- Here_to_Help15

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- Here_to_Help15

##### 1 Attachment

- Here_to_Help15

@amistre64 now can you help me :D

- Here_to_Help15

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## More answers

- Here_to_Help15

- amistre64

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

- Here_to_Help15

Thats where it went gmergme4gtm5tmpt5p4 < Confusing i dont understand it

- amistre64

i dont know what that means

- Here_to_Help15

Exactly....

- Here_to_Help15

I dont know what the question means either

- amistre64

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

- Here_to_Help15

wait i know what it means now please continue :)

- Here_to_Help15

Its the starting point :)

- amistre64

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?

- Here_to_Help15

Wow so many questions and yes right!

- amistre64

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

- Here_to_Help15

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

- amistre64

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?

- amistre64

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

- Here_to_Help15

:)

- amistre64

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?

- amistre64

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.

- Here_to_Help15

Would i get y by it self? and x?

- amistre64

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

- amistre64

0 = x+6 , solve for x
0 = y+4 , solve for y

- Here_to_Help15

-6=x -4=y ?

- amistre64

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

- Here_to_Help15

Yes

- amistre64

|dw:1433281584913:dw|

- amistre64

so in relation to the xy plane, we now know where it originates at ...
our range is ... hmm, would you agree its a circle?

- Here_to_Help15

Hmm i believe so

- amistre64

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?

- Here_to_Help15

\[2\sqrt{13} ?\]

- amistre64

thats not a distance formula, thats a number

- Here_to_Help15

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

- amistre64

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

- amistre64

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

- Here_to_Help15

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

- amistre64

at this point, yes :)
what is our distance formula?

- Here_to_Help15

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

- amistre64

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

- Here_to_Help15

Ok

- Here_to_Help15

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

- Here_to_Help15

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

- freeG13

what is the question?

- amistre64

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

- Here_to_Help15

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

- Here_to_Help15

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

##### 1 Attachment

- amistre64

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

- amistre64

|dw:1433282849596:dw|

- Here_to_Help15

\(\huge\tt\color{#f9bec7}{6!}\)

- Here_to_Help15

Sorry openstudy is super laggy

- Here_to_Help15

I knew it was 6 just it wouldnt let me type it

- Here_to_Help15

Ok so its 6 now what?

- amistre64

we have a range of 6 (miles kilometers feet?) in all directions from the origin

- amistre64

what is our questions asking for?

- Here_to_Help15

Thats the centre of the the signal

- Here_to_Help15

\[(x-h)^{2} + (y-k)^{2} =r ^{2}\]

- Here_to_Help15

I could have used this formula as well :)

- Here_to_Help15

Basically go left 6 and down 4 from the origin and the signal would be 6 units

- amistre64

of course you could have, but then we would not have had much to study ;)

- Here_to_Help15

lol :) ok i know i am the biggest pain in the behind but can we do a couple more? Not like this (multiple choice)

- amistre64

not at the moment, i have a nother question that needs my attention at the moment.

- Here_to_Help15

##### 1 Attachment

- Here_to_Help15

I will attend it

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