answer:Let ( x ) be the list price of the item. 1. **Alice's Selling Price and Commission:** - Alice sells the item for ( x - 15 ). - Her commission is ( 15% ) of ( x - 15 ), mathematically written as ( 0.15(x - 15) ). 2. **Bob's Selling Price and Commission:** - Bob sells the item for ( x - 25 ). - His commission is ( 25% ) of ( x - 25 ), which is ( 0.25(x - 25) ). 3. **Equating Commissions:** - Equate Alice's commission to Bob's commission: [ 0.15(x - 15) = 0.25(x - 25) ] 4. **Solving the Equation:** - Simplifying and solving for ( x ): [ 0.15x - 2.25 = 0.25x - 6.25 ] [ 0.15x - 0.25x = -6.25 + 2.25 ] [ -0.10x = -4 ] [ x = frac{-4}{-0.10} = 40 ] 5. **Conclusion:** - The list price of the item is ( 40 ). The final answer is boxed{textbf{(C) } 40}boxed{}

answer:Since the ellipse C: frac{x^2}{a^2} + frac{y^2}{4} = 1 (a > 2), we have b=2 and c= sqrt{a^2-4}. Given that P is a point on the ellipse and angle F_1PF_2 = 60^circ, with F_1 and F_2 being the left and right foci, we have |F_1P| + |PF_2| = 2a and |F_1F_2| = 2sqrt{a^2-4}, thus |F_1F_2|^2 = (|PF_1| + |PF_2|)^2 - 2|F_1P||PF_2| - 2|F_1P|cdot|PF_2|cos 60^circ = 4a^2 - 3|F_1P|cdot|PF_2| = 4a^2 - 16, therefore |F_1P|cdot|PF_2| = frac{16}{3}. Hence, the area of triangle PF_1F_2 is S_{triangle PF_1F_2} = frac{1}{2}|F_1P|cdot|PF_2|sin 60^circ = frac{1}{2} times frac{16}{3} times frac{sqrt{3}}{2} = frac{4sqrt{3}}{3}. Therefore, the correct answer is boxed{C}. According to the problem, in triangle F_1PF_2 with angle F_1PF_2 = 60^circ and |F_1P| + |PF_2| = 2a, we find |F_1F_2| = 2sqrt{a^2-4}. Using the cosine theorem, we can calculate the value of |F_1P|cdot|PF_2|, and thus find the area of triangle PF_1F_2. This problem examines the simple properties of an ellipse, the application of the cosine theorem, and the formula for the area of a triangle, making it a medium-level question.

answer:Let's denote the salaries for the months of January, February, March, April, and May as J, F, M, A, and May respectively. According to the information given: (J + F + M + A) / 4 = 8000 => J + F + M + A = 8000 * 4 => J + F + M + A = 32000 ...(1) (F + M + A + May) / 4 = 8700 => F + M + A + May = 8700 * 4 => F + M + A + May = 34800 ...(2) We also know that J = 3700 ...(3) Now, let's subtract equation (1) from equation (2) to find the salary for May: (F + M + A + May) - (J + F + M + A) = 34800 - 32000 => May - J = 2800 => May - 3700 = 2800 => May = 3700 + 2800 => May = 6500 So, the person earned Rs. boxed{6500} in the month of May.

answer:1. **Assertion and Initial Conditions:** Let ( P(x, y) ) denote the assertion ( |f(x+y)| geq |f(x) + f(y)| ). - For ( P(0, 0) ): [ |f(0+0)| geq |f(0) + f(0)| implies |f(0)| geq |2f(0)| ] This implies ( |f(0)| geq 2|f(0)| ). The only solution to this inequality is ( f(0) = 0 ). 2. **Odd Function Property:** - For ( P(x, -x) ): [ |f(x + (-x))| geq |f(x) + f(-x)| implies |f(0)| geq |f(x) + f(-x)| ] Since ( f(0) = 0 ), we have: [ 0 geq |f(x) + f(-x)| ] This implies ( f(x) + f(-x) = 0 ), hence ( f(-x) = -f(x) ). Therefore, ( f ) is an odd function. 3. **Analyzing ( P(x+y, -x) ) and ( P(x+y, -y) ):** - For ( P(x+y, -x) ): [ |f((x+y) + (-x))| geq |f(x+y) + f(-x)| implies |f(y)| geq |f(x+y) - f(x)| ] - For ( P(x+y, -y) ): [ |f((x+y) + (-y))| geq |f(x+y) + f(-y)| implies |f(x)| geq |f(x+y) - f(y)| ] 4. **Summing the Inequalities:** - Squaring both sides of the inequalities: [ f(y)^2 geq (f(x+y) - f(x))^2 implies f(y)^2 geq f(x+y)^2 - 2f(x+y)f(x) + f(x)^2 ] [ f(x)^2 geq (f(x+y) - f(y))^2 implies f(x)^2 geq f(x+y)^2 - 2f(x+y)f(y) + f(y)^2 ] - Adding these inequalities: [ f(x)^2 + f(y)^2 geq f(x+y)^2 - 2f(x+y)f(x) + f(x)^2 + f(x+y)^2 - 2f(x+y)f(y) + f(y)^2 ] [ 2(f(x)^2 + f(y)^2) geq 2f(x+y)^2 - 2f(x+y)(f(x) + f(y)) ] Dividing by 2: [ f(x)^2 + f(y)^2 geq f(x+y)^2 - f(x+y)(f(x) + f(y)) ] Rearranging: [ f(x+y)(f(x) + f(y)) geq f(x+y)^2 ] This implies: [ f(x+y) cdot (f(x) + f(y)) geq f(x+y)^2 ] Since ( f(x+y) neq 0 ), we can divide both sides by ( f(x+y) ): [ f(x) + f(y) geq f(x+y) ] 5. **Conclusion:** - If ( f(x+y) geq 0 ): [ f(x) + f(y) geq f(x+y) geq 0 ] Thus, ( |f(x) + f(y)| = f(x) + f(y) geq f(x+y) = |f(x+y)| ). - If ( f(x+y) leq 0 ): [ f(x) + f(y) leq f(x+y) leq 0 ] Thus, ( |f(x) + f(y)| = -(f(x) + f(y)) leq -f(x+y) = |f(x+y)| ). Therefore, the equality ( |f(x+y)| = |f(x) + f(y)| ) always holds. The final answer is ( boxed{ |f(x+y)| = |f(x) + f(y)| } ).