representation on number line

The direct answer is option A: \([-6,-1)\cup[4,9]\). Set difference means keep the first set and remove every number belonging to the second set. Start with the closed interval \([-6,9]\), which includes both -6 and 9. Remove \([-1,4)\): this removed interval includes -1, includes all numbers up to 4, but does not include 4 itself. The portion left on the left is from -6 through numbers just before -1, so it is \([-6,-1)\): -6 stays, while -1 is removed. The portion left on the right begins at 4 and goes through 9, so it is \([4,9]\): 4 stays because it was excluded from the removed interval, and 9 stays because it belongs to the original interval. Option A is therefore correct. Option B, \([-6,-1]\cup(4,9]\), wrongly keeps -1 and wrongly removes 4. Option C excludes -6 and includes 4 only up to but not including 9, so both endpoint choices are wrong. Option D is a union that overlaps the removed middle part and does not represent subtraction at all. On a number line, shade the original interval, erase the section from -1 including -1 up to but not including 4, and observe the two remaining pieces. Memory cue: in set difference, inspect the brackets at both ends of the part being removed; an included removed endpoint disappears, while an excluded one remains.

No number can be both greater than or equal to ( 3 ) and less than ( 3 ). For AND, the intersection is empty.

In ( x\le -2 ), ( -2 ) is closed, and in ( x>10 ), ( 10 ) is open. OR forms the union of separate parts.

Solving gives ( -19<-5x\le -4 ), then signs reverse to ( \frac{19}{5}>x\ge \frac{4}{5} ). In order, write ( \frac{4}{5}\le x<\frac{19}{5} ).

The first part gives values less than ( -9 ), and the second gives ( -9 ) and larger values. Together they form the whole number line.

The first interval includes ( 4 ), and the second begins after ( 4 ). Together they form ( 1<x\le 12 ).

The direct answer is option C: \(x>7\), shown by an open point at 7 and shading to the right. Start with \((13-3x)/(-4)>2\). Multiply both sides by \(-4\); because it is negative, reverse the inequality: \(13-3x<-8\). Subtract 13: \(-3x<-21\). Divide by \(-3\), reversing the sign again: \(x>7\). The endpoint is open because 7 does not satisfy the original strict inequality: it gives \(2>2\), which is false. Option A wrongly includes 7 and points right. Option B has the wrong direction and excludes the wrong side. Option C has both the correct open endpoint and correct right-hand direction. Option D includes 7 and points left. The important warning is that dividing or multiplying by a negative number reverses the inequality sign.

An open circle at \(-1\) means that \(-1\) is excluded, so the left ray is \(( -\infty,-1)\). A filled circle at \(4\) means that \(4\) is included, so the right ray is \([4,\infty)\). Since the shaded parts are two separate rays, they are joined using union, \(\cup\). Option B reverses the inclusion of both endpoints. Exam tip: use a round bracket for an open circle and a square bracket for a filled circle.

The direct answer is option A: \([1,6)\). Solve the compound inequality as one chain: \(1\le3x-2<16\). Add 2 to all three parts: \(3\le3x<18\). Divide all parts by positive 3, so the signs do not change: \(1\le x<6\). Since 1 is included, use a closed endpoint at 1. Since 6 is excluded by the strict sign \(<\), use an open endpoint at 6. Option A, \([1,6)\), matches exactly. Option B reverses which endpoint is included. Option C is the interval before dividing by 3 and therefore gives values for \(3x\), not for \(x\). Option D has both the wrong interval and wrong endpoint inclusion. A useful check is to test \(x=2\): it satisfies the original inequality, while \(x=6\) gives \(16<16\), false. Memory cue: adding or dividing by a positive number keeps signs; equality makes a closed endpoint.

The first interval stops before ( 5 ), and the second begins after ( 5 ). Therefore, there is no common part.

Absolute value cannot be less than ( 0 ), and ( |x+4|=0 ) gives ( x=-4 ). So only one closed dot appears.

Solving gives ( -5<-2x\le 7 ), then signs reverse to ( \frac{5}{2}>x\ge -\frac{7}{2} ). In order, write ( -\frac{7}{2}\le x<\frac{5}{2} ).

In the interval \((2,8]\), 2 is excluded, so it has an open endpoint, while 8 is included, so it has a closed endpoint. The notation \(\setminus\{5\}\) removes 5 from the interval, so an open hole must be marked at 5. Therefore, option A is correct. Exam tip: round brackets indicate open endpoints, whereas square brackets indicate closed endpoints.

The first inequality gives ( x\le -3 ), and the second gives ( x>10 ). For OR, take the union of both parts.

The first inequality gives ( x>3 ), and the second gives ( x\le 7 ). For both conditions, the common part is ( 3<x\le 7 ).

The two intervals give all real numbers except ( 0 ). The open hole in the middle shows the removed point.

Both sides have square brackets, so both boundary values are included. Included boundaries are shown by closed dots.

The first interval does not include ( 2 ), and the second starts at ( 2 ). Therefore, there is no common point.

The direct answer is option C: only the closed point 3. An intersection contains numbers that belong to both intervals. In \((-4,3]\), numbers are greater than -4 and up to and including 3. In \([3,9)\), numbers start at and include 3, then go below 9. The only common number is 3, and both intervals include it, so the result is \(\{3\}\), shown by a closed dot. Option A, \((-4,9)\), is a broad union-like interval and includes numbers belonging to only one interval. Option B, \((3,9)\), excludes 3 and contains numbers not in the first interval. Option C is correct. Option D is wrong because the intersection is not empty. Memory cue: for intersection, keep the overlap; when endpoints meet and both brackets are square, keep the endpoint.

( |4x+1|\le 13 ) gives ( -13\le 4x+1\le 13 ). Thus ( -\frac{7}{2}\le x\le 3 ), not the listed interval.

The complement of the given union gives values less than ( -1 ) and values greater than ( 4 ) up to ( 9 ). The point ( 9 ) remains because the second part excludes ( 9 ).

Simplifying gives ( 9-6x\le -9 ), so ( -6x\le -18 ) and ( x\ge 3 ). Reverse the sign when dividing by a negative.

The round bracket excludes \( \frac{1}{4} \), and the rightward arrow gives larger values. Therefore \( x>\frac{1}{4} \).

A filled circle shows that the boundary point belongs to the solution set, so it is used with \(\le\) or \(\ge\). An open circle is used for \(<\) or \(>\). Exam tip: check for equality before drawing the endpoint.

The expression \( A\setminus B \) means elements in \(A\) that are not in \(B\). Here \(A=(-\infty,6] \) and \(B=(1,6]\). So we remove all numbers greater than \(1\) up to \(6\), including \(6\). Therefore all numbers \(x\le 1\) remain. Hence the result is \( (-\infty,1] \).