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1 Boolean Algebra Chapter 2 3 Z Aliyazici Basic Identities of Boolean Algebra 1 X0 = X 2 X1 = X 3 X1 = 1 4 X0 = 0 5 XX = X 6 XX = X 7 XX' = 1 8Click here👆to get an answer to your question ️ Let y'(x) y(x)g'(x) = g(x)g'(x) , y(0) = 0, x∈ R , where f'(x) denotes df(x)/dx and g(x) is a given non constant differentiable function on R with g(0) = g(2) = 0 Then the value of y(2) isG(x,y)=xyx'y' is in DNF But h(x,y,z)=xyx'y'z is not in DNF because xy is not a minterm of size 3 Definition (Conjunctive Normal Form) A Boolean function/expression is in Conjunctive Normal Form (CNF), also called maxterm canonical form, if the function/expression is a product of maxterms

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4Vp *>j © j ¤Q;3P'Zj3@BADZVCRZ3±5@?x < C G=xcy Gcy=x answer= GCY=X THANK YOU you're welcome Advertisement Advertisement New questions in Mathematics Step 1 Identify the LCD Step 2 Transform the equation into quadratic equation Step 3 Solve for the roots using any methodThe CDC AZ Index is a navigational and informational tool that makes the CDCgov website easier to use It helps you quickly find and retrieve specific information

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When looking at the basic macroeconomy, we need to know what components make up GDP (Gross Domestic Product) The most basic equation for representing GDP is the following Y=CIGNX where Y is GDP C is consumer spending I is investment G is government spending and NX is net exports This means the GDP of an economy (or the total value of all of the final outputs), is equalFunction f(x,y,z) subject to a constraint (or side condition) of the form g(x,y,z) =k Assumptions made the extreme values exist ∇g≠0 Then there is a number λ such that ∇ f(x 0,y 0,z 0) =λ ∇ g(x 0,y 0,z 0) and λ is called the Lagrange multiplier• Constant Multiple Rule g(x)=c·f(x)theng0(x)=c·f0(x) • Power Rule f(x)=x n thenf 0 (x)=nx n−1 • Sum and Difference Rule h(x)=f(x)±g(x)thenh 0 (x)=f 0 (x)±g 0 (x)

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Title COSEXindd Author christopherdinardo Created Date 007 PM g''(x) = − e−x − (e−x − xe−x) ∴ g''(x) = − 2e−x xe−x Similarly the third derivative g(3)(x) = 2e−x e−x − xe−x ∴ g(3)(x) = 3e−x − xe−x So it looks like clear pattern is forming, but let us just check by looking at the fourth derivative;Proof lnexy = xy = lnex lney = ln(ex ·ey) Since lnx is onetoone, then exy = ex ·ey 1 = e0 = ex(−x) = ex ·e−x ⇒ e−x = 1 ex ex−y = ex(−y) = ex ·e−y = ex 1 ey ex ey • For r = m ∈ N, emx = e z }m { x···x = z }m { ex ···ex = (ex)m • For r = 1 n, n ∈ N and n 6= 0, ex = e n n x = e 1 nx n ⇒ e n x = (ex) 1 • For r rational, let r = m n, m, n ∈ N

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N ¹ y f1 _NM È á h# Í o _# ,X Ï* > = 0 0 < ;The Algebra of Functions Like terms, functions may be combined by addition, subtraction, multiplication or division Example 1 Given f ( x ) = 2x 1 and g ( x ) = x2 2x – 1 find ( f g ) ( x ) and ( f g ) ( 2 )

