translations, Mathematics

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Once we get out of the review, we are not going to be doing a lot with Taylor series, but they are a fine method to get us back into the swing of dealing with power series. Through

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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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i need somehelp i am not the sharpest in the pack so plz help me thank you i hope you do

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