Solve the inequality, and express the solutions in terms of intervals whenever possible.
6/3x + 2 ≥ 0
PRE CALC HELP!!!!?
2017-01-17 12:29 pm
回答 (10)
2017-01-17 12:40 pm
6 / (3x + 2) ≥ 0
3x + 2 > 0
3x > -2
x > -2/3
(When 3x + 2 = 0, the fraction 6/0 is undefined.)
3x + 2 > 0
3x > -2
x > -2/3
(When 3x + 2 = 0, the fraction 6/0 is undefined.)
2017-01-23 6:53 pm
phuck you all idiots
2017-01-20 11:45 am
is it 6/(3x + 2) ≥ 0 ?
2017-01-19 12:23 pm
6 / (3x +2) > 0
This happens whenever 3x +2 > 0
x > -2/3
6 / (3x + 2) = 0
This cannot happen.
This happens whenever 3x +2 > 0
x > -2/3
6 / (3x + 2) = 0
This cannot happen.
2017-01-19 7:26 am
6/3x simplifies to 2x...2x >= -2....x = -1
2017-01-17 11:39 pm
Presentation is unclear.
Brackets required.
Brackets required.
2017-01-17 11:12 pm
6/ (3x) + 2 ≥ 0
multiply both sides by 3x
6 + 6x ≥ 0
subtract 6 from both sides
6x ≥ -6
divide both sides by 6
x ≥ -1
[-1, infinity) is the solution
multiply both sides by 3x
6 + 6x ≥ 0
subtract 6 from both sides
6x ≥ -6
divide both sides by 6
x ≥ -1
[-1, infinity) is the solution
2017-01-17 2:04 pm
6/3x + 2 ≥ 0 means (6/3)x + 2 ≧ 0 .
2x + 2 ≧ 0
2x ≧ -2
x ≧ -1 ---> [-1 , ∞)
If you want to mean 6/(3x) + 2 ≧ 0 , it means 6/(3x) ≧ -2 , clearly x ≠ 0 .
Case1 . x > 0
6 ≧ -2*(3x)
6 ≧ -6x
-1 ≦ x
It must be x > 0 , so x > 0 .
Case2. x < 0
6 ≦ -2*(3x)
6 ≦ -6x
-1 ≧ x
It must be 0 < x , so x ≦ -1 .
Therefore , x ≦ -1 or x > 0 . ---> (-∞ , -1] ∪ (0 , ∞)
2x + 2 ≧ 0
2x ≧ -2
x ≧ -1 ---> [-1 , ∞)
If you want to mean 6/(3x) + 2 ≧ 0 , it means 6/(3x) ≧ -2 , clearly x ≠ 0 .
Case1 . x > 0
6 ≧ -2*(3x)
6 ≧ -6x
-1 ≦ x
It must be x > 0 , so x > 0 .
Case2. x < 0
6 ≦ -2*(3x)
6 ≦ -6x
-1 ≧ x
It must be 0 < x , so x ≦ -1 .
Therefore , x ≦ -1 or x > 0 . ---> (-∞ , -1] ∪ (0 , ∞)
2017-01-17 1:36 pm
I will do it different that above since you did not have parenthesis.
6/x + 2 >/= 0 ................. subtract 2
..... 6/x >/= - 2 .............. multiply by x .............. x not= 0
......... 6 >/= -2x ............. divide by ( - 2)
....... -3 >/= x....... OR
........ x </= - 3 ............... OR ...... x = ( - oo , -3 ] ...... ANSWER
6/x + 2 >/= 0 ................. subtract 2
..... 6/x >/= - 2 .............. multiply by x .............. x not= 0
......... 6 >/= -2x ............. divide by ( - 2)
....... -3 >/= x....... OR
........ x </= - 3 ............... OR ...... x = ( - oo , -3 ] ...... ANSWER
2017-01-19 3:32 am
6/3x + 2 ≥ 0
6 + 6x ≥ 0
6x ≥ -6
x ≥ -1
6 + 6x ≥ 0
6x ≥ -6
x ≥ -1
收錄日期: 2021-04-18 15:58:47
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20170117042924AAGrfEd