answer:1. **Define the Parameters:** Let ( S ) be the distance between points ( A ) and ( B ). The speeds of the two people are given as follows: - Person 1: ( v_1 = 6 , mathrm{m/s} ) - Person 2: ( v_2 = 4 , mathrm{m/s} ) The time after the turn until the second person arrives back is given as: - ( t_2 = 10 , mathrm{min} = 600 , mathrm{s} ) 2. **Express the Total Distance ( S ) in Terms of ( t_1 ):** The first person walks towards the second person at ( A ). At the moment of meeting, each has traveled a distance of ( v_1 t_1 ) and ( v_2 t_1 ) respectively. So: [ S = v_1 t_1 + v_2 t_1 ] [ S = (6 , mathrm{m/s}) t_1 + (4 , mathrm{m/s}) t_1 ] [ S = 10 t_1 , mathrm{m} ] 3. **Calculate the Distance Covered by Person 2 When Person 1 Reaches Their Starting Point:** After meeting, Person 1 turns back and covers a distance ( v_1 t_1 ). Person 2 continues to walk, so the total distance ( S ) from the initial point to where they meet and beyond back to ( B ) must account for Person 2's continued motion. Therefore, the path can also be described by: [ S = v_2 t_1 + v_2 t_1 + v_2 t_2 ] [ S = 2 v_2 t_1 + v_2 t_2 ] [ S = 2 (4 , mathrm{m/s}) t_1 + (4 , mathrm{m/s}) (600 , mathrm{s}) ] [ S = 8 t_1 + 2400, mathrm{m} ] 4. **Equate the Two Expressions for ( S ) and Solve for ( t_1 ):** Now, we equate the two expressions for ( S ): [ 10 t_1 = 8 t_1 + 2400 ] 5. **Isolate ( t_1 ):** Subtract ( 8t_1 ) from both sides: [ 10 t_1 - 8 t_1 = 2400 ] [ 2 t_1 = 2400 ] 6. **Solve for ( t_1 ):** [ t_1 = frac{2400}{2} ] [ t_1 = 1200 , mathrm{s} ] Convert seconds to minutes: [ t_1 = frac{1200 , mathrm{s}}{60} = 20 , mathrm{min} ] # Conclusion: [ boxed{20 , text{мин}} ]

answer:The error in the Python program is a syntax error: a colon (:) after the expression in the line `result = a + b:`. The corrected version of the code is: ```python def add_numbers(a, b): result = a + b return result num1 = 5 num2 = 7 print(add_numbers(num1, num2)) ``` Alternatively, you can simplify the function to directly return the sum without storing it in a variable: ```python def add_numbers(a, b): return a + b num1 = 5 num2 = 7 print(add_numbers(num1, num2)) ```

answer:In Python, when multiple base classes have the same method name but different definitions, the conflict is resolved using the Method Resolution Order (MRO). The MRO is an algorithm used by Python to resolve the order in which to search for methods in a class's inheritance graph. In the example you provided: ```python class A: def method(self): print(class A method) class B: def method(self): print(class B method) class C(A, B): pass ``` The MRO for class C would be: C -> A -> B -> object. This means that Python will first look for the method in class C, then in class A, then in class B, and finally in the built-in object class. So, in this case, when you call `C().method()`, Python will use class A's method. ```python c = C() c.method() # prints: class A method ``` This is because class A is listed before class B in the definition of class C's inheritance. If you swap the order of A and B in the inheritance list, class B's method would be used instead: ```python class C(B, A): pass c = C() c.method() # prints: class B method ``` It's worth noting that Python's MRO is based on the C3 linearization algorithm, which ensures that the order of method resolution is predictable and consistent, even in the presence of multiple inheritance and diamond inheritance patterns.

answer:Here's a complete implementation of the LRU Cache class in Python using a dictionary for the cache and a doubly linked list to track the order of usage: ```python class Node: def __init__(self, key, value): self.key = key self.value = value self.prev = None self.next = None class LRUCache: def __init__(self, capacity: int): self.capacity = capacity self.cache = {} self.head = Node(0, 0) self.tail = Node(0, 0) self.head.next = self.tail self.tail.prev = self.head def _remove(self, node): prev_node = node.prev next_node = node.next prev_node.next = next_node next_node.prev = prev_node def _add(self, node): prev_node = self.tail.prev prev_node.next = node self.tail.prev = node node.prev = prev_node node.next = self.tail def get(self, key: int) -> int: if key in self.cache: node = self.cache[key] self._remove(node) self._add(node) return node.value return -1 def put(self, key: int, value: int) -> None: if key in self.cache: self._remove(self.cache[key]) node = Node(key, value) self._add(node) self.cache[key] = node if len(self.cache) > self.capacity: lru_node = self.head.next self._remove(lru_node) del self.cache[lru_node.key] ``` In this implementation, the `LRUCache` class has the following properties and methods: * `capacity`: The maximum number of key-value pairs the cache can hold. * `cache`: A dictionary that maps keys to nodes in the doubly linked list. * `head` and `tail`: The head and tail nodes of the doubly linked list. * `_remove(node)`: Removes the given node from the doubly linked list. * `_add(node)`: Adds the given node to the front of the doubly linked list (i.e., just before the tail node). * `get(key)`: Retrieves the value associated with the given key and moves the corresponding node to the front of the cache. * `put(key, value)`: Inserts or updates the key-value pair and moves the corresponding node to the front of the cache. If the cache is full, it removes the least recently used pair. Note that this implementation uses 0-based indexing for keys and values, and returns -1 if a key is not found in the cache. You can modify these conventions according to your specific requirements.