👍 Correct answer to the question The identity (x^2 y^2)^2 = (x^2 y^2)^2 (2xy)^2 can be used to generate Pythagorean triples What Pythagorean triple could be generated using x = 8 and y = 3?Write a(x)/ b(x) in the form q(x) r(x)/ b(x), where a(x), b(x), q(x), and r(x) areYou can put this solution on YOUR website!
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The identity (x^2 y^2)^2=(x^2-y^2)^2 (2xy)^2 can be used
The identity (x^2 y^2)^2=(x^2-y^2)^2 (2xy)^2 can be used-Answer 2 📌📌📌 question The identity (x^2y^2)^2 = (x^2y^2)^2 (2xy)^2 can be used to generate pythagorean triples what pythagorean triple could be generated using x=8 and y=3 the answers to estudyassistantcom The identity (x^2 y^2)^2 = (x^2 – y^2)^2 (2xy)^2 can be used to generate Pythagorean triples What Pythagorean triple could be generated using x = 8 and y Skip to Content
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When adding and multiplying like terms students may initially confuse x x as x2 instead of 2x factors and can be used to factor polynomials of any power The degree of a polynomial will identity (x 2 y 2) 2 = (x 2 y 2) 2 (2xy) 2 can be used to(x – y) 2 = x 2 – 2xy y 2 LHS = RHS Hence, proved x 2 – y 2 = (x y) (x – y) By taking RHS and multiplying each term (x y) (x – y) = x 2 – xy xy – y 2 (x y) (x – y) = x 2 – y 2 Or x 2 – y 2 = (x y) (x – y) LHS = RHS Hence proved In the same way, you can prove the other above given algebraic identities Problems on Algebraic Identities Problem Solve (x 3) (x – 3) using algebraic identities Solution By the algebraic identity, x 2 How do you use Implicit differentiation find #x^2 2xy y^2 x=2# and to find an equation of the tangent line to the curve, at the point (1,2)?
1 6 x 2 4 y 2 9 z 2 1 6 x y 1 2 y z 2 4 x z Medium View solution Factorise 4 x 4 9 y 4 6 x 2 y 2 Easy View solution Factorise the following, using the identity a 2Write a(x)/ b(x) in the form q(x) r(x)/ b(x), where a(x), b(x), q(x), and r(x) are polynomials with the degree of r(x) less• See x4 – y4 as (x2)2 – (y2)2, thus recognizing it as a difference of squares that can be factored as (x2 – y2)(x2 y2) • In the equation x2 2x 1 y2 = 9, see an opportunity to rewrite the first three terms as (x1)2, thus recognizing the equation of a circle with radius 3 and center (−1, 0)
By using formula (xy)^2 = x^2 y^2 2xy we can solve this problem Now x^2 y^2 = (xy)^2 2xy Simply put the given values in the above equation & you will get the answer 28 The following identity can be used to find Pythagorean triples, where the expressions x2−y2, 2xy, and x2y2 represent the lengths of three sides of a right triangle;Believe me this can't be simplified It can be expanded and multiplied out (x^(2)2xyy)^2 becomes x^44 x^3 y4 x^2 y^22 x^2 y4 x y^2y^2 and then you still have to multiply by (xy) and that can become x^5(5 x2) x^3 y(8 x^26 x1) x y^2(2 x1)^2 y^3 or that can become
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PreAlgebra Simplify (2xy^22x^3x^2y) (2x^2y2xy^2y^3) (2xy2 2x3 − x2y) − (−2x2y 2xy2 − y3) ( 2 x y 2 2 x 3 x 2 y) ( 2 x 2 y 2 x y 2 y 3) Simplify each term Tap for more steps Apply the distributive propertyFor example, the polynomial identity (x^2 y^2)^2 = (x^2 – y^2)^2 (2xy)^2 can be used to generate Pythagorean triples I can prove polynomial identitiesAnswer to Verify the identity \bigtriangledown ^2 \vec \text v = \ grad \ div \vec \text v \ curl \ curl \vec \text v for the vector field \vec
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The identity (x^2 y^2)^2 = (x^2 y^2)^2 (2xy)^2 can be used to generate Pythagorean triples What Pythagorean triple could Mathematics, 0640 yo2lo15 The identity (x^2 y^2)^2 = (x^2 y^2)^2 (2xy)^2 can be used to generate Pythagorean triples$\psi = x^2(y2)^2\lambda(x^2y^21)$ Establish Lagrange Equations $$\psi_x = 2x\lambda(2x)=0 $$ $$\psi_y = 2y4\lambda(2y)=0$$ $$\psi_\lambda = x^2y^21=0$$The polynomial identity (x2 y2)2 = (x2 – y2)2 (2xy)2 can be used to generate Pythagorean triples D Rewrite rational expressions AAPRD6 Rewrite simple rational expressions in different forms;
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An algebraic identity is an equality that holds for any values of its variables For example, the identity ( x y) 2 = x 2 2 x y y 2 (xy)^2 = x^2 2xy y^2 (x y)2 = x2 2xyy2 holds for all values of x x x and y y y Since an identity holds for all values of its variables, it is possible to substitute instances of one side of theFor example, the polynomial identity (x 2 y 2) 2 = (x 2 y 2) 2 (2xy) 2 can be used to generate Pythagorean triples Authors National Governors Association Center for Best Practices, Council of Chief State School Officers Title CCSSMathContentHSAAPRC4 Prove Polynomial Identities And Use Them To DescribeFor example, the polynomial identity (x^2 y^2 )^ 2 = (x^2 – y^ 2 )^ 2 (2xy)^2 can be used to generate Pythagorean triples AAPRC5 () Know and apply the Binomial Theorem for the expansion of (x y)^n in powers of x and y for a positive integer n, where x and y are any numbers, with coefficients determined for example by Pascal's Triangle
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The Pythagorean triple Identity is (x 2 y 2) 2 = (x 2 y 2) 2 (2xy) 2, where c = x 2 y 2, a = x 2 y 2, and b = 2xy With x = 3 and y = 5 a = 3 2 5 2 = 16, (this is 16, by the rules x should be greater than y (x > y), in this case √(x 2 y 2) 2 = a) b = 2*3*5 = 30 c = 3 2 5 2For cases where equality is shown using numerical values, (name) will use algebraic operations (eg distributive property, collecting like terms) to prove that the polynomials are equivalentMAFS912AAPR22 In this video, students will use the Polynomial Remainder Theorem to determine whether a linear expression is a factor of a polynomial expression • Internet connection • Speakers/headphones • Computer • Scientific calculator (if necessary) Dividing Polynomials • MAFS912AAPR22 This tutorial can be used to help
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