representation on number line

The interval notation is determined independently at each endpoint. An open circle at 1 means that 1 is excluded, so the interval must begin with a parenthesis. A closed circle at 8 means that 8 is included, so the interval must end with a square bracket. Shading between the points means that every number greater than 1 and less than or equal to 8 is selected. Therefore the interval is (1, 8]. Option A includes 1 incorrectly, option B excludes 8 incorrectly, and option D reverses both endpoint symbols. Hence option C is correct.

The strict inequalities x<6 and x>6 together include every real number except 6. Therefore, the number line has an open point at 6 and is shaded on both sides. Option B is incorrect because it includes 6. Exam tip: use open endpoints for < and >, and closed endpoints for ≤ and ≥.

The direct answer is option A: only a closed point at -14. The interval [-14,-14] has the same number at both ends, so it contains exactly one real number: -14. Square brackets mean that the endpoint is included, and because both endpoints coincide, the interval collapses to a single point. Option A is therefore correct: draw one filled or closed dot at -14. Option B, an open circle, would represent exclusion of -14, as in an interval with round brackets, so it is wrong. Option C, the whole line, would be represented by (-∞,∞), not this one-point interval. Option D, no solution, is also wrong because -14 is included. A useful check is to ask which values lie between -14 and -14: only -14 itself. Remember that [a,a] represents the single value a.

(\le) includes the boundary (\frac{5}{6}) and takes numbers to the left. Therefore ((-\infty,\frac{5}{6}]) is correct.

An open circle means that 2 is excluded, while a ray to the right represents values greater than 2. Hence, \(x>2\) is correct. \(x\geq2\) would require a filled circle at 2. Exam tip: rightward shading means greater values.

The notation (-9, -3] describes all real numbers greater than -9 and less than or equal to -3. The round bracket at -9 means that -9 itself is excluded, while the square bracket at -3 means that -3 is included. Among the integers greater than -9, the first one encountered is -8, and it also satisfies -8 ≤ -3. Therefore, -8 is the smallest integer in the interval, so option B is correct. Option A is rejected because the left endpoint is open. Option C is in the interval but is not the smallest member. Option D is smaller than -9 and lies outside the interval.

(12) is not included, so the greatest integer before it is (11). Do not take the open right endpoint as the answer.

The open point at \(-4\) shows that this boundary value is excluded from the solution set, while the closed point at \(3\) shows that \(3\) is included. Therefore, \(x=-4\) is not a solution in this interval. Exam tip: an open circle represents < or >, whereas a closed circle represents ≤ or ≥.

Parentheses exclude endpoints and square brackets include endpoints. This creates the endpoint difference on the number line.

The inequality (x>9) includes all numbers greater than 9, while (x<9) includes all numbers less than 9. Their union contains every real number except 9. Since both inequalities are strict, 9 is marked with an open circle on the number line. Exam tip: In x>9 or x<9, the boundary value 9 is not included.

An open circle means that \(-20\) is not included in the solution set. Shading to the left represents all numbers less than \(-20\). Therefore, the inequality is \(x < -20\). In contrast, \(x \le -20\) would include \(-20\), so it would use a closed circle. Exam tip: an open circle corresponds to \(<\) or \(>\), while a closed circle corresponds to \(\le\) or \(\ge\).

The interval (-∞, ∞) is the standard interval notation for the set of all real numbers. Infinity is not an endpoint that can be included or excluded like an ordinary number; the notation simply indicates that the interval extends without bound in both directions. Consequently, every real-number point on the number line satisfies x ∈ (-∞, ∞), so the complete number line must be shaded. Option A would represent only numbers extending toward negative infinity, and option B would represent only numbers extending toward positive infinity. Option D would represent an empty set, which this interval certainly is not. Hence option C is correct.

The interval \((-15,-7)\) has parentheses at both ends, so neither \(-15\) nor \(-7\) is included. Therefore, the correct inequality is \(-15<x<-7\). Option A is incorrect because it includes both endpoints. 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,-4)\cup(11,\infty)) is correct.

A closed interval with equal endpoints includes only that number. Therefore one closed point is drawn at (\frac{4}{5}).

The closed circle includes (-\frac{5}{2}), and greater numbers lie to the right. Therefore (x\ge -\frac{5}{2}) is correct.

