## anonymous one year ago medal and fan

1. anonymous

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

hint: two vectors are orthogonal each other, when their dot product or scalar product is equal to zero

3. anonymous

parallel ?

4. Michele_Laino

what is: $u \cdot v = ..?$

5. anonymous

10, -9

6. anonymous

no 0, 0?

7. Michele_Laino

exact! we have: $u \cdot v = 0$ please the dot product is a real number not a vector

8. Michele_Laino

so, what can you conclude?

9. anonymous

so neither

10. Michele_Laino

if dot product=0---> the two vector are orthogonal by definition

11. Michele_Laino

vectors*

12. Michele_Laino

then?

13. anonymous

orthogonal

14. Michele_Laino

that's right!

15. anonymous

thank you ! i have 3 more questions...

16. Michele_Laino

ok!

17. anonymous

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

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

-54

20. Michele_Laino

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

162 486??

22. Michele_Laino

more precisely: $\Large \begin{gathered} {a_5} = 3{a_4} = - 162 \hfill \\ {a_6} = 3{a_5} = - 486 \hfill \\ \end{gathered}$

23. anonymous

thank you :)

24. Michele_Laino

:)

25. anonymous

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

27. Michele_Laino

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

And I have no idea hoe to continue

29. anonymous

how

30. Michele_Laino

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

oops..it is left limit, not right limit

32. anonymous

ok..

33. anonymous

limit 1 ???

34. Michele_Laino

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

I am so confused.... so sorry

36. Michele_Laino

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

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

right*