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kittiwitti1

  • one year ago

I DON'T GET IT How is either method wrong? - (the plus minus sign wasn't an option in their symbols bank?) http://prntscr.com/8ft0nc http://prntscr.com/8ft0se

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  1. imqwerty
    • one year ago
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    \[\sqrt{4-(2\sin \theta)^2}\]\[\sqrt{4-4\sin^2\theta} \]\[\sqrt{4(1-\sin^2\theta)}\]and 1-sin^2(theta)=cos^2theta so we get \[\sqrt{4\cos^2\theta}\]=\[\pm2\cos \theta\]

  2. kittiwitti1
    • one year ago
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    I got that. The system isn't taking the answer in any form...

  3. imqwerty
    • one year ago
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    wait :D i didn see this - \[0<\theta <90\]

  4. kittiwitti1
    • one year ago
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    I tried these combinations\[\pm\sqrt{4\cos^{2}\theta}\]\[\sqrt{4\cos^{2}\theta},-\sqrt{4\cos^{2}\theta}\]yes? what does this mean?\[0\lt\theta\lt90?\]

  5. imqwerty
    • one year ago
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    that means that the angle theta lies between 0degrees and 90 deegrees

  6. kittiwitti1
    • one year ago
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    Yes ? I don't know what that applies to my answer lol, don't know how to

  7. imqwerty
    • one year ago
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    hav u tried this- \[2\cos \theta\]

  8. imqwerty
    • one year ago
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    tried that as the answer?

  9. kittiwitti1
    • one year ago
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    http://prntscr.com/8ft6av?

  10. imqwerty
    • one year ago
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    ?

  11. kittiwitti1
    • one year ago
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    I have tried the positive version already. I have only one attempt left. However, I solved by hand and the answer I got was\[\pm2\cos\theta\]

  12. kittiwitti1
    • one year ago
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    I mean\[\pm2\left|\cos\theta\right|\]

  13. imqwerty
    • one year ago
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    yea

  14. kittiwitti1
    • one year ago
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    I don't get it. See, I got this one right without issues. http://prntscr.com/8ft75p

  15. kittiwitti1
    • one year ago
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    Maybe they are negative?

  16. kittiwitti1
    • one year ago
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    I can't risk the attempt though, got to make sure. :S

  17. imqwerty
    • one year ago
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    :) if they r not accepting positive then no scope for negative

  18. imqwerty
    • one year ago
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    maybe theres some error in their server

  19. kittiwitti1
    • one year ago
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    "No scope"?? -Confused- Yes maybe. One time I put space after a comma and they told me it was wrong. Someone on here said "take the space out" and I got it right after that... =_=

  20. imqwerty
    • one year ago
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    hahaha :D these things r annoying have u tried it like this- http://prntscr.com/8ft8mg i mean that space between theta nd that sec or tan or whateva

  21. kittiwitti1
    • one year ago
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    Sadly I tried it before, doesn't register as a trig function if I do that

  22. imqwerty
    • one year ago
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    have tried any such question before ...a question in which the answer was in the form of a trigo expression?

  23. kittiwitti1
    • one year ago
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    que?

  24. hartnn
    • one year ago
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    i would try \( 2| \cos \theta| \) for 1st and \(3\tan \theta \) for 2nd

  25. hartnn
    • one year ago
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    i mean \(3 |\tan \theta |\)

  26. kittiwitti1
    • one year ago
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    @hartnn first one does not accept second one I cannot try, I must be sure - only one submit attempt left ^^; sorry!

  27. hartnn
    • one year ago
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    its definitely not \(\pm\) the answer to a \(\sqrt{...}\) is always a positive

  28. imqwerty
    • one year ago
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    yes maybe it doesn't accept because of that bracket u put :P http://prntscr.com/8ftchw

  29. kittiwitti1
    • one year ago
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    The coding comes with the bracket. @imqwerty I cannot remove that. I have tried that.

  30. kittiwitti1
    • one year ago
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    @hartnn I was taught that there are two types of signs to a square root? I mean (-)(-)=+ right?

  31. imqwerty
    • one year ago
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    u must consult ur teacher

  32. kittiwitti1
    • one year ago
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    HAHAHAHHA my teacher is dumb she won't help me, I've tried. She sucks at technology and give us online homework LOL

  33. hartnn
    • one year ago
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    if already not tried, \(3 |\tan(\theta)|\) is worth trying for 2nd. don't forget to keep the fingers crossed :P

  34. kittiwitti1
    • one year ago
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    ehhh. I have done that?

  35. imqwerty
    • one year ago
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    lol

  36. imqwerty
    • one year ago
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    is this flvs? maybe u get the solution to this issue on google :)

  37. hartnn
    • one year ago
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    if x^2 = a then x = \(\pm \sqrt a\) thats because \((\sqrt a)^2 \\ and \\ (-\sqrt a)^2 \\ \text{both equal a}\)

  38. hartnn
    • one year ago
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    and \(\sqrt a\) is just one of them, and it can never be negative

  39. kittiwitti1
    • one year ago
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    ????????

  40. kittiwitti1
    • one year ago
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    Ah my question key die from pressing too much LOL x_x!

  41. hartnn
    • one year ago
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    was just trying to explain why sqrt a can never be negative

  42. imqwerty
    • one year ago
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    so u got the 1st one wrong ....did it tell u the correct answer afterwards?

