💡To determine the value of ( m^n ) given that ( 3a^{2}b^{n-1} ) and ( -frac{1}{2}a^{m+3}b^2 ) are like terms, we need to ensure that the exponents of the corresponding variables match.1. **Matching the exponents of ( a ):** [ m + 3 = 2 ] Solving for ( m ): [ m = 2 - 3 = -1 ]2. **Matching the exponents of ( b ):** [ n - 1 = 2 ] Solving for ( n ): [ n = 2 + 1 = 3 ]3. **Calculating ( m^n ):** [ m^n = (-1)^3 = -1 ]Therefore, the value of ( m^n ) is (-1).

💡Okay, so I have this function ( g(x) ) defined piecewise. It means that depending on the value of ( x ), the function behaves differently. Specifically, if ( x ) is less than 5, ( g(x) ) is ( x^3 + 1 ). But if ( x ) is 5 or greater, ( g(x) ) is simply ( x + 7 ). The problem is asking for ( g(g(g(9))) ). That means I need to apply the function ( g ) three times, starting with the number 9. I think the best way to approach this is to work from the inside out. So first, I'll find ( g(9) ), then use that result to find ( g(g(9)) ), and finally use that to find ( g(g(g(9))) ).Starting with ( g(9) ). Since 9 is greater than or equal to 5, I should use the second part of the piecewise function, which is ( x + 7 ). So, ( g(9) = 9 + 7 = 16 ). That seems straightforward.Next, I need to find ( g(g(9)) ), which is ( g(16) ). Again, 16 is greater than or equal to 5, so I'll use the same part of the function. Therefore, ( g(16) = 16 + 7 = 23 ). So far, so good.Now, the last step is to find ( g(g(g(9))) ), which is ( g(23) ). Checking the condition, 23 is definitely greater than or equal to 5, so once more, I'll use ( x + 7 ). Calculating that, ( g(23) = 23 + 7 = 30 ).Wait, let me double-check each step to make sure I didn't make a mistake. Starting with ( g(9) ): 9 is greater than 5, so ( 9 + 7 = 16 ). Correct. Then ( g(16) ): 16 is also greater than 5, so ( 16 + 7 = 23 ). That's right. Finally, ( g(23) ): 23 is greater than 5, so ( 23 + 7 = 30 ). Yep, that all adds up.I was a bit worried that maybe one of these numbers would be less than 5, which would require using the ( x^3 + 1 ) part of the function. But since all the intermediate results (9, 16, 23) are greater than 5, I didn't have to worry about that. If, for example, ( g(9) ) had resulted in a number less than 5, then the next step would have used the cubic function. But in this case, everything stayed above 5, so it was just a matter of adding 7 each time.I think I'm confident with the result. Each step was straightforward, and since all the values were above 5, the function was consistently ( x + 7 ). So, after applying ( g ) three times, starting from 9, I end up with 30.

