representation on number line

The round bracket at 5 means that 5 is not included. The interval contains all real numbers less than 5, so it represents x < 5.

The square bracket at 2 means that 2 is included, and the interval extends to the right. Thus, it represents all real numbers greater than or equal to 2: \(x\ge 2\).

The governing concept is interpretation of a compound inequality as an interval. The statement 0≤x≤6 means that x must be at least 0 and at most 6. Because both symbols are ≤ rather than <, the endpoints 0 and 6 are included. Every number between them, including 3, also satisfies both conditions: 0≤3 and 3≤6. The value 7 fails the upper condition because 7 is greater than 6, so it lies outside the interval [0,6]. Therefore option D is not included. On a number line, the solution is shown by a solid segment from 0 to 6 with filled dots at both endpoints; 7 is to the right of that segment. Options A and B are included boundary values, and option C is an included interior value.

The inequality x > −4 includes all real numbers greater than −4 but excludes −4 itself. Therefore, its interval notation is (−4, ∞); the parenthesis shows that −4 is not included.

\(x\le 0\) includes every number less than 0 as well as 0 itself. Hence, the interval extends leftward up to 0, with a square bracket at 0: \(( -\infty,0]\). A parenthesis is always used with \(-\infty\), since infinity is not a real number. Option D excludes 0, so it represents \(x<0\). Exam tip: For \(\le\) or \(\ge\), include the boundary value using a square bracket.

The compound inequality \(2\le x<5\) means \(2\le x\) and \(x<5\). Thus, \(x\) may be equal to 2, but it cannot be equal to 5, so option C is correct. Option D represents values outside this interval. Exam tip: \(\le\) includes the endpoint, whereas \(<\) excludes it.

The governing ideas are the meaning of a strict inequality and the stated domain of x. Since x is a natural number, use the school convention that natural numbers are 1, 2, 3, and so on. The condition x < 4 permits only natural numbers smaller than 4, namely 1, 2, and 3. The number 4 is excluded because the inequality is strict, not ≤. Option B includes 0, which is not included under this convention; options C and D contain values that are at least 4. Therefore option A gives the complete solution set. If a textbook defines natural numbers to include 0, the convention should be stated, but here the supplied choices clearly use the positive-natural-number convention.

A closed dot means that 4 is included, while rightward shading represents numbers greater than 4. Hence \(x \geq 4\) is correct. In \(x>4\), 4 is excluded. Exam tip: use \(\geq\) or \(\leq\) for a closed dot.

The inequality \(x<-2\) means that \(x\) can be any real number smaller than \(-2\). On a number line, smaller numbers lie to the left of \(-2\). Since the sign is '<', \(-2\) itself is not included, so an open circle is used. Exam tip: with '<' or '>', the boundary point is excluded.

The inequality \(x\ge 2\) means that \(x\) can be 2 or any value greater than 2. Thus, 2 is included in the solution set, so a square bracket is used at 2, and the values extend to the right up to infinity. Therefore, the correct interval is \([2,\infty)\). Option A excludes 2 and represents \(x>2\). Exam tip: Use a square bracket for an included endpoint, but always use a round bracket with \(\infty\).

The governing concept is interpreting a compound inequality by reading both parts together. In 3 < x ≤ 8, the first relation says that x is strictly greater than 3, so 3 is excluded. The second says that x is less than or equal to 8, so 8 is included. Combining them gives the interval 3 < x ≤ 8. Therefore option A is correct. On a number line this would be shown with an open circle at 3, a closed circle at 8, and the segment between them shaded. Option B reverses both relationships, option C keeps only an excluded boundary value, and option D ignores the upper bound condition.

The governing property is order reversal under multiplication by a negative number. Starting with x ≤ 2, multiply each side by −3. Because −3 is negative, the inequality sign changes from ≤ to ≥: (−3)x ≥ (−3)(2), hence −3x ≥ −6. Therefore option B is correct. Option A performs the multiplication but forgets the reversal. Option C changes the coefficient to positive 3 without justification, and option D changes the right-hand product incorrectly. A numerical check supports the result: x = 2 gives equality, −6 ≥ −6; x = 1 gives −3 ≥ −6, which is true. Thus the transformed inequality is equivalent to the original one.

