A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

What is the compound inequality 8 < or = to 2x-4 < 2

  • This Question is Closed
  1. DanJS
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hmm

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Hmm what?

  3. DanJS
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[8 \le (2x-4) \lt 2\] that right?

  4. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes

  5. DanJS
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    that is a compound inequality

  6. DanJS
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    a<b<c

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Except there is no parentheses

  8. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ya but i have to find the solution to it

  9. DanJS
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so you have 8 is less than or equal to 2x-4 AND 2x-4 is less than 2

  10. DanJS
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh, do you want to just isolate x to get the values for x, solve it

  11. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    My options are: 1. -6 < or = to X < -4 2. -2 < or = to X < -1 3. 0 < or = to X < 5 4. -2 < or = to X < 3

  12. DanJS
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[8 \le (2x-4) \lt 2\] \[8 +4\le (2x) \lt 2 + 4\] \[\frac{ 12 }{ 2 } \le x \lt \frac{ 6 }{ 2 }\] this says \[6 \le x \lt 3\] which is not true, x can not be bigger than 6 AND less than 3 at the same time

  13. DanJS
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    It ca be either OR

  14. DanJS
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you mistyped something in the prob i think

  15. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm confused

  16. DanJS
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the way you put the problem, you cant have solutions for that, you can break it up into \[x \ge 6~~~~OR~~~~x \lt 3\]

  17. DanJS
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    a value can't be both those at the same time,

  18. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I just put it the way that the school has it

  19. DanJS
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    you missing a negative sign, or something,

  20. DanJS
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    put the question down again using the draw tool

  21. DanJS
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the inequality part

  22. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1443228817681:dw|

  23. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Is that what you mean @DanJS ?

  24. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i say its 1

  25. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can you explain why? I am so confused

  26. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    was it really \[-8 \le x-4\lt 2\]

  27. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    or \[-8 \le2 x-4\lt 2\]

  28. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    the second one

  29. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    add 4 all the way across\[-4\leq 2x<6\] then divide all by 2

  30. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so it is \[-2lex <3\]

  31. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah

  32. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[-2 \le x < 3\]

  33. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[-2\leq x<3\]

  34. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thank you so much!

  35. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.