The Identity X2+y22x2 Y22+2xy2 Can Be Used To Generate

Correct answers 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 identity x2+y22x2 y22+2xy2 can be used to generate. One (simple) way let t = x 2 y 2 xy then t xy ≥ 0 since it is the sum of two real squares x 2 y 2 and t xy ≥ 0 since it is the square of the real (x y) since (x y) 2 = x 2 y 2 2xy adding these, we get 2t ≥ 0, therefore t ≥ 0 Share Follow this answer to receive notifications. Consider x^ {2}y^ {2}xy22xy as a polynomial over variable x Find one factor of the form x^ {k}m, where x^ {k} divides the monomial with the highest power x^ {2} and m divides the constant factor y^ {2}y2 One such factor is xy1 Factor the polynomial by dividing it by this factor. We have M(x,y) = y2 2xy and N(x,y) = −x2, which leads to ∂M ∂y = 2x2y and ∂N ∂x = −2x Since ∂M ∂y 6= ∂N ∂x the equation is not exact However, 1 M ∂N ∂x − ∂M ∂y = − 2 y which implies that thereexists an integratingfactor, dependingonly on y, which satisfies 1 µ dµ dy = − 2 y An integrating factor is therefore given by µ(y) = 1 y2 Multiplying through.

Answer 1 📌📌📌 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?.  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)?. X x2 y2 − i y x2 y2, (211) whose real and imaginary parts are graphed in Figure 1 Note that these functions have an interesting singularity at the origin x= y= 0, but are harmonic everywhere else A slightly more complicated example is the function f(z) = z−1 z1 (212).

The answers to estudyassistantscom. Calculus Basic Differentiation Rules Implicit Differentiation. (c) u(x,y) = sinxcoshycosxsinhy Solution (a) If u(x,y) = x2y2, then u xxu yy = 22 = 4 6= 0 Therefore u(x,y) = x2y2 is not a solution (b) If u(x,y) = ln(x2 y2)3/2 = 3 2 ln(x2 y2), then we get u x = 3x x2 y2, u y = 3y x2 y2, u xx = 3 −x2 y2 (x2y2)2, u yy = 3 x2 −y2 (x2 y2)2.

Reverse the roles of x and y and compute the least squares regression line for the new data set x 2 4 6 5 9 y 0 1 3 5 8;.  Factorize the following using appropriate identities (i) 9x2 6xy y2 Solution 9x 2 6xy y 2 = (3x) 2 (2 × 3x × y) y 2 Using formula, a 2 2ab b 2 = (a b) 2 Then, a = 3x b = y 9x 2 6xy y 2 = (3x) 2 (2 × 3x × y) y 2. √ the identity (x^2 y^2)^2=(x^2y^2)^2 (2xy)^2 can be used 30= 6* (xy) So, (xy)= 30÷6 = 5 So, (xy)= 5 14K views Get stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us!Math Algebra 2 Use the identity (x^2y^2)^2=(x^2−y^2)^2(2xy)^2 to determine the sum of the squares of two numbers if the difference of the squares of the numbers.

 👍 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?. Find dy/dx x^2y^2=2xy Differentiate both sides of the equation Differentiate the left side of the equation Tap for more steps Differentiate Tap for more steps By the Sum Rule, the derivative of with respect to is Differentiate using the Power Rule which states that is where Evaluate Tap for more steps Differentiate using the chain rule, which states that is where and Tap. 2 The Bivariate Normal Distribution has a normal distribution The reason is that if we have X = aU bV and Y = cU dV for some independent normal random variables U and V,then Z = s1(aU bV)s2(cU dV)=(as1 cs2)U (bs1 ds2)V Thus, Z is the sum of the independent normal random variables (as1 cs2)U and (bs1 ds2)V, and is therefore normalA very important property of.

 I am trying to solve the equation $$ (x^2y^2)y' 2xy = 0 $$ I have rearranged to get $$ y' = f(x,y) $$ where $$ f(x,y) = \frac{2xy}{x^2y^2} $$ From here I tried to use a trick that I learned Stack Exchange Network Stack Exchange network consists of 178 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn,. (a) u(x,y) = x2 y2;.  (xy)^2 = x^2 2xy y^2 (x y)2 = x2 2xyy2 holds for all values of x x and y y Since an identity holds for all values of its variables, it is possible to substitute instances of one side of the equality with the other side of the equality For example, because of the identity above, we can replace any instance of (xy)^2 (x y)2 with.