In the interval \((-3,7]\), the square bracket after 7 shows that 7 belongs to the solution set. Hence, a filled dot is marked at x=7 on the number line, whereas -3 is shown with an open dot. Exam tip: a square bracket indicates inclusion, while a parenthesis indicates exclusion.

The open circle excludes (0), and smaller numbers lie to the left. Hence the interval is ((-\infty,0)).

The direct answer is option D: \(2\le x<9\). A closed endpoint means that endpoint is included, so it uses \(\le\). An open endpoint means that endpoint is excluded, so it uses \(<\). The shaded part is between 2 and 9. Since 2 is closed, write \(x\ge2\); since 9 is open, write \(x<9\). Combining them gives \(2\le x<9\). Option A, \(2<x<9\), wrongly excludes 2. Option B, \(2\le x\le9\), wrongly includes 9. Option C, \(2<x\le9\), excludes 2 and includes 9, so both endpoints are reversed. Option D correctly matches the two endpoint markings. On a number line, a filled or closed dot means included and an empty or open circle means excluded. Memory trick: closed circle goes with the inequality having an equal sign; open circle goes with a strict inequality.

A number-line endpoint carries inclusion information. An open circle at −1 means that −1 is excluded, so the inequality must be strict there: x > −1. A closed circle at 6 means that 6 is included, so the inequality must use x ≤ 6. Combining the two conditions gives −1 < x ≤ 6, represented by the interval (−1, 6]. Therefore option C is correct. Option B incorrectly excludes 6, option A incorrectly includes −1 and excludes 6, and option D incorrectly includes both endpoints. The direction of shading between the endpoints confirms that all values in this bounded interval are intended.

Interval notation records whether each endpoint belongs to the represented set. A round parenthesis means the endpoint is open and excluded, whereas a square bracket means the endpoint is closed and included. Since a is marked open, the interval contains numbers greater than a but not a itself. Since b is marked closed, b is included, so the upper condition is less than or equal to b. Combining these conditions gives a < x ≤ b, written as (a,b]. Hence option B is correct. Option A includes both endpoints and is wrong at a; option C includes a but excludes b; option D excludes both endpoints. The order a < b ensures this is a valid nonempty interval.

The endpoint symbols in an inequality determine whether the endpoint belongs to the solution set. The symbol ≤ includes equality, so x = −4 is allowed and −4 must be shown by a closed, filled circle. The symbol < excludes equality, so x = 2 is not allowed and 2 must be shown by an open, unfilled circle. Every value between −4 and 2 is included, so the interval is [−4, 2), with the segment between the endpoints shaded. Therefore option B is correct. Option A reverses the meanings of the two signs, option C excludes both endpoints, and option D includes both endpoints. The governing idea is closed for inclusive and open for strict inequalities.

Solve the inequality step by step: 3x − 2 ≥ 7. Adding 2 to both sides gives 3x ≥ 9. Dividing by the positive number 3 preserves the inequality sign, so x ≥ 3. Because equality is included, the value 3 belongs to the solution set and is represented by a closed circle. Values greater than 3 also satisfy the inequality, so the ray is shaded to the right. Therefore option A is correct. An open circle would incorrectly exclude 3, and shading left would represent values less than 3. The inequality sign would reverse only if we multiplied or divided by a negative number, which does not occur here.

The governing concept is the boundary, or critical, point of a linear inequality. In x < 5, the variable is compared directly with the constant 5, so the solution region changes at x = 5. Values such as 4 and 0 lie to the left and satisfy the inequality, whereas values such as 6 lie to the right and do not. Because the sign is strictly less than, the point 5 itself is not included and would be shown by an open circle; however, it remains the boundary point. Thus option C, 5, is correct. The numbers 0, 1, and −5 are possible solution values, not the point where the direction changes.

The governing concept is interval notation and endpoint inclusion. The interval [0,∞) contains every real number from 0 onward. The square bracket at 0 means that 0 itself is included, while the direction toward infinity means all numbers greater than 0 are also included. Therefore the equivalent inequality is x≥0, so option C is correct. Option A, x>0, would exclude 0 and would be represented by a parenthesis at 0. Options B and D describe numbers on the nonpositive or negative side of the number line, which is the opposite direction. Infinity is not an actual endpoint, so no bracket is placed there.