x+2 > 10 solves to x > 8 after we subtract 2 from both sides
So set A is the set of real numbers that are larger than 8. The value 8 itself is not in set A. The same can be said about 5 as well. 
Set B is the set of values that are larger than 5 since 2x > 10 turns into x > 5 after dividing both sides by 2. The value x = 5 is not in set B since x > 5 would turn into 5 > 5 which is false. The values x = 6, x = 8, and x = 9 are in set B.
----------------
Summarizing everything, we can say...
5 is not in set A. True
5 is in set B. False
6 is in set A. False
6 is not in set B. False
8 is not in set A. True
8 is in set B. True
9 is in set A. True
9 is not in set B. False
 
        
                    
             
        
        
        
6a-2(a-1)
6a-2a-2
4a-2
I think it is 4a-2
        
             
        
        
        
Answer:
Kyle saves 40 percent
$50 is 100% of what he had
$20 would be 40%
50= 100
40= 80
30= 60
20= 40
10= 20 
do what u will with that info I'm bad at wording things
 
        
             
        
        
        
Sqrt[x] = -x
   
x= x ^ 2  
x - x^2 = 0
-x(x-1)=0
x(x-1)=0
x = 1 or x = 0
sqrt 1 isn not equal to 0
x= 0
 
        
        
        
9514 1404 393
Answer:
   -3 ≤ x ≤ 19/3
Step-by-step explanation:
This inequality can be resolved to a compound inequality:
   -7 ≤ (3x -5)/2 ≤ 7
Multiply all parts by 2.
   -14 ≤ 3x -5 ≤ 14
Add 5 to all parts.
   -9 ≤ 3x ≤ 19
Divide all parts by 3.
   -3 ≤ x ≤ 19/3
_____
<em>Additional comment</em>
If you subtract 7 from both sides of the given inequality, it becomes ...
   |(3x -5)/2| -7 ≤ 0
Then you're looking for the values of x that bound the region where the graph is below the x-axis. Those are shown in the attachment. For graphing purposes, I find this comparison to zero works well. 
__
For an algebraic solution, I like the compound inequality method shown above. That only works well when the inequality is of the form ...
   |f(x)| < (some number) . . . . or ≤
If the inequality symbol points away from the absolute value expression, or if the (some number) expression involves the variable, then it is probably better to write the inequality in two parts with appropriate domain specifications:
   |f(x)| > g(x)   ⇒   f(x) > g(x) for f(x) > 0; or -f(x) > g(x) for f(x) < 0
Any solutions to these inequalities must respect their domains.