Write the inequality in interval notation. Then graph the interval.
x ≤ - 6 or x > 0
Interval notation: (-∞, -6] U (0, ∞)
Graph: 
Solve and graph the equation.
│x│= 7
To solve the equation │x│= 7, we need to consider two cases:
Case 1: x is positive.
In this case, │x│ is equal to x. So the equation becomes x = 7. Solving for x, we have x = 7.
Case 2: x is negative.
In this case, │x│ is equal to -x. So the equation becomes -x = 7. Solving for x, we have x = -7.
Therefore, the solutions to the equation │x│= 7 are x = 7 and x = -7.
Graph:
The graph of │x│= 7 consists of two horizontal lines, one at y = 7 and the other at y = -7, intersecting the x-axis at x = -7 and x = 7.
│3t│= 24
To solve the equation │3t│= 24, we need to consider two cases:
Case 1: 3t is positive.
In this case, │3t│ is equal to 3t. So the equation becomes 3t = 24. Solving for t, we have t = 8.
Case 2: 3t is negative.
In this case, │3t│ is equal to -3t. So the equation becomes -3t = 24. Solving for t, we have t = -8.
Therefore, the solutions to the equation │3t│= 24 are t = 8 and t = -8.
Graph:
The graph of │3t│= 24 consists of two vertical lines, one at x = 8 and the other at x = -8, intersecting the y-axis at y = 24 and y = -24.
│x + 9│≥ 4
To solve the inequality │x + 9│≥ 4, we need to consider two cases:
Case 1: x + 9 is positive or zero.
In this case, │x + 9│ is equal to x + 9. So the inequality becomes x + 9 ≥ 4. Solving for x, we have x ≥ -5.
Case 2: x + 9 is negative.
In this case, │x + 9│ is equal to -(x + 9). So the inequality becomes -(x + 9) ≥ 4. Solving for x, we have x ≤ -13.
Therefore, the solutions to the inequality │x + 9│≥ 4 are x ≥ -5 or x ≤ -13.
Graph:
The graph of │x + 9│≥ 4 consists of two half-lines, one going to the right from x = -5, and the other going to the left from x = -13.
│x│+ 7 = 13
To solve the equation │x│+ 7 = 13, we need to consider two cases:
Case 1: x is positive.
In this case, │x│ is equal to x. So the equation becomes x + 7 = 13. Solving for x, we have x = 6.
Case 2: x is negative.
In this case, │x│ is equal to -x. So the equation becomes -x + 7 = 13. Solving for x, we have x = -6.
Therefore, the solutions to the equation │x│+ 7 = 13 are x = 6 and x = -6.
Graph:
The graph of │x│+ 7 = 13 consists of two vertical lines, one at x = 6 and the other at x = -6, intersecting the y-axis at y = 13.
-3 │x│ = -9
To solve the equation -3│x│ = -9, we need to consider two cases:
Case 1: x is positive or zero.
In this case, │x│ is equal to x. So the equation becomes -3x = -9. Solving for x, we have x = 3.
Case 2: x is negative.
In this case, │x│ is equal to -x. So the equation becomes -3(-x) = -9. Simplifying, we have 3x = -9. Solving for x, we have x = -3.
Therefore, the solutions to the equation -3│x│ = -9 are x = 3 and x = -3.
Graph:
The graph of -3│x│ = -9 consists of a vertical line at x = 3 and another vertical line at x = -3, intersecting the y-axis at y = -9.
3│x-1│=6
To solve the equation 3│x-1│=6, we need to consider two cases:
Case 1: x-1 is positive or zero.
In this case, │x-1│ is equal to x-1. So the equation becomes 3(x-1) = 6. Solving for x, we have x = 3.
Case 2: x-1 is negative.
In this case, │x-1│ is equal to -(x-1). So the equation becomes 3(-(x-1)) = 6. Simplifying, we have -3(x-1) = 6. Distributing the negative sign, we get -3x+3 = 6. Subtracting 3 from both sides, we have -3x = 3. Dividing by -3, we get x = -1.
Therefore, the solutions to the equation 3│x-1│=6 are x = 3 and x = -1.
