answer:To find the value of n, we need to set up an equation using the definition of the average (arithmetic mean) of a set of numbers. The average is the sum of the numbers divided by the number of numbers. Let's add up the given numbers and the unknown number (n + 1), and then divide by the total number of numbers, which is 4, to get the average. We know the average is 20, so we can set up the equation as follows: (3 + 16 + 33 + (n + 1)) / 4 = 20 Now, let's solve for n: (52 + (n + 1)) / 4 = 20 Multiply both sides by 4 to get rid of the denominator: 52 + (n + 1) = 80 Now, subtract 52 from both sides: n + 1 = 80 - 52 n + 1 = 28 Finally, subtract 1 from both sides to solve for n: n = 28 - 1 n = 27 So, the value of n is boxed{27} .

answer:To represent the classes and grades in a coordinate format where the first number represents the grade and the second number represents the class, we follow the given format. Given that left(7,8right) represents Class 8 of Grade 7, we understand that the first number is the grade and the second number is the class. Therefore, to represent Class 6 of Grade 8, we simply place the grade (8) as the first number and the class (6) as the second number in the coordinate format. This gives us: [ left(8,6right) ] Hence, Class 6 of Grade 8 can be represented as boxed{left(8,6right)}.

answer:Given an infinite sequence (a_0, a_1, a_2, ldots) of real numbers that satisfies the condition: [ a_n= left| a_{n+1} - a_{m+2}right| quad text{for every } n geq 0 ] with (a_0) and (a_1) being positive and distinct. We will analyze whether this sequence can be bounded. We claim that the sequence ( {a_n} ) must be unbounded. 1. **Understanding the Condition:** The condition can be rewritten and analyzed as: [ a_n = left| a_{n+1} - a_{m+2} right| quad text{for every } n geq 0 ] This helps us understand that the sequence ( {a_n} ) involves absolute differences. 2. **Sequence Values:** By iterating the condition, we recognize that: [ a_{n+1}=a_n+a_{n-1} quad text{or} quad a_{n+1}=a_n-a_{n-1} ] Depending on if (a_{n+1}) is greater than or less than (a_{n}). 3. **Case Enumeration:** Let us consider different cases and analyze their implications. - If ( a_{n} < a_{n-1} ): Then, by using the condition of the sequence: [ a_{n+1} > max {a_n, a_{n-1} } ] This implies that the new term ( a_{n+1} ) is strictly greater than both previous terms. 4. **Obtaining a Strictly Increasing Subsequence:** Remove all ( a_{n} ) where ( a_{n} < a_{n-1} ). Define a new subsequence: [ b_m ] where ( b_m ) is strictly increasing. 5. **Difference Analysis:** To complete the proof, demonstrate that: [ b_{m+1} - b_{m} geq b_{m} - b_{m-1} quad text{for all } m geq 2 ] - Case (i): If ( a_{n+1} > a_n ): [ b_m = a_{n+1} quad text{and} ] [ b_{m-1} geq a_{n-1} quad text{(since } b_{m-1} text{ is either } a_{n-1} text{ or } a_n text{)} ] Then: [ b_{m+1} - b_m = a_{n+2} - a_{n+1} ] [ a_{n+2} = a_n + a_{n+1} ] [ a_{n+2} - a_{n+1} = a_n ] [ b_{m} - a_{n-1} geq b_{m} - b_{m-1} ] - Case (ii): If ( a_{n+1} < a_n ): [ b_m = a_n ] [ b_{m-1} geq a_{n-1} ] Then: [ b_{m+1} - b_m = a_{n+2} - a_n ] [ a_{n+2} = a_{n}-a_{n+1} ] [ a_{n+2} - a_n = -a_{n+1} ] [ b_m - a_{n-1} geq b_{m} - b_{m-1} ] In both cases, the difference increase holds. # Conclusion: The sequence ( { a_n } ) is unbounded. Therefore, the sequence cannot be bounded. blacksquare

answer:1. **Understanding Euler's Formula for Polyhedra on a Sphere**: Euler's formula states that for a polyhedron (or equivalently, a planar graph), the relationship between the number of vertices (V), edges (E), and faces (F) is given by: [ V - E + F = 2 ] This can be rephrased for faces as: [ F = E - V + 2 ] 2. **Counting Vertices (V)**: Given (n) great circles in general position on the sphere (meaning that at most two circles intersect at the same point), each pair of circles intersects at exactly 2 points. Therefore, the total number of intersection points (vertices (V)) can be calculated by choosing 2 circles from (n) and noting that each pair provides 2 vertices. Thus: [ V = binom{n}{2} times 2 = n(n-1) ] 3. **Counting Edges (E)**: Each vertex is where two circles intersect, and each such intersection contributes 2 edges (the intersecting circles split each other into arcs). Thus, given each vertex is of degree 4 (each intersection splits into 4 edges under proper consideration), the total number of edges (E) can simply be considered as: [ E = 2V = 2n(n-1) ] 4. **Calculating Faces (F)**: By Euler's formula mentioned above: [ F = E - V + 2 ] Substituting the expressions we derived for (E) and (V): [ F = 2n(n-1) - n(n-1) + 2 ] Simplifying the expression: [ F = n(n-1) + 2 ] Therefore, the number of regions the (n) great circles divide the sphere into is: [ F = n^2 - n + 2 ] 5. **Verification for Base Case (n = 1)**: Consider a single great circle (when ( n = 1 )): - It clearly divides the sphere into (2) regions. - Checking the formula: [ F = 1^2 - 1 + 2 = 2 ] Thus, the formula holds true. 6. **Conclusion**: The number of regions that (n) great circles divide the sphere into is indeed: [ boxed{n^2 - n + 2} ]