💡Alright, so I'm trying to figure out how Ali-Baba can win this game against the robber. The game involves a pile of 2017 diamonds, and each player can split one pile into two smaller piles on their turn. The person who can't make a move loses, and the winner takes all the diamonds. Okay, so Ali-Baba wants to end up with all 2017 diamonds. First, I need to understand the mechanics of the game better. When you split a pile, you're essentially increasing the number of piles by one. So, if you start with one pile, after one move, you have two piles, then three, and so on. The game ends when all piles are of size one because you can't split a pile of one diamond anymore. So, starting with one pile of 2017 diamonds, the game will end when there are 2017 piles of one diamond each. That means the total number of moves that can be made is 2016 because each move increases the number of piles by one, and you need to go from 1 pile to 2017 piles. Now, the key here is to determine who makes the last move. Since there are 2016 moves in total, if Ali-Baba goes first, he will make the 1st, 3rd, 5th, ..., 2015th moves, and the robber will make the 2nd, 4th, 6th, ..., 2016th moves. That means the robber would make the last move, which would be the 2016th move, leaving Ali-Baba without a move and thus losing. But wait, if Ali-Baba doesn't go first, and instead lets the robber go first, then the robber would make the 1st, 3rd, 5th, ..., 2015th moves, and Ali-Baba would make the 2nd, 4th, 6th, ..., 2016th moves. In this case, Ali-Baba would make the last move, which is the 2016th move, leaving the robber without a move and thus winning the game. So, it seems like the strategy here is about forcing the opponent to make the last move. Since the total number of moves is even (2016), if Ali-Baba goes second, he can make the last move and win. If he goes first, he'll end up making the first move and the robber will make the last move, causing Ali-Baba to lose. But is there a way for Ali-Baba to ensure he always makes the last move regardless of who starts? Or is it solely dependent on who starts first? It seems like it's dependent on the number of moves and who starts. Since 2016 is an even number, if Ali-Baba starts second, he can make the last move. Let me think about smaller numbers to verify this. Suppose there's a pile of 2 diamonds. If Ali-Baba goes first, he splits it into two piles of 1 each, and the robber can't move, so Ali-Baba wins. But wait, that's only one move. So, if there's 2 diamonds, the number of moves is 1, which is odd. So, if Ali-Baba goes first, he makes the last move and wins. If there are 3 diamonds, starting with one pile, the first player can split it into 1 and 2. Then the second player can split the pile of 2 into 1 and 1, making three piles. Now, the first player can't move, so the second player wins. So, with 3 diamonds, the second player can win if both play optimally. Wait, so with 2 diamonds, the first player wins; with 3 diamonds, the second player can win. So, it alternates depending on the number of diamonds. Let's see 4 diamonds. Starting with one pile, the first player can split it into 1 and 3. Then the second player can split the 3 into 1 and 2. Then the first player can split the 2 into 1 and 1. Now, there are four piles, and the second player can't move, so the first player wins. Hmm, so with 4 diamonds, the first player can win. So, it seems like when the number of diamonds is a power of 2, the first player can win, but when it's not, the second player can win? Wait, 2 is a power of 2, and the first player wins. 4 is a power of 2, and the first player wins. But 3 is not, and the second player wins. But 2017 is not a power of 2. The closest power of 2 is 2048, which is larger than 2017. So, does that mean that the second player can win? But in the case of 3 diamonds, the second player can win, but in 2017 diamonds, it's a much larger number. Wait, maybe I'm overcomplicating it. The total number of moves is 2016, which is even. So, if Ali-Baba goes second, he can make the last move and win. If he goes first, the robber will make the last move and win. So, the key is to go second. But in the problem statement, it says "Ali-Baba is playing a game with one of the robbers." It doesn't specify who starts first. So, maybe Ali-Baba can choose to go second. If he can force the robber to go first, then he can make the last move and win. Alternatively, if the robber starts first, Ali-Baba can respond accordingly to ensure he makes the last move. So, the strategy is to let the robber make the first move, and then Ali-Baba can always respond in a way that ensures he makes the last move. But how exactly can Ali-Baba ensure that? Maybe by mirroring the robber's moves or something like that. But in this game, since you can split any pile into any two, it's not straightforward mirroring. Wait, maybe it's about parity. Since the total number of moves is even, if Ali-Baba goes second, he can always make the last move. So, the key is to go second. Therefore, Ali-Baba should let the robber start first, and then he can make the last move and win. So, in conclusion, Ali-Baba should let the robber make the first move, and then he can always respond in a way that ensures he makes the last move, thereby winning the game and taking all 2017 diamonds.

