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How many integers satisfy the inequality |10(x+1)/x^2+2x+3|=1? Solution) first thing thats not an inequality,and second thing its very easy if thats the question.the LHS = |10/x^2 + 10/x + 2x + 3| = 110/x^2 will also be an integer only if x^2<10.....i.e. x<3.hence x= 0,+1,-1,+2,-2,+3,-3.but the only value that holds the equality is -1.hence the no. of INTEGERS is only 1.
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Prove the subsequent Boolean expression: (x∨y) ∧ (x∨~y) ∧ (~x∨z) = x∧z Ans: In the following expression, LHS is equal to: (x∨y)∧(x∨ ~y)∧(~x ∨ z) = [x∧(x∨ ~y)] ∨ [y∧(x∨
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Q. Illustrate Field Properties of Numbers? Ans. What the associative law of addition states is this: for any numbers a, b, and c,
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