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anonymous

  • 5 years ago

need help solving this inequality please. x >-7

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  1. anonymous
    • 5 years ago
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    ...isn't that already solved? You know the values of x. That's about as simple as it gets

  2. anonymous
    • 5 years ago
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    it's a question on an exam prep....

  3. anonymous
    • 5 years ago
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    hm... anything more to it than just "solve..."?

  4. anonymous
    • 5 years ago
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    So what are the values of x that satisfy the relation x > -7

  5. anonymous
    • 5 years ago
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    like choices? Otherwise i think im at a loss

  6. myininaya
    • 5 years ago
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    (-7,infinity)

  7. anonymous
    • 5 years ago
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    how would I graph it....thats the other part of the question...sorry guys

  8. anonymous
    • 5 years ago
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    \[x \in (-7, \infty)\] Graph the relation? or the solutions?

  9. anonymous
    • 5 years ago
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    on a number line, draw it with an open circle on -7, with an arrow to the right

  10. anonymous
    • 5 years ago
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    Thanks Bwah!

  11. anonymous
    • 5 years ago
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    on a cartesian plane, a vertical line at x=-7, dotted (non inclusive). shade to the right

  12. myininaya
    • 5 years ago
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    x=-7 is a vertical line so since we have x>-7 and not x>=-7 then you draw a broken vertical line at x=7 and since we have x>-7 shade everything to the right 0f -7

  13. anonymous
    • 5 years ago
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    no prob, glad to help

  14. anonymous
    • 5 years ago
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    how about 3-10x>33

  15. anonymous
    • 5 years ago
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    well, the process is the same as any equation u just have an inequality sign at the end

  16. anonymous
    • 5 years ago
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    Depends on which dimensions he has to include in the graph. In \(R^1\) it's just a number line. In \(R^2\) it is a field comprised of all (x,y) pairs such that \(y\in (-\infty, \infty) \text{ and } x \in (-7, \infty)\) . In \(R^3\) it would be a boxy thing, and so on.

  17. anonymous
    • 5 years ago
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    I think I am lost on the process :)

  18. anonymous
    • 5 years ago
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    How would you solve 3-10x = 33?

  19. anonymous
    • 5 years ago
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    -10x>30 x<-3 (if i recall correctly, u switch the sign if u divide or multiply by a negative number)

  20. anonymous
    • 5 years ago
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    subtract 3 from both sides?

  21. anonymous
    • 5 years ago
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    Indeed. You can do the same thing to inequalities.

  22. anonymous
    • 5 years ago
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    Then what would you do?

  23. anonymous
    • 5 years ago
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    divide by 10......I think I am seeing the light :)

  24. anonymous
    • 5 years ago
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    Yep. The only rule is that if you divide or multiply by a negative you have to change the direction of the inequality. \(-x < 5 \implies x > 5\)

  25. anonymous
    • 5 years ago
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    Ack

  26. anonymous
    • 5 years ago
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    \[-x < 5 \implies x > -5\] rather

  27. anonymous
    • 5 years ago
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    Thanks so very much!

  28. myininaya
    • 5 years ago
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    I don't know if you have ever though of like if we have 5>2 then and we multiply both sides by negative one or divide by negative one then it becomes -5<-2 since -2 is bigger than -5

  29. myininaya
    • 5 years ago
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    any any negative number

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