answer:(1) The derivative of the function f(x)= frac{x-a}{(x+a)^{2}} is f′(x)= frac{3a-x}{(x+a)^{3}}, where x neq -a. The slope of the tangent at the point (0,f(0)) is therefore frac{3}{a^{2}}. Since the tangent is parallel to the line y=3x-2, which has a slope of 3, we equate them: frac{3}{a^{2}}=3. Solving this equation yields a=pm1. [boxed{a=pm1}] (2) For any x_1 in the domain, the existence of x_2 satisfying f(x_2) < f(x_1) implies that f(x) has no minimum when x neq -a. If a=0, f(x)= frac{1}{x} which clearly has no minimum value. Thus, this case is satisfied. For a>0, the derivative of f(x) is f′(x)= frac{3a-x}{(x+a)^{3}}. f(x) is decreasing on (-infty, -a), increasing on (-a, 3a), and decreasing again on (3a, +infty). Hence, f(x) has a local maximum at x=3a. When x>a, f(x)>0; when x<a, f(x)<0. For x_1<a and x_2 neq -a, the condition can always be met. When x_1<-a, f(x) is decreasing on (-infty, -a), and x_1 < x_1+ frac{1}{2}|x_1+a| < -a, hence f(x_1) > f(x_1 + frac{1}{2}|x_1+a|), and thus there exists x_2=x_1+ frac{1}{2}|x_1+a|, such that f(x_2) < f(x_1). Similarly, when -a<x_1<a, let x_2=x_1 - frac{1}{2}|x_1+a|, which also satisfies f(x_2) < f(x_1). Therefore, when a>0, the condition f(x_2) < f(x_1) is always met. For a<0, f(x) is decreasing on (-infty, 3a), increasing on (3a, a), and decreasing on (-a, +infty), and f(x) has a local minimum at x=3a. f(x)_{text{min}} = f(3a), and when x_1 = 3a, there exists no x_2 such that f(x_2) < f(x_1). Therefore, the range of a is [0, +infty). [boxed{a in [0, +infty)}]

answer:Let's denote Johnny's current age as J. According to the problem, J = 8. Johnny says that in 2 years, he will be twice as old as he was a certain number of years ago. Let's denote the number of years ago as X. In 2 years, Johnny's age will be J + 2. According to Johnny's statement, at that time, he will be twice as old as he was X years ago. So, the age he was X years ago is (J - X). The equation based on Johnny's statement is: J + 2 = 2 * (J - X) Now we know that J = 8, so we can substitute J with 8 in the equation: 8 + 2 = 2 * (8 - X) Solving for X: 10 = 16 - 2X 2X = 16 - 10 2X = 6 X = 6 / 2 X = 3 So, Johnny was referring to boxed{3} years ago.

answer:Solution: (Ⅰ) Since cos 2A + 2sin^2(pi + B) + 2cos^2left(frac{pi}{2} + Cright) - 1 = 2sin Bsin C, therefore sin^2B + sin^2C - sin^2A = sin Bsin C. By the Law of Sines, we have b^2 + c^2 - a^2 = bc. By the Law of Cosines, we get cos A = frac{b^2 + c^2 - a^2}{2bc} = frac{1}{2}. Since 0 < A < pi, therefore A = frac{pi}{3}. boxed{A = frac{pi}{3}} (Ⅱ) Since a^2 = b^2 + c^2 - 2bccos A = 16 + 25 - 2 times 4 times 5 times frac{1}{2} = 21, therefore a = sqrt{21}. By the Law of Sines, frac{a}{sin A} = frac{b}{sin B}, we get frac{sqrt{21}}{sin frac{pi}{3}} = frac{4}{sin B}. Solving this, we find sin B = frac{2sqrt{7}}{7}. boxed{sin B = frac{2sqrt{7}}{7}}

answer:1. First, solve for y in terms of x: [ y = frac{200 - 7x}{4} ] 2. Express xy in terms of x: [ xy = x cdot frac{200 - 7x}{4} = frac{200x - 7x^2}{4} ] This is a quadratic expression in terms of x, and it opens downwards (since the coefficient of x^2 is negative). 3. Find the vertex of the parabola to determine the maximum value of xy: [ x = -frac{b}{2a} = -frac{200}{2 times -7} = frac{100}{7} approx 14.29 ] Since x must be an integer, we test x = 14 and x = 15. 4. Check for integer values of y: - For x = 14, y = frac{200 - 7 times 14}{4} = frac{102}{4} = 25.5 (not an integer). - For x = 15, y = frac{200 - 7 times 15}{4} = frac{95}{4} = 23.75 (not an integer). Testing further, we find: - For x = 13, y = frac{200 - 7 times 13}{4} = frac{109}{4} = 27.25 (not an integer). - For x = 12, y = frac{200 - 7 times 12}{4} = frac{116}{4} = 29 (an integer). 5. Calculate xy for x = 12, y = 29: [ xy = 12 times 29 = boxed{348} ]