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BabyNath24
Group Title
Solve the inequality: x^2  5x + 6 ≥ 0.
A. x ≤ 2 or x ≥ 3
B. 2 ≤ x ≤ 3
C. x ≤ or x ≥ 6
D. x ≤ 6 or x ≥ 1
 7 months ago
 7 months ago
BabyNath24 Group Title
Solve the inequality: x^2  5x + 6 ≥ 0. A. x ≤ 2 or x ≥ 3 B. 2 ≤ x ≤ 3 C. x ≤ or x ≥ 6 D. x ≤ 6 or x ≥ 1
 7 months ago
 7 months ago

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arbershabani97 Group TitleBest ResponseYou've already chosen the best response.0
(x3)(x2)= x^25x+6, do you need more help?
 7 months ago

surjithayer Group TitleBest ResponseYou've already chosen the best response.0
\[x ^{2}5x+6>0,x ^{2}5x>6\] adding both sides \[\left( \frac{ 5 }{ 2 } \right)^{2} i.e.,\frac{ 25 }{ 4 }\] \[x ^{2}5x+\frac{ 25 }{4 }>6+\frac{ 25 }{4 }\] \[\left( x\frac{ 5 }{ 2 } \right)^{2}>\frac{ 1 }{ 4 },\left x\frac{ 5 }{2 } \right>\frac{ 1 }{ 2 }\] solve it. \[(\left xa \right>b means xa <b and xa >b)\]
 7 months ago

arbershabani97 Group TitleBest ResponseYou've already chosen the best response.0
(x3)(x2)>=0 (its equal or greater than, but didn't know how to do the sign) x3>=0 and x2>=0 just solve for x and see in which one it is right
 7 months ago

BabyNath24 Group TitleBest ResponseYou've already chosen the best response.0
Oh ok thank you!!
 7 months ago

surjithayer Group TitleBest ResponseYou've already chosen the best response.0
\[correction everywhere write \ge inplace of >\]
 7 months ago
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