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Mathematics 16 Online
OpenStudy (mathmath333):

My question

OpenStudy (ineedhelplz):

?

OpenStudy (mathmath333):

If \(x\ \propto \ y\) and \(y\ \propto \ \dfrac{1}{z}\), then which of the following is correct \(1.)\ x\ \propto \ z \\ 2.)\ x\ \propto \ \dfrac{1}{z}\)

OpenStudy (sparrow2):

what is the sign?it looks like infinity

OpenStudy (mathmath333):

directly proportional/ inversely proportional

OpenStudy (ineedhelplz):

I got 2

OpenStudy (mathmath333):

how

OpenStudy (ineedhelplz):

x(n) X(z) Let y(n) = x(-n) ∞ ∞ ∞ Then Y(z) = ∑ x(-n)z-n = ∑ x(r) z+r = ∑ x(r) (z-1 ) -1 = X (z-1 ) n = -∞ r = -∞ r = -∞

OpenStudy (sparrow2):

yeah that;s right

OpenStudy (mathmath333):

haha ...

OpenStudy (ineedhelplz):

?

OpenStudy (ineedhelplz):

Did I mess up?

OpenStudy (sparrow2):

i guess it doesn't need that kind of calculation or whatever :D

OpenStudy (mathmath333):

i understand nothing , i m confusede

OpenStudy (ineedhelplz):

;-;

OpenStudy (ineedhelplz):

Well thats my work

OpenStudy (mathmath333):

@sparrow2 did u understand his work, if yes, please pass it to me

OpenStudy (sparrow2):

using only words :D: if x increases then y increases... if y increases then z decreases so we have if x increasis then z deacreases which is in the second

OpenStudy (ineedhelplz):

I'm so sorry I couldn't explain better

OpenStudy (mathmath333):

but the answer given is the first one

OpenStudy (sparrow2):

oh then i don;t know

OpenStudy (mathmath333):

in this video the guy explains it , but i can't understand https://www.youtube.com/watch?v=BM7L9dlbS6w

OpenStudy (mathmath333):

from 1;00

OpenStudy (mathmath333):

please dont mistake this post as any openstudy feedback and start unnecessary converstion

OpenStudy (anonymous):

the answer is 2489

OpenStudy (mathmath333):

@jigglypuff314

OpenStudy (freckles):

i was thinking using the transitive property thingy

OpenStudy (freckles):

the only thing i guess is we don't know if the operation has that property

OpenStudy (freckles):

do we know anything about that operation?

OpenStudy (freckles):

oh watching the video

OpenStudy (freckles):

ok so that one symbol means directly proportional to... doesn't this means we can write the equations x=ky and y=a/z where k and a are constants so replacing first y with a/z we obtain x=k(a/z) or x=L/z So x is inversely proportion to z

OpenStudy (freckles):

but you said and the video says that is incorrect

OpenStudy (freckles):

I guess the directly proportional thing doesn't mean what it used to mean in algebra

OpenStudy (mathmath333):

which answer is correct according to u

OpenStudy (freckles):

the one that is incorrect according to the video but I haven't watched the whole video just to the part where he says the one I chose is incorrect

OpenStudy (mathmath333):

u mean \(x \propto \dfrac1z\) this is correct ?

OpenStudy (freckles):

yep did you understand my reasoning above? like I read it as this we are given x=ky and y=a/z then this means x=k(a/z)=ka/z=L/z so given x=ky and y=a/z then x=L/z

OpenStudy (mathmath333):

yea i got ur reasoning

OpenStudy (mathmath333):

u use substitution

OpenStudy (freckles):

yes

OpenStudy (mathmath333):

thnks

OpenStudy (freckles):

he continues to talk about x directly prop to y and y directly prop to 1/z I don't get how x/(y*z) is a constant from this

OpenStudy (ineedhelplz):

Thats what I said at the beginning @mathmath333 :P

OpenStudy (freckles):

if x/y is a constant and yz is a constant then I can say \[\frac{\frac{x}{y}}{yz} \text{ is a constant } \text{ or that } \frac{yz}{\frac{x}{y}} \text{ is a constant } \\ \text{ simplifying } \\ \frac{x}{y^2 z} \text{ and } \frac{y^2 z}{x} \text{ are constants }\]

OpenStudy (freckles):

http://www.themathpage.com/alg/variation.htm this site agrees with us it says varies directly is a transitive relation

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