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 Despite the fact that the same variables appear in every term, this expression does not factorize into one term There is no common factor, and no common bracket EAch term can be factored by differemce of squares xy(x^2y^2) yz(y^2z^2) zx(z^2x^2) =xy(xy)(xy) yz(yz)(yz) zx(zx)(zx) The only other option would be to multiply out the brackets and try aC a^10 a^5 1;Example 2 if x = 10 and y is 4 (10 4) 2 = 10 2 2·10·4 4 2 = 100 80 16 = 36 The opposite is also true 25 a 4a 2 = 5 2 2·2·5 (2a) 2 = (5 2a) 2 Consequences of the above formulas

 Transcript Ex 42, 9 By using properties of determinants, show that 8 (x&x2&yz@y&y2&zx@z&z2&xy) = (x – y) (y – z) (z – x) (xy yz zx) Solving LHS 8 (𝑥&𝑥^2&𝑦𝑧@𝑦&𝑦^2&𝑧𝑥@𝑧&𝑧^2&𝑥𝑦) Applying R1→ R1 – R2 = 8 (𝑥−𝑦&𝑥^2−𝑦^2&𝑦𝑧−𝑥𝑧@𝑦&𝑦^2&𝑧𝑥@𝑧&𝑧^2On the right side of the yaxis (b) 5 =0 q=p 2 q= p 2 x y 0 4 −5 q 16 x = lnt, y = p t, t 1 Solution (a) x = lnt =) t = ex =) y = p ex = ex=2, x 0 (b) x 1 0 y 22 Describe the motion of a particle with position (x;y) as t varies in the given interval x = cos2 t;Solution First note that if z2XY, then z= xywith x2Xand y2Y, so z= xy ab Hence a bis an upper bound for X Y Next, let cbe any upper bound for X Y Suppose for a contradiction that c0 By Exercise 51, there exists x2Xsuch that a 2

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Then the random variable Z = min{X,Y } is also exponentially distributed True Here Z is a derived random variable defined as Z = min{X,Y } We can obtain the PDF of Z by first determining its CDF and then taking the derivative The CDF of Z is given by FZ(z) = P(Z ≤ z) = P(min{X,Y } ≤ z)(b) Verify XY'ZXY'Z'X'Y'Z = XY' Y'Z algebraically (c) Write the dual of the Boolean Expression (B'C)A (d) Obtain a simplified form for a Boolean Expression F(U,V,W,Z,) = ∑(0,2,3,4,7,9,10,13,14,15) (e) Draw the logic circuit for a half adder (Out of Syllabus now) 02 6

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