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anonymous

  • one year ago

Please help.... If x>0, and two sides of a certain triangle have lengths 2x+1 and 3x+4 respectively, which of the following could be the length of the third side of the triangle?

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  1. anonymous
    • one year ago
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    The third side of the triangle must have a length greater than zero and less than the sum of the other two sides.

  2. anonymous
    • one year ago
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    1. 4x+5 2. x+2 3. 6x+1 4. 5x+6 5. 2x+17

  3. anonymous
    • one year ago
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    i know its 1, 3 and 5 but why is 3 an answer?

  4. welshfella
    • one year ago
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    adding the 2 given sides gives 5x + 5

  5. ali2x2
    • one year ago
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    Note x>0 says 3x+4 > 2x+1. So, the possibilities are those which lie between the following sum and difference: (3x+4) + (2x+1) = 5x+5 (3x+4) - (2x+1) = x+3

  6. anonymous
    • one year ago
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    If you add the length of the two sides together, you get \(5x+5\). Set that equal to \(6x+1\) and solve for \(x\). Is \(x > 0\) ?

  7. ali2x2
    • one year ago
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    so its A, B

  8. anonymous
    • one year ago
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    but look at B if you use 10 then it also doesn't work out

  9. anonymous
    • one year ago
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    "COULD BE"

  10. anonymous
    • one year ago
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    13<61<55

  11. anonymous
    • one year ago
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    it COULD BE if, for example, \(x=1\)

  12. anonymous
    • one year ago
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    \[5x + 5 = x+2\]\[4x = -3\]Can that be?

  13. anonymous
    • one year ago
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    hmmm i see wow so i'd have to be extra extra careful..... this question is a tricky one

  14. anonymous
    • one year ago
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    i have to add that this is, in my opinion ,a really foolish question that confuses the issue

  15. welshfella
    • one year ago
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    for arguments sake let x = 2 5x + 5 = 15 6x + 1 = 13 so 3rd side is less than sum of the other 2 6x +1 can be the third side.

  16. anonymous
    • one year ago
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    Satellite i 1000000% agree and the only reason it's such a dumb question is because it's a GRE question.

  17. anonymous
    • one year ago
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    yeah standardized tests strike again

  18. anonymous
    • one year ago
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    okay got it thanks everyone =)

  19. phi
    • one year ago
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    I would do it this way: from 2x+1 and 3x+4 , we must be > x+3 and < 5x+5 1. x+3 < 4x+5 < 5x+5 4x>x and 5>3 and 4x<5x and 5=5. this is OK 2. x+3 < x+2 < 5x+5 for all x>0, x+3 is bigger than x+2. NOT legal 3. x+3 <6x+1 < 5x+5 x+3< 6x+1 means 2 < 5x, 2/5 < x and 6x+1 < 5x + 5 means x< 4. so : 2/5 < x < 4 works. OK 4. x+3 <5x+6 < 5x+5 for all x>0 5x+6 > 5x+5. NOT legal 5. x+3 < 2x+17< 5x+5 2x>x and 17>3 . also 2x+17<5x+5 means 12<3x, 4<x so for x>4 this is OK

  20. phi
    • one year ago
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    when I checked (for example) x+3 < 4x+5 I said it is obviously true for x>0 but we could analyze it by simplifying the relation to -2 < 3x (add -x to both sides, add -5 to both sides) -2/3 < x and it is true for all x > -2/3. because x>0, this will always be true.

  21. phi
    • one year ago
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    and if we wanted to analyze x+3 < x+2 add -x to both sides 3<2 and we see this is not true. thus we could simplify each relation and make sure it is true for some x

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