Solution (x 4 – 1) is of the form Identity III where a = x 2 and b = 1 So we have, (x 4 – 1) = (x 2) 2 – 1 2) = (x 2 1)(x 2 – 1) The factor (x 2 – 1) can be further factorised using the same Identity III where a = x and b = 1 So, (x 4 – 1) = (x 2 1)((x) 2 –(1) 2) = (x 2 1)(x 1)(x – 1) Eample 3 Factorise 16x 2 4y 2 9z 2 – 16xy 12yz – 24zx using standard. MGSE912AAPR4 Prove polynomial identities and use them to describe numerical relationships For example, the polynomial identity (x^2 y^2)^2 = (x^2 – y^2)^2 (2xy)^2 can be used to generate Pythagorean triples Identities V Last updated at by Teachoo Identity V is (a b c) 2 = a 2 b 2 c 2 2ab 2bc 2ca Let us prove it Proof (a b c) 2 = ( (a b) c) 2 Using (x y) 2 = x 2 y 2 2xyAnd x>y. Consider the function f (x, y) = x 2 y 2 f (x, y) = x 2 y 2 from Example 69 Figure 611 shows the level curves of this function overlaid on the function’s gradient vector field The gradient vectors are perpendicular to the level curves, and the magnitudes of the vectors get larger as the level curves get closer together, because closely grouped level curves indicate the graph is steep.

The answers to estudyassistantcomYes, we can call it as an identity For example Let us take x = 2 and y = 1 in x 3 − y 3 = ( x − y ) ( x 2 x. Interchanging x and y corresponds geometrically to reflecting the scatter plot in a 45degree line Reflecting the regression line for the original data the same way gives a line with the equation y ^ = 1346 x − 3600. From this example we see that the method have the following steps 1 Write y(x) = X n=0 ∞ a n xn (19) 2 Substitute into the equation and determine a n A recurrence relation – a formula determining a n using a i, i.

= df dr r r Note that f(r) is spherically symmetrical and the resultant vector field is radial out of a sphere The significance of grad 66 • We know that the total differential and grad are defined as dU = ∂U ∂x dx ∂U ∂y dy ∂U ∂z dz & ∇U = ∂U ∂x ˆı ∂U ∂y ˆ ∂U. Factor y^{2}2xyx^{2} In general, when x^{2}bxc is a perfect square, it can always be factored as \left(x\frac{b}{2}\right)^{2} \sqrt{\left(yx\right)^{2}}=\sqrt{0} Take the square root of both sides of the equation yx=0 yx=0 Simplify y=x y=x Subtract x from both sides of the equation y=x The equation is now solved Solutions are the same Examples Quadratic. X 2y z is the distance from the origin O to the point P OA OB AP x y z A(x,y,z) B(x,y,z) a R r O Figure 53 Generating Function for Legendre Polynomials a = OA r = OB φ = C R = C √ r2 a2 − 2cos−1 θ 9 Through substitution we can write φ = C r 1− 2xt t2 −1/2 where t = a r,x= cosθ Therefore φ ≡ C r g(x,t) We introduce the angle θ between the vectors OA and OP and write R2.

Add − x 2 y 2 x 2 y 2 and x 2 y 2 x 2 y 2 Add x 2 x 2 x 2 x 2 and 0 0 Simplify each term Tap for more steps Multiply x 2 x 2 by x 2 x 2 by adding the exponents Tap for more steps Use the power rule a m a n = a m n a m a n = a m n to combine exponents Add 2 2 and 2 2. 2 Pauli spin matrices The Pauli spin matrices, σx, σy, and σz are defined via S~= ~s~σ () (a) Use this definition and your answers to problem 131 to derive the 2×2 matrix representations.  Saikiran Reddy and Kwasi F give excellent solutions to this The answer, dy/dx =1 might make us think about the question a bit For x^2y^2=2xy, we get (by differentiating implicitly), dy/dx =1 That's the same as the derivative of a linear function with slope, 1 Hmmmmm Let's see If we have x^2y^2=2xy The we must also have x^22xy y^2=0 Factoring gets us.

