answer：Alright, so I've got this problem here about a factory with three workshops: A, B, and C. They produce the same kind of product, but in different quantities—120 from A, 80 from B, and 60 from C. The question is about determining if there's a significant difference in the quality of their products using a stratified sampling method. They took a sample of size n, and they drew 3 products from workshop C. I need to figure out what n is.Hmm, okay. Let me recall what stratified sampling is. From what I remember, stratified sampling is a method where the population is divided into subgroups, or strata, and then a random sample is taken from each stratum. The idea is that each subgroup is represented in the sample, which can lead to more accurate results compared to simple random sampling.So, in this case, the workshops A, B, and C are the strata. Each workshop has a different production quantity, so the sample size from each workshop should be proportional to their production quantity, right? That way, each workshop is fairly represented in the sample.Let me write down the given quantities:- Workshop A: 120 products- Workshop B: 80 products- Workshop C: 60 productsTotal production is 120 + 80 + 60. Let me calculate that: 120 + 80 is 200, plus 60 is 260. So, the total number of products is 260.Now, they used stratified sampling to draw a sample of size n. They took 3 products from workshop C. I need to find n.Since it's stratified sampling, the number of samples taken from each workshop should be proportional to the size of each workshop relative to the total production.So, the proportion for workshop C is 60 out of 260. Let me write that as a fraction: 60/260. I can simplify that fraction. Both numerator and denominator are divisible by 20, so 60 ÷ 20 is 3, and 260 ÷ 20 is 13. So, the proportion is 3/13.If the sample size from workshop C is 3, then 3 is equal to the proportion of workshop C times the total sample size n. So, mathematically, that would be:3 = (3/13) * nI need to solve for n. Let me rearrange the equation:n = 3 / (3/13) = 3 * (13/3) = 13Wait, so n is 13? Let me check that again. If the proportion is 3/13, and 3 is the number from C, then n must be 13. That seems right.But let me think about it another way to make sure. If n is 13, then the number of samples from each workshop should be:- From A: (120/260) * 13- From B: (80/260) * 13- From C: (60/260) * 13 = 3Calculating each:For A: 120/260 simplifies to 12/26, which is 6/13. So, 6/13 * 13 = 6. So, 6 samples from A.For B: 80/260 simplifies to 8/26, which is 4/13. 4/13 * 13 = 4. So, 4 samples from B.For C: As given, 3 samples.Adding them up: 6 + 4 + 3 = 13. That matches n. So, that seems consistent.Alternatively, I can think about the ratio of the sample sizes. Since the total production is 260, and they took a sample of size n, the sampling fraction is n/260.But since it's stratified, each workshop's sample size is the total sample size multiplied by their proportion in the population.So, for workshop C, sample size is n * (60/260) = 3.So, n = 3 * (260/60) = 3 * (13/3) = 13. Yep, same result.Wait, another way: 60/260 is the proportion, which is 3/13. So, 3 is 3/13 of n, so n is 13.I think that's solid. So, n is 13.But just to make sure, let me consider if there's another interpretation. Sometimes, in stratified sampling, people might use equal sample sizes from each stratum, but that's not the case here because the workshops have different sizes. So, proportional allocation is the standard approach unless stated otherwise.If they had used equal sample sizes, they might have taken the same number from each workshop, but since the sample from C is 3, and C is the smallest workshop, that would mean taking 3 from each, which would be 9 total. But that's not proportional. So, the question specifies stratified sampling, which usually implies proportional allocation.Therefore, I think n is 13.**Final Answer**The value of ( n ) is boxed{13}.

answer：To determine the position of (-20) in the sequence ({(-1)^{n+1}n(n+1)}), I start by setting the general term equal to (-20):[(-1)^{n+1}n(n+1) = -20]Since the term is negative, the exponent (n+1) must be even, which means (n) is odd. I can rewrite the equation as:[n(n+1) = 20]Expanding and rearranging gives a quadratic equation:[n^2 + n - 20 = 0]Using the quadratic formula:[n = frac{-1 pm sqrt{1 + 80}}{2} = frac{-1 pm 9}{2}]This yields two solutions:[n = frac{8}{2} = 4 quad text{and} quad n = frac{-10}{2} = -5]Since (n) must be a positive integer, the valid solution is (n = 4). Therefore, (-20) is the 4th term of the sequence.

answer：Okay, so I have this problem where I need to find the maximum value of the function ( f(x) = 3 sin x + 4 cos x ). Hmm, I remember that for functions like this, which are combinations of sine and cosine, there's a method to find their maximum and minimum values. Let me try to recall how that works.First, I think it has something to do with rewriting the function in a different form. Maybe using the amplitude-phase form? I remember something like ( A sin(x + phi) ) or ( A cos(x + phi) ), where ( A ) is the amplitude, which would give the maximum value. Yeah, that sounds right. So, if I can express ( 3 sin x + 4 cos x ) as a single sine or cosine function, then the amplitude would be the maximum value.How do I do that? I think it involves using the identity for combining sine and cosine terms. Let me write down the general form: ( a sin x + b cos x = R sin(x + phi) ), where ( R = sqrt{a^2 + b^2} ) and ( phi ) is some phase shift. Is that correct? I think so. So, in this case, ( a = 3 ) and ( b = 4 ). Therefore, ( R = sqrt{3^2 + 4^2} = sqrt{9 + 16} = sqrt{25} = 5 ). So, the amplitude ( R ) is 5. That would mean the maximum value of the function is 5, since the sine function oscillates between -1 and 1, so multiplying by 5 would make it oscillate between -5 and 5. Therefore, the maximum value is 5.Wait, let me make sure I did that correctly. Maybe I should verify by expanding ( R sin(x + phi) ) and see if it matches ( 3 sin x + 4 cos x ). So, expanding ( R sin(x + phi) ) gives ( R sin x cos phi + R cos x sin phi ). Comparing this to ( 3 sin x + 4 cos x ), we can set up the equations:1. ( R cos phi = 3 )2. ( R sin phi = 4 )We already found ( R = 5 ), so plugging that in:1. ( 5 cos phi = 3 ) => ( cos phi = 3/5 )2. ( 5 sin phi = 4 ) => ( sin phi = 4/5 )So, ( phi ) is an angle whose cosine is 3/5 and sine is 4/5. That makes sense because ( (3/5)^2 + (4/5)^2 = 9/25 + 16/25 = 25/25 = 1 ), so it's a valid angle. Therefore, the function can indeed be written as ( 5 sin(x + phi) ), and since the sine function has a maximum of 1, the maximum value of ( f(x) ) is 5*1 = 5.Alternatively, I remember another method using calculus. Maybe I can take the derivative of ( f(x) ) and find its critical points to determine the maximum. Let me try that approach as a check.The function is ( f(x) = 3 sin x + 4 cos x ). Taking the derivative with respect to x:( f'(x) = 3 cos x - 4 sin x )To find critical points, set the derivative equal to zero:( 3 cos x - 4 sin x = 0 )Let me solve for x:( 3 cos x = 4 sin x )Divide both sides by ( cos x ) (assuming ( cos x neq 0 )):( 3 = 4 tan x )So,( tan x = 3/4 )Therefore, ( x = arctan(3/4) + kpi ), where k is an integer. So, the critical points occur at these x-values.Now, to