Question

@Luigi0210

Answer 1

I need yhuuu....to explain how to do this XD

Like, don't do it for me, just give me an example XD

Answer 2

http://i.imgur.com/BkTdd6M.jpg

Answer 3

Do you know the magnitude of a vector formula?

Answer 4

uhhh...good question ;) let me look at my notes..

Answer 5

\[|V|=\sqrt{x^2+y^2}\]right?

Answer 6

so it will always be positive? or no..

Answer 7

Yess, now plug them in :P

Answer 8

One second XD need new sheet of paper..

Answer 9

wait...what?

Answer 10

I'm given 2 ordered pairs..

I'm so confused lol

Answer 11

or is it the difference between them?

Answer 12

Yea, your formula would probably make more sense this way xD

\(\Large |PQ|=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2} \)

Answer 13

Ohhh, that makes more sense...

Answer 14

So it's basically the distance formula?

Answer 15

If that's what it is, then yea xD

I don't know the distance formmula /.\

Answer 16

Wait, what? The math genius doesn't know the distance formula?!?!?

And I got sqrt 2 lol

Answer 17

Well \(\sqrt{2} \) isn't an option xD

And I'm a math noob, get it right ;)

Answer 18

lol

Answer 19

oh, sqrt 2 is 1.41 lol

Answer 20

so (-1, 1) would be correct

Answer 21

You found the ordered pair already? Nicee

Answer 22

That was easy lol

Answer 23

The ordered pair that represents it is just the difference between the 2 ordered pairs...Now that I know how to do it, these 2 dimensional ones are easy lol

Answer 24

Wait, so to find the pair you do \(b-a \) right?

Answer 25

Pretty sure, but if you did the distance formula, you would just simplify it down to \[\sqrt{x^2+y^2}\] Which, is easy XD

Answer 26

As you can tell, I enjoyed the distance formula...it was easy XD

Answer 27

Hold up on the second one...I'm trying something XD

Answer 28

The second one is
<1, -2, 4>, 4.58

XD

That one was easy, too...if it's right

Answer 29

you just add to the distance formula...\[\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\]and you'd just simplify it to \[\sqrt{x^2+y^2+z^2}\]I hope I'm doing this correctly lol

Answer 30

Psssh, and you said you needed help xD

Answer 31

I did lol

Answer 32

you seem to have done it all already tho ;P

Answer 33

math is confusing to me until I learn it :P

Answer 34

I have plenty more xD

Answer 35

Bring it ;)

But once you get to integrals, then we'll have real fun <3

Answer 36

lol

Answer 37

I'll just post the first one...2 dimensional one XD I should be able to figure out the other one if I get this one XD

Answer 38

http://imgur.com/Kgx56i8

Answer 39

\(\Large u= 2 \color{red}{w}-\color{purple}{v} \)

\(\large \color{red}{w}=<-2, 4> \)

\(\large \color{purple}{v}=<3, 1> \)

Answer 40

XD

Answer 41

http://imgur.com/bFfWriz

Answer 42

\(\Large <1-(-2), 5-3, -2-6 > \)

\(\Large <3i, 2j, -8k > \)

Answer 43

weird lol

Answer 44

Here's my source if you're curious xP
http://www.algebra.com/algebra/homework/complex/Complex_Numbers.faq.question.229691.html

Answer 45

This cross product is confusing, but it is similar to determinants

Answer 46

so it is...

|a2b3-a3b2|
|a3b1-a1b3|
|a1b2-a2b1|

a1=-1
b1=-3
a2=2
b2=-1
a3=4
b3=5

right? lol

Answer 47

|(2)(5)-(4)(-1)|
|(4)(-3)-(-1)(5)|
|(-1)(-1)-(2)(-3)|

I think LOL

Answer 48

|10+4|
|-12+5|
|1+6|

Therefore, we get <14, -7, 7>

Answer 49

Same thing I got, way to go Haley ;P

Answer 50

XD yay

Answer 51

How do I find the direction and angle of a resultant vector?