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mathmath333

  • one year ago

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align}& \normalsize \text{Let U be the set of all triangles in a plane.}\hspace{.33em}\\~\\ & \normalsize \text{If A is the set of all triangles with at}\hspace{.33em}\\~\\ & \normalsize \text{least one angle different from 60°, what is A′?}\hspace{.33em}\\~\\ \end{align}}\)

  2. vera_ewing
    • one year ago
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    A' is the set of all triangles with all angles 60°.

  3. mathmath333
    • one year ago
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    how

  4. vera_ewing
    • one year ago
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    A is the set of all triangles with at least one different angle from 60°.

  5. vera_ewing
    • one year ago
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    So that means that A' is the set of all equilateral triangles.

  6. mathmath333
    • one year ago
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    lol u wrote the same thing

  7. mathmath333
    • one year ago
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    i neeed A compliment

  8. mathmath333
    • one year ago
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    \(\Huge A'\)

  9. ParthKohli
    • one year ago
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    Jesus. I just typed everything out and it removed it.

  10. ParthKohli
    • one year ago
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    OK, so \(A'\) *is* the set of equilateral triangles.

  11. mathmath333
    • one year ago
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    type it on notepad first then copy

  12. Empty
    • one year ago
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    Or you could use a better site such as www.peeranswer.com

  13. ParthKohli
    • one year ago
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    \(A \cup A' = T\) where \(T\) is the set of all triangles. Thus, \(A'\) is the set of all triangles that do not satisfy the criterion for \(A\) (that at least one angle should be different from 60). The only way to violate that? Have none of them different from 60, which pretty much means that all of them are 60.

  14. mathmath333
    • one year ago
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    no i have faith in OP.com

  15. ParthKohli
    • one year ago
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    Read that as "I have faith in OP's mom"

  16. imqwerty
    • one year ago
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    OP.com??????

  17. mathmath333
    • one year ago
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    OPenstudy.com

  18. imqwerty
    • one year ago
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    ohhhhh :O

  19. mathmath333
    • one year ago
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    lol

  20. ganeshie8
    • one year ago
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    negation of "at least one" is "none" so the negation of "at least one angle different from 60" is "none of the angles different from 60", which is same as "all angles are 60"

  21. mathmath333
    • one year ago
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    hmm this is confujing

  22. mathmath333
    • one year ago
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    Fill in the blanks to make each of the following a true statement : (i) A ∪ A′ = . . . (ii) φ′ ∩ A = . . . (iii) A ∩ A′ = . . . (iv) U′ ∩ A = . . .

  23. ganeshie8
    • one year ago
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    whats the negation of below statement : at least of the gifts is a car

  24. mathmath333
    • one year ago
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    car is not in the gifts

  25. ganeshie8
    • one year ago
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    right, another whats the negation of below statement : at least of the gifts is not a car

  26. mathmath333
    • one year ago
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    Their might be car as gift more than once.

  27. ganeshie8
    • one year ago
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    try again

  28. mathmath333
    • one year ago
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    at least of the gifts is a car

  29. ganeshie8
    • one year ago
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    ``` at least of the gifts is not a car ``` means, there exist one or more gifts that are not cars

  30. ganeshie8
    • one year ago
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    whats the negation of that ?

  31. mathmath333
    • one year ago
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    this_>? at least of the gifts is a car

  32. ganeshie8
    • one year ago
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    nope

  33. ganeshie8
    • one year ago
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    negation of "at least one" is "none"

  34. mathmath333
    • one year ago
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    i mean negation of which sentence

  35. ganeshie8
    • one year ago
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    ``` at least of the gifts is not a car ``` negation of above statement would be : ``` none of the gifts is not a car ```

  36. ganeshie8
    • one year ago
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    that means, there are no gifts that are not cars, which is same as saying that all the gifts are cats

  37. mathmath333
    • one year ago
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    ok i see my english is poor

  38. ganeshie8
    • one year ago
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    it has nothing to do with english, it is logic, you just need to see it once

  39. mathmath333
    • one year ago
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    oh, hmm ok i have one morre question Fill in the blanks to make each of the following a true statement : (i) A ∪ A′ = . . . (ii) φ′ ∩ A = . . . (iii) A ∩ A′ = . . . (iv) U′ ∩ A = . . .

  40. ganeshie8
    • one year ago
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    \(A'\) is everything that is not \(A\) therefore, combining\(A\) with \(A'\) gives you everything : \[A\cup A' = U\]

  41. ganeshie8
    • one year ago
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    may be lets work this with venn diagrams

  42. mathmath333
    • one year ago
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    |dw:1438798363617:dw|

  43. ganeshie8
    • one year ago
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    Yes, that looks good. can you show me A and A' in above venn diagram

  44. mathmath333
    • one year ago
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    |dw:1438798432076:dw|

  45. ganeshie8
    • one year ago
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    as expected, you can see that combining A and A' covers the entire U

  46. mathmath333
    • one year ago
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    ok

  47. ganeshie8
    • one year ago
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    what about \(A\cap A'\) ? is there anything common between A and A' ?

  48. mathmath333
    • one year ago
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    nothing common

  49. ganeshie8
    • one year ago
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    since there is nothing common, we say it equals `emptyset`, \(\varphi\)

  50. ganeshie8
    • one year ago
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    \[A\cap A' = \varphi\]

  51. mathmath333
    • one year ago
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    ok

  52. mathmath333
    • one year ago
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    this one \(\emptyset′ ∩ A\) =

  53. ganeshie8
    • one year ago
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    whats the compliment of emptyset ?

  54. ganeshie8
    • one year ago
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    make a guess

  55. mathmath333
    • one year ago
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    \( \emptyset\)

  56. ganeshie8
    • one year ago
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    try again

  57. ganeshie8
    • one year ago
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    by definition, complement is different from the original

  58. ganeshie8
    • one year ago
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    emptyset and universal set are complements of each other

  59. ganeshie8
    • one year ago
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    \(\varphi' = U\) \(U' = \varphi\)

  60. mathmath333
    • one year ago
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    ok i see

  61. ganeshie8
    • one year ago
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    so \(\emptyset′ ∩ A=U\cap A = ?\)

  62. mathmath333
    • one year ago
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    A

  63. ganeshie8
    • one year ago
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    Yep, try the last one

  64. mathmath333
    • one year ago
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    \[\emptyset\]

  65. ganeshie8
    • one year ago
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    Yes \(U′ ∩ A = \varphi \cap A = \varphi\)

  66. mathmath333
    • one year ago
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    thnx

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