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anonymous

  • one year ago

Derive the trigonometric addition formula for sine :

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  1. anonymous
    • one year ago
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    \[\sin(a+b)=sinacosb+cosasinb\]

  2. anonymous
    • one year ago
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    also, derive these identities using the addition or subtraction formulas for sine or cosine : \[sinasinb=\frac{ 1 }{ 2 }(\cos(a-b)-\cos(a+b))\]

  3. IrishBoy123
    • one year ago
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    easiest i know is: \( e^{i \ a} = (cos a + i \ sin a)\) \( e^{i \ b} = (cosb + i \ sin b)\) \( e^{i \ a} . e^{i \ b} = e^{i \ (a+b)} = \) then multiply out and equate the real and complex bits for the second bit, just churn the formula. eg expand the RHS using the cos (a+b) and cos (a-b).

  4. IrishBoy123
    • one year ago
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    if you post some stuff, i'm sure you will get all the help you need :p

  5. anonymous
    • one year ago
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    I've never seen that first formula before . I'm even more confused

  6. freckles
    • one year ago
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    you can derive cos(a-b)=cos(a)cos(b)+sin(a)sin(b) using the distance formula then you can use cos(a-b)=cos(a)cos(b)+sin(a)sin(b) along with cofunction identities to show sin(a+b)=sin(a)cos(b)+sin(b)cos(a)

  7. anonymous
    • one year ago
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    what are cofunction identities ?

  8. anonymous
    • one year ago
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    This is an algebra 2 class and they threw in a trig section

  9. freckles
    • one year ago
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    you are asking to derive the formula sin(a+b)=sin(a)cos(b)+sin(b)cos(a) right?

  10. anonymous
    • one year ago
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    yes

  11. freckles
    • one year ago
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    hmmm maybe you don't remember the name of the identity but you remember that cos(90-A)=sin(A) and sin(90-A=cos(A)? If not you easily show these by drawing right triangles and comparing the above ratios.

  12. anonymous
    • one year ago
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    the notes didn't name the identities and I'm sure I learned that . Trigonometry is just really hard for me to understand

  13. IrishBoy123
    • one year ago
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    hi @jxaf sorry for the confusion, i should have asked first what you knew. do you think you could follow this proof, with some guidance. https://gyazo.com/be936132eb2a67f24dbd6841f0cf94a3

  14. anonymous
    • one year ago
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    Um, that looks really complex. Do you want a picture of my notes to see what I learned ?

  15. IrishBoy123
    • one year ago
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    yeah, go for it!

  16. anonymous
    • one year ago
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    okay it might take a bit cause my computer is really slow

  17. IrishBoy123
    • one year ago
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    take your time.

  18. IrishBoy123
    • one year ago
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    well i've drawn the shape for the proof anyways just in case it comes in useful. i think it's more straightforward that it might appear. |dw:1440346130714:dw|

  19. yamyam70
    • one year ago
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    Try searching , sum and difference trigonometric identities, that's the name of the first one

  20. anonymous
    • one year ago
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    these are my notes

  21. anonymous
    • one year ago
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    |dw:1440346974875:dw|

  22. anonymous
    • one year ago
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    you guys keep doing the same triangle and I haven't seen that :/

  23. anonymous
    • one year ago
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    Yeah , apex doesn't teach that

  24. anonymous
    • one year ago
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    they probably want you to just memorize and expect to learn.

  25. IrishBoy123
    • one year ago
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    @jxaf i'll happily have a read of your notes once i get back, am popping out

  26. anonymous
    • one year ago
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    https://archive.org/details/TeachYourselfTrigonometry_94

  27. anonymous
    • one year ago
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    idk why they would when I'm taking an algebra 2 class and @IrishBoy123 okay (:

  28. anonymous
    • one year ago
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    your identities are covered in that really old teach yourself trig book beginning from pp 98

  29. anonymous
    • one year ago
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    it is the same supplementary textbook I used in 6th grade

  30. anonymous
    • one year ago
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    you learned trig in the 6th grade ?

  31. anonymous
    • one year ago
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    they teach this early as 6th grade in some parts of Asia

  32. freckles
    • one year ago
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    after seeing your notes...I'm unsure the question is to actually derive the sine sum formula you know derive this formula: sin(a+b)=sin(a)cos(b)+sin(b)cos(a) are you sure it says derive that formula?

