anonymous
  • anonymous
medal and fan
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • 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
Michele_Laino
  • Michele_Laino
hint: two vectors are orthogonal each other, when their dot product or scalar product is equal to zero
anonymous
  • anonymous
parallel ?

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Michele_Laino
  • Michele_Laino
what is: \[u \cdot v = ..?\]
anonymous
  • anonymous
10, -9
anonymous
  • anonymous
no 0, 0?
Michele_Laino
  • Michele_Laino
exact! we have: \[u \cdot v = 0\] please the dot product is a real number not a vector
Michele_Laino
  • Michele_Laino
so, what can you conclude?
anonymous
  • anonymous
so neither
Michele_Laino
  • Michele_Laino
if dot product=0---> the two vector are orthogonal by definition
Michele_Laino
  • Michele_Laino
vectors*
Michele_Laino
  • Michele_Laino
then?
anonymous
  • anonymous
orthogonal
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
thank you ! i have 3 more questions...
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • 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
Michele_Laino
  • 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
anonymous
  • anonymous
-54
Michele_Laino
  • Michele_Laino
correct! \[\Large \begin{gathered} {a_1} = - 2 \hfill \\ {a_2} = 3{a_1} = 3 \times \left( { - 2} \right) = - 6 \hfill \\ {a_3} = 3{a_2} = 3 \times \left( { - 6} \right) = - 18 \hfill \\ {a_4} = 3{a_3} = 3 \times \left( { - 18} \right) = - 54 \hfill \\ {a_5} = 3{a_4} = ... \hfill \\ {a_6} = 3{a_5} = ... \hfill \\ \end{gathered} \]
anonymous
  • anonymous
162 486??
Michele_Laino
  • Michele_Laino
more precisely: \[\Large \begin{gathered} {a_5} = 3{a_4} = - 162 \hfill \\ {a_6} = 3{a_5} = - 486 \hfill \\ \end{gathered} \]
anonymous
  • anonymous
thank you :)
Michele_Laino
  • Michele_Laino
:)
anonymous
  • 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
anonymous
  • anonymous
Michele_Laino
  • Michele_Laino
Using the definition of right limit and left limit, I think that we can write this: \[\Large \mathop {\lim }\limits_{x \to 2 - } f\left( x \right) = 5,\quad \mathop {\lim }\limits_{x \to 2 + } f\left( x \right) = - 2,\]
anonymous
  • anonymous
And I have no idea hoe to continue
anonymous
  • anonymous
how
Michele_Laino
  • 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
Michele_Laino
  • Michele_Laino
oops..it is left limit, not right limit
anonymous
  • anonymous
ok..
anonymous
  • anonymous
limit 1 ???
Michele_Laino
  • Michele_Laino
I think that y=1 is a value of your function, namely your function can be defined piecewise as below: \[\Large f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {5,\quad x < 2} \\ {1,\quad x = 2} \\ { - 2,\quad x > 2} \end{array}} \right.\]
anonymous
  • anonymous
I am so confused.... so sorry
Michele_Laino
  • 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
Michele_Laino
  • Michele_Laino
nevertheless righ limit and left limit exist, and we have: \[\Large \mathop {\lim }\limits_{x \to 2 - } f\left( x \right) = 5,\quad \mathop {\lim }\limits_{x \to 2 + } f\left( x \right) = - 2,\]
Michele_Laino
  • Michele_Laino
right*

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