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Exercise 1.3 · Q76

Q.If A={3,5,7,9,11,12}A = \{3, 5, 7, 9, 11, 12\}, determine the truth value of: ∃x∈A\exists x \in A such that 3x+8>403x + 8 > 40

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Solving 3x+8>403x+8>40 gives 3x>323x>32, i.e. x>10.67x>10.67. Among A's elements, x=11x=11 gives 3(11)+8=41>403(11)+8=41>40 ✓ and x=12x=12 gives 3(12)+8=44>403(12)+8=44>40 ✓. Since at least one element satisfies the inequality, the exi …

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