Answer:
The solution of |3x-9|≤15 is [-2;8] and the solution |2x-3|≥5 of is (-∞,2] ∪ [8,∞)
Step-by-step explanation:
When solving absolute value inequalities, there are two cases to consider.
Case 1: The expression within the absolute value symbols is positive.
Case 2: The expression within the absolute value symbols is negative.
The solution is the intersection of the solutions of these two cases.
In other words, for any real numbers a and b,
- if |a|> b then a>b or a<-b
- if |a|< b then a<b or a>-b
So, being |3x-9|≤15
Solving: 3x-9 ≤ 15
3x ≤15 + 9
3x ≤24
x ≤24÷3
x≤8
or 3x-9 ≥ -15
3x ≥-15 +9
3x ≥-6
x ≥ (-6)÷3
x ≥ -2
The solution is made up of all the intervals that make the inequality true. Expressing the solution as an interval: [-2;8]
So, being |2x-3|≥5
Solving: 2x-3 ≥ 5
2x ≥ 5 + 3
2x ≥8
x ≥8÷2
x≥8
or 2x-3 ≤ -5
2x ≤-5 +3
2x ≤-2
x ≤ (-2)÷2
x ≤ -2
Expressing the solution as an interval: (-∞,2] ∪ [8,∞)
I’m not 100% sure but I think it’s y=5x+20
2/5 van not be simiplified because 2 and 5 dont have a common factor lower than them
Answer:
Hans has done a mistake in the second step.
The correct answer is:
5x^2+13x-6=0
5x^2+15x-2x-6x=0 - 5. -6=-30 -30=15. -2
5x(1+3x)-2x(1+3x)=0
(5x-2x)(1+3x)=0
5-2x=0 1+3x=0
x=5/2 x=-1/3
Answer:
True or False
2. When graphing inequalities, a dashed line is used for < or > signs.
True or False
3. If you were to graph x > 6, the circle would be closed and to the left.
True or False
4. On a number line, if you were to graph an equation with a less than sign or a great than sign, < or > respectively, you would use an open circle
True or False
