can someone please help me with algebra 2? im really stuck on this question..:/ picture posted..
We can rule out direct variation. Now we need to check if it is indirect variation or misc. The general form for indirect variation is \[y = \frac{ k }{ x }\] where k is a constant. Check if all four pairs give the same value of k.
wait...what? how do i find that out?
theres a pattern? every time that y goes up one, then x goes up 5.. is that what u mean?
Figure out what \(x\cdot y\) is in each case.
In layman terms.. Direct variation between x and y implies if x increases; then y increases.(and vice-versa) Indirect variation between x and y implies If x increases; then y decreases (and vice-versa)
ooh! 10 times 1/2 is 5.. so x goes up 5?
so its definitely direct variation right?
@LastDayWork how do i write an equation for the function?
Find xy for each case (as wio suggested) If you get xy = constant ; for all pairs of (x,y) Then it is indirect variation. & the equation will be xy = constant I guess, you can easily find the value of the constant ;)
can u help me?
Yea..tell me where you are stuck..
finding xy.. i dont know how to do that.. its probably really simple huh? lol:)
@LastDayWork ?
The table gives you the value of x and y. Consider each column individually. I am doing one as an example. x = 10 y = 1/2 Thus constant (k) = x*y = 10 * (1/2) = 5 This may look unnecessary, as you could easily comprehend "...every time that y goes up one, then x goes up 5.." But, let's do this as a formality ;)
You will have to show the same for other three pairs..
so... (k) = x*y=10*(1/2)=5 is the equation for the function?
Yea.. xy = 5 but you have to show that other pairs also satisfy this equation.
ok.. but the other pairs would be the same?
I don't quiet get your question..
ok. well every other xy pair would also equal 5 right? because each time y goes up 1, then x goes up 5.. umm...how can i show that the other pairs would also satisfy the equation xy=5?
Simply multiply them.. x = 15 y = (1/3) xy = 15 * (1/3) = 5 Repeat these steps..
oh ok!! then after they are all done, then is that the equation for the function?
@LastDayWork
Er..yea Were you lookn' for something complex ?? :P
oh maybe.. thanks so much!:)
You're welcome :)
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