representation on number line

((-\infty,-1]) means (x\le -1), and ([4,\infty)) means (x\ge 4). The symbol (\cup) means the union of both parts.

The governing concept is translating endpoint symbols and shading into interval notation. A closed circle at -3 means that -3 is included, so the left endpoint requires a square bracket. An open circle at 3 means that 3 is excluded, so the right endpoint requires a parenthesis. Shading between the two marked points means that every real number between them is included. The solution is therefore -3 ≤ x < 3, written as the interval [-3, 3). Hence option D is correct. Option A excludes both endpoints, option B includes both, and option C excludes -3 but includes 3. Each of those alternatives contradicts at least one of the endpoint markings shown on the number line.

The conditions \(x<0\) and \(x>0\) represent all real numbers to the left and right of 0, respectively. Their union is \(( -\infty,0)\cup(0,\infty)\), which does not include 0. In option B, 0 is included, so it represents all real numbers. Exam tip: With strict inequalities \(<\) or \(>\), the boundary point is not included.

A closed interval [a, b] contains every number from a through b, including both endpoints. When the two endpoints are equal, as in [12, 12], there is no length between them and the only possible member is 12 itself. On a number line this set is represented by one closed point at 12, because the square brackets indicate inclusion. It is a singleton set, not an interval containing nearby numbers and not an empty set. Therefore option A is correct. An open circle would exclude 12, the whole line contains infinitely many other values, and “no solution” would mean that the set had no elements at all.

(\ge) includes the boundary (-\frac{5}{2}) and takes numbers to the right. Therefore it starts with a square bracket.

In \(x\le 3\), \(x=3\) is included because \(3\le 3\) is true. Therefore, a closed circle is drawn at 3 on the number line. An open circle is used only when the boundary value is excluded, as in \(x<3\). Exam tip: use a closed circle for \(\le\) and \(\ge\), and an open circle for \(<\) and \(>\).

(-4) is not included, so the first integer after it is (-3). On the number line, do not count an open endpoint as a member.

(4) is not included, so the greatest integer before it is (3). Do not take the open right endpoint as the answer.

An open circle on a number line means that endpoint is excluded. Since the circle at 0 is open, (x=0) is not in the solution set. The shaded interval can be written as \(0<x\leq 6\); 6 is included because it has a closed circle. Exam tip: use < or > for an open circle and \(\leq\) or \(\geq\) for a closed circle.

Parentheses exclude endpoints and square brackets include endpoints. This changes the endpoints on the number line.

The inequality \(x>2\) contains every value greater than 2, while \(x<2\) contains every value less than 2. Their union includes every real number except \(x=2\), because 2 satisfies neither inequality. Hence, 2 is left out. Exam tip: In strict inequalities \(>\) and \(<\), the boundary value is excluded and is shown by an open circle on the number line.

An open circle shows that 20 is not included in the solution, so the inequality must be strict. Shading to the right represents all numbers greater than 20. Therefore, the correct inequality is \(x>20\). In \(x\geq 20\), 20 would be included and shown with a closed circle. Exam tip: rightward shading means ‘greater than’, while an open circle means ‘not equal to’.

The direct answer is option C: the entire number line. The interval (-∞,∞) means every real number, with no lower or upper restriction. Infinity is not an endpoint that can be reached; it simply indicates that the set continues without limit in both directions. Option A, only on the left, represents a set extending toward negative infinity but does not include all positive numbers. Option B, only on the right, similarly misses negative numbers. Option C is correct because every point on the real number line belongs to the interval. Option D, nowhere, would represent an empty solution set, but this interval is the opposite of empty: it contains all real numbers. On a drawing, there is no endpoint circle; shade the complete line from left to right. Memory cue: both infinities together mean ‘all real numbers.’

In the interval \((-7,-2)\), parentheses at both ends mean that neither \(-7\) nor \(-2\) is included. Therefore, \(x\) must be greater than \(-7\) and less than \(-2\), giving \(-7<x<-2\). Option C is incorrect because it includes \(-7\). Exam tip: Use \(<\) or \(>\) for parentheses, and \(\le\) or \(\ge\) for square brackets.

Outward shading shows two separate parts and open circles exclude the boundaries. Hence ((-\infty,-1)\cup(4,\infty)) is correct.

A closed interval with equal endpoints includes only that number. Therefore one closed point is drawn at (-6).

The open circle excludes (-8), and the closed circle includes (2). Therefore the correct inequality is (-8<x\le 2).

The square bracket shows that \(-\frac{1}{2}\) is included, and the values extend toward \(-\infty\). Hence \(x\le -\frac{1}{2}\) is correct.

The strict sign (>) does not include the boundary, and greater numbers lie to the right. So use an open circle with rightward direction.

The sign (\le) includes (14) and smaller numbers are on the left. Use a closed circle for an included boundary.

The governing concept is reading both the endpoint symbol and the direction of shading. A closed or filled circle at 16 means that 16 itself belongs to the solution set, so the inequality must include equality. Shading to the right represents numbers greater than 16. Combining these facts gives x ≥ 16. Option A points to values less than 16 and excludes the endpoint. Option D points in the correct general direction but excludes 16 because it uses a strict greater-than sign. Option B includes 16 but shades to the left, representing x ≤ 16. Therefore option C is the only representation consistent with both the filled endpoint and rightward shading.

The square bracket shows that (9) is included and values extend toward \(-\infty\). Hence it is \(x\le 9\).

The governing concept is the graphical representation of a compound strict inequality. The statement -12 < x < -5 means that x must be greater than -12 and less than -5 simultaneously. Since both inequality signs are strict less-than signs, neither endpoint is included. Thus both -12 and -5 must be marked with open circles. The word ‘and’ requires the common region satisfying both conditions, which is the shaded segment between the two points. Therefore the interval is (-12, -5), and option A is correct. Option B incorrectly includes both endpoints. Option C includes -12 and does not describe the complete required marking, while option D represents values less than -12 rather than values lying between the endpoints.

The open circle excludes (-13) and the right side has greater numbers. Hence the interval is ((-13,\infty)).

In \(x>3\), the number 3 is not included in the solution set, so an open point is marked at 3 and the ray extends to the right. A filled point would represent \(x\geq3\), where 3 is included. Exam tip: use an open point for \(>\) or \(<\), and a filled point for \(\geq\) or \(\leq\).