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anonymous
 one year ago
Solve the inequality for x given that (x6)/(x3)(x9)<0
a) (inf,6)U(6,9)
b) (inf,3)U(6,inf)
c) (inf,6)U(6,inf)
d) (inf,3)U(6,9)
e) (inf,3)U(6,9)
anonymous
 one year ago
Solve the inequality for x given that (x6)/(x3)(x9)<0 a) (inf,6)U(6,9) b) (inf,3)U(6,inf) c) (inf,6)U(6,inf) d) (inf,3)U(6,9) e) (inf,3)U(6,9)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is not as hard as it looks changes sign at 3,6,9

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0divide number line up in to four intervals pick some interval and see if it is positive or negative there there, then you will know all

anonymous
 one year ago
Best ResponseYou've already chosen the best response.03,6,9 would all be open circles right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah but you don't need no circles, you need intervals

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(\infty, 3),(3,6),(6,9),(9\infty)\] are your intervals check on one then alternate

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0somehow i got that (inf,3)U(3,6)U(9,inf)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have no idea what im doing wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you have two adjacent intervals \((\infty,3)\) and \((3,6)\) that can't be right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it changes sign at 3 so if it is positive on one interval it must be negative on the other did you check on any one interval? with some practice you won't have too, but it doesn't hurt

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i did a little math wrong and forgot that i'm looking for negatives, not positives

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what interval did you check?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i got that (inf,3) was negative, (3,6) was positive, (6,9) was negative, and (9,inf) was positive

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i would also point out that one only of the above answers is possible, even if you didn't check anything

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah at the risk of repeating myself you only have to check one if you know it is negative on \((\infty,3)\) then it must be \[\huge ,+,,+\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the correct answer is (inf,3)(6,9), right?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0The graph confirms the answer you got https://www.desmos.com/calculator/ipckz8kxhy the portion on the interval (inf, 3) is below the x axis the portion on the interval (6,9) is below the x axis everything else is either on the x axis or above it
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