💡Okay, so I have this expression to evaluate: (left(2c^c - (c+1)(c-1)^cright)^c) where (c = 4). Hmm, let me try to break this down step by step. First, I need to substitute (c = 4) into the expression. That should make it easier to handle.Alright, substituting (c = 4), the expression becomes:[left(2 cdot 4^4 - (4 + 1)(4 - 1)^4right)^4]Okay, now I need to calculate each part inside the parentheses first. Let me tackle the exponents first because they can get big quickly.Starting with (4^4). I remember that (4^4) means (4 times 4 times 4 times 4). Let me compute that:[4 times 4 = 16][16 times 4 = 64][64 times 4 = 256]So, (4^4 = 256). Got that down.Next, I need to compute ((4 - 1)^4). That simplifies to (3^4). Let me calculate that:[3 times 3 = 9][9 times 3 = 27][27 times 3 = 81]So, (3^4 = 81). Good.Now, looking back at the expression, I have (2 cdot 4^4), which is (2 times 256). Let me do that multiplication:[2 times 256 = 512]Alright, so (2 cdot 4^4 = 512).Next, I have ((4 + 1)(4 - 1)^4), which is (5 times 81). Let me compute that:[5 times 81 = 405]So, ((4 + 1)(4 - 1)^4 = 405).Now, putting it all back into the expression inside the parentheses:[512 - 405]Let me subtract those:[512 - 405 = 107]Okay, so the expression inside the parentheses simplifies to 107. Now, the entire original expression is this result raised to the power of (c), which is 4. So, I need to compute (107^4).Hmm, (107^4) is a bit of a big number. Let me break it down. I know that (107^2) is:[107 times 107]Let me compute that. 100 times 100 is 10,000, but since it's 107, I need to account for the extra 7s.Calculating (107 times 107):[100 times 100 = 10,000][100 times 7 = 700][7 times 100 = 700][7 times 7 = 49]Adding those up:[10,000 + 700 + 700 + 49 = 11,449]So, (107^2 = 11,449).Now, I need to compute (11,449^2) to get (107^4). That's going to be a large number. Let me see if I can compute that step by step.First, write it as:[11,449 times 11,449]This is a bit complex, but I can use the standard multiplication method.Let me write it out:``` 11449 x11449 --------```I'll multiply 11449 by each digit of 11449, starting from the right, and then add the results appropriately.But this might take a while. Maybe there's a smarter way. I remember that ((a + b)^2 = a^2 + 2ab + b^2). Maybe I can use that identity here.Let me consider 11,449 as (11,000 + 449). Then:[(11,000 + 449)^2 = 11,000^2 + 2 times 11,000 times 449 + 449^2]Let's compute each term:1. (11,000^2 = 121,000,000)2. (2 times 11,000 times 449) - First, compute (11,000 times 449): - (10,000 times 449 = 4,490,000) - (1,000 times 449 = 449,000) - Adding them: (4,490,000 + 449,000 = 4,939,000) - Then multiply by 2: (4,939,000 times 2 = 9,878,000)3. (449^2) - Let's compute (449 times 449): - (400 times 400 = 160,000) - (400 times 49 = 19,600) - (49 times 400 = 19,600) - (49 times 49 = 2,401) - Adding them up: (160,000 + 19,600 + 19,600 + 2,401 = 201,601)Now, add all three terms together:[121,000,000 + 9,878,000 + 201,601]First, add (121,000,000 + 9,878,000):[121,000,000 + 9,878,000 = 130,878,000]Then add (201,601):[130,878,000 + 201,601 = 131,079,601]Wait, that doesn't seem right because when I computed (107^2) earlier, I got 11,449, and squaring that should give me (107^4). But according to this, I have 131,079,601. Hmm, let me check my calculations again because I might have made a mistake.Wait, when I broke down (449^2), I think I might have miscalculated. Let me recalculate (449 times 449).Calculating (449 times 449):Let me do it step by step:[449 times 449 = (400 + 49)(400 + 49)]Using the formula ((a + b)^2 = a^2 + 2ab + b^2):[400^2 + 2 times 400 times 49 + 49^2]Compute each term:1. (400^2 = 160,000)2. (2 times 400 times 49 = 2 times 19,600 = 39,200)3. (49^2 = 2,401)Now, add them together:[160,000 + 39,200 = 199,200][199,200 + 2,401 = 201,601]Okay, so that part was correct. Then, going back to the earlier step:[121,000,000 + 9,878,000 = 130,878,000][130,878,000 + 201,601 = 131,079,601]Hmm, but I think I might have made a mistake in the initial breakdown. Let me try