Asked by Ummm...
                Ok, I am horrible at math so I am going to need someone's help.....
1. Solve.
|x+3|<5
a. -8<x<2 ****
b. -2<x<8
c.3<x<5
d.-3<x<5
2. Solve.
|-3n| - 2 = 4
a. 2 or -2 ****
b. 3 or -3
c. 4 or -4
d. 6 or -6
3. Solve.
|f| - 2/3=5/6
a. 5/6 or -5/6
b. 2/3 or -2/3
c. 1 1/2 or -1 1/2
d. no solution
4. Solve.
-2 |x - 5|≤ -10
a. x ≤ -5 or x ≥ 5
b. -5 ≤ x ≤ 5
c. x ≤ 0 or x ≥ 10
d. -10 ≤ x ≤ 10
5. The ideal circumference of a woman's basketball is 28.75 in. The actual circumference may vary from the ideal by at most 0.25 in. What are the acceptable circumferences for a woman's basketball? Let the variable,c, represent circumference.
a. c ≤ 28.5 in. or c ≥ 29 in.
b. c < 28.5 in. or c > 29 in.
c. 28.5 in. ≤ c ≤ 29 in.
d. 28.5 < c < 29 in.
Please help!!
            
        1. Solve.
|x+3|<5
a. -8<x<2 ****
b. -2<x<8
c.3<x<5
d.-3<x<5
2. Solve.
|-3n| - 2 = 4
a. 2 or -2 ****
b. 3 or -3
c. 4 or -4
d. 6 or -6
3. Solve.
|f| - 2/3=5/6
a. 5/6 or -5/6
b. 2/3 or -2/3
c. 1 1/2 or -1 1/2
d. no solution
4. Solve.
-2 |x - 5|≤ -10
a. x ≤ -5 or x ≥ 5
b. -5 ≤ x ≤ 5
c. x ≤ 0 or x ≥ 10
d. -10 ≤ x ≤ 10
5. The ideal circumference of a woman's basketball is 28.75 in. The actual circumference may vary from the ideal by at most 0.25 in. What are the acceptable circumferences for a woman's basketball? Let the variable,c, represent circumference.
a. c ≤ 28.5 in. or c ≥ 29 in.
b. c < 28.5 in. or c > 29 in.
c. 28.5 in. ≤ c ≤ 29 in.
d. 28.5 < c < 29 in.
Please help!!
Answers
                    Answered by
            Damon
            
    1. agree
2. agree
3. hint f - 4/6 = 5/6 and -f - 4/6 = 5/6
    
2. agree
3. hint f - 4/6 = 5/6 and -f - 4/6 = 5/6
                    Answered by
            Steve
            
    -2 |x - 5|≤ -10
|x-5| >= 5
That is, x must be at least 5 away from 5
x ≤ 0 or x ≥ 10
or, algebraically,
(x-5)^2 >= 25
x^2 - 10x + 25 >= 25
x^2-10x >= 0
x(x-10) >= 0
so, either both are positive (x >= 10)
or both are negative (x <= 0)
28.75 - 0.25 <= c <= 28.75 + 0.25
28.5 <= c <= 29.0
    
|x-5| >= 5
That is, x must be at least 5 away from 5
x ≤ 0 or x ≥ 10
or, algebraically,
(x-5)^2 >= 25
x^2 - 10x + 25 >= 25
x^2-10x >= 0
x(x-10) >= 0
so, either both are positive (x >= 10)
or both are negative (x <= 0)
28.75 - 0.25 <= c <= 28.75 + 0.25
28.5 <= c <= 29.0
                    Answered by
            Damon
            
    4.
-2 |x - 5|≤ -10
|x-5| >/= 5
ends
(x-5) = 5 is x = 10
-(x-5) = 5 is -x = 0
c
    
-2 |x - 5|≤ -10
|x-5| >/= 5
ends
(x-5) = 5 is x = 10
-(x-5) = 5 is -x = 0
c
                    Answered by
            Damon
            
    circumference between AND including the end points
    
                    Answered by
            playfun
            
    Answer: 
1: A
2: A
3: C
4: C
5: C
    
1: A
2: A
3: C
4: C
5: C
                    Answered by
            oops
            
    playfun is correct i got a 100 !
    
                    Answered by
            girl
            
    playfun is 100% correct.
    
                    Answered by
            Lee
            
    playfun is right, i was on here to check and all the answers were right 
    
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.