hmmm, let's see.
Answer B

why?

D, and E sound like they're out, but let's check.

D) 2 <= |x| <= 5, this translates to |x|>=2 and |x|<=5
x>=2 or x<=-2 and x<=5 or x>=-5 so x is between -5 and -2 or between 2 and 5, not a infinite length line segment.

E) 2<=3x+4<=6
2-4<=3x+4-4<=6-4
-2<=3x<=2
-2/3<=x<=2/3 also not a infinite length line segment.

***on test day, if an option looks as it's not the one focus on the others first, try not to waste your time, but for practise sake i ruled them out***

so, why not A or C
A and C are true for positive and negative numbers.
in A all values greater than 1 and smaller than -1 will funtion here, so the line would break between theese numbers.
in C all values greater than 4 and smaller than -4 will funtion here, so the line would break between theese numbers.

B is the only inequality that has a single infinite line segment, from -infinity to 3.

No the answer is e

moneyman wrote:Which of the following inequalities has a solution set that,when graphed on the number line,is a single line segment of finite length??

x^4>and=1
x^3<and=27
x^2>and=16
2<=|x|<=5
2<=3x+4<=6

x^4>and=1 = X can be 1,2…..any int so not a line segment
x^3<and=27 implies x can be 1,2 or 3 so not a line segment
x^2>and=16 implies x can be 4,5,……any int greater than 4 and negative int can also be the value for x
2<=|x|<=5 implies -2>=x>=2 and -5<=x<=5 so, x has many values
2<=3x+4<=6 implies true for just 1 value of x i.e. x=0 so ……it has fixed length

SO the answer is E

I hope it explained it well

you're right, don't know what i was thinking on that day, any range of numbers squared or cubed will never yield a consecutive set of results, A,B & C out.

and from my own previous answer
E) 2<=3x+4<=6
2-4<=3x+4-4<=6-4
-2<=3x<=2
-2/3<=x<=2/3 also not a infinite length line segment.

but since we're looking for a finite line segment. I need to take more care with reading what is asked