find whether these points are maxima or minima, I can use the second derivative test or evaluate the function at these points.Let me compute the second derivative:( f''(x) = -3 sin x - 4 cos x )At ( x = arctan(3/4) ), let's compute ( f''(x) ). But maybe it's easier to evaluate ( f(x) ) at this point.Alternatively, since I know the function is periodic and smooth, the maximum will occur at one of these critical points. So, let's compute ( f(x) ) at ( x = arctan(3/4) ).But wait, maybe I can express ( sin x ) and ( cos x ) in terms of the tangent. Since ( tan x = 3/4 ), we can imagine a right triangle where the opposite side is 3 and the adjacent side is 4, so the hypotenuse is 5. Therefore, ( sin x = 3/5 ) and ( cos x = 4/5 ).Therefore, plugging into ( f(x) ):( f(x) = 3 sin x + 4 cos x = 3*(3/5) + 4*(4/5) = 9/5 + 16/5 = 25/5 = 5 )So, the function reaches 5 at this critical point. Since the sine function has a maximum of 1, and we've scaled it by 5, this makes sense. Therefore, the maximum value is indeed 5.Wait, but just to be thorough, is this the only critical point? Or are there others? Well, the function is periodic, so the critical points repeat every ( pi ) radians, but the maximum value will be the same at each of these points. So, regardless of where you are on the x-axis, the maximum value of the function is 5.Alternatively, another way to think about this is using vectors. The function ( 3 sin x + 4 cos x ) can be thought of as the dot product of two vectors: ( (3, 4) ) and ( (sin x, cos x) ). The maximum value of this dot product occurs when the two vectors are in the same direction. The maximum value is then the magnitude of the vector ( (3, 4) ), which is ( sqrt{3^2 + 4^2} = 5 ). So, that's another way to see it.So, whether I use the amplitude-phase method, calculus, or vector interpretation, I arrive at the same conclusion: the maximum value of ( f(x) ) is 5.Just to recap, the steps I took were:1. Recognize that the function is a combination of sine and cosine.2. Recall that such functions can be rewritten in the form ( R sin(x + phi) ) with ( R = sqrt{a^2 + b^2} ).3. Compute ( R ) as 5.4. Verify by expanding the expression and comparing coefficients.5. Use calculus by taking the derivative, finding critical points, and evaluating the function at those points.6. Confirm the result using the vector dot product interpretation.Each method consistently gives the maximum value as 5, so I'm confident that's the correct answer.**Final Answer**The maximum value of the function is boxed{5}.

answer：Okay, so I have this physics problem here, and I need to figure out the tension in the string and the force F acting on the board. Let me try to visualize the setup first.There's a board with mass M = 4 kg on a horizontal table. On top of this board, there's a block with mass m = 1 kg. They're connected by a light string that goes over a pulley. So, the block is probably hanging off the edge of the board, right? And there's a horizontal force F applied to the board. The board is accelerating with an acceleration a = g/5, which is 2 m/s² since g is 10 m/s². The coefficient of friction is 0.2 for both the board-block interface and the board-table interface. Alright, so first, I need to find the tension in the string. Let me think about the forces acting on each object.Starting with the block on top of the board. The block has mass m = 1 kg. It's connected to the board via a string over a pulley, so the tension in the string will affect its motion. Since the board is moving, the block is also moving, but relative to the board, is it sliding? Hmm, the problem doesn't specify, but since there's friction mentioned between the block and the board, I think the block might be sliding on the board or maybe it's just about to slide. Wait, the board is moving with acceleration a = 2 m/s². So, if the block is on the board, what's its acceleration? If the string is connected to the pulley, then the block is being pulled by the tension, but also, the board is moving. Hmm, maybe the block is moving with the same acceleration as the board? Or is it moving with a different acceleration because of the tension?Wait, the pulley is probably fixed, right? So, if the board is moving to the right, the pulley is fixed, so the string would cause the block to move either up or down. Wait, hold on, the block is on the board, so if the board is moving to the right, the pulley is fixed, so the string would cause the block to move to the left relative to the board? Or maybe the block is moving with the board?Wait, this is getting confusing. Let me try to draw a free-body diagram.First, the board: mass M = 4 kg. It's on a table, so normal force N1 from the table. There's a horizontal force F applied to the board. The board is moving, so friction is involved. The coefficient of friction between the board and the table is μ = 0.2, so the frictional force on the board due to the table is μ*N1. Since the board is on the table, N1 = M*g = 4*10 = 40 N. So frictional force f1 = μ*N1 = 0.2*40 = 8 N.Also, the board has the block on top of it, so the block exerts a normal force on the board. The mass of the block is m = 1 kg, so the normal force N2 = m*g = 10 N. The friction between the block and the board is μ*N2 = 0.2*10 = 2 N. So, the board has friction from the table (8 N) and friction from the block (2 N). So total frictional force opposing the motion of the board is 8 + 2 = 10 N.Wait, is that right? So, the board is being pushed by force F, but it's also experiencing friction from the table and from the block. So, the net force on the board is F - f1 - f2 = M*a, where f1 is friction from the table, f2 is friction from the block.But wait, the block is on the board, so the friction between the block and the board is acting on the block, not directly on the board. Hmm, maybe I need to consider the forces on the block separately.Let me think again. The board is moving with acceleration a = 2 m/s². The block is on the board. So, if the block is not sliding, it would have the same acceleration as the board. But if the tension in the string is pulling it, maybe it's moving with a different acceleration.Wait, the string is connected over a pulley, so the tension T would cause the block to accelerate. But since the block is on the board, which is also accelerating, the net acceleration of the block relative to the ground would be a combination of the board's acceleration and the acceleration due to the tension.Wait, I think I need to consider the block's motion separately. Let's analyze the forces on the block.The block has mass m = 1 kg. The forces acting on it are:1. Tension T upwards (since it's connected over a pulley, assuming the pulley is fixed, the tension would pull the block towards the pulley, which is probably to the left if the board is moving to the right).2. Frictional force f_block from the board. Since the block is on the board, if the board is moving to the right with acceleration a, and if the block tends to stay in place due to inertia, there would be a frictional force acting on the block to the right, opposing the relative motion.Wait, but the block is connected by a string, so the tension might cause it to move. Hmm, this is getting a bit tangled.Alternatively, maybe the block is being pulled by the tension, which causes it to accelerate relative to the board. So, the block's acceleration relative to the board is different from the board's acceleration.Wait, perhaps it's better to consider the block and the board as separate systems and write equations for both.Let me start with the block.For the block:Forces acting on it:- Tension T upwards (assuming the pulley is fixed, so the tension is pulling the block towards the pulley, which is to the left if the board is moving to the right).