Graph:
The graph of 3│x-1│=6 consists of two vertical lines, one at x = 3 and the other at x = -1, intersecting the y-axis at y = 6.
│2n-4│+19=3
To solve the equation │2n-4│+19=3, we need to consider two cases:
Case 1: 2n-4 is positive or zero.
In this case, │2n-4│ is equal to 2n-4. So the equation becomes 2n-4+19=3. Solving for n, we have 2n+15=3. Subtracting 15 from both sides, we have 2n=-12. Dividing by 2, we get n=-6.
Case 2: 2n-4 is negative.
In this case, │2n-4│ is equal to -(2n-4). So the equation becomes -(2n-4)+19=3. Simplifying, we have -2n+4+19=3. Combining like terms, we get -2n+23=3. Subtracting 23 from both sides, we have -2n=-20. Dividing by -2, we get n=10.
Therefore, the solutions to the equation │2n-4│+19=3 are n=-6 and n=10.
Graph:
The graph of │2n-4│+19=3 consists of two vertical lines, one at n = -6 and the other at n = 10, intersecting the y-axis at y = 3.
6│2y-2│-2=40
To solve the equation 6│2y-2│-2=40, we need to consider two cases:
Case 1: 2y-2 is positive or zero.
In this case, │2y-2│ is equal to 2y-2. So the equation becomes 6(2y-2)-2=40. Simplifying, we have 12y-12-2=40. Combining like terms, we get 12y-14=40. Adding 14 to both sides, we have 12y=54. Dividing by 12, we get y=4.5.
Case 2: 2y-2 is negative.
In this case, │2y-2│ is equal to -(2y-2). So the equation becomes 6(-(2y-2))-2=40. Simplifying, we have -12y+12-2=40. Combining like terms, we get -12y+10=40. Subtracting 10 from both sides, we have -12y=30. Dividing by -12, we get y=-2.5.
Therefore, the solutions to the equation 6│2y-2│-2=40 are y=4.5 and y=-2.5.
Graph:
The graph of 6│2y-2│-2=40 consists of two horizontal lines, one at y = 4.5 and the other at y = -2.5, intersecting the x-axis at x = 40.
│x+9│≥3
To solve the inequality │x+9│≥3, we need to consider two cases:
Case 1: x+9 is positive or zero.
In this case, │x+9│ is equal to x+9. So the inequality becomes x+9≥3. Solving for x, we have x≥-6.
Case 2: x+9 is negative.
In this case, │x+9│ is equal to -(x+9), which means that we need to solve -(x+9)≥3. Simplifying, we get -x-9≥3. Adding 9 to both sides, we have -x≥12. Dividing by -1 (which reverses the inequality), we get x≤-12.
Therefore, the solutions to the inequality │x+9│≥3 are x≥-6 or x≤-12.
Graph:
The graph of │x+9│≥3 consists of two half-lines. The first half-line extends to the right from x = -6, and the second half-line extends to the left from x = -12.
│2x-9│≥3
To solve the inequality │2x-9│≥3, we need to consider two cases:
Case 1: 2x-9 is positive or zero.
In this case, │2x-9│ is equal to 2x-9. So the inequality becomes 2x-9≥3. Solving for x, we have 2x≥12, and dividing by 2, we get x≥6.
Case 2: 2x-9 is negative.
In this case, │2x-9│ is equal to -(2x-9), which means that we need to solve -(2x-9)≥3. Simplifying, we get -2x+9≥3. Subtracting 9 from both sides, we have -2x≥-6. Dividing by -2 (which reverses the inequality), we get x≤3.
Therefore, the solutions to the inequality │2x-9│≥3 are x≥6 or x≤3.
Graph:
The graph of │2x-9│≥3 consists of two intervals. The first interval is from x = 6 to positive infinity, and the second interval is from negative infinity to x = 3.
│3x│-9=3
To solve the equation │3x│-9=3, we need to consider two cases:
Case 1: 3x is positive or zero.
In this case, │3x│ is equal to 3x. So the equation becomes 3x-9=3. Solving for x, we get x=4.
Case 2: 3x is negative.
