💡Okay, so I have this problem where a company produces two types of barrels, A and B. Each product requires different amounts of materials A and B, and each gives a different profit. The company wants to maximize their daily profit without exceeding the daily material consumption limits. Let me try to figure this out step by step.First, let me list down the given information:- **Product A:** - Material A required: 1 kg - Material B required: 2 kg - Profit per barrel: 300 yuan- **Product B:** - Material A required: 2 kg - Material B required: 1 kg - Profit per barrel: 400 yuan- **Daily material constraints:** - Material A: 12 kg - Material B: 12 kgSo, the company can use up to 12 kg of each material daily. I need to find how many barrels of A and B they should produce to maximize their profit.Hmm, this sounds like a linear programming problem. I remember that linear programming involves setting up inequalities based on constraints and then finding the maximum or minimum of an objective function. In this case, the objective function would be the profit, which we want to maximize.Let me define variables for the number of barrels produced:Let ( x ) = number of barrels of product A produced daily.Let ( y ) = number of barrels of product B produced daily.Now, I need to write the constraints based on the materials.For Material A:Each barrel of A uses 1 kg, and each barrel of B uses 2 kg. The total used can't exceed 12 kg.So, the constraint is:( 1x + 2y leq 12 )For Material B:Each barrel of A uses 2 kg, and each barrel of B uses 1 kg. The total used can't exceed 12 kg.So, the constraint is:( 2x + 1y leq 12 )Also, since the number of barrels can't be negative:( x geq 0 )( y geq 0 )Now, the objective function is the total profit, which is:( text{Profit} = 300x + 400y )We need to maximize this profit.So, summarizing the problem:Maximize ( 300x + 400y )Subject to:1. ( x + 2y leq 12 )2. ( 2x + y leq 12 )3. ( x geq 0 )4. ( y geq 0 )I think the next step is to graph these inequalities to find the feasible region, and then evaluate the profit function at each corner point of this region to find the maximum.Let me try to graph the constraints.First, for the constraint ( x + 2y leq 12 ):If ( x = 0 ), then ( 2y = 12 ) => ( y = 6 ).If ( y = 0 ), then ( x = 12 ).So, this line connects the points (0,6) and (12,0).Next, for the constraint ( 2x + y leq 12 ):If ( x = 0 ), then ( y = 12 ).If ( y = 0 ), then ( 2x = 12 ) => ( x = 6 ).So, this line connects the points (0,12) and (6,0).Now, considering both constraints along with ( x geq 0 ) and ( y geq 0 ), the feasible region is a polygon bounded by these lines and the axes.I need to find the intersection points of these two lines to determine all the corner points of the feasible region.To find the intersection of ( x + 2y = 12 ) and ( 2x + y = 12 ):Let me solve these two equations simultaneously.From the first equation:( x = 12 - 2y )Substitute into the second equation:( 2(12 - 2y) + y = 12 )( 24 - 4y + y = 12 )( 24 - 3y = 12 )( -3y = 12 - 24 )( -3y = -12 )( y = 4 )Now, substitute ( y = 4 ) back into ( x = 12 - 2y ):( x = 12 - 2(4) = 12 - 8 = 4 )So, the intersection point is (4,4).Therefore, the corner points of the feasible region are:1. (0,0) – Origin2. (0,6) – From the first constraint when x=03. (4,4) – Intersection point4. (6,0) – From the second constraint when y=0Wait, but hold on. When I look at the second constraint, when y=0, x=6. But the first constraint when y=0 is x=12, but since the second constraint is more restrictive, the feasible region is actually bounded by x=6 on the x-axis.Similarly, for the y-axis, the first constraint allows y=6, while the second constraint allows y=12, but since the first constraint is more restrictive, the feasible region is bounded by y=6 on the y-axis.So, the feasible region is a quadrilateral with vertices at (0,0), (0,6), (4,4), and (6,0).Wait, but actually, when I plot both lines, I think the feasible region is a polygon with these four points.But let me double-check.At (0,0): satisfies all constraints.At (0,6): Check second constraint: 2(0) + 6 = 6 ≤12, which is true.At (4,4): Check both constraints:1. 4 + 2(4) = 12 ≤12 ✔️2. 2(4) + 4 = 12 ≤12 ✔️At (6,0): Check first constraint: 6 + 2(0) =6 ≤12 ✔️So, yes, these four points form the feasible region.Now, I need to calculate the profit at each of these corner points.1. At (0,0):Profit = 300(0) + 400(0) = 0 yuan.2. At (0,6):Profit = 300(0) + 400(6) = 0 + 2400 = 2400 yuan.3. At (4,4):Profit = 300(4) + 400(4) = 1200 + 1600 = 2800 yuan.4. At (6,0):Profit = 300(6) + 400(0) = 1800 + 0 = 1800 yuan.So, comparing these profits: 0, 2400, 2800, 1800.The maximum profit is 2800 yuan at the point (4,4).Therefore, the company should produce 4 barrels of product A and 4 barrels of product B daily to achieve the maximum profit of 2800 yuan.Wait, but just to make sure I didn't make any mistakes in calculations.Let me verify the constraints at (4,4):Material A: 1*4 + 2*4 = 4 + 8 = 12 kg ✔️Material B: 2*4 + 1*4 = 8 + 4 = 12 kg ✔️Yes, both materials are exactly used up, which makes sense since it's the intersection point of the two constraints.So, the calculations seem correct.I think that's the solution.**Final Answer**The maximum profit the company can obtain is boxed{2800} yuan.

