anonymous
  • anonymous
What is the solution set for -x^2+9>?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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.
chestercat
  • chestercat
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anonymous
  • anonymous
\[-x^2+9>\]
anonymous
  • anonymous
Yea but do you know what the solution set is?
anonymous
  • anonymous
> than what? ...

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anonymous
  • anonymous
Oops. Sorry Kreshnik. >0
lgbasallote
  • lgbasallote
go @Kreshnik ! hehe
anonymous
  • anonymous
that's better ;) .... we have: \[\Large -x^2+9>0\] \[\Large 9-x^2>0\quad \longrightarrow \[3^2-x^2>0\]\] now we use this formula... \[\Large a^2-b^2=(a-b)(a+b)\] and we have.. \[\Large 3^2-x^2>0 \longrightarrow (3-x) (3+x )>0\] from here we have two parts 3-x>0 or 3+x>0 so we get that... -x>-3 or x>-3 multiply left inequality by -1 and swap the sight > into < (everytime you multiply\divide by a negative number in inequalities you change sign !) and finally we have: x<3 or x>-3 so \[\LARGE x\in (-\infty ,3) \cup (-3, +\infty)\] from here we get that... \[\LARGE x\in (-3,-2,-1,0,1,2,3)\]
anonymous
  • anonymous
Wow you're great Kreshnik. I really appreciate all of your help.
anonymous
  • anonymous
@lgbasallote :P am I right or not ? :P
anonymous
  • anonymous
@mathsux4real Glad to help ;) (I got to go to school, I'm out ! see you later ;) )
anonymous
  • anonymous
Cool, see ya around!
lgbasallote
  • lgbasallote
yes :D very good job @Kreshnik *thumbs up* ;D
anonymous
  • anonymous
bye ! ;) @lgbasallote thanks ;)
anonymous
  • anonymous
actually it should be... \[\LARGE x\in (-\infty,3)\cap(-3,+\infty)\] the rest is ok.

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