- Frictional force f_block from the board. Since the block is on the board, if the board is moving to the right, and the block tends to stay in place, the frictional force on the block would be to the right, opposing the relative motion.Wait, but if the block is being pulled by tension to the left, then the frictional force would be to the right, opposing the motion caused by the tension.So, the net force on the block is T - f_block = m*a_block.But what's a_block? The block's acceleration relative to the ground. Since the board is moving to the right with acceleration a, and the block is being pulled to the left by tension, the block's acceleration relative to the ground would be a_block = a_board - a_tension_effect.Wait, maybe it's better to consider the block's acceleration relative to the board.Let me denote a_block_ground as the acceleration of the block relative to the ground, and a_board as the acceleration of the board.Then, the acceleration of the block relative to the board is a_block_board = a_block_ground - a_board.But since the block is on the board, if the block is moving with acceleration a_block_ground, and the board is moving with a_board, then the relative acceleration is a_block_board.But I think I'm complicating it. Let me try to write the equations.For the block:Sum of forces in the horizontal direction: T - f_block = m*a_block.Where f_block is the frictional force between the block and the board. Since the block is on the board, f_block = μ*m*g = 0.2*1*10 = 2 N.So, T - 2 = 1*a_block.But what's a_block? Is it the same as the board's acceleration? Or is it different?Wait, the board is moving with acceleration a = 2 m/s². If the block is on the board, and the tension is pulling it, the block could be moving with a different acceleration. But since the string is connected over a pulley, the tension would cause the block to accelerate in the opposite direction of the board's motion.Wait, maybe the block is moving with acceleration a_block = a + a_pulley_effect.Wait, I'm getting confused. Maybe I need to consider the entire system.Alternatively, perhaps the block is moving with the same acceleration as the board, but that seems unlikely because the tension is pulling it.Wait, no, if the block is on the board, and the board is moving, the block would tend to stay in place relative to the ground unless acted upon by another force. So, the tension is the force that would cause the block to accelerate relative to the board.So, the block's acceleration relative to the ground is a_block = a_board - a_tension_effect.Wait, maybe I should think in terms of relative acceleration.Let me denote:a_board = 2 m/s² to the right.a_block = acceleration of the block relative to the ground.Then, the acceleration of the block relative to the board is a_block - a_board.But since the block is connected by a string over a pulley, the tension would cause the block to accelerate in the opposite direction of the board's motion.Wait, perhaps the block's acceleration relative to the board is such that the string doesn't slip, so the acceleration of the block relative to the board is equal to the acceleration caused by the tension.Wait, maybe I should consider the block's acceleration as a_block = a_board - a, where a is the acceleration due to tension.But I'm not sure. Maybe it's better to write the equations for both the board and the block.For the board:Forces acting on it:- Applied force F to the right.- Frictional force from the table: f1 = μ*M*g = 0.2*4*10 = 8 N to the left.- Frictional force from the block: f2 = μ*m*g = 0.2*1*10 = 2 N to the left.- Tension T: Wait, is the tension acting on the board? Or is the tension only acting on the block?Wait, the string is connected to the block and goes over the pulley, so the tension is acting on the block, not directly on the board. So, the board only experiences F, f1, and f2.So, the net force on the board is F - f1 - f2 = M*a.So, F - 8 - 2 = 4*2.F - 10 = 8.So, F = 18 N.Wait, but that seems too straightforward. But wait, is that correct? Because the tension in the string is also affecting the block, which is on the board, so maybe the tension affects the frictional force on the board.Wait, no, the tension is acting on the block, not directly on the board. So, the frictional force on the board is due to the block's weight, which is m*g, so f2 = μ*m*g = 2 N, regardless of the tension.So, the net force on the board is F - f1 - f2 = M*a.So, F - 8 - 2 = 4*2 => F - 10 = 8 => F = 18 N.Hmm, but then the tension in the string is another part. So, for the block, the forces are tension T to the left, frictional force f_block to the right, and the net force is m*a_block.But what's a_block? Is it the same as the board's acceleration? Or is it different?Wait, if the block is on the board, and the board is moving to the right with acceleration a, then if the block is not sliding, it would also have acceleration a to the right. But the tension is pulling it to the left, so maybe the block is actually moving to the left relative to the board, meaning its acceleration relative to the ground is a_block = a_board - a_tension_effect.Wait, but how do we find a_block?Alternatively, maybe the block is moving with acceleration a_block = a - (T/m). Wait, not sure.Wait, let's write the equation for the block.For the block:Sum of forces = m*a_block.So, T - f_block = m*a_block.But what's a_block? It's the acceleration of the block relative to the ground.But the board is moving to the right with acceleration a = 2 m/s², so if the block is moving to the left relative to the board, its acceleration relative to the ground would be a_block = a_board - a_relative.Wait, maybe it's better to consider the relative acceleration.Let me denote a_block_relative = a_block - a_board.But since the block is being pulled by tension, the relative acceleration would be due to the tension and friction.So, for the block:T - f_block = m*a_block_relative.But a_block_relative = a_block - a_board.So, T - 2 = 1*(a_block - 2).But we also know that the block is connected to the board via the string, so the movement of the block is related to the movement of the board.Wait, perhaps the length of the string is constant, so the acceleration of the block relative to the board is equal to the acceleration caused by the tension.Wait, I'm getting stuck here. Maybe I should consider that the block's acceleration relative to the ground is equal to the board's acceleration minus the acceleration caused by the tension.Alternatively, perhaps the block is moving with the same acceleration as the board, but that doesn't make sense because the tension is pulling it.Wait, maybe the block is moving with acceleration a_block = a - (T/m). Hmm, not sure.Wait, let me think differently. The board is moving to the right with acceleration a = 2 m/s². The block is on the board, so if there