1.x<5 and x ≥ 0
[0, 5) what will be the plotting points on a number line.
2. 4x-3<-3 and 4x+2>-14
(-4,0) what will be the plotting points on a number line.
3.2(3x-3)≥ 6 and 2(3x-4)less than sign or eaual to -8
Solution is false what will be the plotting points on a number line.
4. 4x-3\le 2x+7\:and\:5x\:-2\:>2x-17
(-5, 5] what will the plotting points on the number line.
5. \left(0\le \:2x+4\right)\quad \mathrm{and}\quad \left(2x+4<14\right) what will the plotting points on a the numberline.
2. Graph 4x-3 = -3, and 4x+2 = -14.
4x-3 = -3. X = 0.
4x+2 = -14. X = -4.
Plotting points: X = -4, and X = 0.
All points BETWEEN -4 and 0 should satisfy both inequalities.
3. Graph 2(3x-3) = 6, and 2
93x-4) = -8
3. Graph 2(3x-3) = 6, and 2(3x-4) = -8.
2(3x-3) = 6, X = 2.
2(3x-4) = -8, X = 0.
4. Sorry! but this one makes no sense to me.
To find the plotting points on a number line for these inequalities, we need to solve the inequalities and express the solutions as intervals on the number line. Here's how you can do it step by step:
1. x < 5 and x ≥ 0:
- To solve the inequality x < 5, we note that x can take any value less than 5.
- To solve the inequality x ≥ 0, we note that x can take any value greater than or equal to 0.
- Combining these two, we find that x can take any value between 0 (including 0) and 5 (excluding 5), which is represented as [0, 5).
2. 4x-3 < -3 and 4x+2 > -14:
- To solve 4x - 3 < -3, we add 3 to both sides: 4x < 0.
- Divide by 4 on both sides: x < 0.
- To solve 4x + 2 > -14, we subtract 2 from both sides: 4x > -16.
- Divide by 4 on both sides: x > -4.
- Combining the two inequalities, we find that x can take any value between -4 (excluding -4) and 0 (excluding 0), which is represented as (-4, 0).
3. 2(3x-3) ≥ 6 and 2(3x-4) ≤ -8:
- Simplify the inequalities:
- 6x - 6 ≥ 6
- 6x - 8 ≤ -8
- Simplifying further, we have:
- 6x ≥ 12
- 6x ≤ 0
- Dividing both inequalities by 6:
- x ≥ 2
- x ≤ 0
- The solution x ≥ 2 ∩ x ≤ 0 is not possible as these are contradictory statements. Therefore, the solution is false, and there is no plotting point on the number line.
4. 4x - 3 ≤ 2x + 7 and 5x - 2 > 2x - 17:
- Simplify the inequalities:
- 4x - 2x ≤ 7 + 3
- 5x - 2x > -17 + 2
- Simplifying further, we have:
- 2x ≤ 10
- 3x > -15
- Dividing both inequalities by 2 and 3 respectively:
- x ≤ 5
- x > -5
- Combining the inequalities, we find that x can take any value between -5 (excluding -5) and 5 (including 5), which is represented as (-5, 5].
5. 0 ≤ 2x + 4 and 2x + 4 < 14:
- To solve 0 ≤ 2x + 4, we subtract 4 from both sides: -4 ≤ 2x.
- Dividing both sides by 2: -2 ≤ x.
- To solve 2x + 4 < 14, we subtract 4 from both sides: 2x < 10.
- Dividing both sides by 2: x < 5.
- Combining the two inequalities, we find that x can take any value between -2 (including -2) and 5 (excluding 5), which is represented as [-2, 5).