Y(x) = a0 ex 2 (18) where a0 can take any value – recall that the general solution to a first order linear equation involves an arbitrary constant!. X2y2z2, so ∂r/∂x = x/r and similarly for y,z ⇒ ∇∇U = df dr xˆıyˆ zkˆ r!. X2 y2 z2 (d) f(x,y,z) = 1 p x2 y2 z2 Quiz Choose the Laplacian of f(r) = 1 rn where r = p x2 y2 z2 (a) − 1 rn2 (b) n rn2 (c) n(n−1) rn2 (d) n(n5) rn2 The equation ∇2f = 0 is called Laplace’s equation This is an important equation in science From the above exercises and quiz we see that f = 1 r is a solution of Laplace’s.

 x to the power of 2 y to the power of 2= 3x y to the power of 3 a2b2 cd Reply eyi at 908 pm Which expression is eqiviluant to (7^8) (7^3) Reply yo mama at 3 pm I just want to solve if 4/5 and 6/10 are equivalent BoY WhY Reply Yeri at 1217 am What expressions are equivalent to 5\4?. The identity (x^2 y^2)^2=(x^2y^2)^2 (2xy)^2 can be used to generate Satyajeetdamekar004 satyajeetdamekar004 Math Secondary School answeredUse the identity (x2y2)2=(x2?y2)2(2xy)2 to determine the sum of the squares of two numbers if the difference of the squares of the numbers is 5 and the product of the numbers is 6The following identity can. NCERT Solutions Class 9 Maths Chapter 2 – Polynomials Exercise 25 are given here These NCERT Maths solutions are created by our subject experts which makes it easy for students to learn The students use it for reference while solving the exercise problems The fifth exercise in Polynomials Exercise 25 discusses the Algebraic Identities.

Answer (1 of 5) Divide throughout by x^2 dx (y^2/x^2 dx 2y/x dy) = 0 The term within parentheses is recognizable as the exact differential d(y^2/x) So the antiderivative can be taken easily and the solution is x C y^2/x = 0. Videos, solutions, examples, and lessons to help High School students learn to prove polynomial identities and use them to describe numerical relationships For example, the polynomial identity (x2 y2)2 = (x2 – y2)2 (2xy)2 can be used to generate. (b) u(x,y) = ln(x2 y2)3/2;.

If (x, y, z) (x, y, z) is a point in space, then the distance from the point to the origin is r = x 2 y 2 z 2 r = x 2 y 2 z 2 Let F r F r denote radial vector field F r = 1 r 2 〈 x r, y r, z r 〉 F r = 1 r 2 〈 x r, y r, z r 〉 The vector at a given position in space points in the direction of unit radial vector 〈. コレクション the identity (x^2 y^2)^2=(x^2y^2)^2 (2xy)^2 can be used X and y are positive integers;Example 5532 For the function f(x,y)=xy3, calculate the gradient at the point (2,1) and estimate z if x =1and y =2 Let's begin my finding the partial derivatives at (2, 1) • fx(x,y)=y3, which implies fx(2,1) = 1 • fy(x,y)=3xy2, which implies fy(2,1) = 6 Now, we may plug into. Stack 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.

Extended Keyboard Examples Upload Random Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, music WolframAlpha brings expertlevel knowledge. Correct answers 3 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?. 442 Use the tangent plane to approximate a function of two variables at a point 443 Explain when a function of two variables is differentiable 444 Use the total differential to approximate the change in a function of two variables In this section, we consider the problem of finding the tangent plane to a surface, which is analogous to finding the equation of a tangent line to a curve.

Inverse of x iy is the complex number (−x) i(−y), the multiplicative identity is 1 and the multiplicative inverse of the non–zero complex number xiy is the complex number uiv, where u = x x2 y2 and v = −y x2 y2 (If xiy 6= 0, then x 6= 0 or y 6= 0, so x2 y2 6= 0) From equations 51 and 52, we observe that addition and multiplication of complex numbers is performed just as.  Ex 25, 3Factorise the following using appropriate identities(i) 9x2 6xy y29x2 6xy y2 = 32 x2 6xy (y)2 = (3x)2 6xy (y)2 = (3x)2 2 (3x) (y) (y)2 Using Identity (a b)2 = a2 b2 2abWhere a = 3x , b = y= (3x y)2= (3x y) (3x y) (टीचू) Maths;.