answer:First, let's calculate the time it will take for Tina to clean the remaining 29 keys. Since it takes 7 minutes to clean one key, the time to clean 29 keys would be: 29 keys * 7 minutes/key = 203 minutes Next, we need to calculate the total drying time. Since each key needs 10 minutes to dry and she has 30 keys in total (including the one she already cleaned), the drying time would be: 30 keys * 10 minutes/key = 300 minutes Now, let's calculate the total break time. Tina takes a break after every 5 keys cleaned. Since she has 30 keys, she will take a break 6 times (30 keys / 5 keys per break). The total break time would be: 6 breaks * 3 minutes/break = 18 minutes Finally, we add the time it will take for Tina to complete her assignment: 20 minutes Adding all these times together gives us the total time Tina will spend: Cleaning time + Drying time + Break time + Assignment time 203 minutes + 300 minutes + 18 minutes + 20 minutes = 541 minutes Therefore, it will take Tina a total of boxed{541} minutes to clean the remaining keys, wait for them to dry, take planned breaks, and finish her assignment.

answer:Since a_{n+1} = frac{2a_n}{a_n + 2}, we have frac{1}{a_{n+1}} = frac{1}{a_n} + frac{1}{2}, which implies frac{1}{a_{n+1}} - frac{1}{a_n} = frac{1}{2}. Therefore, left{frac{1}{a_n}right} forms an arithmetic sequence with the first term frac{1}{2} and common difference frac{1}{2}. Thus, frac{1}{a_n} = frac{1}{2} + (n-1) times frac{1}{2} = frac{n}{2}. Hence, a_n = frac{2}{n}. So, the answer is: boxed{frac{2}{n}}. By transforming the recursive relation through taking the reciprocal, and according to the definition of an arithmetic sequence, we find that left{frac{1}{a_n}right} is an arithmetic sequence. Using the general formula for an arithmetic sequence, we find frac{1}{a_n}, and further derive a_n. This problem tests the ability to construct a new sequence to find the general term and the general formula of an arithmetic sequence.

answer:To show that the first player can always win in the game of Kayles, we will consider two cases based on the parity (odd or even) of the number of pins. The strategy involves mirroring the moves of the second player after the first player's initial move. Let's detail the steps: 1. **Case 1: Odd number of pins** - Suppose the total number of pins, ( n ), is odd. - The first player knocks over the middle pin. The position of the middle pin is given by ( leftlceil frac{n}{2} rightrceil ). For example, if there are 5 pins (which is odd): [ 1 2 3 4 5 ] The first player knocks over the 3rd pin: [ 1 2 _ 4 5 ] 2. **Case 2: Even number of pins** - Suppose the total number of pins, ( n ), is even. - The first player knocks over the two middle pins. The positions of the two middle pins are ( frac{n}{2} ) and ( frac{n}{2} + 1 ). For example, if there are 6 pins (which is even): [ 1 2 3 4 5 6 ] The first player knocks over the 3rd and 4th pins: [ 1 2 _ _ 5 6 ] 3. **Mirroring strategy** - After the first player's initial move, the remaining pins will be in two symmetrical sections (left and right). - The strategy for the first player is now to mirror the moves of the second player. If the second player knocks over a pin or two adjacent pins on one side, the first player makes a move that exactly mirrors it on the other side. This strategy works because it ensures that the first player always has a move until the very end. The symmetrical removal of pins guarantees that the first player can always respond to any move made by the second player by making a corresponding move on the mirrored side. - For instance, if after the first player's initial move, the second player knocks over pin 1 on the left section, then the first player knocks over the corresponding pin in the mirrored position (which is the last pin on the right section). By maintaining this mirroring strategy, the first player forces the second player to make the last move in each symmetrical section. Since the game ends when the last pin(s) are knocked over, the second player is forced to knock over pins in the final move and thus loses. # Conclusion: The strategy of removing the middle pin(s) initially and then mirroring the second player's moves guarantees the first player always has a winning strategy. [ boxed{text{The first player can always win.}} ]

answer:Given that in triangle triangle ABC, we have the equation |2sin A-1|+|frac{sqrt{2}}{2}-cos B|=0. Let's solve this step by step: 1. Since the sum of two absolute values equals zero, it implies that each absolute value must be zero individually. Therefore, we have: - |2sin A-1|=0 - |frac{sqrt{2}}{2}-cos B|=0 2. Solving each equation separately: - For |2sin A-1|=0, we get 2sin A-1=0 Rightarrow sin A=frac{1}{2}. - For |frac{sqrt{2}}{2}-cos B|=0, we get frac{sqrt{2}}{2}-cos B=0 Rightarrow cos B=frac{sqrt{2}}{2}. 3. From sin A=frac{1}{2}, we can determine that angle A=30^{circ}, because sin 30^{circ}=frac{1}{2}. 4. From cos B=frac{sqrt{2}}{2}, we can determine that angle B=45^{circ}, because cos 45^{circ}=frac{sqrt{2}}{2}. 5. Knowing that the sum of angles in a triangle equals 180^{circ}, we can find angle C as follows: [ angle C = 180^{circ} - angle A - angle B = 180^{circ} - 30^{circ} - 45^{circ} = 105^{circ} ] Therefore, the value of angle C in triangle triangle ABC is boxed{105^{circ}}.