representation on number line

The governing concept is translating endpoint marks and shading on a number line into interval notation. An open dot means that the endpoint is excluded from the solution set, whereas a closed dot means that it is included. Since -10 has an open dot, the interval must begin with a parenthesis at -10. Since -4 has a closed dot, the interval must end with a square bracket at -4. The shaded portion includes all numbers between these endpoints, so the set is (-10,-4]. The order is correct because -10 is less than -4. Therefore option B is correct. Option A incorrectly includes -10 and excludes -4; option C excludes both endpoints; and option D includes both. The endpoint symbols, not merely the shaded length, determine the brackets.

The governing concept is interpreting a union of intervals and an excluded point. The interval (-∞,2) contains every real number less than 2, and (2,∞) contains every real number greater than 2. Their union therefore covers the entire real number line except the single value 2. Because both intervals use a parenthesis at 2, the value 2 is not included. The equivalent statement is x≠2, so option B is correct. Option A identifies only the missing point rather than the shaded solution set. Options C and D include 2 and cover only one side or one half-line, so they cannot represent both portions of the union.

The governing concept is endpoint inclusion in a union of intervals. The first interval, [-3,4), contains values up to but not including 4. The second interval, (4,10], begins just after 4 and also excludes 4. Although the number line is shaded on both sides of 4, neither interval contains the point itself. Therefore the graph has an open hole at 4, making option B correct. A closed dot would mean that 4 belongs to at least one interval, which is false here. Options C and D are incomplete because shading occurs on both sides, not only on one side.

The intersection contains numbers that belong to both intervals simultaneously. The interval (−∞, 4) contains every number less than 4, but it excludes 4 because the endpoint is open. The interval [−3, 8] contains all numbers from −3 to 8, including −3 and 8. The common portion therefore starts at −3 and continues up to, but not including, 4. The left endpoint is closed because −3 is included in the second interval and is certainly less than 4. The right endpoint is open because 4 is excluded from the first interval. Consequently, the intersection is [−3, 4), which is option B. The other options either use an incorrect endpoint symbol or include numbers not common to both intervals.

The intersection symbol ∩ means that a number must belong to both intervals. The first interval contains every real number less than or equal to 2, while the second contains numbers from -5 through values less than 7. Their common portion begins at -5 and ends at 2. Both endpoints are included: -5 is included by [-5, 7), and 2 is included by (-∞, 2]. Hence the intersection is [-5, 2], so option B is correct. Option A wrongly excludes -5; option C includes values greater than 2; option D includes numbers outside the common region.

The governing concept is that the symbol ≠ excludes exactly one value while allowing every other real number. Therefore, x may be less than 2 or greater than 2, but it cannot equal 2. On a number line, an excluded endpoint is marked by an open circle or hole, so 2 must be left unfilled. Because values on both sides of 2 are allowed, both the left and right portions of the number line are shaded. In interval notation, the solution is (-∞, 2) ∪ (2, ∞). Hence option B is correct. Option A incorrectly includes only the excluded value. Option C includes 2 and shows only values to the right, while option D excludes 2 but represents only values to the left.

In interval notation, (7, 7) denotes all real numbers x satisfying 7 < x < 7. The governing idea is that an open interval contains numbers strictly between its endpoints, but there is no real number strictly greater than 7 and strictly less than 7 at the same time. The endpoint 7 is also excluded because the parentheses are open. Consequently, the interval has no point and its solution set is the empty set, written as ∅. Option C is therefore correct. Option A, {7}, would include the single value 7, but that value is not allowed in an open interval. Options B and D are one-sided intervals and contain many values that fail one of the two required inequalities.

The union symbol ∪ combines every value belonging to either interval. The first interval covers -2 through 5, including both endpoints. The second covers 5 through values less than 8; it includes 5 but excludes 8. Because both intervals meet at 5 and 5 is included, there is no gap between them. The combined coverage therefore starts at -2, includes -2, continues through 5, and ends just before 8. Its simplified form is [-2, 8), so option C is correct. Options A and B retain only one original interval, while D excludes -2 and incorrectly includes 8.

A horizontal number line increases from left to right, so points to the left of -12 represent numbers smaller than -12. The shading therefore describes x values less than -12. The open dot is equally important: it means the boundary value -12 is excluded, so equality is not allowed. Combining the direction of shading with the type of endpoint gives x < -12, making option C correct. Option A has the correct left-side direction but uses ≤ and would require a closed dot at -12. Options B and D describe numbers to the right of -12, and D also includes the boundary. Thus neither the direction nor the endpoint in those options matches the stated diagram.