representation on number line

An open circle shows that the endpoint is not part of the solution set. It is generally used for strict inequalities, < or >. A filled circle represents an included endpoint, so option B is incorrect. Exam tip: remember open circle = < or >, filled circle = ≤ or ≥.

The inequality \(x>21\) represents all numbers greater than 21. Greater numbers lie to the right of 21 on a number line, so an open circle is marked at 21 and the arrow extends rightward. A leftward arrow would represent \(x<21\). Exam tip: For \(>\) or \(\ge\), shade or draw the ray to the right.

The inequality x ≥ -18 contains -18 because the symbol ≥ means ‘greater than or equal to.’ It also contains every real number larger than -18, so the solution extends indefinitely to the right. In interval notation, an included finite endpoint is written with a square bracket, while infinity always uses a parenthesis because infinity is not an actual number that can be included. Therefore the set is [-18, ∞). Option A and option D describe values extending leftward, and option C wrongly excludes -18. Thus option B is correct.

Since \(3\le x\) includes \(3\), we use a closed bracket at \(3\). Since \(x\le 11\) includes \(11\), we use a closed bracket at \(11\). So the interval is \([3,11]\).

A closed or filled point means that the marked endpoint is included in the solution set. Because the diagram contains only one closed point at -4 and no shaded ray or interval extending to other numbers, the set contains exactly one value. Therefore the solution is \(x=-4\), making option C correct. A leftward ray would represent \(x<-4\), and a rightward ray would represent \(x>-4\); neither is shown. The statement \(x\ne -4\) would include every other number and would normally be represented by an open point at -4 with both directions shaded. Thus the visual information describes a singleton set, not an inequality interval.

In ((-\infty,-16)), (-16) is not included and smaller numbers are taken. So an open circle and left ray are drawn.

(-9) has a square bracket so it is included and (-2) has a parenthesis so it is excluded. This decides the endpoints.

In (x\ne 7), all real numbers except (7) are included. The excluded value is shown by an open circle.

Since \(-6\) is not included, an open bracket \((\) is used at \(-6\). Since \(4\) is included, a closed bracket \(]\) is used at \(4\). Therefore, the interval is \((-6,4]\). In \([-6,4)\), \(-6\) is included and \(4\) is excluded, so it is not correct. Exam tip: use a square bracket for an included endpoint and a round bracket for an excluded endpoint.

In ([-\frac{3}{2},2)), (-\frac{3}{2}) is included but (2) is not. Therefore (-\frac{3}{2}) is a correct member.

The first part is the left side including (5), and the second part is the right side after (5). Together they give the whole number line.

The direct answer is option B, (-∞,-7)∪(12,∞). The condition x<-7 describes every number strictly less than -7, so its interval is (-∞,-7); the round bracket at -7 shows that -7 is excluded. The condition x>12 describes numbers strictly greater than 12, giving (12,∞); 12 is also excluded. Since the word ‘and’ here asks to show both separate regions together on the number line, the union symbol ∪ joins them. Option A, (-7,12), contains numbers between the endpoints, which is the opposite region. Option B correctly shows both outer parts and excludes both endpoints. Option C uses square brackets and incorrectly includes -7 and 12. Option D, [-7,12], shows the middle closed interval, not either strict inequality. Exam cue: < and > use round brackets; separate pieces use union.

\(10>x\) means that \(x\) is less than 10. Writing the variable on the left gives \(x<10\). \(x\le 10\) is incorrect because equality at 10 is not included in the original inequality. Exam tip: when the two sides of an inequality are interchanged, reverse the inequality sign.

The governing concept is the relationship between shading and a solution set on a number line. Shading identifies the real numbers that satisfy the given equation or inequality. If no point and no interval is shaded at all, then no real number satisfies the condition. The solution set is therefore the empty set, written as ∅ or {}. Option A, all real numbers, would require the entire number line to be shaded. Option C, only 1, would require a marked point at 1, and option D, only negative numbers, would require shading on the negative half-line. Since none of those graphical features is present, the correct interpretation of no shading is no solution, or the empty set. Hence option B is correct.

