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A function f from A to B assigns exactly one element of B to each element of A f(a) = b ≡ f assigns b ∈ B to a ∈ A f A → B ≡ f is a function from A to B A is called the domain of f B is called the codomain or range of f Example Functions R×R → R What about minus, subtraction, multiplication, division, exponentiationI f h f g f e N W P Q S ` S Y O d Z U N M S Y O N X W S U M U < F B ;That is, basic questions about whether or not an orthogonal series expansion such as (7) is actually possible were ignored

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=> A' (CC' ) AB' C => A'AB'C (xx'=1) => A'B'C X=A' y=B'C = LHS Hence Proved Question 2 Write the Boolean Expression for the result of the Logic Circuit as shown below Аnswer F = (uv')(uw)(vw') Question 3 Derive a Canonical POS expression for a Boolen function F, represented by the following truth tableA private research university with more than 16,000 students from around the world, the University of Miami is a vibrant and diverse academic community focused on teaching and learning, the discovery of new knowledge, and service to the South Florida region and beyondX O b ta in C o p y O f P o lic e R e p o rt x I n s p e c t V e h ic le x C o m p le te R e p a ir A u th o riz a tio n F o rm A n d B o d y D a m a g e I n s p e c tio n S h e e t x B rin g / T o w V e h ic le T o I S D B o d y S h o p 3 R E C E I V E / R E V I E W REVIEW CONDUCT/ ENSURE PICK UP/ COMPLETE I N S P E C

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/ 0 1 1 23 4 5 6 7 8 9 6 ;2D9 The accompanying picture shows the level curves of a function w = f(x,y) The value of w on each curve is marked A unit distance is given a) Draw in the gradient vector at A b) Find a point B where w = 3 and ∂w/∂x = 0 c) Find a point C where w = 3 and ∂w/∂y = 0 d) At the point P estimate the value of ∂w/∂x and ∂w/∂y The next two letters are likely C & U There are at least 2 patterns in these letters FIrst is the list of vowels that make up every other letter Then the letters alternate from the ends of

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1 Fractions Let a,b,c, and d be numbers (a) You can break up a fraction from a sum in the numerator, but not in the denominator ab c = a c b c but a bc 6= a b a c (b) Cancellation of the c here requires that it appears in each additive term of the numerator and denominator cacb cd = c(ab) cd = ab d but cab cd 6= ab dDefinition An element y in B is called a complement of an element x in B if xy=1 and xy=0 Theorem 2 For every element x in B, the complement of x exists and is unique Proof Existence Let x be in B x' exists because ' is a unary operation X' is a complement of xH o ^ n m l k ?

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