#(xy)^3=(xy)(xy)(xy)# Expand the first two brackets #(xy)(xy)=x^2xyxyy^2# #rArr x^2y^22xy# Multiply the result by the last two brackets #(x^2y^22xy)(xy)=x^3x^2yxy^2y^32x^2y2xy^2# #rArr x^3y^33x^2y3xy^2#Our online expert tutors can answer this problem Get stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us!Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history
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Expand log x^(2)y^(3)z
Expand log x^(2)y^(3)z-Given xyz=12Squaring it we getx2 y2 z2 2(xy yzzx) =144Therefore xy yzzx =24Also 1/x1/y 1/z =36Implies (xy yzzx)/xyz =36⇒ xyz = (xyyzzx)/36 = 24/36 = 2/3We know (x3 y3 z3) = (xyz)(x2 y2 z2 −xy−yz −zx)3xyz⇒ x3 y3 z3 = (12)(96−24)2 = 12∗722 = 8663 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 x
Let X Y Z Be Real Numbers Satisfying X Y Z 3 X 2 Y 2 Z 2 5
If we replace y with (− y) the expression changes to (x − y) 3 So to find the expansion of ( x − y ) 3 , we can replace y with ( − y ) in ( x y ) 3 = x 2 3 x 2 y 3 x y 2 y 3 This is the required expansion for ( x − y ) 3A Taylor Series is an expansion of some function into an infinite sum of terms, where each term has a larger exponent like x, x 2, x 3, etc Example The Taylor Series for e x e x = 1 x x 2 2!⋅(x)3−k ⋅(y)k ∑ k = 0 3
Expand (xy)^3 (x y)3 ( x y) 3 Use the Binomial Theorem x3 3x2y3xy2 y3 x 3 3 x 2 y 3 x y 2 y 3X, y, z are fixed during the integrations, a Taylor's series expansion in the source point coordinates x ', y ', z ' about (0,0,0) provides an approximation of the source coordinate dependence of D (;Expand 1/12*((xyz)^6 2(x^6y^6z^6) 2(x^3y^3z^3)^2 4(x^2y^2z^2)^3 3(xyz)^2(x^2y^2z^2)^2) Extended Keyboard;
X(Y Z) = (XY)Z X (Y Z) = (X Y)Z X(X Y) = X X (XY) = X X (Y Z) = (X Y)(X Z) X(Y Z) = (XY)(XZ) XX = 1 X X = 0 We will use the first nontrivial Boolean Algebra A = {0,1} This adds the law of excluded middle if X 6=0 then X = 1 and if X 6=1 then X = 0The volume is the integral of the function 1 over the ellipsoid You can evaluate it by changing variables to X = x 2y, Y = x− 2y z, Z = 3 z In the new coordinates, one now has Resolution of the singularities of the curve X^2Y^2 Y^2Z^2 X^2Z^2=0 Resolution of the singularities of the curve X 2Y 2 Y 2Z 2 X 2Z 2 = 0Since (3x z) is in parentheses, we can treat it as a single factor and expand (3x z)(2x y) in the same manner as A(2x y) This gives us If we now expand each of these terms, we have Notice that in the final answer each term of one parentheses
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The calculator can also make logarithmic expansions of formula of the form ln ( a b) by giving the results in exact form thus to expand ln ( x 3), enter expand_log ( ln ( x 3)) , after calculation, the result is returned The calculator makes it possible to obtain the logarithmic expansionIn elementary algebra, the binomial theorem (or binomial expansion) describes the algebraic expansion of powers of a binomialAccording to the theorem, it is possible to expand the polynomial (x y) n into a sum involving terms of the form ax b y c, where the exponents b and c are nonnegative integers with b c = n, and the coefficient a of each term is a specific positiveSolve x y z = x 3 y 3 z 3 = 8 in Z First I tried to transform this equation, substituting x = 8 − y − z So I end up with x 3 y 3 z 3 = 8 ( 8 − y − z) 3 y 3 z 3 = 8 Using Wolfram Alpha I expanded this equation and tried to factorize it so finally I got ( z − 8) ( y 2 y ( z − 8) − 8 z) = 168
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Log 2 (8) log 2 ( x4) – log 2 (5) = 3 4 log 2 ( x) – log 2 (5) Each log now finally contains only one thing, and the first log term has been simplified to a numerical value, so this expression is fully expanded Then my final answer is 3 4log 2 ( x) – log 2 (5) Content Continues BelowConsider 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!3 Answers3 For the nontrivial interpretation, you're looking for nonnegative solutions of a b c = n (each of these corresponds to a term x a y b z c ) Code each of these solutions as 1 a 0 1 b 0 1 c, for example ( 2, 3, 5) would be coded as Now it should be easy to see why the answer is ( n 2 n)
Let X Y Z Be Real Numbers Satisfying X Y Z 3 X 2 Y 2 Z 2 5
X Y Z 3 Expand Novocom Top
Simplify X Y Z Whole Square Minus Brainly In If X Y Z Is Equal To 10 And Square 40 Find Exact Brainly In X Y Z Whole Square Brainly In X Y Whole Square Brainly In Expand Using Ideny 3 X Minus 2 Y Z Whole Square Brainly In X Square Minus Y Whole Cube Z Brainly In If X Y Z Are All Diffe Real Numbers And One Upon Minus Whole Square 1 ThatA Find the coefficient of x 6 y 3 z 5 in the multinomial expansion of (x y z) 14 b Find the coefficient of x 6 y 11 in the binomial expansion of (x y) 17 Who are the experts?This calculator can be used to expand and simplify any polynomial expression
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X Y Z 3 Expand Novocom Top
X Y Z Whole Square Is Equal To If x 1 p y q z r and xyz then if x 1 p y q z r and xyz then y and z are positive integers such x 3 y z is equal x 3 y z is equal Formula For X Y Z 2 Brainly In Expand Form Of X Y Z Whole Square Brainly In Simplify X Y Z Whole Square Minus Brainly In If X Y Z Is Equal To 10 And Square 40 FindX1 1meter length wiring harness for extruder motor/Xaxis motor/xaxis end stop5!) Thus, (xyz) 10 = ∑(10!) / (P1
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