The open interval (8, 8) is defined as the set of real numbers x satisfying 8 < x < 8. These two strict inequalities cannot be true at the same time, because a number cannot be both greater than 8 and less than 8. The endpoints are identical, so there is no distance or region between them on the number line. Moreover, the round brackets exclude the endpoint 8 itself, so even 8 is not included. Thus the interval contains no real number and is represented by the empty set, written as ∅. Option A is incorrect because 8 is excluded; option B includes only numbers to the right of 8; and option D includes every real number. Therefore option C is correct.

When the entire number line is shaded, every real number is included in the solution set. Hence, the correct notation is \(x\in \mathbb{R}\). The inequalities \(x>0\) and \(x<0\) represent only one side of the number line, while \(x=0\) represents just one point. Exam tip: A fully shaded number line corresponds to the solution set \(\mathbb{R}\).

In (x<-\frac{9}{4}), the boundary is excluded and smaller numbers are included. Hence ((-\infty,-\frac{9}{4})) is correct.

In \(-1\le x<13\), the symbol \(\le\) includes \(-1\). However, the strict symbol \(<\) excludes \(13\), so \(x=13\) is not allowed. Therefore, \(13\) is the endpoint that is not included. Exam tip: Use a closed dot for \(\le\) or \(\ge\), and an open dot for \(<\) or \(>\) on a number line.

The inequality x ≥ 4 includes 4 because the symbol ≥ permits equality. It also includes every real number greater than 4, so the solution extends indefinitely to the right of 4 on the number line. In interval notation, inclusion of a finite endpoint is shown with a square bracket; therefore the left endpoint is written as [4. The symbol infinity is not a real number and is never included as an endpoint, so a round bracket is used at infinity. Option A has the correct rightward direction but excludes 4. Options C and D describe values extending to the left of 4 and therefore represent the wrong inequality. Hence [4, ∞), option B, is correct.

For a strict inequality such as \(x>7/3\), the boundary value \(7/3\) is not part of the solution set. On a number line, an excluded boundary is represented by an open circle, also called an empty circle. The line or ray would then be shaded to the right because the permitted values are greater than \(7/3\). Therefore option B, “Open circle,” is correct. A closed circle is used for \(\ge\) or \(\le\), where equality is included. A double closed circle has no standard role in this representation, and the inequality certainly has a boundary point even though that point itself is excluded.

The square bracket includes (\frac{2}{3}), and the parenthesis excludes (\frac{8}{3}). So the first endpoint is closed and the second is open.

The governing idea is interpreting a strict inequality on a number line. The statement x > −10 means that x must be greater than −10, so the endpoint −10 is excluded and is represented by an open circle. The solution region extends to the right. Test the choices: −12 and −15 are less than −10 and lie to the left, while −10 itself is the excluded boundary. The number −9 is greater than −10 and lies to the right, so it belongs to the shaded region. Remember that among negative numbers the one closer to zero is larger; therefore −9 > −10. Hence option C is the only value satisfying the inequality.

The shaded portion lies between 18 and 25, so x cannot be less than 18 or greater than 25. A closed endpoint means that the corresponding boundary value is included in the solution. Since both 18 and 25 are shown with closed endpoints, both values must be included. The resulting compound inequality is 18 ≤ x ≤ 25, and its interval notation is [18, 25]. Option A uses strict inequalities and excludes both endpoints. Option B includes 18 but excludes 25, while option C excludes 18 but includes 25. Neither matches two closed endpoints. Therefore option D is the only inequality that accurately represents the shaded segment.

In (x\le -22), (-22) is included and smaller numbers are taken. So a closed circle with a left arrow is used.

The union symbol ∪ means that a number may belong to either interval. The first interval, (-∞, -6], contains every real number less than or equal to -6; the square bracket includes -6. The second interval, [10, ∞), contains every real number greater than or equal to 10; the square bracket includes 10. Thus the graph has two separate shaded rays: one extending left from -6 and another extending right from 10. Option A incorrectly uses strict inequalities and excludes both boundary values. Options C and D describe the central interval between -6 and 10, which is not included at all. Therefore the equivalent inequality is x ≤ -6 or x ≥ 10, making option B correct.