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brittany_lundgren

  • 2 years ago

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

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

  2. inkyvoyd
    • 2 years ago
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    *first isolate v^2

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

  4. brittany_lundgren
    • 2 years ago
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    ok

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

  6. brittany_lundgren
    • 2 years ago
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    ok

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

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

  9. brittany_lundgren
    • 2 years ago
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    ok im following you

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

  11. brittany_lundgren
    • 2 years ago
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    ok im waiting :)

  12. inkyvoyd
    • 2 years ago
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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?

  13. inkyvoyd
    • 2 years ago
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    @brittany_lundgren ?

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

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

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

  17. inkyvoyd
    • 2 years ago
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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?

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

  19. inkyvoyd
    • 2 years ago
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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.

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

  21. brittany_lundgren
    • 2 years ago
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    okay, im following you so far!

  22. inkyvoyd
    • 2 years ago
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    Good. :)

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

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

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

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

  27. inkyvoyd
    • 2 years ago
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    Ok. In general, or just for this equation?

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

  29. inkyvoyd
    • 2 years ago
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    Okk.

  30. inkyvoyd
    • 2 years ago
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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.

  31. brittany_lundgren
    • 2 years ago
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    okay so what would I do then?

  32. inkyvoyd
    • 2 years ago
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    What's in front of v?

  33. brittany_lundgren
    • 2 years ago
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    1.Solve S = 4v2 for v.

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

  35. brittany_lundgren
    • 2 years ago
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    4

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

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

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

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

  40. inkyvoyd
    • 2 years ago
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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".

  41. brittany_lundgren
    • 2 years ago
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    okay, I'm sooo confused!!! :'(

  42. inkyvoyd
    • 2 years ago
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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.

  43. brittany_lundgren
    • 2 years ago
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    OMG!!! dumb right now :(

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

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

  46. inkyvoyd
    • 2 years ago
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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"

  47. brittany_lundgren
    • 2 years ago
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    ok,im following so far.

  48. inkyvoyd
    • 2 years ago
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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.

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

  50. inkyvoyd
    • 2 years ago
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    Ok. Now simplify the rigt side.

  51. brittany_lundgren
    • 2 years ago
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    S=v^2

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

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

  54. inkyvoyd
    • 2 years ago
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    yess.

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

  56. brittany_lundgren
    • 2 years ago
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    S/4=v

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

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

  59. inkyvoyd
    • 2 years ago
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    That's ok.

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

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

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

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

  64. inkyvoyd
    • 2 years ago
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    There's a lot going on right here.

  65. inkyvoyd
    • 2 years ago
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    Does everything make sense?

  66. brittany_lundgren
    • 2 years ago
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    ima little lost there

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

  68. brittany_lundgren
    • 2 years ago
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    nope, im confused whole step.......

  69. inkyvoyd
    • 2 years ago
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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.

  70. brittany_lundgren
    • 2 years ago
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    damn, what time is it there?

  71. inkyvoyd
    • 2 years ago
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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

  72. brittany_lundgren
    • 2 years ago
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    dang........

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

  74. inkyvoyd
    • 2 years ago
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    Time for a cup of coffee. thanks, @nbouscal .

  75. nbouscal
    • 2 years ago
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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 \[ v=\pm\sqrt{\frac{S}{4}} \]

  76. brittany_lundgren
    • 2 years ago
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    @nbouscal brb imma go get something to drink

  77. brittany_lundgren
    • 2 years ago
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    ok, im following so far.

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

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

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

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

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

  83. brittany_lundgren
    • 2 years ago
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    2

  84. brittany_lundgren
    • 2 years ago
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    @inkyvoyd @nbouscal

  85. inkyvoyd
    • 2 years ago
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    Ok, then, what do we get?

  86. inkyvoyd
    • 2 years ago
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    (use draw if you have to)

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

  88. inkyvoyd
    • 2 years ago
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    @brittany_lundgren , do you get it?

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