Question

Find the angle between the given vectors to the nearest tenth of a degree. (2 points)

u = <2, -4>, v = <3, -8>

3.0°
6.0°
-7.0°
16.0°

Answer 1

@misty1212

Answer 2

HI!!

Answer 3

find the dot product first

Answer 4

6 and 32

Answer 5

lol the dot product is a number, not a choice of numbers

Answer 6

38

Answer 7

ok then find the absolute value of each vector

Answer 8

how

Answer 9

\[||<a,b>||=\sqrt{a^2+b^2}\]

Answer 10

but u just said it was one # XD

Answer 11

oh my

Answer 12

we have a long way to go don't we?

Answer 13

i mean it depends how much time u want to waste with me ill forget it tomorrow lol

Answer 14

\[\cos(\theta)=\frac{u\cdot v}{|u|\times |v|}\]

Answer 15

what is that

Answer 16

cosine theta and then i draw a blank lol

Answer 17

the numerator is the dot product, you got 38

Answer 18

the denominator is the product of the absolute value of the vectors

Answer 19

ok can u give me the numerator please

Answer 20

\[|<2,-4>=\sqrt{2^2+4^2}=\sqrt{20}\]

Answer 21

do the same for \(v\)

Answer 22

i still dont understand lol

Answer 23

like i get what u did to get here but what is v

Answer 24

the other vector

Answer 25

what is the other vector

Answer 26

you wrote it , not me

Answer 27

i have 38 and i dont knwo what is after

Answer 28

please i want to go to sleep can i please just have the asnwer

Answer 29

38 is the numerator
you need the denominator too

Answer 30

im gonna cry please i have like 1 minute left on my quiz

Answer 31

this is what you are going to use \[\cos(\theta)=\frac{u\cdot v}{|u|\times |v|}\]
we got \(u\cdot v=38\)

Answer 32

i need a good grade if not ill get beat

Answer 33

jk that wont happen lol

Answer 34

dont u get the same thing as the denominator???

Answer 35

http://www.wolframalpha.com/input/?i=arccos%2819%2Fsqrt%28365%29%29

Answer 36

looks like \(6^\circ\)

Answer 37

so thats the asnwer?!?!?!