Question

H!E!L!P! M!E! P!L!E!A!S!E!

Answer 1

@Hero

Answer 2

@amistre64 now can you help me :D

Answer 3

@Hero

Answer 4

@campbell_st

Answer 5

tell me your thoughts, how do we define an origin?

Answer 6

Thats where it went gmergme4gtm5tmpt5p4 < Confusing i dont understand it

Answer 7

i dont know what that means

Answer 8

Exactly....

Answer 9

I dont know what the question means either

Answer 10

then you might need to google what an origin is ... otherwise we will not be speaking on the same terms

Answer 11

wait i know what it means now please continue :)

Answer 12

Its the starting point :)

Answer 13

we want to find the origin of the signal, how do we define an origin?

what is the origin on our conventional graphing system? we start at 0,0 right?

Answer 14

Wow so many questions and yes right!

Answer 15

this is a study session, how can we study if im the only one that does the thinking?

Answer 16

Sorry its just this is our final unit and im just coming to terms with these problems :(

Answer 17

consider the form:

u^2 + v^2 = 36

the origin is at u=0, v=0

now compare that to your form where u = x+6, and v=y+4

how do we relate the origin to your problem?

Answer 18

this is our study session, when you decide to study is not my concern :) but we will still study it regardless

Answer 19

:)

Answer 20

u = x+6, and v=y+4

our origin is u=0, v=0 therefore, in relation to x and y

0 = x+6 and 0 = y+4

what are our values of x and y in relation to the origin?

Answer 21

you shold already know that, im not here to teach you the entirety of algebra. they cover solving for a single variable way before they cover this stuff.

Answer 22

Would i get y by it self? and x?

Answer 23

x is the name of a variable, there is nothing special in a name

Answer 24

0 = x+6 , solve for x

0 = y+4 , solve for y

Answer 25

-6=x -4=y ?

Answer 26

good, so in relation to our xy plane, the origin is at: -6,-4

right?

Answer 27

Yes

Answer 28

so in relation to the xy plane, we now know where it originates at ...

our range is ... hmm, would you agree its a circle?

Answer 29

Hmm i believe so

Answer 30

well, the circle equation is just the distance from a central points, the radius measures the distance from the center of a circle

what is our distance formula?

Answer 31

\[2\sqrt{13} ?\]

Answer 32

thats not a distance formula, thats a number

Answer 33

Wait do we have 2 points or is it that one that i solved for ?

Answer 34

we have 2 points, an origin (-6,-4), and all (or any) points (x,y) from it

Answer 35

what is the distance between the points (-6,-4) and (x,y)?

Answer 36

If i told you im a little lost would you leave me ?

Answer 37

at this point, yes :)

what is our distance formula?

Answer 38

lol wait are you serious and do you want me to write the formula?

Answer 39

i want you to write the formula for distance, this will help us determine the range

Answer 40

Ok

Answer 41

\[d=\sqrt{(x _{2}}-x _{1})^{2} + (y _{2}-y _{1})^{2}\] = ?

Answer 42

Sorry im not so good with that equation button but everything is squared

Answer 43

what is the question?

Answer 44

good, but lets square each side

(x2-x1)^2 + (y2-y1)^2 = d^2

let -6,-4 be the point x1,y1

and let x,y be some other point

(x-(-6))^2 + (y-(-4))^2 = d^2

how does this compare to

(x+6)^2 + (y+4)^2 = 36

Answer 45

Thanks but @freeG13 i will show you if @amistre64 gives up on me :) but i doubt he will

Answer 46

Are the exact thing
its just the bottom is the simplified version

Answer 47

so, d^2 = 36 then, what is d?