representation on number line

In (x<5), (5) is not included so a parenthesis is used at (5). Always use a parenthesis with infinity.

The sign (\ge) includes (0), and greater numbers are to the right. So it is a closed circle with a right ray.

The governing concept is interpreting an endpoint and direction on a number line. An open circle at 6 means that 6 is excluded, so the inequality must be strict rather than inclusive. A ray extending to the left represents values smaller than 6, because numbers decrease as one moves left. Combining these facts gives x < 6, so option C is correct. Option D has the correct direction but includes 6, which would require a closed circle. Options A and B describe values to the right of 6 and therefore conflict with the ray’s direction; B also includes the endpoint. The graph can be written in interval notation as (-∞, 6), confirming that 6 is not included.

In ([-3,2)), (-3) is included but (2) is not. The open endpoint is placed at the number not included.

Interval notation records whether each endpoint is included. The inequality \(1<x\) is strict, so the value 1 is excluded and must be written with a parenthesis. The condition \(x\le 8\) includes 8, so the right endpoint uses a square bracket. Combining both conditions gives the interval \((1,8]\), which is option C. On a number line, this is shown by an open circle at 1, a closed circle at 8, and a shaded segment between them. Option A incorrectly includes 1, option B incorrectly excludes 8, and option D includes 1 while excluding 8. Thus the endpoint symbols directly follow the inequality signs.

The governing concept is combining endpoint inclusion with the direction of shading. A closed circle at -5 means that -5 belongs to the solution set, so the inequality must include equality. Shading to the right represents numbers greater than -5 because values increase from left to right on the number line. Therefore the inequality is x ≥ -5, making option D correct. Option C has the correct rightward direction but excludes -5 and would need an open circle. Options A and B point left, representing values less than -5; B includes the endpoint but has the wrong direction. In interval notation, the graph is [-5, ∞), which also confirms inclusion of -5 and extension toward larger values.

In (x\le 4), (4) is included so a square bracket is used at (4). Smaller numbers extend toward (-\infty).

Parentheses mean that boundary numbers are not included. Thus both (-4) and (3) have open circles.

In the interval \([0,5]\), square brackets show that both endpoints, 0 and 5, are included. Hence, \(x\) is not less than 0 and not greater than 5, so \(0\le x\le 5\). In option C, 5 is excluded, so it does not represent \([0,5]\). Exam tip: use \(\le\) for square brackets and \(<\) for round brackets.

In \(x>-7\), the solution consists of values greater than \(-7\), but \(-7\) itself is not included. Therefore, an open circle is drawn at \(-7\), and the ray extends to the right. Although \(0\) and \(7\) satisfy the inequality, they are not boundary points. Exam tip: use an open circle for \(>\) or \(<\), and a closed circle for \(\ge\) or \(\le\).

The sign (<) shows smaller numbers, and smaller numbers are on the left of the number line. Identifying direction is the first step.

The sign (\ge) includes the boundary and greater numbers lie to the right. So a closed circle with a right ray is drawn.

An open circle means that endpoint is not included in the solution set. Since the shading lies between 2 and 6 and both endpoints are open, \(x\) must be greater than 2 and less than 6: \(2<x<6\). Options C and D include one endpoint, so they are incorrect. Exam tip: use \(<\) or \(>\) for an open circle, and \(\le\) or \(\ge\) for a filled circle.

The interval ([4,\infty)) includes (4) and all larger numbers. Hence a closed circle with a right ray is correct.

(-6) is included so a square bracket is used, and (1) is not included so a parenthesis is used. Read brackets to decide endpoints.

(x=3) means that x has exactly one value, 3. Therefore, it is shown by a closed point at 3 on the number line. An open circle is usually used for an excluded boundary value, such as in x > 3 or x < 3. Rays represent inequalities with many solutions. Exam tip: for '=', mark only the stated number with a point.

An open circle means the boundary value is not part of the solution. In such questions, check the type of circle first.

The governing concept is endpoint inclusion on a number line. A closed, or filled, circle means that the boundary value itself belongs to the solution set. This notation is used with inclusive inequalities such as x ≤ b or x ≥ b. For example, x ≤ b contains b as well as every number smaller than b, so the point b must be filled. An open circle would instead indicate that b is excluded, as in x < b or x > b. The direction of shading tells which side is selected, but the closed circle specifically communicates inclusion. Therefore option A is correct; option B describes an open circle, while options C and D do not follow merely from the presence of a closed endpoint.

The interval ((3,\infty)) does not include (3) and contains larger numbers. So it uses an open circle and a right ray.

The governing rule is that a strict inequality excludes its endpoint, whereas a non-strict inequality includes it. In -8 < x ≤ -3, the sign < beside -8 means that -8 is not part of the solution, so an open circle is placed at -8. The sign ≤ beside -3 means that -3 is included, so a closed circle is placed at -3. The interval is therefore (-8, -3], with the segment between the two endpoints shaded. Option A incorrectly includes -8, option C reverses the endpoint markings, and option D excludes both endpoints. Hence option B gives the exact number-line representation.

A parenthesis gives a strict inequality and a square bracket gives an inclusive inequality. Therefore (-2<x\le 4) is correct.

The governing concept is conversion between an inequality and interval notation. The inequality x > 0 selects every real number greater than zero, but it does not select zero itself. Therefore the endpoint 0 is written with a parenthesis, not a square bracket. The solution continues without bound toward positive infinity, and infinity is always written with a parenthesis because it is not an actual real endpoint. Thus the interval is (0, ∞), making option B correct. Option A incorrectly includes zero, while options C and D describe negative numbers, with D also including zero. The inequality sign and endpoint notation must agree.

In (x\ne 2), all real numbers except (2) are included. The excluded point is shown by an open circle.

In \(-4\le x\), reversing the order of the two sides gives \(x\ge -4\). It means that \(x\) is equal to or greater than \(-4\). On a number line, mark a closed point at \(-4\) and shade to the right. The closest distractor, \(x>-4\), is incorrect because it excludes \(-4\) itself. Exam tip: \(\le\) and \(\ge\) include equality, so use a closed point on the number line.

In (x\le 10), the boundary is (10), and smaller numbers are on the left. Equality means the point is closed.