were no tension, the block would move with the same acceleration due to friction. But since there's tension pulling it to the left, the block's acceleration relative to the ground would be less than a.So, the net acceleration of the block is a_block = a - (T/m - f_block/m).Wait, maybe not. Let's write the equation properly.For the block:Sum of forces in the horizontal direction: T - f_block = m*a_block.Where f_block = 2 N.So, T - 2 = 1*a_block.But what's a_block? It's the acceleration of the block relative to the ground.But the board is moving to the right with acceleration a = 2 m/s², so if the block is moving to the left relative to the board, its acceleration relative to the ground would be a_block = a_board - a_relative.Wait, but without knowing a_relative, it's hard to proceed.Alternatively, maybe the block's acceleration relative to the board is such that the tension causes it to accelerate at a certain rate.Wait, perhaps the key is that the block is moving with the same acceleration as the board, but that seems contradictory because the tension is pulling it.Wait, maybe the block is moving with the same acceleration as the board, but in the opposite direction? No, that can't be because the board is moving to the right, and the tension is pulling the block to the left, so the block would be moving to the left relative to the board, but its absolute acceleration would be a_block = a_board - a_tension_effect.Wait, I'm going in circles. Maybe I should consider that the block's acceleration is a_block = a - (T/m - f_block/m). Hmm, not sure.Wait, let me try to write the equations for both the board and the block.For the board:F - f1 - f2 = M*a.Where f1 = 8 N, f2 = 2 N, M = 4 kg, a = 2 m/s².So, F - 8 - 2 = 4*2 => F - 10 = 8 => F = 18 N.So, that's the first part, F = 18 N.But wait, is that correct? Because the tension in the string is also affecting the block, which is on the board, so maybe the frictional force on the board is not just f1 + f2, but also something else.Wait, no, the frictional force on the board is due to the block's weight, which is m*g, so f2 = μ*m*g = 2 N, regardless of the tension. The tension is acting on the block, not on the board.So, the net force on the board is F - f1 - f2 = M*a.So, F = M*a + f1 + f2 = 4*2 + 8 + 2 = 8 + 8 + 2 = 18 N.So, F = 18 N.Okay, so that's part 2 answered.Now, for part 1, the tension in the string.For the block:Sum of forces: T - f_block = m*a_block.We need to find T, so we need to find a_block.But what's a_block? It's the acceleration of the block relative to the ground.But the block is on the board, which is moving to the right with acceleration a = 2 m/s². If the block is being pulled by tension to the left, its acceleration relative to the ground would be a_block = a_board - a_tension_effect.Wait, but how do we find a_block?Alternatively, maybe the block is moving with the same acceleration as the board, but that can't be because the tension is pulling it.Wait, perhaps the block is moving with acceleration a_block = a - (T/m - f_block/m).Wait, let me think again.The block is subject to tension T to the left and friction f_block = 2 N to the right.So, the net force on the block is T - 2 = m*a_block.But the block is on the board, which is moving to the right with acceleration a = 2 m/s². So, if the block is moving to the left relative to the board, its absolute acceleration would be a_block = a_board - a_relative.But without knowing a_relative, it's hard to proceed.Wait, maybe the key is that the block is moving with the same acceleration as the board, but that can't be because the tension is pulling it. Alternatively, the block's acceleration is such that the tension and friction balance out to give it an acceleration relative to the board.Wait, perhaps the block's acceleration relative to the board is a_relative = (T - f_block)/m.So, a_relative = (T - 2)/1 = T - 2.But the board is moving to the right with acceleration a = 2 m/s², so the block's absolute acceleration is a_block = a_board - a_relative = 2 - (T - 2) = 4 - T.But wait, that seems odd because if T increases, a_block decreases, which makes sense because more tension would pull the block more to the left, reducing its acceleration relative to the ground.But we also know that the block is connected to the board via the string, so the movement of the block is related to the movement of the board.Wait, maybe the length of the string is such that the block's movement is directly related to the board's movement.Wait, if the board moves to the right, the pulley is fixed, so the block would move to the left. So, the block's acceleration is in the opposite direction of the board's acceleration.So, if the board is accelerating to the right with a = 2 m/s², the block is accelerating to the left with some acceleration, say, a_block.But how is a_block related to the board's acceleration?Wait, perhaps the block's acceleration is equal in magnitude to the board's acceleration, but in the opposite direction. So, a_block = -a = -2 m/s².But that might not be the case because the tension and friction would affect it.Wait, let's test this idea.If a_block = -2 m/s², then:T - 2 = 1*(-2) => T - 2 = -2 => T = 0.But that can't be right because the tension can't be zero if the block is moving.So, that idea is wrong.Alternatively, maybe the block's acceleration is such that the tension causes it to accelerate at a certain rate, while the board is accelerating at 2 m/s².Wait, perhaps the block's acceleration is a_block = a - (T/m - f_block/m).Wait, not sure.Alternatively, maybe the block's acceleration is a_block = a - (T/m).But that doesn't account for friction.Wait, let's go back to the equation for the block:T - f_block = m*a_block.We have T - 2 = 1*a_block.But we need another equation to relate a_block to the board's acceleration.Wait, since the block is on the board, and the board is moving to the right with acceleration a = 2 m/s², the block's acceleration relative to the ground is a_block = a_board - a_relative.But a_relative is the acceleration of the block relative to the board, which is caused by the tension and friction.So, a_relative = (T - f_block)/m = (T - 2)/1 = T - 2.Therefore, a_block = a_board - a_relative = 2 - (T - 2) = 4 - T.So, a_block = 4 - T.But from the block's equation:T - 2 = a_block.So, substituting a_block = 4 - T into this equation:T - 2 = 4 - T.So, T - 2 = 4 - T.Adding T to both sides:2T - 2 = 4.Adding 2 to both sides:2T = 6.So, T = 3 N.Okay, so the tension in the string is 3 N.Let me check if this makes sense.If T = 3 N, then from the block's equation:3 - 2 = 1*a_block => a_block = 1 m/s².But the board is moving with a = 2 m/s² to the right, so the block's acceleration relative to the board is a_relative = a_block - a_board = 1 - 2 = -1 m/s², meaning the block is accelerating to the left relative to the board at 1 m/s².Which makes sense because the tension is pulling it to the left, causing it to slow down relative to the board.So, the tension is 3 N.And earlier, we found F = 18 N.So, that seems consistent.Wait, but let me double-check the board's equation.F - f1 - f2 = M*a.F - 8 - 2 = 4*2 => F - 10 = 8 => F = 18 N.Yes, that's correct.And for the block:T - f_block = m*a_block => 3 - 2 = 1*1 => 1 = 1, which checks out.So, the tension is 3 N, and the force F is 18 N.I think that's the answer.

