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anonymous

  • one year ago

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  1. anonymous
    • one year ago
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    Determine whether the vectors u and v are parallel, orthogonal, or neither. (5 points) u = <10, 0>, v = <0, -9> Parallel Orthogonal Neither 10. Find the first six terms of the sequence. (5 points) a1 = -2, an = 3 • an-1 -6, -18, -54, -162, -486, -1458 -2, -6, -18, -54, -162, -486 -2, -6, -3, 0, 3, 6 0, 3, -6, -3, 0, 3

  2. Michele_Laino
    • one year ago
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    hint: two vectors are orthogonal each other, when their dot product or scalar product is equal to zero

  3. anonymous
    • one year ago
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    parallel ?

  4. Michele_Laino
    • one year ago
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    what is: \[u \cdot v = ..?\]

  5. anonymous
    • one year ago
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    10, -9

  6. anonymous
    • one year ago
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    no 0, 0?

  7. Michele_Laino
    • one year ago
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    exact! we have: \[u \cdot v = 0\] please the dot product is a real number not a vector

  8. Michele_Laino
    • one year ago
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    so, what can you conclude?

  9. anonymous
    • one year ago
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    so neither

  10. Michele_Laino
    • one year ago
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    if dot product=0---> the two vector are orthogonal by definition

  11. Michele_Laino
    • one year ago
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    vectors*

  12. Michele_Laino
    • one year ago
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    then?

  13. anonymous
    • one year ago
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    orthogonal

  14. Michele_Laino
    • one year ago
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    that's right!

  15. anonymous
    • one year ago
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    thank you ! i have 3 more questions...

  16. Michele_Laino
    • one year ago
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    ok!

  17. anonymous
    • one year ago
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    Find the first six terms of the sequence. (5 points) a1 = -2, an = 3 • an-1 -6, -18, -54, -162, -486, -1458 -2, -6, -18, -54, -162, -486 -2, -6, -3, 0, 3, 6 0, 3, -6, -3, 0, 3

  18. Michele_Laino
    • one year ago
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    question #10 we can write this: \[\Large \begin{gathered} {a_1} = - 2 \hfill \\ {a_2} = 3{a_1} = 3 \times \left( { - 2} \right) = - 6 \hfill \\ {a_3} = 3{a_2} = 3 \times \left( { - 6} \right) = - 18 \hfill \\ \end{gathered} \] please continue

  19. anonymous
    • one year ago
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    -54

  20. Michele_Laino
    • one year ago
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    correct! \[\Large \begin{gathered} {a_1} = - 2 \hfill \\ {a_2} = 3{a_1} = 3 \times \left( { - 2} \right) = - 6 \hfill \\ {a_3} = 3{a_2} = 3 \times \left( { - 6} \right) = - 18 \hfill \\ {a_4} = 3{a_3} = 3 \times \left( { - 18} \right) = - 54 \hfill \\ {a_5} = 3{a_4} = ... \hfill \\ {a_6} = 3{a_5} = ... \hfill \\ \end{gathered} \]

  21. anonymous
    • one year ago
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    162 486??

  22. Michele_Laino
    • one year ago
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    more precisely: \[\Large \begin{gathered} {a_5} = 3{a_4} = - 162 \hfill \\ {a_6} = 3{a_5} = - 486 \hfill \\ \end{gathered} \]

  23. anonymous
    • one year ago
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    thank you :)

  24. Michele_Laino
    • one year ago
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    :)

  25. anonymous
    • one year ago
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    Use the given graph to determine the limit, if it exists. https://lss.brainhoney.com/Resource/24831188,0/Assets/46212_50f97695/0908b_g28_q2a.gif

  26. anonymous
    • one year ago
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  27. Michele_Laino
    • one year ago
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    Using the definition of right limit and left limit, I think that we can write this: \[\Large \mathop {\lim }\limits_{x \to 2 - } f\left( x \right) = 5,\quad \mathop {\lim }\limits_{x \to 2 + } f\left( x \right) = - 2,\]

  28. anonymous
    • one year ago
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    And I have no idea hoe to continue

  29. anonymous
    • one year ago
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    how

  30. Michele_Laino
    • one year ago
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    here you have to apply the definition of left limit and right limit. For example, in the case of right limit, if we pick a neighborhood U of the point y=5, we can find a left neighborhood V of the point x=2, such that for each point x_0, in V except x=2, the value of the function f(x_0) belongs to the neighborhood U

  31. Michele_Laino
    • one year ago
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    oops..it is left limit, not right limit

  32. anonymous
    • one year ago
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    ok..

  33. anonymous
    • one year ago
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    limit 1 ???

  34. Michele_Laino
    • one year ago
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    I think that y=1 is a value of your function, namely your function can be defined piecewise as below: \[\Large f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {5,\quad x < 2} \\ {1,\quad x = 2} \\ { - 2,\quad x > 2} \end{array}} \right.\]

  35. anonymous
    • one year ago
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    I am so confused.... so sorry

  36. Michele_Laino
    • one year ago
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    no worries!! :) your function is not continue at x=2, so y=1 can not be its limit as x goes to 2

  37. Michele_Laino
    • one year ago
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    nevertheless righ limit and left limit exist, and we have: \[\Large \mathop {\lim }\limits_{x \to 2 - } f\left( x \right) = 5,\quad \mathop {\lim }\limits_{x \to 2 + } f\left( x \right) = - 2,\]

  38. Michele_Laino
    • one year ago
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    right*

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