For each vector u 2 V, the norm (also called the length) of u is deflned as the number kuk= p hu;ui If kuk = 1, we call u a unit vector and u is said to be normalized For any nonzero vector v 2 V, we have the unit vector v^ = 1 kvk v This process is called normalizing v Let B = u1;u2;;un be a basis of an ndimensional inner product space VFor vectors u;v 2 V, write
The identity (x^2 y^2)^2=(x^2-y^2)^2 (2xy)^2 can be used to generate pythagorean triples-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 factorUse the identity (x^2y^2)^2=(x^2−y^2)^2(2xy)^2 to determine the sum of the squares of two numbers if theChứng minh (x^2y^2)^2(2xy)^2=(xy)^2(xy)^2 chúng minh đẳng thức (x^2y^2)^2(2xy)^2=(xy)^2(xy)^2 Theo dõi Vi phạm YOMEDIA Toán 8 Bài 3 Trắc nghiệm Toán 8 Bài 3 Giải bài tập Toán 8 Bài 3 Trả lời (1) Ta có (x 2 y 2) 2(2xy) 2The algebraic identities for class 9 consist of identities of all the algebraic formulas and expressions You must have learned algebra
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Pythagorean triples are given by the formula x^2y^2, 2xy, and x^2y^2 Use the formulas for the Pythagorean triples to find a right triangle with leg lengths of 16 and an odd number Show all of your work for full creditAnswer (1 of 2) Let's solve this by graphing because I'm feeling a bit lazy with the latex My apologies You can see by the graph that we have 3 solutions for x and y We have (0,2),(0,\sqrt{2}), and (0,\sqrt{2}) The last two solutions solve the
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