representation on number line

An open circle means that 2 is not included in the solution set, while shading to the left represents all values less than 2. Therefore, the inequality is \(x<2\). For \(x\leq 2\), the point at 2 would be a filled circle. Exam tip: open circle means the endpoint is excluded; filled circle means it is included.

In \(x<0\), all numbers less than 0 are included, but \(0\) itself is not included in the solution. Therefore, an open circle is drawn at \(0\), with the line or arrow extending to the left. A closed circle would be used for \(x\leq 0\), because 0 is included there. Exam tip: use an open circle for \(<\) or \(>\), and a closed circle for \(\leq\) or \(\geq\).

Because (3\le x), (3) is included, and because (x<9), (9) is not included. So (3) is closed and (9) is open.

The governing concept is testing possible values in a strict one-variable inequality. For x > -2, a value belongs to the solution set only when it is strictly greater than -2; equality is not sufficient. Check each option: -3 is less than -2, so it fails; -2 equals -2, so it is excluded; -1 is greater than -2, so it satisfies the inequality; and -5 is less than -2, so it fails. Therefore option C, -1, is correct. On a number line the graph has an open circle at -2 and shading to the right, and -1 lies in that shaded region. If the inequality had been x ≥ -2, the endpoint -2 would also have been included, but the strict symbol > excludes it.

Since (5) is not included, a parenthesis is used, and smaller numbers extend toward (-\infty). Hence ((-\infty,5)) is correct.

(x\le 2) includes all numbers up to (2), and (x>2) includes all numbers after (2). Together they form the whole number line.

Square brackets include endpoints and parentheses exclude endpoints. This creates the open or closed endpoint difference on the number line.

The parenthesis excludes (-5), while the square bracket includes (-1). Therefore (-5) is not included.

The governing concept is the meaning of the set of real numbers, denoted by ℝ. Every point on the ordinary number line corresponds to a real number, including negative numbers, zero, and positive numbers. Therefore, the statement x ∈ ℝ places no further restriction on x: every real value is allowed. Its solution set is the whole real line, written in interval notation as (-∞, ∞). A correct number-line representation must consequently shade the complete line. Option B represents only x > 0, option C represents only x < 0, and option D represents the singleton set {0}. Since none of those options contains all real numbers, option A is the only correct representation.

The direct answer is option D, -2. The inequality x≥-1 means x may be -1 or any number greater than -1. The endpoint -1 is included because the symbol ≥ means greater than or equal to. Check each choice directly: option A, -1, satisfies -1≥-1, so it belongs to the solution. Option B, 0, is greater than -1, so it belongs. Option C, 2, is also greater than -1, so it belongs. Option D, -2, is less than -1, so it does not satisfy the inequality and is not in the solution. On a number line, use a filled dot at -1 and shade to the right. A common mistake is to use an open circle for ≥; an open circle is used only when the endpoint is excluded, such as >.

((-\infty,\infty)) represents the whole real number line. Infinity is not an endpoint, so parentheses are used with it.

For \(x< -4\), the value \(-4\) is not included, so we use a parenthesis at \(-4\). All numbers less than \(-4\) extend to \(-\infty\). Hence the interval is \(( -\infty,-4)\).

Strict inequalities exclude the endpoints and the middle part is the solution. Therefore both endpoints remain open.

The governing concept is the interpretation of a degenerate closed interval. In [7, 7], the lower endpoint and upper endpoint are both 7. The square brackets indicate that the endpoint is included, so the interval means 7 ≤ x ≤ 7. The only real number satisfying both inequalities is x = 7; hence the interval is the singleton set {7}. On a number line, this set is shown by one filled or closed point at 7, with no shaded segment extending to either side. Option A would describe the empty set. Options C and D would describe rightward and leftward rays, respectively, and therefore contain infinitely many numbers rather than just 7. Thus option B is correct.

In ((7,7)), no number can be greater than (7) and less than (7) at the same time. So it is an empty set.

In \(-3\le x<0\), \(-3\) has \(\le\), so it is included and is shown by a closed dot on the number line. Zero has \(<\), so it is excluded and is shown by an open dot. Hence, the interval is \([-3,0)\). Option A incorrectly excludes \(-3\) as well. Exam tip: use a square bracket for \(\le\) or \(\ge\), and a round bracket for \(<\) or \(>\).

The sign (\le) includes equality, so the boundary (-6) is in the solution. An included boundary is shown by a closed circle.

The direct answer is option C: \([-2,5)\). Interval notation records whether each endpoint is included. A closed dot means the endpoint belongs to the solution, so it uses a square bracket. An open dot means the endpoint is excluded, so it uses a round parenthesis. Here \(-2\) is closed, so the interval starts with \([\). The number 5 is open, so the interval ends with \) . Therefore the interval is \([-2,5)\), meaning \(-2\le x<5\). Option A, \((-2,5)\), excludes both endpoints and therefore wrongly excludes \(-2\). Option B, \([-2,5]\), includes 5 even though its point is open. Option C correctly includes \(-2\) and excludes 5. Option D, \((-2,5]\), does the opposite: it excludes \(-2\) and includes 5. The order of the brackets matters, because it tells exactly what happens at each boundary. Memory cue: square bracket means “included”; round bracket means “not included.”

Two graphical rules must be applied together. First, a closed circle at 8 means that 8 is included, so the inequality must contain ≤ or ≥ rather than a strict sign. Second, shading to the left represents numbers smaller than 8. Combining these facts gives x ≤ 8: it includes 8 and every real number less than 8. Thus option B is correct. Option A has the correct direction but uses an open endpoint, so it excludes 8. Option D includes 8 but shades to the right, and option C both excludes 8 and selects the right side. The endpoint marking and shading direction must both agree.

An open circle means (-6) is not included, and a closed circle means (2) is included. Therefore the correct inequality is (-6<x\le 2).

The sign (>) excludes the boundary (11) and greater numbers lie to the right. Check the circle type first and then the direction.

(\ge) includes the boundary and greater numbers are on the right. Therefore a closed circle is drawn at (-9).

A number-line representation communicates two facts: the direction of shading identifies smaller or larger values, and the endpoint symbol identifies inclusion. An open circle at 13 means that 13 itself is excluded, so the inequality must use a strict sign rather than ≤ or ≥. Shading to the left represents numbers smaller than 13. Combining these facts gives x < 13. Therefore option C is correct. The statement x ≤ 13 would require a closed circle at 13 because equality would be allowed. The right-pointing alternatives describe values greater than 13 and do not match the leftward shading.

The endpoint with a square bracket is included and the endpoint with a parenthesis is excluded. Therefore (-4) is closed.

The inequality \(4<x<12\) means that x must be greater than 4 and less than 12 at the same time. The endpoints 4 and 12 are not allowed because the signs are strict: the statement does not say \(4\le x\) or \(x\le12\). On a number line, an excluded endpoint is shown with an open circle. Every number strictly between the endpoints is included, so that interval is shaded.

Mark an open circle at 4 and another open circle at 12, then shade only the segment between them. Numbers such as 5, 8, and 11 belong to the solution, while 4 and 12 do not. A closed circle would incorrectly include an endpoint, and shading to the right would represent values greater than 4 without the upper restriction. Therefore option C is correct.