answer:To solve this problem, we need to consider the conditions under which rearranging the digits of "124669" results in an even number. An even number ends with 0, 2, 4, or 6. Since 0 is not a digit of the given number, we only consider 2, 4, and 6 as possible units digits. 1. **Case 1: Units digit is 2 or 4** For an even number with the units digit being either 2 or 4, we have two choices for the units digit. After choosing the units digit, we are left with 5 digits, out of which we need to choose 3 digits and arrange them along with the fixed units digit. This can be done in {C}_{2}^{1}{A}_{5}^{3} ways, where {C}_{n}^{r} represents the combination of n items taken r at a time and {A}_{n}^{r} represents the arrangement of n items taken r at a time. Therefore, the calculation goes as follows: {C}_{2}^{1}{A}_{5}^{3} = 2 times 5 times 4 times 3 = 120 2. **Case 2: Units digit is 6** When the units digit is 6, we have already fixed the units digit, and we need to arrange the remaining 5 digits. This can be done in {A}_{5}^{5} ways, which is calculated as: {A}_{5}^{5} = 5 times 4 times 3 times 2 times 1 = 120 3. **Total Different Even Numbers** To find the total number of different even numbers that can be formed, we add the numbers obtained from both cases: 120 + 120 = 240 Therefore, the total number of different even numbers that can be formed by rearranging the digits of "124669" is boxed{240}. Hence, the correct answer is boxed{D}.

answer:Based on the problem description: 1. **Tortoise's Movement:** - Constant pace from start to finish implies a graph that is a straight line rising steadily. 2. **Hare's Movement:** - Initial burst of speed (sharp rise), then a flat line indicating a pause, another sharp rise, a second flat line for the second pause, followed by a final sharp rise up to just before the tortoise's finishing point. 3. **Outcome:** - The tortoise's graph should end slightly higher on the y-axis (finishing first) despite its consistent, lower slope. Review provided graph options: - **Option A:** Does not show siamesed rest periods for the hare. - **Option B:** Correctly shows the hare with two pauses (flat lines), quick starts, and stopping just before reaching the tortoise line at the finish. - **Option C:** Incorrect, does not display any distinct stops for rest. - **Option D:** Similar issues as others with lack of adequate stops. Conclusion: Based on the transformations and adaptations to suit the new problem statement, the best choice is: textbf{Option B} The final answer is boxed{B. The hare has two stops, and the tortoise finishes first.}

answer:(1) The function f(x) is given by f(x)=sqrt{3}cos 2x+2sin (frac{3π}{2}+x)sin (π-x). This simplifies to f(x)=sqrt{3}cos 2x-2cos xsin x and further to f(x)=sqrt{3}cos 2x-sin 2x. This can be written as f(x)=-2sin (2x-frac{π}{3}). From frac{π}{2}+2kπleqslant 2x-frac{π}{3}leqslant frac{3π}{2}+2kπ, we get the increasing interval as [frac{5π}{12}+kπ, frac{11π}{12}+kπ], kin mathbb{Z}. (2) From f(x)=-sqrt{3}, we get sin (2A-frac{π}{3})=frac{sqrt{3}}{2}. Since Ain (0, frac{π}{2}), we have A=frac{π}{3}. Using the cosine rule, we get a^{2}=b^{2}+c^{2}-2bccos A, which simplifies to 9=b^{2}+c^{2}-bc. This implies bcleq 9. Let h be the height on side BC. By the equal area property of triangles, we have frac{1}{2}ah=frac{1}{2}bcsin A. This simplifies to 3h=frac{sqrt{3}}{2}bc. Since bcleq 9, we have hleq frac{3sqrt{3}}{2}. Therefore, the maximum value of h is boxed{frac{3sqrt{3}}{2}}.

answer:Solution: (1) The axis of symmetry for f(x)=x^{2}-2x+2 is x=1, which implies that f(x) is increasing in [1,2], thus, the maximum value of f(x) is f(2)=2, and the minimum value is f(1)=1, hence, we get d(1,2)=2-1=1; (2) If f(x)=x+ frac {m}{x}, and d(1,2)neq |f(2)-f(1)|, it implies that f(x) is not a monotonic function, when m=0, f(x) is an increasing function in [1,2], when m < 0, f(x) is an increasing function in [1,2], when m > 0, since f(x) has an extreme value at x= sqrt {m}, then 1 < sqrt {m} < 2, we get 1 < m < 4, thus, the range of m is (1,4). Therefore, the answers are: boxed{1}, boxed{(1,4)}. (1) By finding the axis of symmetry of f(x), we can determine the monotonicity of f(x) in [1,2] to find the extreme values, which leads to the solution; (2) Based on the problem statement, we know that f(x) is not monotonic in [1,2]. Discussing the cases when m=0, m < 0, and m > 0, and combining with the monotonicity of the hook function, we can find the range of m. This problem examines the method of finding the maximum and minimum values of functions, emphasizing the use of monotonicity of quadratic functions and hook functions, and the idea of discussing cases and computational skills, classified as a medium-level question.