💡First, I recognize that the expression ((1 + 2x)^5) can be expanded using the binomial theorem. The binomial theorem states that ((a + b)^n = sum_{k=0}^{n} binom{n}{k} a^{n-k} b^k).In this case, (a = 1), (b = 2x), and (n = 5). I need to find the coefficient of the (x^2) term, which corresponds to (k = 2) in the expansion.Using the binomial coefficient formula, (binom{5}{2} = 10). Then, I calculate (b^2 = (2x)^2 = 4x^2).Multiplying these together, the coefficient of (x^2) is (10 times 4 = 40).

💡First, I recognize that each term inside the square root is (2^4).Calculating (2^4) gives me 16.Since there are four such terms, I add them together: (16 + 16 + 16 + 16 = 64).Finally, taking the square root of 64 results in 8.

💡First, calculate the number of births per day. Since a baby is born every 6 hours, there are ( frac{24}{6} = 4 ) births per day.Next, note that there are 2 deaths per day.The net population increase per day is ( 4 - 2 = 2 ) people.To find the annual increase, multiply the daily net increase by the number of days in a year: ( 2 times 365 = 730 ) people.Finally, round 730 to the nearest hundred, which is 700.

💡First, I recognize that 80 fish were tagged and released back into the pond. Later, another 80 fish were caught, and we need to determine how many of these were tagged.I know that the proportion of tagged fish in the second catch should reflect the proportion of tagged fish in the entire pond. Given that the total estimated number of fish in the pond is 3200, I can set up a proportion to find the number of tagged fish in the second catch.By setting up the equation T/80 = 80/3200, where T represents the number of tagged fish in the second catch, I can solve for T.Calculating T:T = (80/3200) * 80T = (1/40) * 80T = 2Therefore, approximately 2 fish in the second catch were found to have been tagged.

💡To determine how many zeros the function ( f(x) = sin(log x) ) has on the interval ( 1 < x < e^pi ), we need to find the values of ( x ) for which ( sin(log x) = 0 ).1. **Understanding the Function**: The function ( f(x) = sin(log x) ) involves the composition of the sine function and the natural logarithm. The zeros of this function occur where the argument of the sine function is an integer multiple of ( pi ).2. **Range of ( log x )**: For ( x ) in the interval ( 1 < x < e^pi ), the natural logarithm ( log x ) ranges from ( 0 ) to ( pi ). This is because ( log 1 = 0 ) and ( log(e^pi) = pi ).3. **Zeros of ( sin(t) )**: The sine function ( sin(t) ) is zero at integer multiples of ( pi ), i.e., ( t = kpi ) for integers ( k ).4. **Finding Relevant ( k ) Values**: We need to find integers ( k ) such that ( 0 leq kpi leq pi ). This occurs for ( k = 0 ) and ( k = 1 ).5. **Counting the Zeros**: - For ( k = 0 ), ( log x = 0 ) implies ( x = e^0 = 1 ). However, ( x = 1 ) is not included in the interval ( 1 < x < e^pi ). - For ( k = 1 ), ( log x = pi ) implies ( x = e^pi ). However, ( x = e^pi ) is also not included in the interval ( 1 < x < e^pi ).6. **Conclusion**: Since both ( x = 1 ) and ( x = e^pi ) are excluded from the interval, there are no zeros of ( f(x) = sin(log x) ) within the interval ( 1 < x < e^pi ).Thus, the number of zeros of ( f(x) = sin(log x) ) on the interval ( 1 < x < e^pi ) is ( 0 ).The final answer is (boxed{A}).

