Here_to_Help15
  • Here_to_Help15
H!E!L!P! M!E! P!L!E!A!S!E!
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Here_to_Help15
  • Here_to_Help15
@Hero
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Here_to_Help15
  • Here_to_Help15
@amistre64 now can you help me :D
Here_to_Help15
  • Here_to_Help15
@Hero

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Here_to_Help15
  • Here_to_Help15
@campbell_st
amistre64
  • amistre64
tell me your thoughts, how do we define an origin?
Here_to_Help15
  • Here_to_Help15
Thats where it went gmergme4gtm5tmpt5p4 < Confusing i dont understand it
amistre64
  • amistre64
i dont know what that means
Here_to_Help15
  • Here_to_Help15
Exactly....
Here_to_Help15
  • Here_to_Help15
I dont know what the question means either
amistre64
  • amistre64
then you might need to google what an origin is ... otherwise we will not be speaking on the same terms
Here_to_Help15
  • Here_to_Help15
wait i know what it means now please continue :)
Here_to_Help15
  • Here_to_Help15
Its the starting point :)
amistre64
  • 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
  • Here_to_Help15
Wow so many questions and yes right!
amistre64
  • amistre64
this is a study session, how can we study if im the only one that does the thinking?
Here_to_Help15
  • Here_to_Help15
Sorry its just this is our final unit and im just coming to terms with these problems :(
amistre64
  • 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
  • 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
  • Here_to_Help15
:)
amistre64
  • 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
  • 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
  • Here_to_Help15
Would i get y by it self? and x?
amistre64
  • amistre64
x is the name of a variable, there is nothing special in a name
amistre64
  • amistre64
0 = x+6 , solve for x 0 = y+4 , solve for y
Here_to_Help15
  • Here_to_Help15
-6=x -4=y ?
amistre64
  • amistre64
good, so in relation to our xy plane, the origin is at: -6,-4 right?
Here_to_Help15
  • Here_to_Help15
Yes
amistre64
  • amistre64
|dw:1433281584913:dw|
amistre64
  • 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
  • Here_to_Help15
Hmm i believe so
amistre64
  • 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
  • Here_to_Help15
\[2\sqrt{13} ?\]
amistre64
  • amistre64
thats not a distance formula, thats a number
Here_to_Help15
  • Here_to_Help15
Wait do we have 2 points or is it that one that i solved for ?
amistre64
  • amistre64
we have 2 points, an origin (-6,-4), and all (or any) points (x,y) from it
amistre64
  • amistre64
what is the distance between the points (-6,-4) and (x,y)?
Here_to_Help15
  • Here_to_Help15
If i told you im a little lost would you leave me ?
amistre64
  • amistre64
at this point, yes :) what is our distance formula?
Here_to_Help15
  • Here_to_Help15
lol wait are you serious and do you want me to write the formula?
amistre64
  • amistre64
i want you to write the formula for distance, this will help us determine the range
Here_to_Help15
  • Here_to_Help15
Ok
Here_to_Help15
  • Here_to_Help15
\[d=\sqrt{(x _{2}}-x _{1})^{2} + (y _{2}-y _{1})^{2}\] = ?
Here_to_Help15
  • Here_to_Help15
Sorry im not so good with that equation button but everything is squared
freeG13
  • freeG13
what is the question?
amistre64
  • 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
  • Here_to_Help15
Thanks but @freeG13 i will show you if @amistre64 gives up on me :) but i doubt he will
Here_to_Help15
  • Here_to_Help15
Are the exact thing its just the bottom is the simplified version
1 Attachment
amistre64
  • amistre64
so, d^2 = 36 then, what is d?
amistre64
  • amistre64
|dw:1433282849596:dw|
Here_to_Help15
  • Here_to_Help15
\(\huge\tt\color{#f9bec7}{6!}\)
Here_to_Help15
  • Here_to_Help15
Sorry openstudy is super laggy
Here_to_Help15
  • Here_to_Help15
I knew it was 6 just it wouldnt let me type it
Here_to_Help15
  • Here_to_Help15
Ok so its 6 now what?
amistre64
  • amistre64
we have a range of 6 (miles kilometers feet?) in all directions from the origin
amistre64
  • amistre64
what is our questions asking for?
Here_to_Help15
  • Here_to_Help15
Thats the centre of the the signal
Here_to_Help15
  • Here_to_Help15
\[(x-h)^{2} + (y-k)^{2} =r ^{2}\]
Here_to_Help15
  • Here_to_Help15
I could have used this formula as well :)
Here_to_Help15
  • Here_to_Help15
Basically go left 6 and down 4 from the origin and the signal would be 6 units
amistre64
  • amistre64
of course you could have, but then we would not have had much to study ;)
Here_to_Help15
  • 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
  • amistre64
not at the moment, i have a nother question that needs my attention at the moment.
Here_to_Help15
  • Here_to_Help15
1 Attachment
Here_to_Help15
  • Here_to_Help15
I will attend it

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