Answer by lenny460 (1073) ( Show Source ) You can put this solution on YOUR website!Here is the question What is the coefficient of w˛xłyzł in the expansion of (wxyz) 9 There are 9 4term factors in (wxyz) 9 (wxyz)(wxyz)(wxyz)(wxyz)(wxyz)(wxyz)(wxyz)(wxyz)(wxyz) To multiply it all the way out we would choose 1 term from each factor of 4 terms To get w˛xłyzł, · (xy)^3 (yz)3 (zx)^3 = 3(xy)(yz)(zx) That is it no constraints etc It mentions "This can be done by expanding out the brackets, but there is a more elegant solution" Homework Equations The Attempt at a Solution First of all this only seems to hold in special cases as I have substituted random values for x,y and z and they do not agree
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What is the formula of (x+y+z)^3-Given that x 3 4y 3 9z 3 = 18xyz and x 2y 3z = 0 x 2y = 3z, 2y 3z = x and 3z x = 2y Now `( x 2y )^2/(xy) (2y 3z)^2/(yz) (3x x)^2/(zx · 8x y z = 9, 2x 2y 3z = 22 and x 3y 2z = 15 solve for an ordered tripple asked Mar 6, 14 in ALGEBRA 2 by harvy0496 Apprentice systemofequations
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3 The right hand and left hand limit of the function are respectively 4 If tan A cot A = 2, then the value of t a n 4 A c o t 4 A = 5 If y = 2 x n 1 3 x n ,then x 2 d 2 y d x 2 is 6 If the curves 2 x = y 2 and 2 x y = K intersect perpendicularly, then the value of K 2 isThe author is saying y is not a particular number, that equation holds for every real number So it works for y=3 but also for y = 3 or y = \pi or whateverExpand (xy)^3 full pad » x^2 x^ {\msquare} \log_ {\msquare} \sqrt {\square} \nthroot \msquare {\square} \le \ge
This is the Solution of Question From RD SHARMA book of CLASS 9 CHAPTER POLYNOMIALS This Question is also available in R S AGGARWAL book of CLASS 9 You can FThis preview shows page 3 6 out of 11 pages 33 Find the sumofproducts expansion of the Boolean function f (x y z) that is 1 if and only if exactly two of the three variables have value 1 Ans xyz xyz xyz 34 Find the sumofproducts expansion of the Boolean function f (x y z) that is 1 if and only if either x z 1 and y 0, or x 0 and y z 1This term will be of the form n x 3 y 2 z 3 w for some integer n, so its "coefficient" is n w We can expand ( 2 x 3 y − 4 z w) 9 = ( ( 2 x 3 y − 4 z) w) 9 as ( 2 x 3 y − 4 z) 9 9 w ( 2 x 3 y − 4 z) 8 , so n is equal to 9 times the coefficient of x 3 y 2 z 3 in the expansion of ( 2 x 3 y − 4 z
As you can see for (a b)n contains just n 1 terms Note that we have to keep the sum of powers in each of the combinations of x, y, z to n, so it will be reduced Now replace a and b by x and (y z) respectively So total number of terms should be 1 2 3 ⋯ (n 1) = (n 1)(n 2) 2 Share · Explanation (x −y)3 = (x − y)(x −y)(x −y) Expand the first two brackets (x −y)(x − y) = x2 −xy −xy y2 ⇒ x2 y2 − 2xy Multiply the result by the last two brackets (x2 y2 −2xy)(x − y) = x3 − x2y xy2 − y3 −2x2y 2xy2 ⇒ x3 −y3 − 3x2y 3xy2Obtain The Taylor Expansion Of Z(x, Y) Near (0,0) Upto The Second Order Using The Gradient And Hessian Matrix Z(x, Y) Is Given In Implicit Form As F(x, Y, Z) = Z Cos X X2 – Eyz = 0 This problem has been solved!
