Subtract 18 to Isolate z: Subtract 18 from all parts of the compound inequality to isolate z. 6 ≤ z + 18 < 10 6 - 18 ≤ z + 18 - 18 < 10 - 18 -12 ≤ z < -8Check Solution: Check the solution to ensure no mathematical errors were made. We subtracted 18 from all parts of the inequality, which is the correct operation to isolate z. The resulting inequality -12 ≤ z < -8 is the solution to the original compound inequality.

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Solve for $z$ . $\newline$ 6 \leq z + 18 < 10 $\newline$ Write your answer as a compound inequality with integers. $\newline$ Choices: $\newline$ (A) 24 \leq z < -8 $\newline$ (B) -12 \leq z < -8 $\newline$ (C) -12 \leq z < 28 $\newline$ (D) 24 \leq z < 28

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Math Problems

Algebra 1

Solve compound inequalities

Full solution

Q. Solve for

z

\newline

6 \leq z + 18 < 10

\newline

Write your answer as a compound inequality with integers.

\newline

Choices:

\newline

(A)

24 \leq z < -8

\newline

(B)

-12 \leq z < -8

\newline

(C)

-12 \leq z < 28

\newline

(D)

24 \leq z < 28

Analyze Compound Inequality: Analyze the given compound inequality. $\newline$ The inequality 6 \leq z + 18 < 10 involves two inequalities combined: one is $6 \leq z + 18$ and the other is z + 18 < 10. We need to isolate $z$ in both inequalities by performing the same operation on all parts of the compound inequality.
Subtract $18$ to Isolate z: Subtract $18$ from all parts of the compound inequality to isolate z. $\newline$ 6 \leq z + 18 < 10 $\newline$ 6 - 18 \leq z + 18 - 18 < 10 - 18 $\newline$ -12 \leq z < -8
Check Solution: Check the solution to ensure no mathematical errors were made. $\newline$ We subtracted $18$ from all parts of the inequality, which is the correct operation to isolate $z$ . The resulting inequality -12 \leq z < -8 is the solution to the original compound inequality.