representation on number line

The closed circle includes (7) and the right ray shows numbers greater than (7). Hence the interval is ([7,\infty)).

Parentheses exclude both boundary points from the solution. Therefore both (-8) and (0) have open circles.

In (x\le -15), (-15) is included and smaller numbers are taken. Infinity always uses a parenthesis.

Equality is included on both sides, so both endpoints are included. In this case, square brackets are used at both ends.

A single closed point at 3 means that 3 is included in the solution set and no other value is included. Hence, the solution is \(x=3\). For \(x>3\) or \(x<3\), a ray would extend to the right or left of the point respectively. Exam tip: one isolated closed point represents exactly one value.

In ((-\infty,6)), (6) is not included and smaller numbers are present. So an open circle at (6) and a left ray are drawn.

(5) has a parenthesis, so it is excluded, and (10) has a square bracket, so it is included. This decides the open and closed endpoints.

In (x\ne -3), all real numbers except (-3) are included. The excluded point is shown by an open circle.

Interval notation uses a square bracket when an endpoint belongs to the solution set and a round parenthesis when it does not. The statement says that the interval runs from 1 to 5 and only 1 is included. Therefore 1 receives a square bracket, while 5 receives a parenthesis. The resulting interval is [1, 5), meaning 1 ≤ x < 5. Option A excludes both endpoints, option B includes both, and option D excludes 1 but includes 5. Since those endpoint conditions contradict the wording, option C is the only correct answer. The notation and the corresponding inequality express exactly the same set of real numbers.

The governing concept is endpoint inclusion in interval notation. The interval (0, 8] represents all real numbers x satisfying 0 < x ≤ 8. The parenthesis at 0 means that zero is excluded, while the square bracket at 8 means that eight is included. Testing the choices makes the result clear: 0 fails because it is not greater than 0; 9 fails because it exceeds the upper bound; and -1 fails because it is less than the lower bound. The value 8 satisfies both conditions, since 0 < 8 and 8 ≤ 8. Therefore option C is correct. This example shows why the distinction between a parenthesis and a square bracket is essential when reading an interval.

The symbol (\cup) is used to combine two separate parts. Here both (-2) and (5) are not included.

A filled dot means that the boundary value 3 is included in the solution set, and shading to the left represents all values less than 3. Therefore, the inequality is \(x \leq 3\). For \(x < 3\), the value 3 would not be included, so the dot would be open. Exam tip: a filled dot corresponds to \(\leq\) or \(\geq\), while an open dot corresponds to \(<\) or \(>\).

\((-5<x)\) means \(-5<x\), which is the same as \(x>-5\) by swapping sides of the inequality sign.

An open circle shows that 4 is not included in the solution set, and shading to the right represents numbers greater than 4. Therefore, the inequality is \(x>4\). For \(x\geq 4\), the point at 4 would be shown with a closed circle. Exam tip: right-side shading means ‘greater than’, while left-side shading means ‘less than’.

No number can be greater than (3) and less than (3) at the same time. Therefore ((3,3)) has no member.

A shaded entire number line means that negative numbers, zero, and positive numbers are all included in the solution set. Hence, \(x\in \mathbb{R}\) is correct. \(x>0\) and \(x<0\) represent only one part of the number line, not the whole line. Exam tip: when the complete number line is shaded, the solution set is all real numbers.

The direct answer is option C: \((-6,\infty)\). The inequality \(x>-6\) includes every real number greater than \(-6\), but it does not include \(-6\) itself. In interval notation, an excluded finite endpoint uses a round parenthesis, so the left side is \((-6\). The values continue without an upper limit, represented by \(\infty\), and infinity always uses a round parenthesis because it is not an actual number. Hence the interval is \((-6,\infty)\). Option A, \([-6,\infty)\), wrongly includes \(-6\). Option B, \((-\infty,-6)\), describes numbers less than \(-6\), so it points in the opposite direction. Option C correctly describes all numbers greater than \(-6\). Option D, \((-\infty,-6]\), also describes the wrong side and includes \(-6\). For a quick check, 0 belongs because \(0>-6\), while \(-7\) does not. Memory cue: greater than means move right; strict inequalities use an open endpoint.

The governing concept is interpreting inclusive and strict inequality signs at the endpoints. The condition 2 ≤ x permits x to equal 2, so 2 belongs to the solution set and is shown by a closed circle. The condition x < 6 requires x to be strictly less than 6; equality is not allowed, so 6 is excluded and is shown by an open circle. Combining the two conditions gives the interval [2, 6). Therefore the endpoint that is not included is 6, making option B correct. Option A incorrectly excludes 2. Option C claims that both endpoints are excluded, while option D claims that neither is excluded. Both statements conflict with the inequality symbols and the interval notation.

The inequality \(x\le 0\) includes 0 and every number less than 0. Therefore, 0 has a closed bracket \(]\), and the interval extends left towards \(-\infty\): \(( -\infty, 0 ]\). Option B excludes 0, so it represents \(x<0\). Exam tip: use a closed bracket at an endpoint for \(\le\) or \(\ge\).

The inequality \(x<\frac{3}{2}\) is strict, so \(\frac{3}{2}\) itself is not included in the solution set. Therefore, an open circle is drawn at \(\frac{3}{2}\), with the line shaded to its left. A closed circle would be used for an inequality such as \(x\leq\frac{3}{2}\). Exam tip: Use an open circle for < or >.

In the interval \([\frac{1}{2},\frac{5}{2}]\), square brackets include both boundary values. Hence, closed circles are drawn at both \(\frac{1}{2}\) and \(\frac{5}{2}\), with the segment between them represented. Open circles are used when an endpoint is not included. Exam tip: remember that square brackets mean closed circles, while parentheses mean open circles.

The inequality \(x\le -4\) includes values equal to \(-4\) or less than \(-4\). Since \(-5<-4\), \(-5\) lies in the shaded region. In contrast, \(-3\) is greater than \(-4\), so it is not included. Exam tip: among negative numbers, the number farther left on the number line is smaller.

An open endpoint means that the boundary value is not included in the solution. A strict inequality, shown by < or >, always excludes its boundary. Thus an open circle is used at that endpoint. A closed circle would be used only for an inequality containing equality, such as \(\le\) or \(\ge\).

Here the interval is \(-2<x<1\). The value \(-2\) is excluded because \(x\) must be greater than it, and the value \(1\) is excluded because \(x\) must be less than it. Therefore, the number line has an open circle at both \(-2\) and \(1\), giving a total of two open endpoints. Hence option C follows.

A closed endpoint on a number line means that the endpoint belongs to the solution set and is represented by an inclusive inequality sign. Since both -11 and -2 are shown with closed circles, both boundary values must be included. The shaded portion between them therefore consists of all x satisfying -11 ≤ x ≤ -2. Option A excludes both endpoints, option B includes only -11, and option C includes only -2. None of those matches the statement that both ends are closed. Hence option D is correct. In interval notation, the same solution set is [-11, -2], which confirms the two inclusive inequality signs.

In (x\ge 18), (18) is included and numbers greater than it are taken. So a closed circle with a right arrow is used.