Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- anonymous

need to find the solution to this equation.

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

Get your **free** account and access **expert** answers to this and **thousands** of other questions

- anonymous

need to find the solution to this equation.

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

\[y ^{-1}-22y -_{2}^{1}+12-=0\]

- anonymous

I couldn't get this to look right. it is supposed to be y with the -1/2 as the square root

- anonymous

Is it
\[{1 \over y}-22y^{-1/2}+12=0?\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- myininaya

let u=y^(-1/2) so we have
u^2=y^(-1)

- myininaya

u^2-22u+12=0

- anonymous

I got 12 and 10. is that correct?

- anonymous

or maybe they are negative?

- myininaya

one sec im scanning something

- anonymous

ok, thanks

- myininaya

- anonymous

wow, thanks for doing that! I can see where I maybe slipped up. I am a bit confused as to what the answer is though. Is there still one more step in solving those 2 solutions?

- myininaya

i didn't simplify

- anonymous

for some reason i'm not getting a whole number

- myininaya

i'm not getting a whole number

- anonymous

i just realized that you had 12 in the problem when it's supposed to be 120. Sorry if I typed it wrong.

- anonymous

yep, I did... ahh. sorry.

- myininaya

lol. okay lets take another look at it

- anonymous

Thanks! :)

- myininaya

so we get u=10 and u=12 like you been talking about this whole time but u=y^(-1/2)
y^(-1/2)=10 and y^(-1/2)=12
so y=10^(-2)=1/(10^2)=1/100
so y=12^(-2)=1/(12^2)=1/144
there you basically had it you just needed to solve for y

- myininaya

we got u=12 and 10 because u^2-22u+120=(u-12)*(u-10)=0
so very good

- anonymous

ok I just wasn't sure if they were supposed to stay positive.

- anonymous

Thanks for clarifying!

- myininaya

to get rid of that -1/2 exponent i just raised both sides to -2
because law of exponents tells me if we have (x^{-1/2)^{-2} to multiply the exponents but if i do this to one side i have to do it to the other

- anonymous

okk got it:)

Looking for something else?

Not the answer you are looking for? Search for more explanations.