another approach to compute (11,449 times 11,449).Alternatively, I can use the standard multiplication algorithm:Multiply 11,449 by 11,449:First, multiply 11,449 by 9:[11,449 times 9 = 103,041]Then, multiply 11,449 by 40 (which is the tens place, so shifted one position):[11,449 times 40 = 457,960]Next, multiply 11,449 by 400 (hundreds place, shifted two positions):[11,449 times 400 = 4,579,600]Then, multiply 11,449 by 10,000 (ten-thousands place, shifted four positions):[11,449 times 10,000 = 114,490,000]Now, add all these results together:[103,041 + 457,960 = 561,001][561,001 + 4,579,600 = 5,140,601][5,140,601 + 114,490,000 = 119,630,601]Wait, that's different from the previous result. Hmm, now I'm confused. Which one is correct?Let me check using a calculator approach. Maybe I made a mistake in one of the methods.Alternatively, I can use the fact that (107^4 = (107^2)^2 = 11,449^2). Let me compute (11,449 times 11,449) more carefully.Breaking it down:[11,449 times 11,449 = (11,000 + 449) times (11,000 + 449)]Expanding this:[11,000 times 11,000 + 11,000 times 449 + 449 times 11,000 + 449 times 449]Compute each term:1. (11,000 times 11,000 = 121,000,000)2. (11,000 times 449 = 4,939,000)3. (449 times 11,000 = 4,939,000)4. (449 times 449 = 201,601)Now, add them all together:[121,000,000 + 4,939,000 + 4,939,000 + 201,601]First, add (121,000,000 + 4,939,000 = 125,939,000)Then, add another (4,939,000): (125,939,000 + 4,939,000 = 130,878,000)Finally, add (201,601): (130,878,000 + 201,601 = 131,079,601)Okay, so that seems consistent. Earlier, when I did the standard multiplication, I got 119,630,601, which is different. I must have made a mistake in that method. Maybe I shifted the digits incorrectly.Let me try the standard multiplication again, more carefully.Multiplying 11,449 by 11,449:First, write it as:``` 11449 x11449 --------```Multiply 11449 by 9 (units place):[11449 times 9 = 103,041]Multiply 11449 by 40 (tens place, so shift one position to the left):[11449 times 40 = 457,960]Multiply 11449 by 400 (hundreds place, shift two positions):[11449 times 400 = 4,579,600]Multiply 11449 by 1000 (thousands place, shift three positions):[11449 times 1000 = 11,449,000]Multiply 11449 by 10,000 (ten-thousands place, shift four positions):[11449 times 10,000 = 114,490,000]Now, add all these partial products:``` 103041 457960 4579600 11449000 114490000```Let me add them step by step:First, add 103,041 and 457,960:[103,041 + 457,960 = 561,001]Next, add 4,579,600:[561,001 + 4,579,600 = 5,140,601]Then, add 11,449,000:[5,140,601 + 11,449,000 = 16,589,601]Finally, add 114,490,000:[16,589,601 + 114,490,000 = 131,079,601]Ah, okay, now I get the same result as before: 131,079,601. So, my initial mistake was probably in the shifting or addition during the standard multiplication. Glad I caught that.So, (107^4 = 131,079,601). Therefore, the value of the original expression is 131,079,601.Wait a second, but earlier when I computed (107^2 = 11,449), and then squared that to get (107^4), I got 131,079,601. But when I first tried to compute (11,449 times 11,449) using the standard method, I made a mistake and got a different number, but after correcting it, both methods gave me the same result. So, I think 131,079,601 is the correct value for (107^4).Therefore, putting it all together, the original expression evaluates to 131,079,601.Wait, but in the initial problem, the expression is raised to the power of (c), which is 4. So, after simplifying inside the parentheses, we had 107, and then we raised that to the 4th power, which is 131,079,601.I think that's the final answer. Let me just recap to make sure I didn't miss anything:1. Substitute (c = 4) into the expression.2. Compute (4^4 = 256).3. Compute (3^4 = 81).4. Multiply (2 times 256 = 512).5. Multiply (5 times 81 = 405).6. Subtract: (512 - 405 = 107).7. Raise 107 to the 4th power: (107^4 = 131,079,601).Yes, that seems correct. I think I've covered all the steps and double-checked my calculations to ensure accuracy. So, the evaluated expression is 131,079,601.