💡Alright, so I've got this problem here about crates of peaches. Let me try to wrap my head around it. There are 154 crates, and each crate has between 130 and 160 peaches, inclusive. I need to find the largest integer n such that there must be at least n crates containing the exact same number of peaches. The options are A. 4, B. 5, C. 6, D. 11.Hmm, okay. So, first off, I think this is a problem about the pigeonhole principle. The idea that if you have more pigeons than pigeonholes, at least one pigeonhole must contain more than one pigeon. In this case, the "pigeonholes" are the possible numbers of peaches in a crate, and the "pigeons" are the crates themselves.Let me see. The number of possible peach counts is from 130 to 160, inclusive. So, how many different numbers is that? Let's calculate that. If I subtract 130 from 160, that gives me 30, but since both 130 and 160 are included, I need to add 1. So, 30 + 1 = 31 different possible numbers of peaches.Okay, so there are 31 possible numbers of peaches per crate. Now, there are 154 crates. So, if I think of this as distributing 154 crates among 31 possible numbers, I can use the pigeonhole principle to find the minimum number of crates that must have the same number of peaches.The formula for the pigeonhole principle to find the minimum number in the most even distribution is to divide the total number by the number of categories and then take the ceiling of that. So, in this case, it would be 154 divided by 31.Let me do that division. 31 times 4 is 124, and 31 times 5 is 155. So, 154 divided by 31 is approximately 4.967. Since we're looking for the minimum number that must exist, we take the ceiling of this, which would be 5. So, does that mean the answer is 5?Wait, but let me think again. The question is asking for the largest integer n such that there must be at least n crates with the same number of peaches. So, if the most even distribution gives us that each number has 4 crates, and then there are some leftovers, how does that affect the minimum?Let me break it down. If we have 31 numbers and 154 crates, and we try to distribute the crates as evenly as possible, each number would get 4 crates, right? Because 31 times 4 is 124, and that leaves us with 154 minus 124, which is 30 crates left.So, those 30 crates would have to be distributed among the 31 numbers. That means 30 numbers would have an extra crate, making them 5 crates each, and 1 number would still have 4 crates. So, in this case, the minimum number of crates that must have the same number is 5, because even in the most even distribution, 30 of the numbers would have 5 crates each.But wait, the question is asking for the largest integer n such that there must be at least n crates with the same number. So, does that mean n is 5 because in the worst case, you have 5 crates with the same number? Or is it 4 because there's a possibility that only 4 crates have the same number?I think it's 5 because, regardless of how you distribute them, you must have at least 5 crates with the same number. Because if you try to spread them out as much as possible, you end up with 30 numbers having 5 crates each and 1 number having 4 crates. So, the minimum number that must exist is 5.But let me double-check. If I have 31 numbers and 154 crates, and I want to find the smallest number n such that at least n crates have the same number, I can use the formula:n = floor(total / categories) + 1So, floor(154 / 31) is 4, and then adding 1 gives us 5. So, that confirms it. The answer should be 5.Wait, but the options are A. 4, B. 5, C. 6, D. 11. So, B is 5, which seems to be the answer based on the pigeonhole principle.But let me think about it differently. Suppose I try to minimize the maximum number of crates with the same number. How would I do that? I would spread the crates as evenly as possible among all the numbers. So, each number gets 4 crates, and then I have 30 crates left. Those 30 crates would each go to a different number, making those numbers have 5 crates each. So, in this case, the maximum number of crates with the same number is 5, but the minimum number that must exist is also 5 because there are 30 numbers with 5 crates each.So, regardless of how you distribute them, you must have at least 5 crates with the same number. Therefore, the largest integer n such that there must be at least n crates with the same number is 5.I think I'm confident now that the answer is B. 5.

💡First, I note that McGregor and Floyd were assigned a total of 89 books to read during the week.Floyd finished 32 books, and there are still 23 books left to read.To find out how many books McGregor finished, I subtract the number of books Floyd finished and the number of books still left to read from the total number of books assigned.So, McGregor finished 89 minus 32 minus 23, which equals 34 books.

