medal and fan

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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

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

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