💡Okay, so I have these two logical expressions to prove. The first one is A or (B and C) is equivalent to (A and B) or (A and C). The second one is a bit more complicated: not [(A or not B) or (C and (A or not B))] is equivalent to not A and B. Hmm, let's start with the first one.Alright, for the first problem, I need to show that A ∨ (B ∧ C) is equivalent to (A ∧ B) ∨ (A ∧ C). I think this is related to distributive laws in logic. I remember that in Boolean algebra, there are distributive properties similar to arithmetic, where multiplication distributes over addition, and vice versa. Maybe something similar applies here.So, let's break it down. If I have A ∨ (B ∧ C), I can think of it as A being true, or both B and C being true. On the other hand, (A ∧ B) ∨ (A ∧ C) means either A and B are both true, or A and C are both true. It seems like both expressions are trying to capture scenarios where A is true, or certain combinations of B and C are true.Maybe I can use a truth table to verify this equivalence. Let's list all possible truth values for A, B, and C, and see if both expressions yield the same results.There are 2^3 = 8 possible combinations. Let me create a table:| A | B | C | A ∨ (B ∧ C) | (A ∧ B) ∨ (A ∧ C) ||---|---|---|-------------|-------------------|| T | T | T | T ∨ (T ∧ T) | (T ∧ T) ∨ (T ∧ T) || T | T | F | T ∨ (T ∧ F) | (T ∧ T) ∨ (T ∧ F) || T | F | T | T ∨ (F ∧ T) | (T ∧ F) ∨ (T ∧ T) || T | F | F | T ∨ (F ∧ F) | (T ∧ F) ∨ (T ∧ F) || F | T | T | F ∨ (T ∧ T) | (F ∧ T) ∨ (F ∧ T) || F | T | F | F ∨ (T ∧ F) | (F ∧ T) ∨ (F ∧ F) || F | F | T | F ∨ (F ∧ T) | (F ∧ F) ∨ (F ∧ T) || F | F | F | F ∨ (F ∧ F) | (F ∧ F) ∨ (F ∧ F) |Now, let's compute each column step by step.First row: A=T, B=T, C=T- A ∨ (B ∧ C) = T ∨ (T ∧ T) = T ∨ T = T- (A ∧ B) ∨ (A ∧ C) = (T ∧ T) ∨ (T ∧ T) = T ∨ T = TSecond row: A=T, B=T, C=F- A ∨ (B ∧ C) = T ∨ (T ∧ F) = T ∨ F = T- (A ∧ B) ∨ (A ∧ C) = (T ∧ T) ∨ (T ∧ F) = T ∨ F = TThird row: A=T, B=F, C=T- A ∨ (B ∧ C) = T ∨ (F ∧ T) = T ∨ F = T- (A ∧ B) ∨ (A ∧ C) = (T ∧ F) ∨ (T ∧ T) = F ∨ T = TFourth row: A=T, B=F, C=F- A ∨ (B ∧ C) = T ∨ (F ∧ F) = T ∨ F = T- (A ∧ B) ∨ (A ∧ C) = (T ∧ F) ∨ (T ∧ F) = F ∨ F = FWait, here we have a discrepancy. In the first expression, A ∨ (B ∧ C) is T, but in the second expression, (A ∧ B) ∨ (A ∧ C) is F. That means the two expressions are not equivalent? But that contradicts what I thought earlier.Hold on, maybe I made a mistake in computing the second expression. Let me double-check.In the fourth row: A=T, B=F, C=F- (A ∧ B) = T ∧ F = F- (A ∧ C) = T ∧ F = F- So, (A ∧ B) ∨ (A ∧ C) = F ∨ F = FBut A ∨ (B ∧ C) = T ∨ (F ∧ F) = T ∨ F = TSo, indeed, they are different in this case. That suggests that the original statement might not hold. But the problem says to show that they are equivalent. Maybe I misunderstood the problem or made an error in the truth table.Wait, let me check the problem again. It says:A or (B and C) is equivalent to (A and B) or (A and C)But from the truth table, when A is true, regardless of B and C, the first expression is true, but the second expression depends on whether B or C is true. So, when A is true, the first expression is always true, but the second expression is only true if either B or C is true.Therefore, they are not equivalent. Hmm, that's