representation on number line

The direct answer is option A: x≤-3, with a closed point at -3 and shading to the left. Start with 5-4x≥17. Subtract 5 from both sides: -4x≥12. Divide by -4 and reverse the inequality sign: x≤-3. Because the original sign is ≥, equality is included, so -3 receives a closed point. Values less than -3 are solutions, so shade left. Option A matches every part. Option B incorrectly excludes -3 by using <. Option C shades right, which represents x≥-3 rather than the actual solution. Option D also shades right and uses a strict > sign, so both its direction and endpoint are wrong. A reliable check is x=-4: 5-4(-4)=21≥17, confirming that numbers to the left work. The common mistake is forgetting to reverse the sign when dividing by -4.

The given interval includes ( 2 ) and excludes ( 11 ). In the complement, ( 2 ) is removed and ( 11 ) is included.

A filled point includes \(-2\), so a square bracket is used at \(-2\). A hollow point excludes \(4\), so a parenthesis is used at \(4\). Since the shading is in the two outer rays, the solution is \(( -\infty,-2]\cup(4,\infty)\). Option C reverses the inclusion status of \(-2\) and \(4\). Exam tip: a filled point means included, while a hollow point means excluded.

( |3x-2|\ge 10 ) gives ( 3x-2\le -10 ) or ( 3x-2\ge 10 ). Because equality is included, both boundary points are closed.

The removed part excludes ( -2 ) but includes ( 3 ). Therefore ( -2 ) remains and ( 3 ) is removed.

No number can be both less than or equal to ( 0 ) and greater than ( 0 ). For AND, the intersection is empty.

Subtracting 6 gives \(-18\le -3x<-3\). On dividing by \(-3\), both inequality signs reverse: \(6\ge x>1\). Thus \(1<x\le6\), so the correct interval is \((1,6]\). The interval \([1,6)\) is incorrect because it includes 1 and excludes 6. Exam tip: always reverse inequality signs when multiplying or dividing by a negative number.

Together the two intervals cover the entire number line including ( -5 ). A union includes all shaded parts.

The first interval reaches up to before ( 5 ), and the second includes ( 5 ). Together they form shading from ( -2 ) to ( 12 ) with open endpoints.

Multiplying by ( -5 ) reverses the sign and gives ( 9-2x>-5 ). Then ( -2x>-14 ) gives ( x<7 ).

The square bracket includes \( -\frac{5}{3} \), and the round bracket excludes \( \frac{4}{3} \). The same rule applies to fractional endpoints.

Solving gives ( 6<2x\le 12 ), so ( 3<x\le 6 ). Keep the strict endpoint open and the equality endpoint closed.

The first interval stops before ( 4 ), and the second begins after ( 4 ). Therefore, there is no common part.

Absolute value cannot be less than ( 0 ), and ( |x-8|=0 ) gives ( x=8 ). So only one closed dot appears.

Direct answer: option A is correct. The statement \(x\ne-6\) means x may be every real number except \(-6\). On a number line, shade the entire line because all other numbers are allowed, and mark \(-6\) with an open hole because that single value is excluded. Option A shows exactly this. Option B uses a closed dot, which means \(x=-6\) is included, the opposite of the condition. Option C also includes \(-6\) and keeps only numbers to its right, so it represents \(x\ge-6\), not the required set. Option D excludes \(-6\) correctly with an open point but shades only the left side, representing \(x<-6\); numbers to the right should also be included. The important distinction is: an open point means “not included,” while shading shows the allowed values. Memory cue: “not equal” means the whole number line with one hole.

Solving gives ( -5\le -2x<5 ). Dividing by a negative gives ( \frac{5}{2}\ge x>-\frac{5}{2} ), that is ( -\frac{5}{2}<x\le \frac{5}{2} ).

Multiplying by ( -2 ) reverses the sign and gives ( x\le 2 ). Reverse the sign when multiplying or dividing by a negative number.

The intersection \( (-\infty,1] \cap (-4,6) \) keeps only the numbers that satisfy both conditions. From \( (-\infty,1] \), we have \(x\le 1\). From \( (-4,6) \), we have \( -4<x<6\). Combining gives \( -4<x\le 1\), which in interval notation is \( (-4,1] \).

The integers between ( \frac{1}{2} ) and ( \frac{11}{2} ) are from ( 1 ) to ( 5 ). For integer solutions, mark only separate points.

The closed dot includes ( -8 ), and the open dot excludes ( 1 ). Decide inequality signs from endpoint types.

Given \(-6x+8<20\), subtract 8 from both sides to get \(-6x<12\). Dividing by \(-6\) reverses the inequality sign, so \(x>-2\). Since the inequality is strict, \(-2\) is shown with an open circle, and values greater than \(-2\) are shaded to the right. Option C has the correct direction but wrongly includes \(-2\) with a closed circle. Exam tip: always reverse an inequality when multiplying or dividing by a negative number.

Multiplying all three parts of \( -4\le \frac{x-3}{2}<5 \) by the positive number \(2\) gives \( -8\le x-3<10 \). Adding \(3\) throughout gives \( -5\le x<13 \). Thus, \(-5\) is included while \(13\) is excluded, so the interval is \(\boxed{[-5,13)}\). In option A, the inclusion of the endpoints is reversed. Exam tip: inequality signs do not reverse when multiplying or dividing by a positive number.

In the complement, ( -3 ) becomes included and ( 6 ) is excluded. Open and closed endpoints switch in a complement.

The inequality \(|x-6|<9\) means that \(x\) is less than 9 units away from 6. Hence, \(-9<x-6<9\). Adding 6 to all three parts gives \(-3<x<15\), so the required interval is \((-3,15)\). Option \([-3,15]\) is incorrect because a strict inequality \(<\) does not include the endpoints \(-3\) and \(15\). Exam tip: Use open circles and open interval brackets for \(<\) or \(>\).

( |2x-5|\ge 11 ) gives ( 2x-5\le -11 ) or ( 2x-5\ge 11 ). Since equality is included, boundary points are closed.