answer:We begin by converting the decimal 0.abc into a fraction: [ 0.abc = frac{abc}{1000} = frac{1}{z}. ] Now, z must be a divisor of 1000, and z leq 15. The divisors of 1000 within this range are 1, 2, 4, 5, 8, 10. Let's consider each case: 1. z = 2: [ abc = 500 Rightarrow a+b+c = 5+0+0 = 5. ] 2. z = 4: [ abc = 250 Rightarrow a+b+c = 2+5+0 = 7. ] 3. z = 5: [ abc = 200 Rightarrow a+b+c = 2+0+0 = 2. ] 4. z = 8: [ abc = 125 Rightarrow a+b+c = 1+2+5 = 8. ] 5. z = 10: [ abc = 100 Rightarrow a+b+c = 1+0+0 = 1. ] Thus, the largest possible value of a+b+c among the valid values of z is boxed{8}.

answer:(1) We first find the derivative of g(x) with respect to x, which is g'(x)=3kx^{2}-1. * When kleqslant 0, we have g'(x)=3kx^{2}-1leqslant 0, which means that g(x) is monotonically decreasing in the interval (1,2), and thus does not satisfy the given condition. * When k > 0, we have g(x) decreasing in the interval (0, sqrt { frac{1}{3k}}) and increasing in the interval (sqrt { frac{1}{3k}},+∞). Since g(x) is not monotonic in the interval (1,2), we have 1 < sqrt { frac{1}{3k}} < 2, which gives us frac{1}{12} < k < frac{1}{3}. Combining both cases, we conclude that the range of k is frac{1}{12} < k < frac{1}{3}. (2) From the given inequality f(x)geqslant g(x)+x+2, we have kleqslant frac{(x-2)e^{x}}{x^{3}}. Let h(x)=frac{(x-2)e^{x}}{x^{3}}. Then, the derivative of h(x) is h'(x)=frac{(x^{2}-4x+6)e^{x}}{x^{4}}, which is always positive. Therefore, h(x) is monotonically increasing in the interval [1,+∞). The minimum value of h(x) in this interval is h(1)=-e. Hence, we have kleqslant -e, which implies that the maximum value of k is boxed{-e}.

answer:To solve the problem of choosing four edges in a cube such that any two among those four chosen edges have no common point, we need to consider the geometric properties of the cube and the constraints given. 1. **Case 1: The four edges are pairwise parallel.** - A cube has 12 edges, and these edges can be grouped into 3 sets of 4 parallel edges each. - For each set of parallel edges, we can choose 4 edges such that no two edges share a common vertex. - Since there are 3 sets of parallel edges, there are (3) ways to choose this set of edges. 2. **Case 2: Two pairs of parallel edges from opposite faces, and any two edges from different faces are skew.** - A cube has 6 faces, and these faces can be grouped into 3 pairs of opposite faces. - For each pair of opposite faces, we can choose 2 parallel edges from one face and 2 parallel edges from the opposite face. - Each face has 2 sets of parallel edges, so for each pair of opposite faces, there are (2 times 2 = 4) ways to choose the edges. - Since there are 3 pairs of opposite faces, there are (3 times 4 = 12) ways to choose this set of edges. Combining both cases, we get: - Case 1: (3) ways - Case 2: (12) ways Therefore, the total number of ways to choose four edges such that any two among those four chosen edges have no common point is: [ 3 + 12 = 15 ] The final answer is (boxed{15})

answer:To find the dividend, we can use the formula: Dividend = (Divisor × Quotient) + Remainder Given that the divisor is 17, the quotient is 9, and the remainder is 5, we can plug these values into the formula: Dividend = (17 × 9) + 5 Dividend = 153 + 5 Dividend = 158 Therefore, the dividend is boxed{158} .