Question

1. each number line below shows a set of numbers.

(a)

Explanation:

Step1: Determine the scale

From 0 to 5, there are 5 units (since 5 - 0 = 5) and the number of intervals between 0 and 5 is 5 (counting the small ticks). Wait, actually, looking at the number line, between 0 and 5, how many ticks? Let's see: from 0, the first tick, then second, then third, then the dot, then to 5. Wait, maybe the interval between each small tick is 1? Wait, no, let's check the positions. The first dot: let's count the ticks. From 0, each small tick is 1 unit? Wait, 0 to 5: the distance from 0 to 5 is 5, and if we look at the number of intervals between 0 and 5, let's see: 0, then 1, 2, 3 (the dot), 4, 5? Wait, no, the dot is at position 3? Wait, then the next dot: between 10 and 15. Let's see, 10 to 15: 5 units. The dot is at 12? Wait, 10, 11, 12 (the dot), 13, 14, 15. Wait, no, maybe the scale: from 0 to 5, there are 5 intervals? Wait, no, the number line has 0, then some ticks, then a dot, then to 5, then more ticks, then 10, then more ticks, then a dot, then to 15. Let's count the distance between 0 and 5: the number of small intervals. From 0 to the first dot: let's see, 0, then 1, 2, 3 (the dot), so that's 3 units from 0? Wait, no, maybe each small tick is 1 unit. So the first dot is at 3 (since 0 + 3 = 3), and the second dot is at 12 (since 10 + 2 = 12? Wait, no, 10 to 15: 5 units, with the dot at 12? Wait, 10, 11, 12 (dot), 13, 14, 15. So the first dot is at 3, the second at 12. And the line is between these two dots, including the dots (since they are filled). So the inequality is \( 3 \leq x \leq 12 \).

Wait, let's re-examine:

• The first dot: count the ticks from 0. Each small tick is 1 unit. So 0, 1, 2, 3 (dot). So that's 3.

• The second dot: from 10, count the ticks: 10, 11, 12 (dot). So that's 12.

• The line is between these two dots, including them (since the dots are filled), so the set of numbers is all x such that \( 3 \leq x \leq 12 \).

Step1: Identify the left endpoint

The left dot is at position 3 (counting from 0, each small tick is 1, so 0 + 3 = 3). So the lower bound is 3, with a closed circle (so inclusive, \( \leq \)).

Step2: Identify the right endpoint

The right dot is at position 12 (counting from 10: 10 + 2 = 12? Wait, no, 10, 11, 12: so 12. Closed circle, so inclusive, \( \leq \)).

Step3: Write the inequality

Since the line is between 3 and 12, including both endpoints, the inequality is \( 3 \leq x \leq 12 \).

Answer:

\( 3 \leq x \leq 12 \)