💡Alright, so I've got this problem about a chocolate bar shaped like an equilateral triangle with side length ( n ). It's divided into smaller equilateral triangles with side length 1. Two players take turns breaking off a triangular piece and eating it, passing the rest to the opponent. The person who gets the last piece, which is a single small triangle, wins. If you can't make a move, you lose.Okay, let me try to wrap my head around this. First, the chocolate bar is an equilateral triangle, so all sides are equal, and it's divided into smaller equilateral triangles. That means the number of small triangles is ( frac{n(n+1)}{2} ) because that's the formula for the number of triangles in a triangular number arrangement.Now, each player can break off a triangular piece. So, on each turn, a player can remove a smaller equilateral triangle from the remaining chocolate. The key here is that after each break, the remaining chocolate is still a polygon, but it's no longer a perfect triangle—it becomes something else, maybe a trapezoid or another shape.The goal is to be the one who takes the last piece, which is a single small triangle. So, if you're forced to receive the last piece, you win. If you can't make a move, you lose. That means the strategy revolves around forcing the opponent into a position where they can't make a move or they have to take the last piece.Let me think about smaller cases to get a sense of the pattern.**Case 1: ( n = 1 )**If the chocolate bar is just a single small triangle, the first player takes it and wins. So, for ( n = 1 ), Player 1 wins.**Case 2: ( n = 2 )**The chocolate bar is a larger triangle made up of 3 small triangles. Player 1 can take one small triangle, leaving two small triangles. Then Player 2 can take one of the remaining two, leaving the last one for Player 1, who then wins. Wait, that doesn't make sense because Player 2 would want to win. Maybe Player 1 can take two small triangles at once? But the problem says you can only break off a triangular piece, which would be a single small triangle. So, Player 1 takes one, Player 2 takes one, and Player 1 is left with the last one and wins. Hmm, so for ( n = 2 ), Player 1 still wins.Wait, maybe I'm misunderstanding. If Player 1 takes one triangle, the remaining chocolate is not just two separate triangles but a shape that's still connected. So, Player 2 can take a triangle from the remaining shape, but the remaining shape after Player 1's move is a trapezoid made up of two small triangles. So, Player 2 can take one of those, leaving the last one for Player 1, who then wins. So, again, Player 1 wins.**Case 3: ( n = 3 )**The chocolate bar has 6 small triangles. Player 1 can take one, leaving 5. Then Player 2 can take one, leaving 4. Player 1 takes one, leaving 3. Player 2 takes one, leaving 2. Player 1 takes one, leaving 1, which Player 2 takes and wins. Wait, so in this case, Player 2 wins.But this seems like a simple turn-based game where the number of pieces determines the winner. If the total number of pieces is odd, Player 1 wins; if even, Player 2 wins. But in the case of ( n = 3 ), the total number of pieces is 6, which is even, so Player 2 would win. For ( n = 2 ), total pieces are 3, which is odd, so Player 1 wins. For ( n = 1 ), it's 1, which is odd, Player 1 wins.But wait, the problem isn't just about taking pieces one by one; it's about breaking off a triangular piece, which might change the structure of the remaining chocolate. So, it's not just about the number of pieces but also about the shape.Let me think again. When you break off a triangular piece, you're not just removing a single small triangle; you're altering the shape of the remaining chocolate. So, the game isn't just about reducing the number of pieces but also about controlling the shape to force the opponent into a losing position.Maybe this is similar to the game of Nim, where players take turns removing objects, and the last to remove an object wins. But in this case, the objects are connected in a specific shape, so the moves are more constrained.I need to find a pattern or a mathematical property that determines the winner based on ( n ). Maybe it's related to whether ( n ) is a power of 2 or something like that.Wait, let's think about the number of pieces. For ( n = 1 ), 1 piece. ( n = 2 ), 3 pieces. ( n = 3 ), 6 pieces. ( n = 4 ), 10 pieces. These are triangular numbers: ( T_n = frac{n(n+1)}{2} ).If the total number of pieces is odd or even, that might determine the winner. For ( n = 1 ), odd, Player 1 wins. ( n = 2 ), 3 pieces, odd, Player 1 wins. ( n = 3 ), 6 pieces, even, Player 2 wins. ( n = 4 ), 10 pieces, even, Player 2 wins. ( n = 5 ), 15 pieces, odd, Player 1 wins. Hmm, so it seems like when ( T_n ) is odd, Player 1 wins, and when it's even, Player 2 wins.But wait, is that always the case? Let me check ( n = 4 ). If there are 10 pieces, which