answer:Let f(x)= frac {ln x}{x} (xgeqslant e), then f′(x)= frac {1-ln x}{x^{2}}leqslant 0, therefore the function f(x) is monotonically decreasing on [e,+infty), therefore a= frac {ln 3}{3} > b= frac {ln 5}{5} > c= frac {ln 6}{6}, which means a > b > c. Therefore, the correct answer is: boxed{B}. By setting f(x)= frac {ln x}{x} (xgeqslant e), and then finding f′(x)= frac {1-ln x}{x^{2}}leqslant 0, we can conclude that the function f(x) is monotonically decreasing on [e,+infty). This problem tests the ability to use derivatives to study the monotonicity of functions, as well as reasoning and computational skills, and is considered a medium-level question.

answer:Let's denote the cost price of the bicycle for A as CP_A. A sells the bicycle to B at a profit of 20%. This means that B pays A 120% of CP_A. Let's denote the selling price of A to B as SP_AB. So we have: SP_AB = CP_A + (20/100 * CP_A) SP_AB = CP_A * (1 + 20/100) SP_AB = CP_A * 1.2 Now, B sells the bicycle to C at a profit of 25%. This means that C pays B 125% of SP_AB. Let's denote the selling price of B to C as SP_BC. So we have: SP_BC = SP_AB + (25/100 * SP_AB) SP_BC = SP_AB * (1 + 25/100) SP_BC = SP_AB * 1.25 We are given that C pays Rs. 225 for the bicycle, which is SP_BC. So we have: SP_BC = 225 Now we can substitute SP_AB * 1.25 for SP_BC: SP_AB * 1.25 = 225 We know that SP_AB is 1.2 times CP_A, so we can substitute 1.2 * CP_A for SP_AB: 1.2 * CP_A * 1.25 = 225 Now we can solve for CP_A: 1.2 * 1.25 * CP_A = 225 1.5 * CP_A = 225 CP_A = 225 / 1.5 CP_A = 150 Therefore, the cost price of the bicycle for A is Rs. boxed{150} .

answer:Since the maximum of the quadratic occurs at (2, 7), the vertex formula allows for y = a(x - 2)^2 + 7, where a is negative due to the maximum at that vertex. We know the graph passes through (0, -7), so substituting x = 0 into the vertex form: [ -7 = a(0 - 2)^2 + 7 ] [ -7 = 4a + 7 ] [ -14 = 4a ] [ a = -frac{14}{4} = -3.5 ] Thus, the equation of the quadratic is: [ y = -3.5(x - 2)^2 + 7 ] To find the value of y when x = 5: [ y = -3.5(5 - 2)^2 + 7 ] [ y = -3.5(3)^2 + 7 ] [ y = -3.5(9) + 7 ] [ y = -31.5 + 7 ] [ y = -24.5 ] Therefore, m = boxed{-24.5}.

answer:1. Let's denote the initial total weight of the honey pot (including honey) as ( W ). 2. We are given that the pot without honey weighs 200 grams. Therefore, the weight of the honey initially is: [ W_{honey} = W - 200 ] 3. According to the problem, every time Winnie-the-Pooh tastes the honey, the total weight becomes half of the previous weight. Let's denote the initial total weight as ( W = W_0 ). When Winnie-the-Pooh tastes the honey the first time, the new weight of the pot with honey becomes: [ W_1 = frac{W_0}{2} ] 4. He tastes the honey a second time, and the new weight is: [ W_2 = frac{W_1}{2} = frac{frac{W_0}{2}}{2} = frac{W_0}{4} ] 5. When he tastes the honey a third time, the weight is: [ W_3 = frac{W_2}{2} = frac{frac{W_0}{4}}{2} = frac{W_0}{8} ] 6. Upon the fourth tasting, the weight becomes: [ W_4 = frac{W_3}{2} = frac{frac{W_0}{8}}{2} = frac{W_0}{16} ] 7. At this point, it is stated that there is no honey left. Thus ( W_4 ) should be equal to the weight of the empty pot: [ W_4 = 200 text{ grams} ] 8. To find ( W_0 ), we solve: [ frac{W_0}{16} = 200 ] Hence, [ W_0 = 200 times 16 = 3200 text{ grams} ] 9. The initial weight of the honey is: [ W_{honey} = W_0 - 200 = 3200 - 200 = 3000 text{ grams} ] # Conclusion: [ boxed{3000 text{ grams}} ]