answer:: 1. Given the limit problem: [ lim _{x rightarrow 3} frac{ln left(x^{2}-5 x+7right)}{x-3} ] 2. As (x rightarrow 3), both the numerator (ln (x^2 - 5x + 7)) and the denominator (x - 3) approach zero. Hence, we have an indeterminate form of the type (frac{0}{0}). 3. First, we simplify the expression (x^2 - 5x + 7): [ x^2 - 5x + 7 = 1 + (x^2 - 5x + 6) ] 4. Notice that: [ x^2 - 5x + 6 = (x-2)(x-3) ] 5. Define (z) as: [ z = (x^2 - 5x + 6) = (x-2)(x-3) ] When (x rightarrow 3), (z) approaches 0. 6. We utilize the approximation (ln(1 + z) approx z) for (z rightarrow 0) (see Example 2 or Taylor series expansion): [ ln(x^2 - 5x + 7) = ln(1 + z) sim z text{ as } z rightarrow 0 ] 7. Thus, [ ln(x^2 - 5x + 7) sim x^2 - 5x + 6 ] 8. We then rewrite the original limit using this approximation: [ lim _{x rightarrow 3} frac{ln(x^2 - 5x + 7)}{x-3} sim lim _{x rightarrow 3} frac{x^2 - 5x + 6}{x-3} ] 9. Factorize the numerator (x^2 - 5x + 6) as ((x-2)(x-3)): [ lim _{x rightarrow 3} frac{(x-2)(x-3)}{x-3} ] 10. Cancel the common factor (x - 3) in the numerator and the denominator: [ lim _{x rightarrow 3} (x-2) ] 11. Now, evaluate the limit as (x) approaches 3: [ lim _{x rightarrow 3} (x-2) = 3 - 2 = 1 ] # Conclusion: [ boxed{1} ]

answer:1. **Calculate the percentage of students scoring 95 points**: [ 100% - (15% + 25% + 40%) = 100% - 80% = 20% ] 2. **Determine the median score**: Since 15% + 25% = 40% of students scored below 85 points, and 40% + 20% = 60% scored 85 points or above, the median score is the score at the 50% mark, which is 85 points. 3. **Calculate the mean score**: [ text{Mean} = (15% cdot 60) + (25% cdot 75) + (40% cdot 85) + (20% cdot 95) ] [ text{Mean} = (0.15 cdot 60) + (0.25 cdot 75) + (0.40 cdot 85) + (0.20 cdot 95) ] [ text{Mean} = 9 + 18.75 + 34 + 19 = 80.75 ] 4. **Calculate the difference between the mean and the median**: [ text{Difference} = text{Median} - text{Mean} = 85 - 80.75 = 4.25 approx 4 ] Conclusion with boxed answer: [ 4 ] The final answer is boxed{textbf{(C)} 4}

answer:To find out how many games Alyssa plans to go to next year, we need to add up the games she went to this year and last year, and then subtract that total from the overall total of 39 games. This year, Alyssa went to 11 games. Last year, she went to 13 games. So, the total number of games she went to in the last two years is: 11 (this year) + 13 (last year) = 24 games Now, we subtract this total from the overall total she plans to attend: 39 (total planned) - 24 (already attended) = 15 games Alyssa plans to go to boxed{15} games next year.

answer:: We need to show that the minimum number of players required to satisfy the given conditions is (3k + 3), where ( k = max(a_i) / 6 ). 1. **Step 1: Establishing the lower bound** Suppose we have a player ( A ) who played in (6k) matches. - If (A) was in only one double, then that double played against (6k) other doubles. - Since each player can be in at most 2 doubles, there need to be at least (frac{6k + 1}{2}) different players to accommodate these (6k) doubles. However, this is an overestimation; the correct figure would require at least (6k) different players, which leads to [ 6k + 1 > 3k + 3.] - Suppose (A) was in two doubles. Each of the two doubles must play at least (3k) other doubles, thus involving at least (3k + 3) players in total: (A) himself, his two partners, and at least (3k) opponents. Therefore, the number of players is always at least: [ boxed{3k + 3}.] 2. **Step 2: Showing that a tournament can be arranged with (3k + 3) players** Given ( n ) distinct positive integers ( b_1, b_2, ldots, b_n ) where ( max(b_i) = k ), we need to construct a graph with at least ( 3k + 3 ) players such that: - Each player plays exactly ( a_i = 6b_i ) matches for some ( i ). - For each ( i ), there is at least one player who played exactly ( a_i ) matches. 3. **Step 3: Finding an appropriate graph** Using induction on ( B = b_1 + b_2 + cdots + b_n ): - **Base case (( B = 1 ))**: If ( n=1 ) and ( b_1 = 1 ), then ( k = 1 ). A simple graph with 2 points (one edge) satisfies the conditions. - **Inductive step**: Suppose the statement holds for ( B leq z ). We must show it is valid for ( B = z + 1 ). - If all ( b_i > 1 ), apply the induction hypothesis to ( b_1-1, b_2-1, ldots, b_n-1 ) and then add an extra point connected to each existing ( k ) points. - Suppose ( b_{n-1} = 1 ) and ( b_n = k ), identify ( h ) as the largest of ( b_1, b_2, ldots, b_{n-2} ) (( h < k )). Using induction, form a graph for ( b_1 - 1, b_2 - 1, ldots, b_{n-2} - 1 ) with ( h ) points. Add ( k - h ) isolated points, and finally, another point connected to all ( k ) points, completing the graph construction. 4. **Step 4: Associating points with doubles and matches** Associate each point ( P_i ) of the graph with a set of doubles: ( A_i B_i, B_i C_i, ) and ( C_i A_i ). - Two points connected by an edge imply the respective doubles play against each other. - A point with ( m ) edges indicates each double at that point plays ( 3 ) games per edge, and each player belonging to any double at that point plays ( 6m ) matches in total. Finally, verifying that each graph point meets the conditions: - For a given point with ( m ) edges, any player plays ( 6m ) matches, which meets the requirement ( a_i = 6b_i ). Hence the minimum number of players required to organize the tournament such that each condition is maintained is: [ boxed{3k + 3} ]