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Algebra 18 Online
OpenStudy (anonymous):

What is the inverse of the given function? y = 7x2 - 3.

OpenStudy (zzr0ck3r):

\(y=7x^2-3\) can you solve for \(x\)

OpenStudy (zzr0ck3r):

we cant open those links....

OpenStudy (anonymous):

And how do i solve for x?

OpenStudy (zzr0ck3r):

they ask for user/pass

OpenStudy (anonymous):

All of them?

OpenStudy (zzr0ck3r):

yes

OpenStudy (anonymous):

Well just forget them lol So how do i do this ?

OpenStudy (zzr0ck3r):

\(y=7x^2-3\) add 3 to both sides \(y+3=7x^2\) divide by 7 \(\frac{y+3}{7}=x^2\) square root both sides \(\pm\sqrt\frac{y+3}{7}=x\)

OpenStudy (zzr0ck3r):

now switch the \(x's\) and \(y's\) \(y=\pm\sqrt{\frac{x+3}{7}}\)

OpenStudy (zzr0ck3r):

note that this is not a function

OpenStudy (anonymous):

Umm thanks ? :)

OpenStudy (zzr0ck3r):

lol whats the ? for?

OpenStudy (anonymous):

Cause you kinda did the problem for me /.\ I mean thanks like so much :) but i wanted to like ...do it lol in case i get a problem like it again

OpenStudy (anonymous):

so i know how to solve it:P

OpenStudy (zzr0ck3r):

lol I showed you how to do it,

OpenStudy (zzr0ck3r):

I showed each step and explain each step, if you don't understand a part, ask about it. Else do the same thing....

OpenStudy (zzr0ck3r):

its the same thing if i said 7=3x+8 and asked you to solve for x just now you have a y in there, but thats ok .... just treat it like a number

OpenStudy (anonymous):

ok:)

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