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A compound inequality is graphed, and its graph consists of all real numbers. Which open sentence could have resulted in this solution set? O r < 3 or r>-2 O r < 3 orr 3 orr 3 or r>-2



User Dawa
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The open sentence "r < 3 or r > -2" would result in a compound inequality graph with a solution set of all real numbers.

The open sentence that could have resulted in a compound inequality graph with a solution set of all real numbers is "r < 3 or r > -2".

Let's break it down step by step to understand why this open sentence would produce a solution set of all real numbers.

1. "r < 3" represents all values of r that are less than 3. This includes negative numbers, zero, and positive numbers that are less than 3.

2. "r > -2" represents all values of r that are greater than -2. This includes negative numbers greater than -2, zero, and positive numbers.

When we combine these two inequalities using the "or" operator, we are saying that the solution set includes all values of r that satisfy either "r < 3" or "r > -2".

Since these two conditions cover the entire number line, the solution set includes all real numbers. This means that any value of r will make the compound inequality true.

Therefore the open sentence "r < 3 or r > -2" would result in a compound inequality graph with a solution set of all real numbers.

User Kharda
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