The square bracket at 3 means that 3 is included. The interval contains 3 and every real number less than 3, so it represents x ≤ 3.

The inequality \\(-1\le x<4\\) includes values that are at least \\(-1\\) and less than \\(4\\). Since \\(0\\) lies within this interval, option C is correct. The value \\(4\\) is excluded because the inequality uses \\(<4\\), while \\(5\\) and \\(-2\\) lie outside the interval. Exam tip: an endpoint with \\(\le\\) is included, whereas an endpoint with \\(<\\) is excluded.

The inequality x ≤ 4 includes every real number less than 4 and also 4 itself. These values extend without bound toward negative infinity, which is never included as an endpoint, so the interval is (−∞, 4].

The inequality \(x>6\) contains every real number greater than 6, but it does not include 6 itself. Hence, 6 has a round parenthesis and the interval continues to the right up to infinity: \((6,\infty)\). Option B includes \(6\), so it represents \(x\geq6\). Exam tip: use a round parenthesis at an endpoint for \(>\) or \(<\).

The governing concept is translating interval notation into inequality notation. The interval (−∞, 2] contains every real number less than 2 and also contains 2 itself. The square bracket at 2 is the key: it means the endpoint is included. Therefore the corresponding inequality is x ≤ 2, which is option B. The symbol −∞ is always written with a parenthesis because infinity is not an actual number that can be included as an endpoint. Option A describes the interval (−∞, 2), excluding 2, while option C describes numbers greater than 2 and option D includes 2 but extends to the right. Thus only B represents the stated interval exactly.

Interval notation communicates endpoint inclusion through its brackets. A square bracket means that the endpoint belongs to the set, while a round parenthesis means that it is excluded. In [-4, 3), the square bracket at −4 includes −4, and the parenthesis at 3 excludes 3. Therefore the equivalent double inequality is −4 ≤ x < 3, which is option A. Option B reverses the inclusion status of both endpoints. Option C excludes both endpoints, and option D includes both endpoints. Checking each endpoint separately prevents the common mistake of treating all interval endpoints in the same way. Hence A is the exact translation.

The governing rules for a number-line graph are: a closed circle includes the boundary value, an open circle excludes it, and the shaded direction shows the values being selected. In x≤1, the symbol ≤ includes 1, so a closed circle must be placed at 1. The solution also contains every number smaller than 1, so the line is shaded to the left. Therefore option A is correct. Option B points left but wrongly excludes 1. Option C includes 1 but shades toward numbers greater than 1, representing x≥1. Option D both excludes 1 and points in the wrong direction. Together, inclusion and direction uniquely determine A.

Interval notation must be read using its endpoint symbols. The interval (−∞, 4) contains every real number less than 4. The parenthesis at 4 means that 4 itself is not included; infinity is never treated as an included endpoint either. Therefore the equivalent inequality is x < 4, so option B is correct. Option A incorrectly includes 4, which belongs to the interval [−∞ conceptually not used, 4] only in a different notation; more simply, substituting x = 4 shows it fails the open-end condition. Options C and D describe values at or above 4 and point in the opposite direction. The original question’s option assignment was corrected because x ≤ 4 is not equivalent to the stated open interval.

Subtracting 5 gives 2x ≥ 12; dividing by the positive number 2 gives x ≥ 6. Since 6 is included, the interval is [6, ∞).

The sign (>) does not include the boundary so an open circle is used. In exams check the direction toward greater numbers.

The sign (\le) includes (-2) and smaller numbers lie to the left. Use a closed circle when the endpoint is included.

The governing concept is reading endpoint inclusion and interval direction from a number-line representation. A closed or filled circle means the endpoint is included in the solution set, so both -1 and 4 must use inclusive symbols. Shading between the two points means that x lies from -1 through 4, giving -1 ≤ x ≤ 4. This is the closed interval [-1, 4], and option B is correct. Option A would correspond to open circles at both ends because it excludes both endpoints. Option C would require a closed circle at -1 and an open circle at 4, while option D would require the reverse. Thus the symbols and the shaded direction together identify only B.

A parenthesis endpoint is not included and a square bracket endpoint is included. Therefore (2) is open and (7) is closed.