answer:Let event A represent "not both Archer A and Archer B hit the bullseye"; let event B represent "both Archer A and Archer B hit the bullseye". Then, event B is the complementary event of A. Since P(B) = dfrac{1}{3} times dfrac{1}{2} = dfrac{1}{6}, we have P(A) = 1 - P(B) = 1 - dfrac{1}{6} = dfrac{5}{6}. Therefore, the answer is boxed{dfrac{5}{6}}. This solution involves defining two events and determining that A and B are complementary events; calculating the probability of event B, and using the formula for the probability of complementary events to find the probability of event A. This problem tests the formula for the probability of complementary events and the steps for calculating the probability of an event.

answer:1. First, find t(5): [ t(5) = sqrt{4 cdot 5 + 2} = sqrt{20 + 2} = sqrt{22} ] 2. Evaluate f(5) using the updated t(5): [ f(5) = 7 - 2t(5) = 7 - 2sqrt{22} ] 3. Finally, calculate t(f(5)): [ t(f(5)) = t(7 - 2sqrt{22}) = sqrt{4(7-2sqrt{22}) + 2} = sqrt{28 - 8sqrt{22} + 2} = sqrt{30 - 8sqrt{22}} ] [ boxed{sqrt{30 - 8sqrt{22}}} ]

answer:1. **Define the problem and the cube:** - We are given a cube with side length (1). - We need to place the vertices of a triangle on the faces of the cube such that the shortest side of the triangle is as long as possible. 2. **Identify possible distances between vertices:** - The possible distances between any two vertices of the cube are (1), (sqrt{2}), and (sqrt{3}). - (1): Distance between two adjacent vertices (edge of the cube). - (sqrt{2}): Distance between two vertices on the same face but not adjacent (diagonal of a face). - (sqrt{3}): Distance between two vertices that are not on the same face (space diagonal of the cube). 3. **Analyze the triangle sides:** - A triangle cannot have all three sides equal to (sqrt{3}) because there are not three vertices in the cube that are all (sqrt{3}) apart. - If a triangle has one side of length (sqrt{3}), the other two sides must be (sqrt{2}) and (1) respectively, which does not maximize the shortest side. 4. **Consider a triangle with all sides (sqrt{2}):** - If we place the vertices of the triangle such that each pair of vertices is (sqrt{2}) apart, we get an equilateral triangle with side length (sqrt{2}). - This configuration is possible by placing the vertices on the centers of three adjacent faces of the cube. 5. **Verify the configuration:** - Consider the vertices at ((0,0,0)), ((1,1,0)), and ((0,1,1)). - Distance between ((0,0,0)) and ((1,1,0)) is (sqrt{(1-0)^2 + (1-0)^2 + (0-0)^2} = sqrt{2}). - Distance between ((0,0,0)) and ((0,1,1)) is (sqrt{(0-0)^2 + (1-0)^2 + (1-0)^2} = sqrt{2}). - Distance between ((1,1,0)) and ((0,1,1)) is (sqrt{(1-0)^2 + (1-1)^2 + (0-1)^2} = sqrt{2}). 6. **Conclusion:** - The triangle with all sides of length (sqrt{2}) is the configuration where the shortest side is the longest possible. The final answer is ( boxed{ sqrt{2} } ).

answer:To find the average number of trees planted by the three classes, we first need to find out how many trees each class planted. Class (A) planted 35 trees. Class (B) planted 6 more trees than class (A), so class (B) planted 35 + 6 = 41 trees. Class (C) planted 3 less trees than class (A), so class (C) planted 35 - 3 = 32 trees. Now, we add the number of trees planted by all three classes together and divide by the number of classes to find the average: Total trees planted by all three classes = 35 (A) + 41 (B) + 32 (C) = 108 trees. Average number of trees = Total trees / Number of classes Average number of trees = 108 trees / 3 classes = 36 trees. So, the average number of trees planted by the three classes is boxed{36} trees.