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07-03-2020
Mathematics

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[tex] \frac{u}{3} + 16 \geqslant 14[/tex]

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jdoe0001 jdoe0001

07-03-2020

[tex]\bf \cfrac{u}{3}+16 \geqslant 14\implies \cfrac{u}{3}\geqslant -2\implies u\geqslant -6[/tex]

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