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Lukecrayonz

  • 4 years ago

Vectors (Precalculus)

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  1. Lukecrayonz
    • 4 years ago
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    http://screensnapr.com/v/FfJoiS.png

  2. nenadmatematika
    • 4 years ago
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    a) (4,3) b) (-2,1) c) (-7,1)

  3. Lukecrayonz
    • 4 years ago
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    Explain C please:) and I have a new one for you.

  4. Lukecrayonz
    • 4 years ago
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    http://screensnapr.com/v/rx9TyR.png Just #29

  5. nenadmatematika
    • 4 years ago
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    2*u you get if you multiply its coordinates by 2, 2*v also and than just substract them: first coordinate of u minus first coordinate of v and second minus second and you'll get that result

  6. nenadmatematika
    • 4 years ago
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    v=(\[5/\sqrt{2},5/\sqrt{2}\])

  7. Lukecrayonz
    • 4 years ago
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    How did you get that?;o

  8. Lukecrayonz
    • 4 years ago
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    Nevermind i got it.

  9. nenadmatematika
    • 4 years ago
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    :D

  10. Lukecrayonz
    • 4 years ago
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    No wait D:I didnt.

  11. Lukecrayonz
    • 4 years ago
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    Actually forget it. I don't even need to do these problems, I've been doing the wrong ones!

  12. Lukecrayonz
    • 4 years ago
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    I was wondering why I never learned any of this!

  13. Lukecrayonz
    • 4 years ago
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    http://screensnapr.com/v/Hws0vK.png C of this part :D

  14. nenadmatematika
    • 4 years ago
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    first you find the unit vector (vector which length is 1) in the direction of the given vector by formula \[u _{0}=u/\left| u \right|=(3,3)/\sqrt{(3^2+3^2)}=(1/\sqrt{2},1/\sqrt{2})\] ...than your vector is \[v=5\times(1/\sqrt{2},1/\sqrt{2})=(5/\sqrt{2},5/\sqrt{2})\]

  15. nenadmatematika
    • 4 years ago
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    I must go now, I'll answer you later, good luck :D

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