confusing because the problem states to show that they are equivalent. Maybe I misapplied the distributive law.Wait, actually, in logic, the distributive law is that A ∨ (B ∧ C) is equivalent to (A ∨ B) ∧ (A ∨ C). Is that what the problem is asking? Let me check.No, the problem says A ∨ (B ∧ C) is equivalent to (A ∧ B) ∨ (A ∧ C). But from the truth table, they are not equivalent because in the case where A is true, B and C are false, the first expression is true, but the second is false.So, perhaps the problem is misstated? Or maybe I need to revisit the logic.Alternatively, maybe the problem is to show that A ∨ (B ∧ C) is equivalent to (A ∧ B) ∨ (A ∧ C). But according to the truth table, they are not equivalent.Wait, perhaps I need to use logical equivalences to transform one into the other.Let me try to manipulate A ∨ (B ∧ C).Using distributive law: A ∨ (B ∧ C) ≡ (A ∨ B) ∧ (A ∨ C). So, that's different from (A ∧ B) ∨ (A ∧ C).So, perhaps the problem is miswritten? Or maybe I need to consider that (A ∧ B) ∨ (A ∧ C) is equivalent to A ∧ (B ∨ C), which is another distributive law.Wait, yes, that's right. So, (A ∧ B) ∨ (A ∧ C) ≡ A ∧ (B ∨ C). So, perhaps the problem intended to state that A ∨ (B ∧ C) is equivalent to (A ∨ B) ∧ (A ∨ C), which is the standard distributive law.Alternatively, if the problem is as stated, then it's incorrect because A ∨ (B ∧ C) is not equivalent to (A ∧ B) ∨ (A ∧ C).Given that, maybe I need to proceed with the assumption that the problem has a typo, and it should be A ∨ (B ∧ C) ≡ (A ∨ B) ∧ (A ∨ C). That would make sense with the distributive law.Alternatively, perhaps the problem is correct, and I need to show that A ∨ (B ∧ C) is equivalent to (A ∧ B) ∨ (A ∧ C). But based on the truth table, that's not the case.Wait, let's try to see if they can be equivalent. Maybe by applying some logical equivalences.Starting with A ∨ (B ∧ C). Let's see if we can transform it into (A ∧ B) ∨ (A ∧ C).Using distributive law: A ∨ (B ∧ C) ≡ (A ∨ B) ∧ (A ∨ C). So, that's different from (A ∧ B) ∨ (A ∧ C).Alternatively, let's see if (A ∧ B) ∨ (A ∧ C) can be transformed into A ∨ (B ∧ C).Using distributive law: (A ∧ B) ∨ (A ∧ C) ≡ A ∧ (B ∨ C). So, that's A ∧ (B ∨ C), which is different from A ∨ (B ∧ C).Therefore, unless there's some additional equivalence or unless the problem is misstated, these two expressions are not equivalent.Given that, perhaps the problem intended to state the standard distributive law, which is A ∨ (B ∧ C) ≡ (A ∨ B) ∧ (A ∨ C), or A ∧ (B ∨ C) ≡ (A ∧ B) ∨ (A ∧ C).Given that, maybe I need to proceed with that understanding.Moving on to the second problem: ¬[(A ∨ ¬B) ∨ (C ∧ (A ∨ ¬B))] ≡ ¬A ∧ B.This looks more complex. Let's try to simplify the left-hand side step by step.First, let's denote D = A ∨ ¬B. Then the expression becomes ¬[D ∨ (C ∧ D)].Now, let's simplify D ∨ (C ∧ D). Using distributive law: D ∨ (C ∧ D) ≡ (D ∨ C) ∧ (D ∨ D) ≡ (D ∨ C) ∧ D.Since D ∨ D is just D, so it simplifies to D ∧ (D ∨ C). But D ∧ (D ∨ C) is just D, because if D is true, then D ∧ (D ∨ C) is D, and if D is false, then D ∧ (D ∨ C) is false.Therefore, D ∨ (C ∧ D) ≡ D.So, the original expression ¬[D ∨ (C ∧ D)] ≡ ¬D.But D = A ∨ ¬B, so ¬D ≡ ¬(A ∨ ¬B) ≡ ¬A ∧ B.Therefore, the entire expression simplifies to ¬A ∧ B, which is exactly the right-hand side.So, the second problem is correct.But going back to the first problem, I think there might be a misstatement. Perhaps the intended equivalence was A ∨ (B ∧ C) ≡ (A ∨ B) ∧ (A ∨ C), which is the standard distributive law.Alternatively, if the problem is as stated, then it's incorrect because, as shown in the truth table, when A is true, B and C are false, the two expressions have different truth values.Therefore, unless there's some additional context or unless I'm missing something, I think the first problem might have a typo or misstatement.But assuming that the first problem is intended to be the standard distributive law, then it's correct, and I can proceed to show that.So, to recap:1. Show that A ∨ (B ∧ C) ≡ (A ∨ B) ∧ (A ∨ C).2. Show that ¬[(A ∨ ¬B) ∨ (C ∧ (A ∨ ¬B))] ≡ ¬A ∧ B.I think that's the correct interpretation.For the first problem, using the distributive law, we can directly state that A ∨ (B ∧ C) is equivalent to (A ∨ B) ∧ (A ∨ C).For the second problem, by simplifying the expression step by step, we can show that it reduces to ¬A ∧ B.So, I think that's the way to go.

💡Okay, so I have this problem about an ellipse. Let me try to understand it step by step. The equation of the ellipse is given as (x²/a²) + (y²/b²) = 1, where a > b > 0. The eccentricity e is given as √5 / 5. I need to find the slope of the line FM, where F is the right focus and M is the foot of the perpendicular from the vertex A(0, b) to the right directrix l.First, I remember that for an ellipse, the standard form is (x²/a²) + (y²/b²) = 1, with a being the semi-major axis and b the semi-minor axis. The eccentricity e is defined as e = c/a, where c is the distance from the center to each focus. Also, the relationship between a, b, and c is a² = b² + c².Given that e = √5 / 5, so e = c/a = √5 / 5. That means c = (√5 / 5) * a. I can write this as c = (a√5)/5.Now, since a² = b² + c², I can substitute c into this equation to find b in terms of a. Let's compute c² first:c² = (a² * 5)/25 = (a²)/5.So, a² = b² + (a²)/5. Let's solve for b²:b² = a² - (a²)/5 = (5a² - a²)/5 = (4a²)/5.Therefore, b = (2a)/√5. Hmm, that's b in terms of a. Maybe I can rationalize the denominator: b = (2a√5)/5.Okay, so now I have expressions for c and b in terms of a. Let me note them down:c = (a√5)/5,b = (2a√5)/5.Now, the right focus F is located at (c, 0), since it's the right focus on the x-axis. So, F = (c, 0) = ((a√5)/5, 0).Next, I need to find the right directrix l. For an ellipse, the directrix is given by x = ±a/e. Since we're dealing with the right directrix, it's x = a/e.Given that e = √5 / 5, so 1/e = 5/√5 = √5. Therefore, the right directrix l is the line x = a * √5.Wait, hold on. Let me verify that. The formula for directrix is x = ±a/e. So, since e = c/a, then a/e = a/(c/a) = a²/c. So, x = ±a²/c.Wait, that's different from what I thought earlier. So, actually, the directrix is x = a²/c. Since c = (a√5)/5, then a²/c = a² / [(a√5)/5] = (a² * 5)/(a√5) = (5a)/√5 = a√5.So, yes, the right directrix is x = a√5. So, the line l is x = a√5.Now, we need to draw AM perpendicular to l, where A is the vertex (0, b). So, point A is (0, b). We need to drop a perpendicular from A to the directrix l, which is the vertical line x = a√5. Since l is a vertical line, the perpendicular to it will be a horizontal line. So, the foot of the perpendicular M will have the same y-coordinate as A, which is b, and the x-coordinate will be the same as the directrix, which is a√5.Therefore, the coordinates of M are (a√5, b).Wait, hold on. If l is x = a√5, then to drop a perpendicular from A(0, b) to l, since l is vertical, the perpendicular is horizontal. So, the foot M will have the same y-coordinate as A, which is b, and the x-coordinate will be a√5. So, yes, M is (a√5, b).But wait, in the initial problem statement, it says "draw AM ⊥ l through the vertex A(0, b), with M as the foot of the perpendicular." So, that seems correct.So, now, we have point F at ((a√5)/5, 0) and point M at (a√5, b). We need to find the slope of the line FM.Slope is given by (y2 - y1)/(x2 - x1). So, let's compute the coordinates:F = (c, 0) = ((a√5)/5, 0),M = (a√5, b).So, the slope k of FM is:k = (b - 0)/(a√5 - (a√5)/5) = b / [a√5(1 - 1/5)] = b / [a√5*(4/5)] = (b) / [(4a√5)/5] = (5b)/(4a√5).Simplify this expression. Let's substitute b in terms of a. Earlier, we found that b = (2a√5)/5.So, substituting:k = (5 * (2a√5)/5) / (4a√5) = (2a√5) / (4a√5) = (2√5 a) / (4√5 a) = 2/4 = 1/2.So, the slope is 1/2.Wait, that seems straightforward. Let me double-check my steps.1. Found c = (a√5)/5.2. Found b = (2a√5)/5.3. Directrix l is x = a√5.4. Foot of perpendicular M is (a√5, b).5. Coordinates of F: (c, 0) = ((a√5)/5, 0).6. Calculated slope between F and M: (b - 0)/(a√5 - (a√5)/5) = b / (4a√5/5) = (5b)/(4a√5).7. Substituted b = (2a√5)/5: (5*(2a√5)/5)/(4a√5) = (2a√5)/(4a√5) = 1/2.Yes, that seems correct. So, the slope of FM is 1/2.**Final Answer**The slope of the line ( FM ) is boxed{dfrac{1}{2}}.

💡First, I need to determine how many days John walked the dog in April. April has 30 days, and since John didn't walk the dog on 4 Sundays, he walked the dog for 26 days.Next, I'll calculate John's earnings. There are 4 full weeks in April, resulting in 4 Saturdays. John earns 10 per hour on weekdays and 15 per hour on weekends. For weekdays (Monday to Friday):- There are 5 weekdays in a week.- With 4 full weeks, that's 5 weekdays * 4 weeks = 20 weekdays.- Earnings: 20 days * 10/hour * 1 hour/day = 200.For weekends (Saturday):- With 4 Saturdays in April.- Earnings: 4 days * 15/hour * 1 hour/day = 60.Total earnings for April:- Total earnings = 200 (weekdays) + 60 (weekends) = 260.Now, calculating the expenses:- First expense (books): 50- Second expense (gift): 25- Total expenses before giving money to Kaylee: 50 + 25 = 75.Money left after these expenses:- Money left = 260 - 75 = 185.John gives Kaylee 20% of the remaining money:- Money given to Kaylee = 20% of 185 = 0.20 * 185 = 37.Finally, calculating the money left after all expenses:- Money left after all expenses = 185 - 37 = 148.

💡First, I need to determine the number of men who entered the exhibition hall. Since 7/18 of the 1,260 people were women, I can calculate the number of women and then subtract that from the total to find the number of men.Next, I'll calculate the number of men who wore glasses by applying the given fraction (6/11) to the total number of men.Finally, I'll perform the calculations step by step to arrive at the final answer.