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Therefore, F = m3 m4 m5 m6 m7, which is the same as above when we used term expansion x y z Minterms Notation 0 0 0 x' y' z' m0 0 0 1 x' y' z m1 0 1 0 x' y z' m2 0 1 1 x' y z m3 1 0 0 x y' z' m4 1 0 1 x y' z m5 1 1 0 x y z' m6 1 1 1 x y z m7 Table 39 F = x' y z x y' z x y z' x y z0218 · The number of terms in the expansion of (xyz) n Related questions 0 votes 1 answer If the integers r > 1, n > 2 and coefficients of (3r)th and (r 2)nd terms in the binomial expansion of (1 x)2n are equal, then asked Feb , 18 in Class XI Maths by rahul152Question from Binomial Theorem,cbse,math,class11,ch8,binomialtheorem,exemplar,q26,secb,medium
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See the answer Show transcribed image text Expert AnswerC n are the binomial coefficients in the expansion of (1 x) n then prove that 2 (3 2 − 1) C 1 2 2 (3 2 2 2 − 1) C 2 2 3 (3 3 2 3 − 1) C 3 2 n (3 n 2 n − 1) C n = 2 n (2 3 n − 3 n)The binomial theorem (or binomial expansion) is a result of expanding the powers of binomials or sums of two terms The coefficients of the terms in the expansion are the binomial coefficients
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We know the corollary if abc = 0 then a3 b3 c3 = 3abcUsing the above identity taking a = x−y, b = y−z and c= z−x, we have abc= x−yy−zz −x= 0 then the equation (x− y)3 (y−z)3 (z−x)3 can be factorised as follows(x−y)3 (y−z)3 (z−x)3 = 3(x−y)(y−z)(z−x)Hence, (x−y)3 (y−z)3 (z −x)3 = 3(x−y)(y −z)(z −x)243x 5 810x 4 y 1080x 3 y 2 7x 2 y 3 240xy 4 32y 5 Finding the k th term Find the 9th term in the expansion of (x2y) 13 Since we start counting with 0, the 9th term is actually going to be when k=8 That is, the power on the x will 138=5 and the power on the 2y will be 8Answered 1 year ago · Author has 115 answers and 3342K answer views Solution here, = (Xyz) (Xyz)² = (Xyz) (x²xyxzyxy²yzzxzyz²) = (x³y³z³3x²y3xy²3x²z3z²x3y²z3z²y6xyz) 38K views · View upvotes
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This information can be summarized by the Binomial Theorem For any positive integer n, the expansion of (x y)n is C(n, 0)xn C(n, 1)xn1y C(n, 2)xn2y2 C(n, n 1)xyn1 C(n, n)yn Each term r in the expansion of (x y)n is given by C(n, r 1)xn (r1)yr1 Example Write out the expansion of (x y)75!) Thus, (xyz) 10 = ∑(10!) / (P1(x y) 3 = x 3 3x 2 y 3xy 2 y 3 Example (1 a 2 ) 3 = 1 3 31 2 a 2 31(a 2 ) 2 (a 2 ) 3 = 1 3a 2 3a 4 a 6 (x y z) 2 = x 2 y 2 z 2 2xy 2xz 2yz
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⋅(x)3−k ⋅(y)k ∑ k = 0 3 Total number of terms in the expansion of $(x y z)^{10}$ is $=\,^{1031}C_{31} = \,^{12}C_{2}$ $= \frac{12\times11}{2}= 66$ · y 2 = x 3 ax 2 bx c (this is called the long Weierstrass form It is not strictly needed, but sometimes it's more convenient) It's a basic fact that any elliptic curve can be brought to this form (except if you're working over fields of
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Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, historyExpand Using the Binomial Theorem (xyz)^3 (x y z)3 ( x y z) 3 Use the binomial expansion theorem to find each term The binomial theorem states (ab)n = n ∑ k=0nCk⋅(an−kbk) ( a b) n = ∑ k = 0 n n C k ⋅ ( a n k b k) 3 ∑ k=0 3!Consider the expansion of (x y z) 10 In the expansion, each term has different powers of x, y, and z and the sum of these powers is always 10 One of the terms is λx 2 y 3 z 5 Now, the coefficient of this term is equal to the number of ways 2x′s, 3y′s, and 5z′s are arranged, ie, 10!
