answer:1. **Define the Initial Total Sum of the Grid**: Let the sum of all numbers in the 2021 times 2021 grid be denoted by ( S ). Our aim is to analyze how the value of ( S ) changes as Bob performs his operations on the grid. 2. **Understand Bob's Operation on the Grid**: Each second, Bob will: - Pick a cell with a value of at least 5. - Decrease the value of that cell by 4. - Increment each of the neighboring cells by 1. 3. **Effect of Bob's Operation on the Sum ( S )**: Note that when Bob decreases a cell’s value by 4, he distributes these 4 units to its neighboring cells. Each cell in the grid has at most 4 neighbors. Thus, the net change in ( S ) due to Bob’s operation is calculated as follows: - Let ( v ) be the value of the chosen cell. - After selecting a cell, the value of that cell becomes ( v - 4 ). - Each of its at most 4 neighbors increases by 1. Therefore, the net sum change in one operation is: [ Delta S = -4 + (1 times text{{number of neighbors}}) ] Since each cell has at most 4 neighbors, the change in ( S ) can never be positive, formally: [ Delta S leq 0 ] If Bob picks a boundary cell (with fewer than 4 neighbors), ( Delta S ) becomes negative: [ Delta S < 0 ] This means the total sum ( S ) either stays the same or decreases but never increases. 4. **Finite Number of Configurations**: There are finitely many configurations of the grid with a sum ( S leq text{{initial sum}} ). As ( S ) never increases and can potentially decrease, there are a finite number of times ( S ) can change. 5. **Avoiding Infinite Loops**: To argue that the process must stop, assume by contradiction that Bob could enter a cycle, repeating some sequence of operations indefinitely. For this to occur, the grid must return to the same configuration while preserving the value of ( S ). 6. **Impact on Specific Rows**: Consider the cells in the top row specifically: - For a loop to occur, any change in the values of cells must be reversed eventually. - If Bob ever affects the top row (or any row), those cells’ values must at some point decrease, as they were first picked for having values ≥ 5. This operation would decrease ( S ), contradicting the return to the initial configuration. 7. **Inductive Removal of Rows**: If the top row can thereby not participate in any loop, we can eliminate the top row from consideration, reducing the grid to a ( 2020 times 2021 ) grid. - The same logic applies iteratively: by continuing to remove affected rows, we eventually conclude the grid can be reduced completely, leading to a contradiction. # Conclusion: Given that each operation either preserves or reduces the total sum ( S ) and a repeated configuration cannot occur without decreasing ( S ), the process must ultimately stop. [ boxed{text{Yes}} ]

answer:First, let's convert all measurements to the same unit. Since the dimensions of the rectangular boxes are given in centimeters, we'll convert the dimensions of the wooden box to centimeters as well. The wooden box dimensions in centimeters: 8 m = 800 cm 10 m = 1000 cm 6 m = 600 cm Now, let's calculate how many rectangular boxes can fit in one layer of the wooden box. We do this by dividing the dimensions of the wooden box by the dimensions of the rectangular boxes. Number of rectangular boxes along the length of the wooden box: 800 cm / 4 cm = 200 boxes Number of rectangular boxes along the width of the wooden box: 1000 cm / 5 cm = 200 boxes So, in one layer, the wooden box can carry: 200 boxes (length) * 200 boxes (width) = 40,000 boxes Now, let's calculate how many layers of boxes can fit in the height of the wooden box: 600 cm / 6 cm = 100 layers Therefore, the total number of boxes that can fit in the wooden box is: 40,000 boxes/layer * 100 layers = 4,000,000 boxes However, we also have a weight constraint. Each box weighs 500 grams, which is 0.5 kg. The maximum weight the wooden box can carry is 3000 kg. Let's calculate the maximum number of boxes based on weight: 3000 kg / 0.5 kg/box = 6000 boxes So, the maximum number of boxes that can be carried based on weight is 6000 boxes. Given the constraints, the wooden box can physically fit 4,000,000 boxes, but due to the weight limit, it can only carry 6000 boxes. Therefore, the maximum number of boxes that can be carried in the wooden box, given these constraints, is boxed{6000} boxes.

answer:Set ( y = 0 ) to find when the projectile hits the ground: [ -6.1t^2 + 7t + 10 = 0. ] Multiply both sides by (-10) to eliminate the decimal: [ 61t^2 - 70t - 100 = 0. ] Factoring the quadratic equation: [ (61t + 20)(t - 5) = 0. ] Solving for ( t ), we discard the negative root because time cannot be negative: [ t - 5 = 0, ] [ t = 5. ] Thus, the time when the projectile hits the ground is ( boxed{5} ) seconds.

answer:(1) Since tan alpha=2, we have frac {3sin alpha-2cos alpha}{sin alpha -cos alpha }= frac {3tan alpha-2}{tan alpha -1}= frac {3times2-2}{2-1}=boxed{4}. (2) We will provide two methods to solve this part. Method 1: From frac {sin alpha}{cos alpha }=tan alpha=2, we get sin alpha=2cos alpha. Since sin ^{2}alpha+cos ^{2}alpha=1, we have 5cos ^{2}alpha=1, which implies cos ^{2}alpha= frac {1}{5}. As alpha is an angle in the third quadrant, we know that cos alpha < 0. Therefore, cos alpha=-frac { sqrt {5}}{5}. Method 2: Recall the trigonometric identity cos ^{2}alpha= frac {cos ^{2}alpha}{cos ^{2}alpha +sin ^{2}alpha }= frac {1}{1+tan ^{2}alpha }. Substituting tan alpha=2, we obtain cos ^{2}alpha= frac {1}{1+2^{2}}= frac {1}{5}. Since alpha is an angle in the third quadrant, we have cos alpha < 0. Consequently, cos alpha=-frac { sqrt {5}}{5}. In both methods, we arrive at the same conclusion: cos alpha=-frac { sqrt {5}}{5}. Thus, the final answer is boxed{cos alpha=-frac { sqrt {5}}{5}}.