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brittany_lundgren

1.Solve S = 4v2 for v. Show your work

  • one year ago
  • one year ago

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  1. inkyvoyd
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    Wel, first isolate the term for v^2

    • one year ago
  2. inkyvoyd
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    *first isolate v^2

    • one year ago
  3. inkyvoyd
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    (divide by 4, if you don't know)

    • one year ago
  4. brittany_lundgren
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    ok

    • one year ago
  5. inkyvoyd
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    S/4=v^2

    • one year ago
  6. brittany_lundgren
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    ok

    • one year ago
  7. inkyvoyd
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    Then, take the plus-minus square root of both sides.

    • one year ago
  8. inkyvoyd
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    You know how to simplify \(\Huge \pm\sqrt\frac{S}{4}\) right?

    • one year ago
  9. brittany_lundgren
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    ok im following you

    • one year ago
  10. inkyvoyd
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    Let me show you the whole thing. ONe sec, this may take a minute or two.

    • one year ago
  11. brittany_lundgren
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    ok im waiting :)

    • one year ago
  12. inkyvoyd
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    S = 4v2 \(\Large S=4v^2\) ->Isolate the "\(v^2\)" \(\Large \frac{S}{4}=v^2\) Take the positive negative square root of both sides \(\huge \pm\sqrt{\frac{S}{4}}=v\) \(\huge \pm\frac{\sqrt{S}}{\sqrt{4}}=v\) Can you try from here?

    • one year ago
  13. inkyvoyd
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    @brittany_lundgren ?

    • one year ago
  14. brittany_lundgren
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    I can try........ I dont remember how to do any of ths... I'm sooo lost!

    • one year ago
  15. inkyvoyd
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    Alright, it's ok. Just tell me which steps you don't understand :)

    • one year ago
  16. brittany_lundgren
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    I honestly need help with the whole problem, im soooo confused! :'(

    • one year ago
  17. inkyvoyd
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    Ok, so, from step one, right? Which parts of step one do you not understand? Or is it the problem itself?

    • one year ago
  18. brittany_lundgren
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    It's the problem itself, I just don't understand :'(

    • one year ago
  19. inkyvoyd
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    Ok, what they do is give you an equation. What they want you to do is to "solve for v" This means they want you to put v on one side, and everything else on the other.

    • one year ago
  20. inkyvoyd
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    Note that v can ONLY be on one side. If you have v on both sides, it's a mistake.

    • one year ago
  21. brittany_lundgren
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    okay, im following you so far!

    • one year ago
  22. inkyvoyd
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    Good. :)

    • one year ago
  23. jhonyy9
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    i have solved it before - not is clearly there ?

    • one year ago
  24. inkyvoyd
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    Now, do you remember how we move things around?

    • one year ago
  25. inkyvoyd
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    Like, combine like terms, move things from one side to the other, and take square roots?

    • one year ago
  26. brittany_lundgren
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    ok well can you show me how? Like the steps ??

    • one year ago
  27. inkyvoyd
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    Ok. In general, or just for this equation?

    • one year ago
  28. brittany_lundgren
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    just the steps on how to solve this problem... The problem told me to show my work.......

    • one year ago
  29. inkyvoyd
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    Okk.

    • one year ago
  30. inkyvoyd
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    Remember, we want to put "v" on one side, and everything else on the other. The first step is to get rid of what's in front of v.

    • one year ago
  31. brittany_lundgren
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    okay so what would I do then?

    • one year ago
  32. inkyvoyd
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    What's in front of v?

    • one year ago
  33. brittany_lundgren
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    1.Solve S = 4v2 for v.

    • one year ago
  34. inkyvoyd
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    I mean, what's in front of "v^2"?

    • one year ago
  35. brittany_lundgren
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    4

    • one year ago
  36. inkyvoyd
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    So, how do we move the 4 to the other side?

    • one year ago
  37. brittany_lundgren
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    -4(4-4)=-2(2v-1)

    • one year ago
  38. inkyvoyd
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    Not that complicated. Just divide both sides by 4 :)

    • one year ago
  39. brittany_lundgren
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    Is the answer to the question v=2?

    • one year ago
  40. inkyvoyd
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    Well, we can't actually solve for an exact value of v. We can only express it in terms of "s".

    • one year ago
  41. brittany_lundgren
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    okay, I'm sooo confused!!! :'(

    • one year ago
  42. inkyvoyd
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    Ok. We can't always solve for v (in terms of a number, like 3, or -2). But, we can move everything over to one side and leave v on the other.

    • one year ago
  43. brittany_lundgren
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    OMG!!! dumb right now :(

    • one year ago
  44. inkyvoyd
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    Ok. If you have the equation x-2y=0 you can't find x, because you don't know y.

    • one year ago
  45. brittany_lundgren
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    okay, I understand that, I just don''t understand on how to solve this problem!

