Question

medal and fan

Answer 1

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

Answer 2

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

Answer 3

parallel ?

Answer 4

what is:
\[u \cdot v = ..?\]

Answer 5

10, -9

Answer 6

no 0, 0?

Answer 7

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

Answer 8

so, what can you conclude?

Answer 9

so neither

Answer 10

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

Answer 11

vectors*

Answer 12

then?

Answer 13

orthogonal

Answer 14

that's right!

Answer 15

thank you ! i have 3 more questions...

Answer 16

ok!

Answer 17

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

Answer 18

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

Answer 19

-54

Answer 20

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

Answer 21

162
486??

Answer 22

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

Answer 23

thank you :)

Answer 24

:)

Answer 25

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

Answer 26

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

Answer 27

And I have no idea hoe to continue

Answer 28

how

Answer 29

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

Answer 30

oops..it is left limit, not right limit

Answer 31

ok..

Answer 32

limit 1 ???

Answer 33

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

Answer 34

I am so confused.... so sorry

Answer 35

no worries!! :)
your function is not continue at x=2, so y=1 can not be its limit as x goes to 2

Answer 36

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

Answer 37

right*