answer：Alright, so I've got this problem here about a convex quadrilateral (ABCD). The first thing I notice is that (overrightarrow{BC} = 2 overrightarrow{AD}). Hmm, okay, so vector BC is twice vector AD. That probably means that BC is twice as long as AD and in the same direction, right? So maybe BC is a scaled version of AD.Then there's a point (P) somewhere in the plane such that (overrightarrow{PA} + 2020 overrightarrow{PB} + overrightarrow{PC} + 2020 overrightarrow{PD} = mathbf{0}). That looks like a vector equation involving point (P) and the vertices of the quadrilateral. I remember that in vector problems, sometimes you can use the concept of weighted averages or barycentric coordinates to find the position of a point given such equations.The question is asking for the ratio (frac{t}{s}), where (s) is the area of quadrilateral (ABCD) and (t) is the area of triangle (PAB). So, I need to relate the position of point (P) to the areas involved.Let me try to rewrite the vector equation to find the position vector of (P). The equation is:[overrightarrow{PA} + 2020 overrightarrow{PB} + overrightarrow{PC} + 2020 overrightarrow{PD} = mathbf{0}]I can express each of these vectors in terms of position vectors relative to some origin. Let me denote the position vectors of points (A, B, C, D, P) as (vec{A}, vec{B}, vec{C}, vec{D}, vec{P}) respectively. Then, (overrightarrow{PA} = vec{A} - vec{P}), (overrightarrow{PB} = vec{B} - vec{P}), and so on.Substituting these into the equation:[(vec{A} - vec{P}) + 2020 (vec{B} - vec{P}) + (vec{C} - vec{P}) + 2020 (vec{D} - vec{P}) = mathbf{0}]Let me expand this:[vec{A} - vec{P} + 2020 vec{B} - 2020 vec{P} + vec{C} - vec{P} + 2020 vec{D} - 2020 vec{P} = mathbf{0}]Now, let's collect like terms. The terms with (vec{A}, vec{B}, vec{C}, vec{D}) are straightforward:[vec{A} + 2020 vec{B} + vec{C} + 2020 vec{D}]And the terms with (vec{P}):[- vec{P} - 2020 vec{P} - vec{P} - 2020 vec{P} = (-1 - 2020 -1 -2020) vec{P} = (-4042) vec{P}]So putting it all together:[vec{A} + 2020 vec{B} + vec{C} + 2020 vec{D} - 4042 vec{P} = mathbf{0}]Let me solve for (vec{P}):[4042 vec{P} = vec{A} + 2020 vec{B} + vec{C} + 2020 vec{D}][vec{P} = frac{1}{4042} vec{A} + frac{2020}{4042} vec{B} + frac{1}{4042} vec{C} + frac{2020}{4042} vec{D}]Simplify the coefficients:Notice that (2020 = frac{4040}{2}), so (2020/4042 = 1010/2021). Similarly, (1/4042 = 1/4042). So,[vec{P} = frac{1}{4042} vec{A} + frac{1010}{2021} vec{B} + frac{1}{4042} vec{C} + frac{1010}{2021} vec{D}]Hmm, that seems a bit messy. Maybe I can factor out 1/4042:[vec{P} = frac{1}{4042} (vec{A} + 2020 vec{B} + vec{C} + 2020 vec{D})]Alternatively, I can write it as:[vec{P} = frac{vec{A} + vec{C}}{4042} + frac{2020 (vec{B} + vec{D})}{4042}]Simplify the coefficients:[frac{1}{4042} = frac{1}{2 times 2021}][frac{2020}{4042} = frac{1010}{2021}]So,[vec{P} = frac{vec{A} + vec{C}}{2 times 2021} + frac{1010}{2021} (vec{B} + vec{D})]Hmm, not sure if that helps yet. Maybe I should think about the quadrilateral (ABCD) and the relation (overrightarrow{BC} = 2 overrightarrow{AD}).Let me write that vector equation:[overrightarrow{BC} = 2 overrightarrow{AD}]Expressed in terms of position vectors:[vec{C} - vec{B} = 2 (vec{D} - vec{A})][vec{C} = vec{B} + 2 vec{D} - 2 vec{A}]So, we can express (vec{C}) in terms of (vec{A}, vec{B}, vec{D}). Maybe substitute this into the expression for (vec{P}).From earlier, we have:[vec{P} = frac{vec{A} + vec{C}}{4042} + frac{2020 (vec{B} + vec{D})}{4042}]Substitute (vec{C} = vec{B} + 2 vec{D} - 2 vec{A}):[vec{P} = frac{vec{A} + (vec{B} + 2 vec{D} - 2 vec{A})}{4042} + frac{2020 (vec{B} + vec{D})}{4042}]Simplify the numerator in the first term:[vec{A} + vec{B} + 2 vec{D} - 2 vec{A} = -vec{A} + vec{B} + 2 vec{D}]So,[vec{P} = frac{ -vec{A} + vec{B} + 2 vec{D} }{4042} + frac{2020 (vec{B} + vec{D})}{4042}]Combine the terms:Let me write both terms with denominator 4042:[vec{P} = frac{ -vec{A} + vec{B} + 2 vec{D} + 2020 vec{B} + 2020 vec{D} }{4042}]Combine like terms:- For (vec{A}): (-1 vec{A})- For (vec{B}): (1 + 2020 = 2021 vec{B})- For (vec{D}): (2 + 2020 = 2022 vec{D})So,[vec{P} = frac{ -vec{A} + 2021 vec{B} + 2022 vec{D} }{4042}]Hmm, maybe factor out 2021 from some terms:Wait, 2022 is 2021 + 1. So,[vec{P} = frac{ -vec{A} + 2021 vec{B} + (2021 + 1) vec{D} }{4042}][= frac{ -vec{A} + 2021 vec{B} + 2021 vec{D} + vec{D} }{4042}][= frac{ -vec{A} + 2021 (vec{B} + vec{D}) + vec{D} }{4042}]Not sure if that helps. Maybe another approach.Alternatively, let's think about the quadrilateral (ABCD). Since (overrightarrow{BC} = 2 overrightarrow{AD}), maybe (ABCD) is a trapezoid or something? Or perhaps it's a parallelogram? Wait, in a parallelogram, opposite sides are equal, but here BC is twice AD, so not a parallelogram.Alternatively, maybe we can express the position of (P) in terms of the centroid or something.Wait, the vector equation given is a weighted sum of vectors from (P) to each vertex. It might be helpful to think of this as a weighted centroid.In general, for weights (w_A, w_B, w_C, w_D), the centroid (G) is given by:[vec{G} = frac{w_A vec{A} + w_B vec{B} + w_C vec{C} + w_D vec{D}}{w_A + w_B + w_C + w_D}]Comparing this to our equation, which is:[overrightarrow{PA} + 2020 