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\[y ^{-1}-22y ^{-1 \over 2}+120=0\]

let u=y^(-1/2)

so what does u^2=?

x?

u^2=y^(-1)

ok, thanks. so where do I go from here?

do you know how to factor u^2-22u+120=0

i'm trying to figure out what works for 22 and 120..does anything besides 2 go into them both?

(u-10)(u-12)=0

so will the 2 solutions be positive 10 and 12 then?

y^(-1/2)=10
y^(-1/2)=12
since u=y^(-1/2)

solve for y from both of these equations and you are done

y=10^(-2)
y=12^(-2)
you can evaluate those

do you know law of exponents?

possibly. I'm not sure exactly what that entails. I don't know the tnames for all these math terms

y=10^(-2)=1/(10^2)
y=12^(-2)=1/(12^2)
can you evaluate it now?

do they just end up being 10 and 12?

y=1/(10*10)
y=1/(12*12)
can you evaluate these now?

100 and 144?

I just didn't think it would be that large of number.

y=1/100
y=1/144

you can plug these in to check if you want

oh the 1 does stay on top? wasn't sure about that. Sorry i'm so incompetent at math.

Ok will do. thanks

np how about a medal? lol

i'd say you earned it!

yay! :)

thanks you i will give you one too

yay! Idk if i deserve one but thanks anyways! :)

lol ok deserve it here if we have the function f(n)=2
find f(3)=...

2..or n...er.. uhh

lol thats right

woohoo!

the 2 is right

have a goodnight