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Binomial Expansions Binomial Expansions Notice that (x y) 0 = 1 (x y) 2 = x 2 2xy y 2 (x y) 3 = x 3 3x 3 y 3xy 2 y 3 (x y) 4 = x 4 4x 3 y 6x 2 y 2 4xy 3 y 4 Notice that the powers are descending in x and ascending in yAlthough FOILing is one way to solve these problems, there is a much easier wayYou can put this solution on YOUR website!2301 · Example 15 If x, y, z are different and Δ = 8(x&x2&1x3@y&y2&1y3@z&z2&1z3) = 0 , then show that 1 xyz = 0 Solving ∆ = 8(x&x2&1x3@y&y2&1y3@z&z2&1z3) Here, expanding elements of C3 into two determinants = 8(x&x2&1@y&y2&1@z&z2&1) 8(x&x2&x3@y&y2&y3@z&z2&z3) = 8(x
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∗) (valid for any elements x , y of a commutative ring), which explains the name "binomial coefficient" Another occurrence of this number is in combinatorics, where it gives the number of ways, disregarding order, that k objects can be chosen from among n objects;T = x^5/1 x^3/6 x y^4/24 y^2/2 z^5/1 z^4/24 z^3/6 z^2/2 z 2 You can use the sympref function to modify the output order of a symbolic polynomial Redisplay the polynomial in ascending orderAnswer to Determine the coefficient of the term x^2 y z^3 in the expansion of (x y z)^6 Determine the coefficient of the term
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Taquete Expansor Tipo Z de 3/8" x 2" (INCLUYE TORNILLO) Modelo ANCZ38T Marca ANCLO Código SAT Acceder a mi Cuenta Para obtener su password por primera vez, contacte a su Representante de Ventas Productos Certificados SYSCOM Ver Política de Clasificación Probamos, inspeccionamos3 3 3 logxy logxlogy = Use property 3 to rewrite the multiplication as addition 3 3 = 2logx7logy Use Property 5 to move the exponents out front Thus, ( ) 27 3 3 3 logxy 2logx7logy = Example 2 Use the properties of logarithms to expand 8 3 x log y æ ö ç ÷ ç ÷ Ł ł 1 2 8 8 3 3 x xSolutionShow Solution ( x y z ) 2 = x 2 y 2 z 2 2 (x) (y) 2 (y) (z) 2 (z) (x) = x 2 y 2 z 2 2xy 2yz 2zx Concept Expansion of Formula
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21 · Ex 42, 9 By using properties of determinants, show that 8(x&x2&yz@y&y2&zx@z&z2&xy) = (x – y) (y – z) (z – x)(xy yz zx) Solving LHS 8(𝑥&𝑥^2&𝑦𝑧@𝑦&𝑦^2&𝑧𝑥@𝑧&𝑧^2&𝑥𝑦) Applying R1→ R1 – R2 = 8(𝑥−𝑦&𝑥^2−𝑦^2&𝑦𝑧−𝑥𝑧@𝑦&𝑦^2&𝑧𝑥@𝑧&𝑧^2&𝑥𝑦) Ex 42, 9 By using properties of determinants, s(xyz)^3 (x y z) (x y z) (x y z) We multiply using the FOIL Method x * x = x^2 x * y = xy x * z = xz y * x = xyDe Morgan's laws NAND x · y = x y NOR x y = x · y Redundancy laws The following laws will be proved with the basic laws Counterintuitively, it is sometimes necessary to complicate the formula before simplifying it
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ICS 241 Discrete Mathematics II (Spring 15) d) x = y = z = 0 xy z 122 pg 2 # 3 Find the sumofproducts expansions of these Boolean functionsShare It On Facebook Twitter Email 1 Answer 1 vote answered Sep 17, 19 by Juhy (631k points) selected Sep 18, 19 by Vikash Kumar Best answerMore formally, the number of k element subsets (or k combinations) of an n element set This number can be
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