answer:1. **Understanding the Problem:** Each cell in a (15 times 15) grid contains a natural number not exceeding 4. The sum of the numbers in each (2 times 2) square is 7. We need to determine the maximum possible sum of all the numbers in the entire (15 times 15) grid. 2. **Identifying Constraints:** - Each (2 times 2) square sums to 7. - The maximum number in any cell is 4. 3. **Analyzing Cell Constraints:** Let’s focus on a (2 times 2) square with cells (a, b, c, d) arranged as: [ begin{array}{|c|c|} hline a & b hline c & d hline end{array} ] The condition tells us: [ a + b + c + d = 7 ] Given (a, b, c, d leq 4), the maximum sum of any two neighboring cells (either horizontally or vertically) that doesn’t exceed 8 is crucial. If it exceeds, it would violate the sum constraints within other intersecting (2 times 2) squares. 4. **Dividing the Grid:** We divide the (15 times 15) grid into smaller units for systematic calculation: - There are (49) (2 times 2) squares in a (15 times 15) grid. - There remain additional cells due to the incomplete rows and columns, forming a stripe of width 1 along the right and bottom sides. - This edge consists of 14 dominoes (1 times 2) cells and a single cell in the corner. 5. **Summing Values:** - Sum of numbers in each (2 times 2) square: [ 49 times 7 ] Because there are 49 such squares, the total sum for these squares is: [ 343 ] - Sum of numbers in each (1 times 2) domino. This could have the maximum sum of 5: [ 14 times 5 ] The total sum for these dominoes is: [ 70 ] - Lastly, the corner cell has a number not exceeding 4: [ 4 ] 6. **Adding Up the Sum:** The maximum sum of all the numbers in the 15 x 15 grid is: [ 49 times 7 + 14 times 5 + 4 = 343 + 70 + 4 = 417 ] **Conclusion:** [ boxed{417} ]

answer:# Step-by-Step Solution Part 1: Calculation Given expression: -1^2+sqrt[3]{-64}times(frac{1}{2})^2+|1-sqrt{2}| Step 1: Evaluate each term separately - -1^2 = -1 - sqrt[3]{-64} = -4 (since -4 times -4 times -4 = -64) - (frac{1}{2})^2 = frac{1}{4} - |1-sqrt{2}| = |sqrt{2} - 1| (since absolute value is the same regardless of the order) Step 2: Substitute the evaluated values into the expression - -1^2+sqrt[3]{-64}times(frac{1}{2})^2+|1-sqrt{2}| - = -1 - 4times frac{1}{4} + sqrt{2} - 1 - = -1 - 1 + sqrt{2} - 1 - = sqrt{2} - 3 Final Answer for Part 1: boxed{sqrt{2} - 3} Part 2: System of Equations Given system: left{begin{array}{l}{4x+y=10}{x-2y=-2}end{array}right. Step 1: Multiply the first equation by 2 to eliminate y - 4x + y = 10 becomes 8x + 2y = 20 Step 2: Add the modified first equation to the second equation to eliminate y - (x - 2y) + (8x + 2y) = -2 + 20 - 9x = 18 Step 3: Solve for x - 9x = 18 - x = 2 Step 4: Substitute x = 2 into the first original equation to solve for y - 4(2) + y = 10 - 8 + y = 10 - y = 2 Final Answer for Part 2: boxed{left{begin{array}{l}{x=2}{y=2}end{array}right.}

answer:We compute: 1. f(3) = 3 cdot 3 + 6 = 9 + 6 = 15. 2. g(3) = sqrt{f(3)} - 3 = sqrt{15} - 3. 3. Substituting into f(g(3)), we have h(3) = f(sqrt{15} - 3) = 3(sqrt{15} - 3) + 6 = 3sqrt{15} - 9 + 6 = 3sqrt{15} - 3. The calculated value of ( h(3) ) is boxed{3sqrt{15} - 3}.

answer:To prove that if the number of group meetings ( m ) is more than 1, then it is at least the number of students ( n ), follow these steps: 1. **Reformulation of the Problem:** Given ( n ) distinct elements (students), and a system of ( t ) sets (groups) such that: - No set in the system contains all elements. - Each pair of elements appears together in exactly one set. - If two elements appear in the same set, they do not appear together in any other set. 2. **Initial Observations:** - The system contains at least two sets. If there were only one set, it would contain all elements, contradicting the problem's given condition. - Each element appears in at least two sets. If an element were in only one set, that set would contain all elements, contradicting the problem's conditions. 3. **Key Definitions and Theorems:** - Define the **multiplicity of an element** as the number of sets in which the element appears. - The number of elements in a set is called its **cardinality**. 4. **Sum of Multiplicities Equals Sum of Cardinalities:** - Summing the multiplicities of all elements should equal the sum of the cardinalities of all sets because we count each element once per set it belongs to. 5. **Choosing an Element with Minimum Multiplicity:** - Select an element a_n with the smallest multiplicity ( k_n ). - Denote the sets containing ( a_n ) as A_1, A_2, ldots, A_{k_n}. - Denote remaining sets as A_{k_n+1}, ldots, A_m. 6. **Independent Sets:** - Sets A_1, A_2, ldots, A_{k_n} are independent; no two elements can appear in more than one of these sets. - Therefore, we can label one element from each of these sets as ( a_1, a_2, ldots, a_{k_n} ). 7. **Proof by Contradiction:** - Assume ( m < n ). The element a_n does not belong to sets A_{k_n+1}, ldots, A_m. - By condition ( 5 ), for any element a_i not in A_j, ( k_i geq s_j ). 8. **Summation Inequalities:** - Combining ( s_1, ldots, s_{k_n} leq k_1, ldots, k_{k_n} ) gives [ s_1 + ldots + s_{k_n} leq k_1 + ldots + k_{k_n} ] - Since ( s_{k_n+1}, ldots, s_m leq k_{k_n+1}, ldots, k_m ): [ s_{k_n+1} + ldots + s_m leq k_{k_n+1} + ldots + k_m ] - Summing gives: [ s_1 + ldots + s_m < k_1 + ldots + k_n ] 9. **Contradiction Analysis:** - This contradicts the previous observation that the total multiplicities equal to total cardinalities. Conclusion: Thus, ( m geq n ). Therefore, [ boxed{m geq n} ]