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116 = H > B R G B D g Z F b g g h _ h e h ` d b y m g b \ _ j k b l _ l " K \ B \ Z g J b e k d b", L h f 53, K \I 1 1, F _ o Z g b a Z p b y, _ e _ d l j b nGiven U = {All letters of the alphabet}, A = {c, d, e, f}, and B = {e, f, g, h, k} To start with I am going to assumeSOLUTIONS 1 a F (A,B,C) = A' B' C' A' B' C A B' C' A B' C A B C' A B C Distributive = A'B' (C' C) AB' (C' C) AB (C
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If x,y ∈ A (possibly x = y) then x2 kxy y2 ∈ A for every integer k Determine all pairs m,n of nonzero integers such that the only admissible set containing both m and n is the set of all integers N2 Find all triples (x,y,z) of positive integers such that x ≤ y ≤ z and x3(y3 z3) = 12(xyz 2) N3Jul 04, 07 · We can determine the correct answer choice by substituting numerical values for a and b We could use any two values for a and b, but for simplicity, let's choose a = 1 and b = 2 The function now looks like this f (1 2) = f (1) f (2) f (3) = f (1) f (2) So, we must determine which answer choice has f (3) equal to the sum of f (1) and f (2)Mike Elliott Managing Broker mikeellio@erescompaniescom Energy Real Estate Soluons erescompaniescom CALL FOR OFFERS DEADLINE 5421
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Well, evaluating a function means plugging whatever they gave me in for the argument in the formula This means that I have to plug this character " § " in for every instance of x Here goes f (§) = (§) 2 2 (§) – 1 = § 2 2§ – 1 This is definitely weirdlooking, but itE L IJAH A ND TH E PR O PH ETS O F B A A L M ar c h 1 4 , 2 02 1 // C aleb B o nifay, C PC H ig h S cho ol Directo r ww wcpcda n v illeor g T h e Lo r d s e n t E lija h to c o n f r o n t K in g Ah ab, "Yo u have aban do n ed the Lord's co m m an ds a n d h av e f o llo w ed t h e B aa ls"(R ev ) SE C R E T F E D E R A L B U R E A U O F IN V E S T IG A T IO N b 2 < f y P r e c e d e n c e !
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