  33. anonymous
    • one year ago
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    hold on

  34. freckles
    • one year ago
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    also i did notice an error in your notes

  35. anonymous
    • one year ago
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    I'm sure there is . Apex doesn't make sense when it comes to the note taking

  36. freckles
    • one year ago
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    "page 3: a. The trigonometric sum identity for sine is san(a+b)=sin(a)cos(-b)+cos(a)sin(-b) ...." That should be sin(a+b)=sin(a)cos(b)+cos(a)sin(b)

  37. freckles
    • one year ago
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    sin^2(theta)+cos^2(theta)=1 is true but cos(a+b) doesn't equal sin^2(theta)+cos^2(theta)

  38. anonymous
    • one year ago
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    these are the questions

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  39. freckles
    • one year ago
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    oh and page 1 ... cos(a+b)=cos(a)cos(b)-sin(a)sin(b)

  40. freckles
    • one year ago
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    i guess i'm wrong it is to derive sin(a+b)=sin(a)cos(b)+cos(a)sin(b)

  41. freckles
    • one year ago
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    I wonder if you can use the cos difference identity

  42. freckles
    • one year ago
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    \[\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b) \\ \sin(a+b)=\cos(\frac{\pi}{2}-[a+b])= \cos(\frac{\pi}{2}-a-b) \\ =\cos([\frac{\pi}{2}-a]-[b])\] then plug into the cos difference identity

  43. anonymous
    • one year ago
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    I have no idea lol . I need this packet done by tomorrow and it's stressing me out

  44. freckles
    • one year ago
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    well I don't know what you can use to derive I don't know your tool set for the problem

  45. freckles
    • one year ago
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    but if you can use the cos difference identity then that way I mentioned above is one way to do it

  46. anonymous
    • one year ago
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    the sinacosb-cosasinb ?

  47. freckles
    • one year ago
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    if you aren't allowed to use that then you have to derive also the identity cos(a-b)=cos(a)cos(b)+sin(a)sin(b) and as mentioned earlier one way to do that is using the distance formula

  48. anonymous
    • one year ago
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    I'm ready to give up :/

  49. freckles
    • one year ago
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    just use this then: \[\cos(x-b)=\cos(x)\cos(b)+\sin(x)\sin(b) \\ \sin(a+b)=\cos([\frac{\pi}{2}-a]-b)\]

  50. freckles
    • one year ago
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    replace x with pi/2-a and replace b well with b

  51. anonymous
    • one year ago
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    where's x ?

  52. anonymous
    • one year ago
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    jk i see it

  53. anonymous
    • one year ago
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    is that all or ?

  54. freckles
    • one year ago
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    don't forget to use the cofunction identities

  55. freckles
    • one year ago
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    and then you will have sin(a+b)=sin(a)cos(b)+sin(b)cos(a)

  56. anonymous
    • one year ago
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    what are the cofunction identities

  57. freckles
    • one year ago
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    hmmm... I thought I mentioned two of them earlier...

  58. freckles
    • one year ago
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    cos(pi/2-a)=sin(a) sin(pi/2-a)=cos(a)

  59. anonymous
    • one year ago
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    you probably did , I just forgot trying to understand everything else

  60. anonymous
    • one year ago
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    I'm still confused :/

  61. freckles
    • one year ago
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    ok have you expanded \[\cos((\frac{\pi}{2}-a)-b) \text{ yet using the cosine difference identity ?}\]

  62. anonymous
    • one year ago
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    no ..

  63. freckles
    • one year ago
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    \[\cos(x-b)=\cos(x)\cos(b)+\sin(x)\sin(b)\] I was asking you to replace x with pi/2-a and b can just stay b so you haven't that yet?

  64. anonymous
    • one year ago
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    so it'd be cos(pi/2-a)cos(b)+sin(pi/2-a)sin(b) ?

  65. freckles
    • one year ago
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    right now use the cofunction identities

  66. anonymous
    • one year ago
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    but you just said to substitute x with that ..

  67. freckles
    • one year ago
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    \[\sin(a+b) \\ =\cos(\frac{\pi}{2}-(a+b)) \\ =\cos(\frac{\pi}{2}-a-b) \\ =\cos([\frac{\pi}{2}-a]-[b]) \\ =\cos(\frac{\pi}{2}-a)\cos(b)+\sin(\frac{\pi}{2}-a)\sin(b) \] one last step and you are done the last step is to apply cofunction identities

  68. yamyam70
    • one year ago
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    @jxaf You can do it! :)

  69. freckles
    • one year ago
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    "but you just said to substitute x with that .." I was saying you did that part right... but you have one last step

  70. freckles
    • one year ago
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    remember I just say above cos(pi/2-a)=? and sin(pi/2-a)=?

  71. anonymous
    • one year ago
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    @yamyam70 thank you lol @freckles yeah , I do

  72. freckles
    • one year ago
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    so you have completed the problem right? because you replaced cos(pi/2-a) with sin(a) and sin(pi/2-a) with cos(a)?

  73. anonymous
    • one year ago
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    yes

  74. anonymous
    • one year ago
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    what about the next problem

  75. freckles
    • one year ago
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    try expanding cos(a-b) and cos(a+b)

  76. anonymous
    • one year ago
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    the cos(a-b) would be cosacosb+sinasinb ?

  77. anonymous
    • one year ago
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    and cos(a+b) would be cosacosb-sinasinb ?

  78. freckles
    • one year ago
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    what happens when you subtract that last from the first one you wrote that is can you simplify: [cos(a)cos(b)+sin(a)sin(b)]-[cos(a)cos(b)-sin(a)sin(b)]

  79. anonymous
    • one year ago
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    it becomes 0 ?