is even, Player 2 would win. But does the game actually reduce to just taking pieces one by one? Or does the structure of the chocolate affect the outcome?For example, in ( n = 3 ), if Player 1 takes a corner triangle, the remaining shape is a trapezoid with 5 small triangles. Then Player 2 can take a triangle from the trapezoid, leaving 4. Then Player 1 takes one, leaving 3. Player 2 takes one, leaving 2. Player 1 takes one, leaving 1 for Player 2, who wins. So, in this case, the number of pieces being even leads to Player 2 winning.But if the number of pieces is odd, like ( n = 2 ), Player 1 can force a win by taking one piece, leaving an even number, and then mirroring Player 2's moves.Wait, but in ( n = 3 ), the number of pieces is 6, which is even, so Player 2 wins. For ( n = 4 ), 10 pieces, even, Player 2 wins. For ( n = 5 ), 15 pieces, odd, Player 1 wins.So, it seems like the parity of the total number of small triangles determines the winner. If ( T_n ) is odd, Player 1 wins; if even, Player 2 wins.But let's test this with ( n = 4 ). If Player 1 takes one piece, leaving 9, which is odd. Then Player 2 can take one, leaving 8, which is even, and so on. Wait, but the number of pieces isn't just being reduced by one each time because the remaining chocolate's shape might allow taking more than one piece at a time.Wait, no, each move only allows taking one triangular piece, right? Because you break off a triangular piece along the grooves, which would be a single small triangle. So, each move reduces the total number of pieces by one. Therefore, the parity of the total number of pieces determines the winner.So, if ( T_n ) is odd, Player 1 wins; if even, Player 2 wins.But let's check ( n = 3 ). ( T_3 = 6 ), which is even, so Player 2 wins. That matches our earlier analysis.Similarly, ( n = 4 ), ( T_4 = 10 ), even, Player 2 wins.( n = 5 ), ( T_5 = 15 ), odd, Player 1 wins.So, the general rule seems to be: if ( frac{n(n+1)}{2} ) is odd, Player 1 wins; if even, Player 2 wins.But let's see if this holds for larger ( n ).( n = 6 ), ( T_6 = 21 ), odd, Player 1 wins.( n = 7 ), ( T_7 = 28 ), even, Player 2 wins.Yes, this seems consistent.But wait, is this always the case? Let me think about ( n = 0 ), which is trivial, but ( T_0 = 0 ), which is even, so Player 2 wins, but there's no chocolate, so Player 1 can't make a move and loses, which aligns with Player 2 winning.Okay, so the key is the parity of the total number of small triangles.Now, ( T_n = frac{n(n+1)}{2} ). So, when is ( frac{n(n+1)}{2} ) odd?Let's analyze ( n(n+1) ). Since ( n ) and ( n+1 ) are consecutive integers, one of them is even, so ( n(n+1) ) is always even. Therefore, ( frac{n(n+1)}{2} ) is an integer.Now, when is ( frac{n(n+1)}{2} ) odd?Let's consider ( n ) modulo 4.Case 1: ( n equiv 0 mod 4 )Then ( n = 4k ), so ( n+1 = 4k + 1 )( n(n+1) = 4k(4k + 1) = 16k^2 + 4k )( frac{n(n+1)}{2} = 8k^2 + 2k ), which is even.Case 2: ( n equiv 1 mod 4 )Then ( n = 4k + 1 ), so ( n+1 = 4k + 2 )( n(n+1) = (4k + 1)(4k + 2) = 16k^2 + 12k + 2 )( frac{n(n+1)}{2} = 8k^2 + 6k + 1 ), which is odd.Case 3: ( n equiv 2 mod 4 )Then ( n = 4k + 2 ), so ( n+1 = 4k + 3 )( n(n+1) = (4k + 2)(4k + 3) = 16k^2 + 20k + 6 )( frac{n(n+1)}{2} = 8k^2 + 10k + 3 ), which is odd.Case 4: ( n equiv 3 mod 4 )Then ( n = 4k + 3 ), so ( n+1 = 4k + 4 )( n(n+1) = (4k + 3)(4k + 4) = 16k^2 + 28k + 12 )( frac{n(n+1)}{2} = 8k^2 + 14k + 6 ), which is even.So, summarizing:- If ( n equiv 0 mod 4 ), ( T_n ) is even.- If ( n equiv 1 mod 4 ), ( T_n ) is odd.- If ( n equiv 2 mod 4 ), ( T_n ) is odd.- If ( n equiv 3 mod 4 ), ( T_n ) is even.Therefore, ( T_n ) is odd when ( n equiv 1 ) or ( 2 mod 4 ), and even otherwise.So, the winner depends on the parity of ( T_n ):- If ( T_n ) is odd, Player 1 wins.- If ( T_n ) is even, Player 2 wins.Thus, the winning condition is:- Player 1 wins if ( n equiv 1 ) or ( 2 mod 4 ).- Player 2 wins if ( n equiv 0 ) or ( 3 mod 4 ).But let me double-check this with some examples.For ( n = 1 ):- ( 1 equiv 1 mod 4 ), Player 1 wins. Correct.For ( n = 2 ):- ( 2 equiv 2 mod 4 ), Player 1 wins. Correct.For ( n = 3 ):- ( 3 equiv 3 mod 4 ), Player 2 wins. Correct.For ( n = 4 ):- ( 4 equiv 0 mod 4 ), Player 2 wins. Correct.For ( n = 5 ):- ( 5 equiv 1 mod 4 ), Player 1 wins. Correct.For ( n = 6 ):- ( 6 equiv 2 mod 4 ), Player 1 wins. Correct.For ( n = 7 ):- ( 7 equiv 3 mod 4 ), Player 2 wins. Correct.Okay, this seems consistent.So, the conclusion is that Player 1 wins if ( n ) is congruent to 1 or 2 modulo 4, and Player 2 wins otherwise.But wait, the problem says "with optimal play." So, does this parity argument hold under optimal play?Yes, because each move reduces the total number of pieces by one, and the parity determines who will be forced to take the last piece.Therefore, the winner is determined by the parity of the total number of small triangles, which depends on ( n ) modulo 4.