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anonymous

  • 4 years ago

2r + 1 < 5 and –2r – 4 ≥ 6 Is it empty ?

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  1. nenadmatematika
    • 4 years ago
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    no, from the first inequality you get r<2 and from the second you get r<=-5 so the answer is r<=-5

  2. anonymous
    • 4 years ago
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    This when there is 'or' but now we are dealing with 'and' => intersection. When you want to graph it will look like : |dw:1328566495750:dw|

  3. anonymous
    • 4 years ago
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    Or is it all real numbers ?

  4. nenadmatematika
    • 4 years ago
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    no the real picture is like this

  5. nenadmatematika
    • 4 years ago
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    |dw:1328566721597:dw|

  6. nenadmatematika
    • 4 years ago
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    |dw:1328566814637:dw|

  7. anonymous
    • 4 years ago
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    oohh! So the intersection is only on point r<=-5 ?

  8. nenadmatematika
    • 4 years ago
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    no, it's the intersection of the two intervals so the intersection area are all numbers from -infinity to -5, including -5

  9. anonymous
    • 4 years ago
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    Okay, what is the solution set ?

  10. nenadmatematika
    • 4 years ago
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    if the question is: ''Is it empty?'' your answer is ''no'' and if they ask you to write the solution your answer is: x\[\epsilon(-\infty,-5]\]

  11. anonymous
    • 4 years ago
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    Okay, thank you!!

  12. nenadmatematika
    • 4 years ago
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    yw

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