`log_(3)x 2log_(3)y3log_(x)z` Step by step solution by experts to help you in doubt clearance & scoring excellent marks in examsClick here to see ALL problems on Exponents Question (xyz)^3 Answer by lenny460 (1073) ( Show Source ) You can put this solution on YOUR website!This video explains how to use the gradient of a function of two variables to determine the maximum rate of a function of two variable from a given point T
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(x+y+z)^3 expand-) rr Adopt the standard subscriptcomponent conventions x' = x 1 ', y' = x 2 ' and z' = x 3 'Find the expansion of (xyz)^{4} 🎉 Announcing Numerade's $26M Series A, led by IDG Capital!
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Solved Expert Answer to A By starting with ( (x y) z) 3, expand (x y z) 3 b Find the term independent of x in the expansion of (x 1 x ?1) 4Solve 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(xyz)^3 (x y z) (x y z) (x y z) We multiply using the FOIL Method x * x = x^2 x * y = xy
To disprove a proposal you need just one case where it does not work Set x = y = x = 1 Left hand side = 3^5 = 243 Right hand side = 3 So the question is answered To find an expression for (x y z)^5 you need Pascal's triangle which evaluates (a b)^n Let b = y z then we can use PascalRewrite (x−y −z)2 ( x y z) 2 as (x−y−z)(x−y−z) ( x y z) ( x y z) Expand (x−y−z)(x−y−z) ( x y z) ( x y z) by multiplying each term in the first expression by each term in the second expression Simplify each term Tap for more steps Multiply x x by x x =x^55x^4z10x^3z^210x^2z^35xz^4z^5 The binomial theorem states that (xy)^nsum_(r=1)^n ""^nC_rx^(nr)y^r therefore (xz)^5=sum_(r=1)^5""^5C_rx^(5r)z^r =""^5C_0x^5""^5C_1x^(51)z^1""^5C_2x^(52)z^2""^5C_3x^(53)z^3""^5C_4x^(54)z^4""^5C_5x^(55)z^5 =x^55x^4z10x^3z^210x^2z^35xz^4z^5
Expand 2 (x y) 3 (x y) full pad » x^2 x^ {\msquare} \log_ {\msquare} \sqrt {\square} \nthroot \msquare {\square} \le \geExpand 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!⋅(x)3−k ⋅(y)k ∑ k = 0 3
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If you have tried to take X Y Z data and graph it three dimensionally inside of Microsoft Excel you know that Excel's graphing capabilities are quite limitedWriting any polynomial directly just requires listing all possible exponent combinations of a^x*b^y*c^z where every term has xyz=3 and the coefficient of each term is (xyz)!/(x!y!z!) This works for any length polynomial and in this case we would have a^3 b^3 c^3 3a^2b 3a^2c 3b^2c 6abc 3ab^2 3ac^2 3bc^2 Explanation Note that (a b)4 = a4 4a3b 6a2b2 4ab3 b4 So we can find the terms of (x y z)4 that only involve 2 of x,y,z by combining the expansions of binomial powers, One way to see that is to think about setting each of x,y,z to zero in turn and expanding the remaining binomial By symmetry, the remaining terms involving all three
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Click here👆to get an answer to your question ️ Using the identity and proof x^3 y^3 z^3 3xyz = (x y z)(x^2 y^2 z^2 xy yz zx)Expand\3(x6) expand\2x(xa) expand\(2x4)(x5) expand\(2x5)(3x6) expand\(4x^23)(3x1) expand\(x^23y)^3 x^33x^23x1 "note that" (xa)^3=x^3(aaa)x^2(aaaaaa)xa^3 (x1)^3toa=1 rArr(x1)^3=x^3(111)x^2(111)x(1)^3 =x^33x^23x1
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(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 agreeFind the expansion of (xyz)^{4} Our Discord hit 10K members!The LibreTexts libraries are Powered by MindTouch ® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot We also acknowledge previous National Science Foundation support under grant numbers ,
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X, 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 (; Find an answer to your question Expand it (Xyz)² = Anishahussain043 Anishahussain043 Math Secondary School Expand it (Xyz)² = 2So 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 ) 3 Let's now use these identities to