overrightarrow{PB} + overrightarrow{PC} + 2020 overrightarrow{PD} = mathbf{0}]Which can be rewritten as:[(vec{A} - vec{P}) + 2020 (vec{B} - vec{P}) + (vec{C} - vec{P}) + 2020 (vec{D} - vec{P}) = mathbf{0}]Which simplifies to:[vec{A} + 2020 vec{B} + vec{C} + 2020 vec{D} - 4042 vec{P} = mathbf{0}][vec{P} = frac{vec{A} + 2020 vec{B} + vec{C} + 2020 vec{D}}{4042}]So, yes, this is like a centroid with weights 1, 2020, 1, 2020 on points (A, B, C, D) respectively.So, (P) is the weighted centroid with weights 1, 2020, 1, 2020.But how does this help me find the area ratio?Maybe I can use the concept that the area of triangle (PAB) relates to the areas of other triangles in the quadrilateral.Alternatively, since we have a weighted centroid, perhaps the areas can be related through the weights.Wait, another thought: maybe express point (P) as a combination of other points in the quadrilateral.Given that (overrightarrow{BC} = 2 overrightarrow{AD}), which we can write as (vec{C} - vec{B} = 2 (vec{D} - vec{A})), so (vec{C} = vec{B} + 2 vec{D} - 2 vec{A}).So, maybe substitute this into the expression for (vec{P}):From earlier, we had:[vec{P} = frac{vec{A} + 2020 vec{B} + vec{C} + 2020 vec{D}}{4042}][= frac{vec{A} + 2020 vec{B} + (vec{B} + 2 vec{D} - 2 vec{A}) + 2020 vec{D}}{4042}]Simplify numerator:[vec{A} + 2020 vec{B} + vec{B} + 2 vec{D} - 2 vec{A} + 2020 vec{D}][= (vec{A} - 2 vec{A}) + (2020 vec{B} + vec{B}) + (2 vec{D} + 2020 vec{D})][= (-vec{A}) + 2021 vec{B} + 2022 vec{D}]So,[vec{P} = frac{ -vec{A} + 2021 vec{B} + 2022 vec{D} }{4042}]Hmm, maybe factor out 2021:[= frac{ -vec{A} + 2021 (vec{B} + vec{D}) + vec{D} }{4042}][= frac{ -vec{A} + 2021 (vec{B} + vec{D}) + vec{D} }{4042}][= frac{ -vec{A} + 2021 vec{B} + 2021 vec{D} + vec{D} }{4042}][= frac{ -vec{A} + 2021 vec{B} + (2021 + 1) vec{D} }{4042}][= frac{ -vec{A} + 2021 vec{B} + 2022 vec{D} }{4042}]I think this is as simplified as it gets. Maybe express this in terms of vectors from point (A) or something.Alternatively, perhaps express (vec{P}) in terms of vectors (vec{AB}) and (vec{AD}), since we have a relation between (overrightarrow{BC}) and (overrightarrow{AD}).Let me denote (vec{AB} = vec{B} - vec{A}) and (vec{AD} = vec{D} - vec{A}). Then, from the given condition:[overrightarrow{BC} = 2 overrightarrow{AD}][vec{C} - vec{B} = 2 (vec{D} - vec{A})][vec{C} = vec{B} + 2 vec{D} - 2 vec{A}]So, we can express (vec{C}) in terms of (vec{A}, vec{B}, vec{D}). Maybe express all position vectors in terms of (vec{A}), (vec{B}), and (vec{D}).So, let's rewrite (vec{P}):[vec{P} = frac{ -vec{A} + 2021 vec{B} + 2022 vec{D} }{4042}]Let me factor out (vec{A}):[= frac{ -vec{A} + 2021 vec{B} + 2022 vec{D} }{4042}][= frac{ -vec{A} + 2021 (vec{B} - vec{A}) + 2022 (vec{D} - vec{A}) + 2021 vec{A} + 2022 vec{A} }{4042}]Wait, that might complicate things. Alternatively, express (vec{P}) as a combination of (vec{A}), (vec{B}), and (vec{D}):[vec{P} = frac{ -vec{A} + 2021 vec{B} + 2022 vec{D} }{4042}][= frac{ -vec{A} }{4042} + frac{2021}{4042} vec{B} + frac{2022}{4042} vec{D}][= -frac{1}{4042} vec{A} + frac{2021}{4042} vec{B} + frac{2022}{4042} vec{D}]Simplify the coefficients:[frac{2021}{4042} = frac{2021}{2 times 2021} = frac{1}{2}][frac{2022}{4042} = frac{2022}{2 times 2021} = frac{1011}{2021}][frac{1}{4042} = frac{1}{2 times 2021}]So,[vec{P} = -frac{1}{2 times 2021} vec{A} + frac{1}{2} vec{B} + frac{1011}{2021} vec{D}]Hmm, that might be helpful. Let me write this as:[vec{P} = frac{1}{2} vec{B} + left( frac{1011}{2021} vec{D} - frac{1}{2 times 2021} vec{A} right)]But not sure yet. Maybe another approach: express (vec{P}) in terms of vectors from (A).Let me denote (vec{AB} = vec{B} - vec{A}) and (vec{AD} = vec{D} - vec{A}). Then, any point can be expressed as (vec{A} + x vec{AB} + y vec{AD}) for some scalars (x, y).So, let's express (vec{P}) in terms of (vec{A}), (vec{AB}), and (vec{AD}):From earlier,[vec{P} = -frac{1}{2 times 2021} vec{A} + frac{1}{2} vec{B} + frac{1011}{2021} vec{D}]Express (vec{B}) and (vec{D}) in terms of (vec{A}), (vec{AB}), and (vec{AD}):[vec{B} = vec{A} + vec{AB}][vec{D} = vec{A} + vec{AD}]Substitute into (vec{P}):[vec{P} = -frac{1}{2 times 2021} vec{A} + frac{1}{2} (vec{A} + vec{AB}) + frac{1011}{2021} (vec{A} + vec{AD})]Expand:[= -frac{1}{4042} vec{A} + frac{1}{2} vec{A} + frac{1}{2} vec{AB} + frac{1011}{2021} vec{A} + frac{1011}{2021} vec{AD}]Combine like terms for (vec{A}):[left( -frac{1}{4042} + frac{1}{2} + frac{1011}{2021} right) vec{A} + frac{1}{2} vec{AB} + frac{1011}{2021} vec{AD}]Compute the coefficient for (vec{A}):First, convert all terms to have denominator 4042:- (-frac{1}{4042})- (frac{1}{2} = frac{2021}{4042})- (frac{1011}{2021} = frac{2022}{4042})So,[-frac{1}{4042} + frac{2021}{4042} + frac{2022}{4042} = frac{ -1 + 2021 + 2022 }{4042} = frac{4042}{4042} = 1]So, the coefficient for (vec{A}) is 1.Therefore,[vec{P} = vec{A} + frac{1}{2} vec{AB} + frac{1011}{2021} vec{AD}]So, in terms of vectors from (A), (P) is located at:[vec{P} = vec{A} + frac{1}{2} vec{AB} + frac{1011}{2021} vec{AD}]That's interesting. So, relative to point (A), point (P) is halfway along (AB) and (frac{1011}{2021}) along (AD).So, if I imagine the quadrilateral (ABCD), with (A) as the origin, (AB) as one side, and (AD) as another side, then (P) is somewhere inside the quadrilateral, halfway between (A) and (B), and almost halfway between (A) and (D) (since (frac{1011}{2021}) is just under 0.5).Wait, actually, (frac{1011}{2021}) is approximately 0.5, but slightly less. Because 2021 divided by 2 is 1010.5, so 1011 is just 0.5 more than half. So, actually, (frac{1011}{2021} = frac{1010.5 + 0.5}{2021} = frac{1}{2} + frac{0.5}{2021}), which is just a tiny bit more than half. So, point (P) is very slightly beyond the midpoint towards (D).But maybe that's not too important right now. The key is that (P) is expressed in terms of vectors from (A), so maybe I can use this to find the area of triangle (PAB).Since (P) is expressed as (vec{A} + frac{1}{2} vec{AB} + frac{1011}{2021} vec{AD}), perhaps I can find the coordinates of (P) relative to (A), (B), and (D), and then compute the area.Alternatively, maybe using barycentric coordinates or area ratios.Wait, another idea: since (P) is expressed in terms of vectors from (A), perhaps I can express the position of (P) as a combination of points (A), (B), and (D), and then use the formula for the area of a triangle given by three points.But first, let's note that the area (s) of quadrilateral (ABCD) can be expressed as the sum of areas of triangles (ABC) and (ADC), or alternatively, as the sum of triangles (ABD) and (BCD). But given the relation (overrightarrow{BC} = 2 overrightarrow{AD}), maybe there's a way to express the area in terms of triangle (ABD) or something.Wait, let's try to find the area of quadrilateral (ABCD). Since (overrightarrow{BC} = 2 overrightarrow{AD}), which is a vector equation, so BC is twice AD in length and direction.So, if I consider vectors, then the length of BC is twice that of AD, and they are parallel.So, quadrilateral (ABCD) has sides (AB), (BC), (CD), (DA), with (BC) parallel to (AD) and twice its length.Wait, that sounds like a trapezoid, but with only one pair of sides parallel. But in a trapezoid, both pairs of sides are not necessarily parallel, just one pair. So, in this case, (AD) and (BC) are parallel, with (BC = 2 AD).So, quadrilateral (ABCD) is a trapezoid with bases (AD) and (BC), where (BC = 2 AD).Therefore, the area of trapezoid (ABCD) is given by the average of the two bases times the height. But since we don't have specific lengths, maybe we can express the area in terms of the area of triangle (ABD) or something.Alternatively, since we have vectors, maybe express the area using cross products.But perhaps it's better to use coordinates. Let me assign coordinates to the quadrilateral to make things more concrete.Let me place point (A) at the origin, so (vec{A} = mathbf{0}). Let me denote (vec{AB} = vec{b}) and (vec{AD} = vec{d}). Then, point (B) is at (vec{b}), point (D) is at (vec{d}), and point (C) can be found using the relation (overrightarrow{BC} = 2 overrightarrow{AD}).Since (overrightarrow{BC} = vec{C} - vec{B} = 2 vec{AD} = 2 vec{d}), so:[vec{C} = vec{B} + 2 vec{d} = vec{b} + 2 vec{d}]So, coordinates of the points:- (A = (0, 0))- (B = vec{b})- (D = vec{d})- (C = vec{b} + 2 vec{d})Now, point (P) is given by:[vec{P} = vec{A} + frac{1}{2} vec{AB} + frac{1011}{2021} vec{AD} = frac{1}{2} vec{b} + frac{1011}{2021} vec{d}]So, in coordinates, (P) is at (left( frac{1}{2} vec{b} + frac{1011}{2021} vec{d} right)).Now, let's compute the area of quadrilateral (ABCD). Since it's a trapezoid with bases (AD) and (BC), and the height is the distance between these two bases.But since we're working with vectors, maybe the area can be expressed using the cross product.The area of trapezoid (ABCD) is equal to the magnitude of (frac{ (vec{AB} times vec{AD}) + (vec{BC} times vec{CD}) }{2}), but I might be mixing things up.Wait, actually, the area of a trapezoid is the average of the two bases times the height. In vector terms, the area can be calculated as the magnitude of the cross product of the diagonals or something else. Maybe it's better to compute it as the sum of two triangles.Quadrilateral (ABCD) can be divided into triangles (ABC) and (ADC). Let's compute the area of each triangle.Area of triangle (ABC):Vectors (AB = vec{b}) and (AC = vec{C} - vec{A} = vec{b} + 2 vec{d}). The area is (frac{1}{2} | vec{AB} times vec{AC} | = frac{1}{2} | vec{b} times (vec{b} + 2 vec{d}) | = frac{1}{2} | vec{b} times vec{b} + 2 vec{b} times vec{d} | = frac{1}{2} | 0 + 2 vec{b} times vec{d} | = | vec{b} times vec{d} | ).Similarly, area of triangle (ADC):Vectors (AD = vec{d}) and (AC = vec{b} + 2 vec{d}). The area is (frac{1}{2} | vec{AD} times vec{AC} | = frac{1}{2} | vec{d} times (vec{b} + 2 vec{d}) | = frac{1}{2} | vec{d} times vec{b} + 2 vec{d} times vec{d} | = frac{1}{2} | - vec{b} times vec{d} + 0 | = frac{1}{2} | vec{b} times vec{d} | ).Therefore, the total area (s) of quadrilateral (ABCD) is:[s = | vec{b} times vec{d} | + frac{1}{2} | vec{b} times vec{d} | = frac{3}{2} | vec{b} times vec{d} |]So, (s = frac{3}{2} | vec{b} times vec{d} |).Now, let's compute the area (t) of triangle (PAB).Points (A), (B), and (P):- (A = (0, 0))- (B = vec{b})- (P = frac{1}{2} vec{b} + frac{1011}{2021} vec{d})The area of triangle (PAB) can be found using the cross product of vectors (AB) and (AP).Vector (AB = vec{b}), vector (AP = vec{P} - vec{A} = frac{1}{2} vec{b} + frac{1011}{2021} vec{d}).The area is (frac{1}{2} | vec{AB} times vec{AP} | = frac{1}{2} | vec{b} times left( frac{1}{2} vec{b} + frac{1011}{2021} vec{d} right) | ).Compute the cross product:[vec{b} times left( frac{1}{2} vec{b} + frac{1011}{2021} vec{d} right) = frac{1}{2} vec{b} times vec{b} + frac{1011}{2021} vec{b} times vec{d}][= frac{1}{2} cdot 0 + frac{1011}{2021} vec{b} times vec{d} = frac{1011}{2021} vec{b} times vec{d}]Therefore, the area (t) is:[t = frac{1}{2} left| frac{1011}{2021} vec{b} times vec{d} right| = frac{1}{2} cdot frac{1011}{2021} | vec{b} times vec{d} | = frac{1011}{4042} | vec{b} times vec{d} |]Simplify (frac{1011}{4042}):Note that 4042 = 2 × 2021, so:[frac{1011}{4042} = frac{1011}{2 times 2021} = frac{1011}{4042}]But 1011 and 2021: let's see if they have a common factor. 2021 ÷ 1011 ≈ 2 with remainder 2021 - 2×1011 = 2021 - 2022 = -1. So, gcd(1011, 2021) is 1, since 2021 = 2×1011 -1, so gcd(1011,1)=1.Therefore, (frac{1011}{4042}) is already in simplest terms.So, (t = frac{1011}{4042} | vec{b} times vec{d} |).Earlier, we found that (s = frac{3}{2} | vec{b} times vec{d} |).Therefore, the ratio (frac{t}{s}) is:[frac{t}{s} = frac{ frac{1011}{4042} | vec{b} times vec{d} | }{ frac{3}{2} | vec{b} times vec{d} | } = frac{1011}{4042} times frac{2}{3} = frac{1011 times 2}{4042 times 3}]Simplify numerator and denominator:Numerator: 1011 × 2 = 2022Denominator: 4042 × 3 = 12126So,[frac{2022}{12126}]Simplify this fraction:Divide numerator and denominator by 2022:2022 ÷ 2022 = 112126 ÷ 2022 = 6 (since 2022 × 6 = 12132, which is 6 more than 12126. Wait, that's not exact. Wait, 2022 × 6 = 12132, which is 6 more than 12126, so actually, 12126 = 2022 × 6 - 6 = 2022 × 5.997, which is not an integer. Hmm, maybe another approach.Wait, let's factor numerator and denominator:2022 = 2 × 3 × 33712126 = 2 × 3 × 3 × 7 × 31Wait, let me check:2022 ÷ 2 = 10111011 ÷ 3 = 337, which is prime.12126 ÷ 2 = 60636063 ÷ 3 = 20212021 ÷ 43 = 47 (since 43 × 47 = 2021)So,2022 = 2 × 3 × 33712126 = 2 × 3 × 43 × 47So, the greatest common divisor (GCD) of 2022 and 12126 is 2 × 3 = 6.Therefore, divide numerator and denominator by 6:2022 ÷ 6 = 33712126 ÷ 6 = 2021So,[frac{2022}{12126} = frac{337}{2021}]Therefore, (frac{t}{s} = frac{337}{2021}).Wait, but let me verify the calculation:We have (t = frac{1011}{4042} | vec{b} times vec{d} |) and (s = frac{3}{2} | vec{b} times vec{d} |).So,[frac{t}{s} = frac{1011}{4042} times frac{2}{3} = frac{1011 times 2}{4042 times 3} = frac{2022}{12126}]As above, simplifying gives (frac{337}{2021}).But let me check 2022 ÷ 6 is 337, since 6 × 337 = 2022.Similarly, 12126 ÷ 6 = 2021, since 6 × 2021 = 12126.Yes, so (frac{2022}{12126} = frac{337}{2021}).Therefore, the ratio (frac{t}{s} = frac{337}{2021}).But let me check if 337 and 2021 have any common factors. 337 is a prime number (since it's not divisible by 2, 3, 5, 7, 11, 13, 17, 19; 19×17=323, 19×18=342>337). 2021 ÷ 337 ≈ 6. So, 337 × 6 = 2022, which is 1 more than 2021, so no, 337 doesn't divide 2021. Therefore, the fraction is in simplest terms.So, the answer is (frac{337}{2021}).But wait, let me double-check my earlier steps because 337 seems a bit arbitrary.Wait, in the area of triangle (PAB), I had:[t = frac{1}{2} times frac{1011}{2021} | vec{b} times vec{d} | = frac{1011}{4042} | vec{b} times vec{d} |]And the area of quadrilateral (ABCD) was:[s = frac{3}{2} | vec{b} times vec{d} |]So, the ratio is:[frac{t}{s} = frac{1011}{4042} div frac{3}{2} = frac{1011}{4042} times frac{2}{3} = frac{1011 times 2}{4042 times 3} = frac{2022}{12126} = frac{337}{2021}]Yes, that seems consistent.Alternatively, maybe there's a simpler way to see this ratio without going through coordinates.Given that (P) is a weighted centroid, the areas might relate through the weights. But since the weights are 1, 2020, 1, 2020, the centroid divides the figure in a certain ratio.But in this case, since the area (t) is just the area of triangle (PAB), and (s) is the area of the entire quadrilateral, the ratio is a fraction determined by the position of (P).Given that (P) is closer to (B) and (D), but in our coordinate system, it's expressed as a combination of (AB) and (AD). The key was expressing (P) in terms of vectors from (A), then computing the cross product areas.So, I think the answer is (frac{337}{2021}).But wait, 337 is a prime number, and 2021 is 43×47, so they don't share any common factors. So, yes, that's the simplest form.Therefore, the ratio (frac{t}{s} = frac{337}{2021}).**Final Answer**boxed{dfrac{337}{2021}}

answer：First, I need to calculate the total amount of food the puppy will eat over the next four weeks, including today.Today, Lily has already fed the puppy 1/2 cup of food.For the first two weeks, starting tomorrow, Lily will feed the puppy 1/4 cup three times a day. So, each day the puppy will eat 3/4 cup. Over two weeks, that's 14 days, resulting in a total of 10.5 cups.For the following two weeks, Lily will feed the puppy 1/2 cup twice a day. Each day, the puppy will eat 1 cup. Over another 14 days, this amounts to 14 cups.Adding today's feeding to the totals for the first and second two weeks, the puppy will consume a total of 25 cups of food over the next four weeks.