answer:Step 1: Combinations for sum of 9 The combinations of two numbers summing to 9 are: - (1,8) - (2,7) - (3,6) - (4,5) - (5,4) - (6,3) - (7,2) - (8,1) Each pair can occur in two ways (one number on each die), giving a total of 8 times 2 = 16 favorable outcomes. Step 2: Total possible outcomes Each die has 8 faces, and each face can show any number from 1 to 8, making the total outcomes for two dice 8 times 8 = 64. Step 3: Probability calculation The probability of obtaining a sum of 9 is the number of favorable outcomes divided by the total number of outcomes: P(text{Sum of 9}) = frac{16}{64} = frac{1}{4} Conclusion: The probability that the sum is 9 is frac{1{4}}. The final answer is boxed{C}

answer:The determinant D' is represented as: [ D' = mathbf{u} cdot (mathbf{v} times mathbf{w}). ] The determinant for the matrix with column vectors mathbf{u} - mathbf{v}, mathbf{v} - mathbf{w}, mathbf{w} - mathbf{u} is: [ (mathbf{u} - mathbf{v}) cdot ((mathbf{v} - mathbf{w}) times (mathbf{w} - mathbf{u})). ] Expanding the cross product: [ (mathbf{v} - mathbf{w}) times (mathbf{w} - mathbf{u}) = mathbf{v} times mathbf{w} - mathbf{v} times mathbf{u} - mathbf{w} times mathbf{w} + mathbf{w} times mathbf{u}. ] Since mathbf{w} times mathbf{w} = mathbf{0}, we simplify to: [ mathbf{v} times mathbf{w} - mathbf{v} times mathbf{u} + mathbf{w} times mathbf{u}. ] Continuing with the dot product: [ (mathbf{u} - mathbf{v}) cdot (mathbf{v} times mathbf{w} - mathbf{v} times mathbf{u} + mathbf{w} times mathbf{u}) = mathbf{u} cdot mathbf{v} times mathbf{w} - mathbf{u} cdot mathbf{v} times mathbf{u} + mathbf{u} cdot mathbf{w} times mathbf{u} - mathbf{v} cdot mathbf{v} times mathbf{w} + mathbf{v} cdot mathbf{v} times mathbf{u} - mathbf{v} cdot mathbf{w} times mathbf{u}. ] Many terms become zero due to orthogonality (i.e., any vector dotted with a cross product involving itself is zero), leaving us with: [ mathbf{u} cdot mathbf{v} times mathbf{w} - mathbf{v} cdot mathbf{w} times mathbf{u}. ] By the scalar triple product properties, both terms are D' but with opposite signs, so: [ D'' = mathbf{u} cdot mathbf{v} times mathbf{w} - mathbf{v} cdot mathbf{w} times mathbf{u} = D' - D' = 0. ] Thus, the determinant of the matrix with column vectors mathbf{u} - mathbf{v}, mathbf{v} - mathbf{w}, mathbf{w} - mathbf{u} is boxed{0}.

answer:**Analysis:** The focus is on the intersection and its operations. By analyzing the given sets S = {x in mathbb{R} | x + 1 geq 2} and T = {-2, -1, 0, 1, 2}, we solve for sets S and T, and then calculate their intersection according to the definition and operational rules of intersections. Solving for S = {x in mathbb{R} | x + 1 geq 2} therefore S = {x in mathbb{R} | x geq 1}, given T = {-2, -1, 0, 1, 2}, thus, S cap T = {1, 2}. Therefore, the correct option is boxed{text{B}}. **Review:** This problem mainly examines the solution methods for quadratic inequalities and the operations of intersections and complements of sets. The operations among sets, including intersection, union, and complement, are common topics in high school exams.

answer:Consider the equilateral tetrahedron. Each face of the tetrahedron is an equilateral triangle. 1. **Observation about the triangles and their angles**: Notice that the faces of the tetrahedron are congruent equilateral triangles, and the sum of the planar angles at each vertex of the tetrahedron is 180^circ. Consequently, these equilateral triangles can be arranged in such a way that they tile the plane seamlessly (consider a plane tiling similar to a honeycomb pattern). 2. **Unwrapping the surface of the tetrahedron**: By unfolding these equilateral triangles and considering their tiling on a plane, each triangle in the tiling corresponds to a face of the tetrahedron. 3. **Placing the tetrahedron on the plane tiling**: Position one of the faces of the tetrahedron (an equilateral triangle) on a triangle in the tiling. If we imagine rolling the tetrahedron along the edges from one triangle to another, it will always land on a triangle in the tiling. Additionally, any point on the plane can be specified by its coordinates obtained by translating along the vectors that define the edges of the triangles, effectively creating a lattice. 4. **Connecting points on the tiling**: If we mark two points K and M on this plane, connecting them with a straight line segment represents the shortest path between them on the plane surface. This line segment can then be mapped back onto the surface of the tetrahedron. 5. **Detailed geometry**: - Place the first ant at point K, and consider translational equivalents of K by vectors defining the side lengths of the triangles. - We form a tiling or a coordinates system where the same triangular lattice structure is applied. 6. **Finding shortest distance on the tetrahedron**: Suppose the second ant is at point M. From the triangle connecting the points K_{1}, K_{2}, K_{3} (lattice points obtained through translations), one of the distances MK_{1}, MK_{2}, or MK_{3} is always less than or equal to the radius of the circumcircle of the face of the tetrahedron. We construct the circumcircle as follows: - The center O of this circumcircle is the same for all faces. - The triangle OK_{1}K_{2}, OK_{1}K_{3}, and OK_{2}K_{3} holds the point M, and the distance to one of the vertices of the triangle from M will be bounded by the side length of the triangle. 7. **Conclusion**: The shortest path connecting points on the surface of the tetrahedron maps to a straight line connecting equivalent points on the planar tiling (through roll-overs). Therefore, considering the largest distance involved (which translates directly to the radius of the circumcircle of one of the equilateral faces of the tetrahedron), the shortest path on the surface of the tetrahedron matches the diameter of the circumscribed circle around the face. # Conclusion: The shortest path is equivalent to the planar distance considering the surface of the tetrahedron, where the maximum distance a point might be from another is along the diameter of the circumscribed circle around a face. Thus, we conclude the shortest path as calculated above. boxed{}