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Algebra Expand (xyz)^2 (x − y z)2 ( x y z) 2 Rewrite (x−y z)2 ( x y z) 2 as (x−yz)(x−yz) ( x y z) ( x y z) (x−y z)(x−yz) ( x y z) ( x y z) Expand (x−yz)(x−yz) ( x y z) ( x y z) by multiplying each term in the first expression by each term in the second expression 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 3xy2Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science
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🎉 Meet students and ask top educators your questionsExample Express the Boolean function F = x y z as a sum of minterms Solution F = x y z = x (y z) AND (multiply) has a higher precedence than OR (add) = x(yy')(zz') (xx')yz expand 1st term by ANDing it with (y y')(z z'), and 2nd term with (x x') = x y z x y z' x y' z x y' z' x y z x' y z = m7 m6 m5 m4 m3Given 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 = 866
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Or, you can group the terms (as in, ((xy) z) n) and apply the Binomial Theorem as usual For example, (xyz) 3 = ((xy) z) 3 = (xy) 3 3(xy) 2 z 3(xy)z 2 z 3 Then, expand the (xy) terms using the Binomial TheoremStack Exchange network consists of 178 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers Visit Stack Exchange (x2y3z)² (x2y3z)²=x²4y²9z²4xy12yz6zx Stepbystep explanation /* We know the algebraic identity (abc)²=a²b²c²2ab2bc2ca */ Here , (x2y3z)² = x²(2y)(3z)² = x²(2y)²(3z)²2×x(2y)2×(2y)(3z)2×(3z)x = x²4y²9z²4xy12yz6zx Therefore, (x
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Get answer The coefficient of `x^2 y^5 z^3` in the expansion of `(2x y 3z)^10` is Step by step solution by experts to help you in doubt clearance & scoring excellent marks in examsB Repeat part a with x y z 3 c Using parts a and b predict the expansion of x y from SCIENCE 101 at Bishop McNally High SchoolWolframAlpha Computational Intelligence Enter what you want to calculate or know about Extended Keyboard Examples Compute expertlevel answers using Wolfram's breakthrough algorithms, knowledgebase and AI technology
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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 3Math, 0850, sahini99 (x y)^3 expandExpand (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 3
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Click here👆to get an answer to your question ️ Expand ( x y 3z )^2Expand (x y z)^10 Extended Keyboard;Expand (− 2 x 5 y − 3 z) 2 using suitable identities Hard Answer (− 2 x 5 y − 3 z) 2 is of the form (a b c) 2 (a b c) 2 = a 2 b 2 c 2 2 a b 2 b c 2 c a where a = − 2 x, b = 5 y, c = − 3 z ∴ (− 2 x 5 y − 3 z) 2 =
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Expand and simplify (p3)(p7) Log Expand Log5(5squareroot x/7) the first 5 is a base how would you expand that?In 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 positive Find an answer to your question 3 Expand (i) (x y z)^2 akshayagkg00gmailcom is waiting for your help Add your answer and earn points
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x 3 y 3 z 3 – 3xyz = (x y z)(x 2 y 2 z 2 – xy – yzzx) = 0(x 2 y 2 z 2 – xy yzzx) (∵ x y z = 0 given) = 0 => x 3 y 3 z 3 = 3xyz Hence proved Ex 25 Class 9 Maths Question 14 Without actually calculating the cubes, find the value of each of the following (i) (12) 3 (7) 3 (5) 3 (ii) (28) 3 (15) 3Get answer The coefficient of x^3 y^4 z ^5 in the expansion of (xy yz xz ) ^6 is Apne doubts clear karein ab Whatsapp par bhi Try it now(x y) 7 = x 7 7x 6 y 21x 5 y 2 35x 4 y 3 35x 3 y 4 21x 2 y 5 7xy 6 y 7 When the terms of the binomial have coefficient(s), be sure to apply the exponents to these coefficients Example Write out the expansion of (2 x 3 y ) 4
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Expand by multiplying each term in the first expression by each term in the second expressionScience (Chemistry) Hello DrBob222 Regarding the answer you gave me thank you very much However when you expand the denominator, I don't get 014 but I get 028Solve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and more
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