In this case, │3x│ is equal to -3x. So the equation becomes -3x-9=3. Solving for x, we get x=-4.
Therefore, the solutions to the equation │3x│-9=3 are x=4 and x=-4.
Graph:
The graph of │3x│-9=3 consists of two vertical lines, one at x=4 and the other at x=-4, intersecting the y-axis at y=3.
Starting from 200 ft away, your friend skates toward you and then passes you. She skates at a constant speed of 20 ft/s. Her distance d from you in feet after t seconds is given by d=│200-20t│. At what times is she 60 ft from you?
To find the times when your friend is 60 ft away from you, we need to solve the equation │200-20t│=60.
We will consider two cases:
Case 1: 200-20t is positive or zero.
In this case, │200-20t│ is equal to 200-20t. So the equation becomes 200-20t=60. Solving for t, we have -20t=-140, which gives t=7.
Case 2: 200-20t is negative.
In this case, │200-20t│ is equal to -(200-20t). So the equation becomes -(200-20t)=60. Simplifying, we have -200+20t=60. Solving for t, we get 20t=260, which gives t=13.
Therefore, your friend is 60 ft away from you at t=7 seconds and t=13 seconds.
Let D{8,11,13}, E{8,10,11,12} and F{7,9,10,11,13}.
D{8, 11, 13} represents a set with elements 8, 11, and 13.
E{8, 10, 11, 12} represents a set with elements 8, 10, 11, and 12.
F{7, 9, 10, 11, 13} represents a set with elements 7, 9, 10, 11, and 13.
List the elements in the set D U E.
The set D U E represents the union of sets D and E, which includes all the elements that are in either set D or set E.
D U E = {8, 10, 11, 12, 13}
Therefore, the elements in the set D U E are 8, 10, 11, 12, and 13.
nope
I apologize for the confusion. Let's go through the process again.
To find the union of sets D and E, we combine all the elements from both sets, removing any duplicate elements.
D = {8, 11, 13}
E = {8, 10, 11, 12}
Combining the elements from both sets, we have: {8, 11, 13, 10, 12}
Therefore, the elements in the set D U E are 8, 11, 13, 10, and 12.
Let B={x | x is an even whole number less than 9} and C{3,6,8,10}. Find the intersection BNC.
To find the intersection of sets B and C, we need to determine the elements that are common to both sets.
B = {x | x is an even whole number less than 9} = {2, 4, 6, 8}
C = {3, 6, 8, 10}
The elements that are common to both sets B and C are 6 and 8. Therefore, the intersection B ∩ C = {6, 8}.
A college survey was taken to determine where students study. Of 144 students surveyed, 75 studied in the library ,55 studied in the student union, 24 studied in both the library and the student union.
Of those interviewed how many did not study in either the library or the student union?
To find the number of students who did not study in either the library or the student union, we need to subtract the students who studied in both places from the total number of students surveyed.
Total number of students surveyed = 144
Number of students who studied in the library = 75
Number of students who studied in the student union = 55
Number of students who studied in both the library and the student union = 24
To find the number of students who did not study in either place, we add the number of students who studied in the library and the number of students who studied in the student union, and then subtract the number of students who studied in both places:
Number of students who did not study in either place = Total number of students surveyed - (Number of students who studied in the library + Number of students who studied in the student union - Number of students who studied in both places)
Number of students who did not study in either place = 144 - (75 + 55 - 24)
Number of students who did not study in either place = 144 - 130
Number of students who did not study in either place = 14
Therefore, 14 students did not study in either the library or the student union.
nope try again
Apologies for the mistake. Let's try again.
To find the number of students who did not study in either the library or the student union, we can use the principle of inclusion-exclusion.
Total number of students surveyed: 144
Number of students who studied in the library: 75
Number of students who studied in the student union: 55
Number of students who studied in both the library and the student union: 24
To find the number of students who studied in either the library or the student union, we sum the number of students in each set and subtract the number of students who studied in both sets:
Number of students who studied in either library or student union = Number of students who studied in the library + Number of students who studied in the student union - Number of students who studied in both
Number of students who studied in either library or student union = 75 + 55 - 24
Number of students who studied in either library or student union =