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anonymous
 one year ago
medal and fan
anonymous
 one year ago
medal and fan

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Determine 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 • an1 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
 one year ago
Best ResponseYou've already chosen the best response.1hint: two vectors are orthogonal each other, when their dot product or scalar product is equal to zero

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1what is: \[u \cdot v = ..?\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1exact! we have: \[u \cdot v = 0\] please the dot product is a real number not a vector

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so, what can you conclude?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1if dot product=0> the two vector are orthogonal by definition

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you ! i have 3 more questions...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Find the first six terms of the sequence. (5 points) a1 = 2, an = 3 • an1 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
 one year ago
Best ResponseYou've already chosen the best response.1question #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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1correct! \[\Large \begin{gathered} {a_1} =  2 \hfill \\ {a_2} = 3{a_1} = 3 \times \left( {  2} \right) =  6 \hfill \\ {a_3} = 3{a_2} = 3 \times \left( {  6} \right) =  18 \hfill \\ {a_4} = 3{a_3} = 3 \times \left( {  18} \right) =  54 \hfill \\ {a_5} = 3{a_4} = ... \hfill \\ {a_6} = 3{a_5} = ... \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1more precisely: \[\Large \begin{gathered} {a_5} = 3{a_4} =  162 \hfill \\ {a_6} = 3{a_5} =  486 \hfill \\ \end{gathered} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Use the given graph to determine the limit, if it exists. https://lss.brainhoney.com/Resource/24831188,0/Assets/46212_50f97695/0908b_g28_q2a.gif

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1Using 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
 one year ago
Best ResponseYou've already chosen the best response.0And I have no idea hoe to continue

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1here 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
 one year ago
Best ResponseYou've already chosen the best response.1oops..it is left limit, not right limit

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I 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
 one year ago
Best ResponseYou've already chosen the best response.0I am so confused.... so sorry

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1no worries!! :) your function is not continue at x=2, so y=1 can not be its limit as x goes to 2

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1nevertheless righ limit and left limit exist, and we have: \[\Large \mathop {\lim }\limits_{x \to 2  } f\left( x \right) = 5,\quad \mathop {\lim }\limits_{x \to 2 + } f\left( x \right) =  2,\]
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