💡Okay, so I have this sequence problem here. Let me try to understand it step by step. The sequence is defined by a₁ = 1, a₂ = 2, and then for n ≥ 1, a_{n+2} = (1 + cos²(nπ/2))a_n + sin²(nπ/2). Part (Ⅰ)① asks me to find a₃, a₄, a₅, and a₆. Hmm, okay. So I need to compute these terms using the given recurrence relation. Let me start with a₃.For a₃, n would be 1 because a_{1+2} = a₃. So plugging into the formula: a₃ = (1 + cos²(π/2))a₁ + sin²(π/2). I know that cos(π/2) is 0, so cos²(π/2) is 0. Similarly, sin(π/2) is 1, so sin²(π/2) is 1. Therefore, a₃ = (1 + 0)*1 + 1 = 1 + 1 = 2.Alright, moving on to a₄. Here, n is 2 because a_{2+2} = a₄. So a₄ = (1 + cos²(2π/2))a₂ + sin²(2π/2). Simplifying, cos²(π) is (-1)² = 1, and sin²(π) is 0. So a₄ = (1 + 1)*2 + 0 = 2*2 = 4.Next, a₅. For this, n is 3 because a_{3+2} = a₅. So a₅ = (1 + cos²(3π/2))a₃ + sin²(3π/2). Cos(3π/2) is 0, so cos² is 0. Sin(3π/2) is -1, so sin² is 1. Therefore, a₅ = (1 + 0)*2 + 1 = 2 + 1 = 3.Now, a₆. Here, n is 4 because a_{4+2} = a₆. So a₆ = (1 + cos²(4π/2))a₄ + sin²(4π/2). Cos(2π) is 1, so cos² is 1. Sin(2π) is 0, so sin² is 0. Thus, a₆ = (1 + 1)*4 + 0 = 2*4 = 8.So, summarizing, I have a₃ = 2, a₄ = 4, a₅ = 3, and a₆ = 8.Moving on to part (Ⅰ)②, I need to prove that the sequence a₁, a₃, a₅, a₇, ..., which are the odd-indexed terms, form an arithmetic sequence. From the values I computed, a₁ = 1, a₃ = 2, a₅ = 3. It seems like each term increases by 1. Let me see if this pattern continues.Looking at the recurrence relation, for the odd terms, let's denote n = 2k - 1, where k is a positive integer. Then, a_{2k+1} = (1 + cos²((2k - 1)π/2))a_{2k - 1} + sin²((2k - 1)π/2).Let me compute cos²((2k - 1)π/2) and sin²((2k - 1)π/2). For any integer k, (2k - 1)π/2 is an odd multiple of π/2, which corresponds to angles like π/2, 3π/2, 5π/2, etc. At these angles, cos is 0, so cos² is 0, and sin is ±1, so sin² is 1. Therefore, the recurrence simplifies to a_{2k+1} = (1 + 0)a_{2k - 1} + 1 = a_{2k - 1} + 1.This shows that each odd-indexed term is 1 more than the previous odd-indexed term. So, starting from a₁ = 1, each subsequent term increases by 1, making it an arithmetic sequence with common difference 1. Therefore, the sequence a₁, a₃, a₅, ... is indeed an arithmetic sequence.Now, part (Ⅱ) introduces bₙ = 1 / [a_{2n - 1} * sqrt(a_{2n + 1}) + a_{2n + 1} * sqrt(a_{2n - 1})]. I need to find Tₙ = b₁ + b₂ + ... + bₙ.First, let me understand what bₙ is. It's the reciprocal of a sum involving a_{2n - 1} and a_{2n + 1}. From part (Ⅰ)②, I know that a_{2n - 1} = n because the arithmetic sequence starts at 1 and increases by 1 each time. Similarly, a_{2n + 1} = n + 1.So, substituting these into bₙ, we have:bₙ = 1 / [n * sqrt(n + 1) + (n + 1) * sqrt(n)].Let me simplify this expression. Notice that both terms in the denominator have sqrt(n) and sqrt(n + 1). Let me factor out sqrt(n) * sqrt(n + 1):Denominator = sqrt(n) * sqrt(n + 1) * [sqrt(n) + sqrt(n + 1)].So, bₙ = 1 / [sqrt(n) * sqrt(n + 1) * (sqrt(n) + sqrt(n + 1))].Hmm, this seems a bit complicated, but maybe I can rationalize it or find a telescoping pattern.Let me consider multiplying numerator and denominator by (sqrt(n + 1) - sqrt(n)):bₙ = [sqrt(n + 1) - sqrt(n)] / [sqrt(n) * sqrt(n + 1) * (sqrt(n) + sqrt(n + 1)) * (sqrt(n + 1) - sqrt(n))].The denominator simplifies because (sqrt(n) + sqrt(n + 1))(sqrt(n + 1) - sqrt(n)) = (n + 1) - n = 1.So, denominator becomes sqrt(n) * sqrt(n + 1) * 1 = sqrt(n(n + 1)).Therefore, bₙ = [sqrt(n + 1) - sqrt(n)] / sqrt(n(n + 1)).But sqrt(n(n + 1)) is sqrt(n) * sqrt(n + 1), so we can split the fraction:bₙ = [sqrt(n + 1) / sqrt(n(n + 1))] - [sqrt(n) / sqrt(n(n + 1))] = 1/sqrt(n) - 1/sqrt(n + 1).Wow, that's a nice simplification! So each bₙ is equal to 1/sqrt(n) - 1/sqrt(n + 1).Therefore, when we sum Tₙ = b₁ + b₂ + ... + bₙ, it's a telescoping series:Tₙ = (1/sqrt(1) - 1/sqrt(2)) + (1/sqrt(2) - 1/sqrt(3)) + ... + (1/sqrt(n) - 1/sqrt(n + 1)).All the intermediate terms cancel out, leaving Tₙ = 1 - 1/sqrt(n + 1).So, the sum Tₙ is equal to 1 minus the reciprocal of the square root of (n + 1).Let me double-check this result. Starting from bₙ, we found that it telescopes, which is a common technique for such sums. The algebra seems correct, and the cancellation makes sense. So, I think this is the right answer.**Final Answer**Ⅰ①: ( a_3 = boxed{2} ), ( a_4 = boxed{4} ), ( a_5 = boxed{3} ), ( a_6 = boxed{8} )Ⅰ②: The sequence ( a_1, a_3, a_5, ldots ) is an arithmetic sequence with common difference 1.Ⅱ: ( T_n = boxed{1 - frac{1}{sqrt{n+1}}} )

💡First, I need to select 1 player out of the 20 to be the point guard. There are 20 possible choices for this role.After selecting the point guard, there are 19 players remaining. I need to choose 4 more players from these 19 to complete the team.The number of ways to choose 4 players out of 19 is calculated using the combination formula, which is (binom{19}{4}).Finally, to find the total number of different lineups, I multiply the number of ways to choose the point guard by the number of ways to choose the remaining 4 players:[ 20 times binom{19}{4} = 20 times 3,876 = 77,520 ]