  80. freckles
    • one year ago
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    well sin(a)sin(b)+sin(a)sin(b) doesn't equal 0 but cos(a)cos(b)-cos(a)cos(b) is 0

  81. anonymous
    • one year ago
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    jk , the subtraction distributes

  82. anonymous
    • one year ago
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    so it'd be 2sinasinb ?

  83. freckles
    • one year ago
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    yes then what is 1/2*2?

  84. anonymous
    • one year ago
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    1

  85. freckles
    • one year ago
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    \[\frac{1}{2}(\cos(a-b)-\cos(a+b)) \\ \frac{1}{2}(\cos(a)\cos(b)+\sin(a)\sin(b)-\cos(a)\cos(b)+\sin(a)\sin(b)) \\ \frac{1}{2}(2 \sin(a)\sin(b)) \\ (\frac{1}{2} \cdot 2) \sin(a)\sin(b)\]

  86. freckles
    • one year ago
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    and yes 1/2*2=1

  87. anonymous
    • one year ago
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    wait why'd you use the half identity thing ?

  88. anonymous
    • one year ago
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    jk

  89. anonymous
    • one year ago
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    I see why . I don't think straight when I'm stressed as you can see

  90. freckles
    • one year ago
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    don't be stressed take a deep breath and enjoy math you should always put aside enough time to think about math so I can see why you are stressed if you have put yourself so close to a deadline (if you have this time) just don't do that again and you might enjoy math

  91. anonymous
    • one year ago
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    I like math , but I've been introduced to trigonometry since geometry and I have never been able to get it and the packet I'm doing was supposed to be due on Friday and I didn't get it done because my mom had major surgery and I was helping her and I just didn't have time to do it but I got an extension

  92. freckles
    • one year ago
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    awww yeah i guess life can get in the way of putting aside time for math hope she is doing better

  93. anonymous
    • one year ago
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    And plus the thing I have to use to get the notes makes no sense to me at all and she is , thank you

  94. IrishBoy123
    • one year ago
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    @jxaf awesome job, so go back and look again at P1, P3a, P8b & P8d in your worksheet. there may be others but those are signs that you are cutting corners and doing things whilst stressed..... well done!

  95. anonymous
    • one year ago
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    Thank you ! @IrishBoy123

  96. freckles
    • one year ago
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    did you want me to show you why the cofunction identities are true @jxaf

  97. freckles
    • one year ago
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    it isn't too long

  98. anonymous
    • one year ago
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    yeah

  99. freckles
    • one year ago
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    okay this is a right triangle (which I have labeled) |dw:1440350795124:dw|

  100. freckles
    • one year ago
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    the sum of the angles in a triangle equal 180 deg

  101. freckles
    • one year ago
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    B+A+90=180 B+A=90

  102. freckles
    • one year ago
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    I can write B in terms of A

  103. freckles
    • one year ago
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    B+A=90 subtract A on both sides B=90-A

  104. freckles
    • one year ago
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    |dw:1440350884800:dw|

  105. freckles
    • one year ago
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    so do you know how to find cos(B) and sin(A) using this right triangle?

  106. anonymous
    • one year ago
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    cos(b) would be a/c sin(a) would be a/c also ?????????????

  107. freckles
    • one year ago
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    and that should be cos(B) and sin(A) respectively right guess what ? that makes cos(B)=sin(A) since they are both equal to a/c but B=90-A so that means cos(90-A)=sin(A)

  108. freckles
    • one year ago
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    this is one of the cofunction identities and you can find 5 other cofunction identities using this right triangle

  109. freckles
    • one year ago
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    \[\cos(B)=\sin(A)=\frac{a}{c} \\ \text{ so } \cos(90-A)=\sin(A) \\ \sin(B)=\cos(A)=\frac{b}{c} \\ \text{ so } \sin(90-A)=\cos(A) \\ \tan(B)=\cot(A)=\frac{b}{a} \\ \text{ so } \tan(90-A)=\cot(A) \\ \csc(B)=\sec(A) =\frac{c}{b}\\ \text{ so } \csc(90-A)=\sec(A) \\ \] you can also from the triangle see that cot(90-A)=tan(A) and sec(90-A)=csc(A)

  110. freckles
    • one year ago
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    by the way all of the above measurements I used were in degrees I hope that was obvious by me choose 90 instead of pi/2

  111. anonymous
    • one year ago
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    No it makes more sense with degrees lol . the whole pi stuff gets me confused . I need the unit circle

  112. anonymous
    • one year ago
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    could you help me with another problem ?

  113. freckles
    • one year ago
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    i will be back in 5 minutes and if you still need help on it you know if someone else hasn't helped you yet i will try to

  114. anonymous
    • one year ago
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    Okay, thank you . Ima make it a different question

  115. anonymous
    • one year ago
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    *

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