    • one year ago
  46. inkyvoyd
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    Ok. Solve for v doesn't mean "get a number value for v". It means, "put just v on one side, and everything else without v on the other"

    • one year ago
  47. brittany_lundgren
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    ok,im following so far.

    • one year ago
  48. inkyvoyd
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    Alright. Since we just want to move everything to the other side, we divide both sides by 4 to get rid of what's in front of v.

    • one year ago
  49. brittany_lundgren
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    ok so S/4=4v^2/4

    • one year ago
  50. inkyvoyd
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    Ok. Now simplify the rigt side.

    • one year ago
  51. brittany_lundgren
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    S=v^2

    • one year ago
  52. inkyvoyd
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    Careful, you can't cancel out the left side...

    • one year ago
  53. brittany_lundgren
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    S/4=v^2

    • one year ago
  54. inkyvoyd
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    yess.

    • one year ago
  55. inkyvoyd
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    Now, what do we do to get rid of v^2? (turn it into v)

    • one year ago
  56. brittany_lundgren
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    S/4=v

    • one year ago
  57. inkyvoyd
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    Careful. You have to take the square root of BOTH sides.

    • one year ago
  58. brittany_lundgren
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    I got confused, just now. I beeen following this whole time and i just now got lost!!! :'(

    • one year ago
  59. inkyvoyd
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    That's ok.

    • one year ago
  60. inkyvoyd
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    S/4=4v^2/4 -> \(\huge \frac{S}{4}=\frac{(4v^2)}{4}\)

    • one year ago
  61. inkyvoyd
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    Then \(\Huge \frac{S}{4}=v^2\)

    • one year ago
  62. inkyvoyd
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    Now, here's the part you got lost. Let me show it in more detail.

    • one year ago
  63. inkyvoyd
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    \(\Huge \pm\sqrt{\frac{S}{4}}=\sqrt{v^2}\)

    • one year ago
  64. inkyvoyd
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    There's a lot going on right here.

    • one year ago
  65. inkyvoyd
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    Does everything make sense?

    • one year ago
  66. brittany_lundgren
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    ima little lost there

    • one year ago
  67. inkyvoyd
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    Ok. Do you know why we have the plus minus sign?

    • one year ago
  68. brittany_lundgren
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    nope, im confused whole step.......

    • one year ago
  69. inkyvoyd
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    I'm really sorry, but I have to go to bed. I'll try to make sure I get people here to help you. (I have to take some tests myself tomorrow, and I can't afford to miss them) @nbouscal , would you be so kind as to help out? THanks.

    • one year ago
  70. brittany_lundgren
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    damn, what time is it there?

    • one year ago
  71. inkyvoyd
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    2:00 AM. I have to go to sleep and get done before 2:00PM, but I have two tests around 1 hour each, :S

    • one year ago
  72. brittany_lundgren
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    dang........

    • one year ago
  73. inkyvoyd
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    ACtually, I'm going to take both right now. No chances.

    • one year ago
  74. inkyvoyd
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    Time for a cup of coffee. thanks, @nbouscal .

    • one year ago
  75. nbouscal
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    Okay, if you have \(v^2=\frac{S}{4}\), and you want to know what \(v\) is, you have to take the square root of both sides. That's the only way to get the left side to just have \(v\) by itself. The problem is, when you square root something, you always have to put a \(\pm\) sign in front, because you don't know whether it's positive or negative. For example, \(2^2=4, (-2)^2=4\), so the square root of 4 could be either 2 or -2. So, we take the square root of both sides and we add a plus-or-minus, and we get \[ v=\pm\sqrt{\frac{S}{4}} \]

    • one year ago
  76. brittany_lundgren
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    @nbouscal brb imma go get something to drink

    • one year ago
  77. brittany_lundgren
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    ok, im following so far.

    • one year ago
  78. inkyvoyd
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    Brittany, I can help too, becaues I just realized I only have one test.

    • one year ago
  79. inkyvoyd
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    \(\huge v=\pm\frac{\sqrt{S}}{\sqrt{4}}\)

    • one year ago
  80. inkyvoyd
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    Remember, \(\Huge \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\)

    • one year ago
  81. inkyvoyd
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    Now, what's the square root of 4?

    • one year ago
  82. inkyvoyd
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    (@nbouscal , I'm going to go ahead and take my scantron before the coffee wears off)

    • one year ago
  83. brittany_lundgren
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    2

    • one year ago
  84. brittany_lundgren
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    @inkyvoyd @nbouscal

    • one year ago
  85. inkyvoyd
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    Ok, then, what do we get?

    • one year ago
  86. inkyvoyd
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    (use draw if you have to)

    • one year ago
  87. nbouscal
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    \(v=\pm\dfrac{\sqrt{S}}{\sqrt{4}}\), \(\sqrt{4}=2\), so \(v=?\)

    • one year ago
  88. inkyvoyd
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    @brittany_lundgren , do you get it?

    • one year ago
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