answer:(1) Since the function f(x)= frac{ax+b}{1+x^2} is defined on (-1,1), is an odd function, and fleft( frac{1}{2} right)=frac{2}{5}, we have the following system of equations: begin{cases} f(0)= frac{a times 0 + b}{1+0^2} = 0, fleft( frac{1}{2} right)= frac{frac{1}{2}a + b}{1+left( frac{1}{2} right)^2} = frac{2}{5}. end{cases} Solving the system, we obtain a=1 and b=0. Therefore, the function is f(x)=frac{x}{1+x^2}. (2) Given that f(x)=frac{x}{1+x^2} is an increasing function on (-1,1) and an odd function, and we have the inequality f(x-1)+f(x) < 0, then f(x-1) < -f(x) = f(-x). Consequently, we have -1 < x-1 < -x < 1. Solving this inequality, we obtain 0 < x < frac{1}{2}. Thus, the solution set for the inequality is boxed{left( 0,frac{1}{2} right)}.

answer:The function h(x) = f(-3x) is defined if and only if -3x is within the domain of f(x). So we require -9 leq -3x leq 6. To find the domain for x, solve these inequalities. Divide each part of the inequality by -3 (and remember to reverse the inequality signs because of division by a negative number): 3 geq x geq -2. Thus, h(x) is defined for x in the interval [-2, 3]. The domain of h(x) is, therefore, boxed{[-2, 3]}.

answer:1. **Define Points and Assumptions:** Consider triangle ABC with points P and Q on sides AB and AC respectively such that angle APC = angle AQB = 45^circ. 2. **Construct New Points:** - Draw the perpendicular from P to AB intersecting BQ at S. - Draw the perpendicular from Q to AC intersecting CP at R. - Let D be a point on BC such that AD perp BC. 3. **Intersection of Extensions:** Extend lines QR and AD to intersect at point E. 4. **Angles and Cyclic Quadrilaterals:** - Observe that angle ABQ = 135^circ - angle BAC = angle ACP, implying points B, P, Q, and C are concyclic. - Therefore, angle APQ = angle ACB due to the inscribed angle theorem. 5. **Evaluating angle AEQ:** - Since AD perp BC, we have angle AEQ = 90^circ - angle EAC = angle ACB, showing that A, P, E, Q are concyclic. - Consequently, angle APE = 180^circ - angle AQE = 90^circ, which implies PE perp AB. 6. **Coincidence of PE and PS:** Given that PE perp AB and PS perp AB, PE and PS must coincide. Thus, PS, AD, and QR concur at a single point E. 7. **Proving SR parallel BC:** With angle BQE = 90^circ - 45^circ = 45^circ = angle CPE, points P, Q, R, S are concyclic, from which angle QPR = angle QSR. - As B, P, Q, C are concyclic, angle QPR = angle QBC. - Therefore, angle QBC = angle QSR, proving that SR parallel BC. # Conclusion: PS, AD, and QR concur at a single point, and SR parallel BC. blacksquare

answer:To find the probability that either event A or event B occurs, we can use the formula for the probability of the union of two events: P(A or B) = P(A) + P(B) - P(A and B) We are given: P(A) = 0.45 (probability that event A occurs) P(B) = 0.4 (probability that event B occurs) P(A and B) = 0.25 (probability that both events A and B occur) Now we can plug these values into the formula: P(A or B) = 0.45 + 0.4 - 0.25 P(A or B) = 0.85 - 0.25 P(A or B) = 0.60 So the probability that either event A or event B occurs is boxed{0.60} or 60%.