  43. kittiwitti1
    • one year ago
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    @imqwerty No, it does not tell me the answer. :[ @hartnn I am still lost, what o-o

  44. kittiwitti1
    • one year ago
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    ...My computer is very laggy right now

  45. imqwerty
    • one year ago
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    :)

  46. hartnn
    • one year ago
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    go for \(3|\tan \theta |\) !

  47. kittiwitti1
    • one year ago
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    ...Fine, I blame you guys if I fail the question lol

  48. kittiwitti1
    • one year ago
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    http://prntscr.com/8ftleb ... @hartnn it's wrong.

  49. kittiwitti1
    • one year ago
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    I don't get iiiit! D:

  50. hartnn
    • one year ago
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    the system should atleast take equivalent answers.. seems to be a faulty system, and an incorrect way to test students! maybe it would have accepted \(|3 \tan \theta |\)

  51. kittiwitti1
    • one year ago
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    The teacher said the answer must be a|cos theta| format

  52. hartnn
    • one year ago
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    it needs to have cos in ti ??

  53. kittiwitti1
    • one year ago
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    no that is an example trig function

  54. hartnn
    • one year ago
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    sec^2 x = 1/cos^2 x

  55. kittiwitti1
    • one year ago
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    LOL

  56. hartnn
    • one year ago
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    3 | tan \(\theta \)| is indeed in that format

  57. kittiwitti1
    • one year ago
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    Yeah idk why they say it's wrong. This question have the right format http://prntscr.com/8ftnwx

  58. kittiwitti1
    • one year ago
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    @hartnn \[\sin10=\cos{?}\]

  59. kittiwitti1
    • one year ago
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    10 degree

  60. kittiwitti1
    • one year ago
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    Wait never mind

  61. kittiwitti1
    • one year ago
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    I got it

  62. hartnn
    • one year ago
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    sin x = cos(90-x) here x = 10 whats 90 -10 ?

  63. hartnn
    • one year ago
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    ok

  64. kittiwitti1
    • one year ago
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    lolol sorry

  65. hartnn
    • one year ago
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    no problem :)

  66. kittiwitti1
    • one year ago
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    But I don't get the formatting of those problem, confusing D:

  67. kittiwitti1
    • one year ago
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    Also: "Use Definition II to explain why it is possible to find an angle θ that will make tan θ as large as we wish" (what?!)

  68. kittiwitti1
    • one year ago
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    @hartnn ? ?

  69. hartnn
    • one year ago
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    whats the Def II? they are asking why tan 90n goes to infinity or -infinity

  70. kittiwitti1
    • one year ago
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    Yes. And I got wrong answer here too LOL very sorry (8 cos 30°)2 I put 96 is wrong

  71. kittiwitti1
    • one year ago
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    ^2 sorry not x2

  72. hartnn
    • one year ago
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    cos 30 = sqrt 3/2 8 cos 30 = 4 sqrt 3 its square 16*3 = ...

  73. kittiwitti1
    • one year ago
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    :o Oh um um um

  74. kittiwitti1
    • one year ago
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    UM.... 48?!

  75. hartnn
    • one year ago
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    yes

  76. kittiwitti1
    • one year ago
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    lol! I got it :D

  77. kittiwitti1
    • one year ago
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    Thank you ! :]

  78. hartnn
    • one year ago
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    welcome ^_^

  79. kittiwitti1
    • one year ago
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    But... what about the weird question? Do I just skip it for now? o-o

  80. hartnn
    • one year ago
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    take 2nd opinion... maybe ask some other expert or you can skip it, if you're not in any hury

  81. kittiwitti1
    • one year ago
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    Okay.

  82. kittiwitti1
    • one year ago
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    @hartnn Sorry question: http://prntscr.com/8ftxg2 ?

  83. hartnn
    • one year ago
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    cos 30 = sqrt 3/2 cos^2 30 = 3/4 4cos^2 30 = 3

  84. hartnn
    • one year ago
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    6 sin 30 = 6/2 =3

  85. hartnn
    • one year ago
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    finally, 3+3 =6

  86. kittiwitti1
    • one year ago
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    o_o Oh.

  87. kittiwitti1
    • one year ago
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    So I should not do it the quadratic formula way yes?

  88. kittiwitti1
    • one year ago
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    Too complex? lol

  89. hartnn
    • one year ago
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    you can, but that not required, because we know the values already

  90. kittiwitti1
    • one year ago
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    ah ok. got it

  91. kittiwitti1
    • one year ago
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    −5 sin 2x if x = 30, how much? I don't know how to do a sin b x

  92. hartnn
    • one year ago
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    x = 30, so 2x = 60 sin 60 = sqrt 3/2 so -5 sqrt 3/2

  93. kittiwitti1
    • one year ago
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    OOOOOOH okay! Thank you :)

  94. kittiwitti1
    • one year ago
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    Umm last one, this really confusing me http://prntscr.com/8fu4ew?

  95. kittiwitti1
    • one year ago
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    ?

  96. kittiwitti1
    • one year ago
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    @hartnn

  97. hartnn
    • one year ago
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    tan is opposite/adjacent so it has to be the 1st one only

  98. kittiwitti1
    • one year ago
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    Ohhh okay my brain is too tired to process right, thank you so much lol

  99. kittiwitti1
    • one year ago
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    I gonna close this now

  100. hartnn
    • one year ago
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    sure, welcome ^_^

  101. kittiwitti1
    • one year ago
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    @hartnn @imqwerty the answer was right EXCEPT for the absolute value... WHAT THE HECK Thanks hartnn I should have listened to you lol

  102. imqwerty
    • one year ago
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    :D

  103. kittiwitti1
    • one year ago
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    @imqwerty what?

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