anonymous
  • anonymous
2r + 1 < 5 and –2r – 4 ≥ 6 Is it empty ?
Mathematics
jamiebookeater
  • jamiebookeater
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nenadmatematika
  • nenadmatematika
no, from the first inequality you get r<2 and from the second you get r<=-5 so the answer is r<=-5
anonymous
  • anonymous
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|
anonymous
  • anonymous
Or is it all real numbers ?

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nenadmatematika
  • nenadmatematika
no the real picture is like this
nenadmatematika
  • nenadmatematika
|dw:1328566721597:dw|
nenadmatematika
  • nenadmatematika
|dw:1328566814637:dw|
anonymous
  • anonymous
oohh! So the intersection is only on point r<=-5 ?
nenadmatematika
  • nenadmatematika
no, it's the intersection of the two intervals so the intersection area are all numbers from -infinity to -5, including -5
anonymous
  • anonymous
Okay, what is the solution set ?
nenadmatematika
  • nenadmatematika
if the question is: ''Is it empty?'' your answer is ''no'' and if they ask you to write the solution your answer is: x\[\epsilon(-\infty,-5]\]
anonymous
  • anonymous
Okay, thank you!!
nenadmatematika
  • nenadmatematika
yw

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