What inequality represents this interval? (-∞, -1) Choices: (A)x > -1 (B)x ≥ -1 (C)x < -1 (D)x ≤ -1

Identify Endpoints: Identify the endpoints of the interval and whether they are included in the interval. The interval is (-∞, -1), which means that the endpoints are -∞ and -1. Since the interval notation uses parentheses, neither endpoint is included in the interval.Determine Direction: Determine the direction of the inequality. Since the interval extends from negative infinity up to but not including -1, the inequality must indicate that x is less than -1.Choose Inequality Symbol: Choose the correct inequality symbol. The inequality must be ＂less than＂ because -1 is not included in the interval. Therefore, the correct symbol is ＂<＂.Write Inequality: Write the inequality that represents the interval. The correct inequality is x < -1, which corresponds to choice (C).

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What inequality represents this interval? $\newline$ $(-\infty, -1)$ $\newline$ Choices: $\newline$ (A) x > -1 $\newline$ (B) $x \geq -1$ $\newline$ (C) x < -1 $\newline$ (D) $x \leq -1$

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Q. What inequality represents this interval?

\newline

(-\infty, -1)

\newline

Choices:

\newline

(A)

x > -1

\newline

(B)

x \geq -1

\newline

(C)

x < -1

\newline

(D)

x \leq -1

Identify Endpoints: Identify the endpoints of the interval and whether they are included in the interval. The interval is $(-\infty, -1)$ , which means that the endpoints are $-\infty$ and $-1$ . Since the interval notation uses parentheses, neither endpoint is included in the interval.
Determine Direction: Determine the direction of the inequality. Since the interval extends from negative infinity up to but not including $-1$ , the inequality must indicate that $x$ is less than $-1$ .
Choose Inequality Symbol: Choose the correct inequality symbol. The inequality must be ＂less than＂ because $-1$ is not included in the interval. Therefore, the correct symbol is ＂<＂.
Write Inequality: Write the inequality that represents the interval. The correct inequality is x < -1, which corresponds to choice $(C)$ .