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anonymous

  • 5 years ago

graph the system of inequalities y>-3 y<- abs(x+2) which two quadrants does the solution lie in? possible answers 2 and 3 1 and 2 1 and 4 3 and 4

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  1. anonymous
    • 5 years ago
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    what is the second inequality?

  2. anonymous
    • 5 years ago
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    y<- abs(x+2)

  3. anonymous
    • 5 years ago
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    abs?absolute?

  4. amistre64
    • 5 years ago
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  5. anonymous
    • 5 years ago
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    absolute so its 1 and 2

  6. anonymous
    • 5 years ago
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    amistre she has to answer 'AND'

  7. amistre64
    • 5 years ago
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    0 < -|2| ; i shaded a wrong area lol

  8. anonymous
    • 5 years ago
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    y < -3 means in quadrants III or IV since below the x axis.

  9. anonymous
    • 5 years ago
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    thats what i was confused

  10. anonymous
    • 5 years ago
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    but the absolute is where i am stuck

  11. anonymous
    • 5 years ago
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    then the second inequality says -x-2<y<x+2

  12. amistre64
    • 5 years ago
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  13. anonymous
    • 5 years ago
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    so has be in quad III

  14. anonymous
    • 5 years ago
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    oh damn

  15. anonymous
    • 5 years ago
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    my mistake. first one says y > -3 sorry

  16. amistre64
    • 5 years ago
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    (0,0) is a false solution to the abs...

  17. anonymous
    • 5 years ago
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    sorry sorry i messed up

  18. anonymous
    • 5 years ago
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    your sleepy aren't you satellite lol

  19. amistre64
    • 5 years ago
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    and I lifted instad of sepereted my graph...

  20. anonymous
    • 5 years ago
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    lol back at square one

  21. anonymous
    • 5 years ago
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    yeah i missed the inequality, my fault. i had it backwards

  22. amistre64
    • 5 years ago
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    this is it lol

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  23. amistre64
    • 5 years ago
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    3 and 4

  24. anonymous
    • 5 years ago
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    3 and 4

  25. anonymous
    • 5 years ago
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    amistre...is it a final attachment :p

  26. amistre64
    • 5 years ago
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    lol..... maybe ;)

  27. amistre64
    • 5 years ago
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    ..ammendment D lol

  28. anonymous
    • 5 years ago
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    no moving towards the final solutions;)

  29. anonymous
    • 5 years ago
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    yes it is 3 and 4. y =- |x+2| upside down v at (-2,0) so y<-|x+2| everything below that. must be in 3 or 4. there is just a tiny piece in 4

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