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Ó ô ^ x Þ ´ á ç G O $ y a X i ^ M { Ó ( x Í $ Ó Ò å ­ ¿ Ä x Ç V b { y w O t s b { "$ 7 ¯ ï ï Ä ¢ ¯ ï ï Ä A q b ¨ µ ¯ ï é w Ô ù w £ ¨ µ ¢ £ l Ó y º G O x ¨ µ ¯ ï é ) " 3), p w G O p b { y C ä ;Fis the same asg q (x)(y)(x i y = x g y) f=g f and g are corestricted 3h(fih &glh) f& f and g are corestrictive 3h(h 1 f & h lg) fAg f and g are cosatisfiable 03x3 t(fx &gx) at t fa f is ultimate CgXgIf~g=n f This notation is suggested by diagrams of fragments of the sortal hierarchy, in which sortals lower down restrict those higher upIf g ( 0) ≠ 0 then put y = 0 in the equation and you get for all other x, g ( x) = 2 First solution g ( x) = 2 Now if g ( 0) = 0 there are other possibilities Choose x = y = 2 and then either g ( 2) = 0 or g ( 2) = 2 If g ( 2) = 0, then either g ( 1) = 0 or g ( 1) = 3

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G(4)(x) = −3e−x − (e−x − xe−x) ∴ g(4)(x) = −4e−xA n d d e nt a l offi ce s , b a n k s a nd c h ild c a re cente rs Ap a r t m e nt s a b ove t h e g ro u n d fl o o r a re a l l owe d At co r n e r t raffi c l ig ht inters e c o n Prop e r t y t a xe s at $ 7 ,5 5 2 ( 21 ) LO C ATI O N D ES C R I P TI O N» & i k x P & Æ Ó ) g 4 z Q y = i k x Ò ò g 4 k n î Ö low high g 4 k Basic Warranty 4 { X Y T g 4 z > ProSupport Plus T a x U & ¹ Ð R þ Ó ) g 4 z ø % 4 { X Y T * Dell Ó g 4 z & MC Addon u c 4 j j k a x U ) i k x 1 ProSupport 4 { X Y T ' o z X Y T Æ Ó ) g 4 z

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0 y = g(t) Make the substitutions x 1 = y, x 2 = y′, x 3 = y″, , x n = y (n−1), and x n′ = y (n) The first n − 1 equations follow thusly Lastly, substitute the x's into the original equation to rewrite it into the nth equation and obtain the system of the form x 1′ = x 2 x 2′ = x 3 x 3′ = x 4 x n1) Thus E(N) = c, the bounding constant, and we can now indeed see that it is desirable to choose our alternative density g so as to minimize this constant c = sup x{f(x)/g(x)} Of course the optimal function would be g(x) = f(x) which is not what we have in mind since the whole point is to choose a different (easy to simulate) alternativeC x4 3x2 3 is irreducible according to Eisenstein's criterion with p = 3 d Consider x5 5x2 1 mod 2, which is x5 x2 1 It is easy to see that this polynomial has no roots in Z 2, and so to prove irreducibility in Z 2 it again suffices to show it has no quadratic factors The only quadratic polynomial in Z 2x that does not have a root in Z 2 is x 2x1 which does not divide x5 x 1

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Advertisement Advertisement FelisFelis FelisFelis Answer The required value is or Stepbystep explanation Consider the provided equation We need to solve the equation for xBounded, G(x) is uniformly continuous But then G(x) = g(x) on (0;1), and so g(x) is also uniformly continuous Failed attempt at a solution ( x h)sin 1 x h sin 1 x sin j j 1 x h 1 x 1 x h jxj sin 1 x h sin 1 x jhj For the rst term, we use the fact that sinA sinB= 2sin A B 2 cos A B 2 ; gx−x=−c Step 3 Factor out variable x x(g−1)=−c Step 4 Divide both sides by g1 x= Hope this helps!

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G (x) g (x 1 ) = g (x) g (x 1 ) g (1) − 2 g (x) g (x 1 ) = g (x) g (x 1 ) 2 − 2 ∵ g (1) = 2 This is valid only for the polynomial ∴ g (x) = 1 ± x n (2) Now g (2) = 5 (Given) ∴ 1 ± 2 n = 5 Using equation (2) ± 2 n = 4, 2 n = 4, − 4 Since the value of 2 n cannot be − V e So, 2 n = 4, n = 2 Now, put n

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