representation on number line

((-1,4)) does not include (-1) and (4), so the complement includes both. Hence the outside regions have closed endpoints.

Solving gives (5-2x\le 3), then (-2x\le -2), and (x\ge 1). The boundary is included, so (1) is closed.

The first interval includes (0), and the second starts after (0). So no gap remains and the result is ([-5,4]).

In \(x\le3\), the endpoint 3 is included, so it is shown by a closed point. Numbers smaller than 3 lie to the left, so that side is shaded. Exam tip: use a closed point for \(\le\) or \(\ge\), not an open point.

The governing concept is intersection, which consists only of numbers common to both sets. The interval (-∞,1] contains every real number less than or equal to 1, so it includes 1. The interval [1,∞) contains every real number greater than or equal to 1, and it also includes 1. A number smaller than 1 fails to belong to the second interval, while a number greater than 1 fails to belong to the first. Consequently, the only common number is 1. The intersection is therefore the singleton set {1}, represented by one closed point at 1 on the number line. Option A is correct; the other choices either include too many points or incorrectly claim that no point is common.

Solving the inequality gives 4x+1>9, so 4x>8 and x>2. The boundary point 2 is not included because the inequality is strict; therefore, an open circle is placed at 2. All values greater than 2 lie to the right, so the right side is shaded. A closed circle would represent an inclusive inequality such as x≥2 or x≤2. Exam tip: use an open circle for > or < and a closed circle for ≥ or ≤.

Solving gives (-5x\le 15) and then (x\ge -3). The inequality sign reverses when dividing by a negative.

The common part is greater than (-2) and less than (4). Both endpoints are excluded in this intersection.

Combine the intervals: \((-\infty,-1)\cup[-1,5)\) includes all \(x\) less than \(5\) and also includes \(-1\) because \(-1\) belongs to \([-1,5)\). Since \(5\) is not included (\(5\) has a parenthesis), the simplified interval is \(( -\infty,5)\).

The governing concept is the complement of an interval in the set of real numbers. In [-4,2), the point -4 is included because of the closed bracket, while 2 is excluded because of the open parenthesis. The complement must contain all real numbers outside this interval. Numbers less than -4 are included, but -4 itself is not, giving (-∞,-4). Numbers greater than or equal to 2 are included, because 2 was excluded from the original interval, giving [2,∞). Therefore the complement is (-∞,-4) ∪ [2,∞), so option B is correct. Option A reverses both endpoint memberships, C gives the original interval, and D wrongly excludes 2.

Solving gives (5\le 2x-3<15) and then (4\le x<9). Therefore (4) is included but (9) is excluded.

In \(-4\le x<3\), \(-4\) is included, but \(3\) is not included. Thus, the integers are \(-4,-3,-2,-1,0,1,2\), giving a total of \(7\) points. Counting \(3\) would be incorrect because the inequality is \(x<3\). Exam tip: Include an endpoint with \(\le\) and exclude an endpoint with \(<\).

\(x>4\) means values greater than \(4\), so \(4\) itself is not included. Hence we use an open circle at \(4\) and shade to the right.

In (\le), the boundary number is included and smaller numbers lie to the left. In exams, a closed circle means included.

The direct answer is option A. The interval \((2,7]\) has two different bracket meanings. A round or open bracket at 2 means 2 is excluded, so the graph must have an open circle at 2. A square or closed bracket at 7 means 7 is included, so the graph must have a filled circle at 7. Every real number between them is included, so shade the segment between the two points. Thus the representation is an open circle at 2, a closed circle at 7, and shading between them. Option A matches exactly. Option B reverses both endpoint types. Option C incorrectly excludes 7 as well as 2. Option D incorrectly includes 2 and also includes 7. In inequality form, the same interval is \(2<x\le7\). Remember that a parenthesis means “not included” and a square bracket means “included”; the bracket closest to each number tells its status.

A closed circle includes (6), and numbers to the right are greater. In exams, always use an open bracket with (\infty).

An open circle excludes the boundary and the left side has smaller numbers. In exams, connect left shading with (<).

A filled circle shows that the endpoint itself belongs to the solution set. Hence, for \(x \le a\) and \(x \ge a\), a filled circle is marked at \(a\). In \(x < a\) or \(x > a\), the endpoint is excluded, so an open circle is used. Exam tip: associate \(\le\) and \(\ge\) with a filled circle, and \(<\) and \(>\) with an open circle.

In ([0,4)), (0) is included but (4) is not included. In exams, exclude the endpoint with an open bracket.

The interval \((-5,1)\) excludes \(-5\) and \(1\). The integers are \(-4,-3,-2,-1,0\). That is \(5\) integers.

Given \(2x+1<9\), subtracting 1 from both sides gives \(2x<8\). Dividing by 2 gives \(x<4\). Thus, 4 is not included, so the number line has an open circle at 4 and is shaded to the left for all values less than 4. Option B represents \(x\le4\), incorrectly including 4. Exam tip: use an open circle for \(<\) or \(>\), and a closed circle for \(\le\) or \(\ge\).

Dividing all parts by (2) gives (-3\le x\le 5). In exams, apply the same operation to every part of a compound inequality.

Multiplying all parts by (2) gives (2<x\le 8). In exams, check carefully whether each endpoint is included.

The integers are (-2,-1,0,1,2,3), so the total is (6). In exams, count both endpoints in a closed interval.

Both circles are open, so both endpoints are excluded. In exams, write an open circle using an open bracket.