💡Alright, so I have this problem: I need to prove that the area of a rectangle inscribed in a triangle doesn't exceed half the area of the triangle. Hmm, okay. Let me try to visualize this. Imagine a triangle, and inside it, somewhere, there's a rectangle touching the sides of the triangle. The rectangle can be of different sizes, but I need to show that no matter how I place it, its area can't be more than half of the triangle's area.First, maybe I should consider a specific type of triangle to make things simpler. Let's start with a right-angled triangle because it might have some properties that are easier to work with. So, suppose I have a right triangle ABC with the right angle at point C. Let's say the legs are AC and BC, and the hypotenuse is AB. Now, if I inscribe a rectangle inside this triangle, one of its sides will lie along the base, say AC, and the opposite side will touch the hypotenuse AB.Let me denote the rectangle as DEFG, where D is on AC, E is on AB, F is on BC, and G is on AC again. Wait, actually, maybe it's simpler to have one vertex at the right angle C. So, let's say the rectangle has one vertex at C, one on AC, one on BC, and one on AB. That might make the calculations easier.So, let's say CD is along AC, and CF is along BC. Then, the rectangle would have sides CD and CF, and the fourth vertex E would be on AB. Let me assign some variables here. Let’s let CD = x and CF = y. Then, the area of the rectangle would be x times y, right?Now, since the rectangle is inside the triangle, the point E must lie on AB. So, I can use similar triangles to relate x and y. The triangle ABC is similar to the triangle formed above the rectangle, which I'll call triangle ADE. Because both are right triangles and share the angle at A, they are similar.So, the ratio of the sides should be the same. In triangle ABC, the base is AC and the height is BC. In triangle ADE, the base is AD and the height is AE. Since they are similar, the ratio of AD to AC should be equal to the ratio of AE to BC. So, AD/AC = AE/BC.But AD is AC minus CD, which is b - x, where b is the length of AC. Similarly, AE is BC minus CF, which is a - y, where a is the length of BC. So, (b - x)/b = (a - y)/a.Cross-multiplying, we get a(b - x) = b(a - y). Simplifying, ab - ax = ab - by. Subtracting ab from both sides, we get -ax = -by, which simplifies to ax = by. So, y = (a/b)x.Okay, so now we have y in terms of x. The area of the rectangle is x times y, which is x*(a/b)x = (a/b)x². So, the area is a quadratic function of x, specifically (a/b)x².To find the maximum area, I can take the derivative of this function with respect to x and set it equal to zero. But wait, since it's a quadratic function, it's a parabola opening upwards if the coefficient is positive, which it is here because a and b are positive lengths. So, actually, this function doesn't have a maximum; it increases as x increases. That doesn't make sense because the rectangle can't extend beyond the triangle.Hmm, maybe I made a mistake in setting up the relationship between x and y. Let me go back. I had (b - x)/b = (a - y)/a, which led to y = (a/b)x. But if I plug y = (a/b)x back into the area, I get (a/b)x², which seems to suggest that as x increases, the area increases without bound, which isn't possible because the rectangle is confined within the triangle.Wait, perhaps I need to consider that as x increases, y decreases, and vice versa. Maybe I should express y in terms of x correctly. Let's see, from (b - x)/b = (a - y)/a, cross-multiplying gives a(b - x) = b(a - y), which simplifies to ab - ax = ab - by. Then, subtracting ab from both sides, -ax = -by, so ax = by, which gives y = (a/b)x. So, that part seems correct.But then, if y = (a/b)x, and the area is x*y = x*(a/b)x = (a/b)x², which is a quadratic function. But in reality, the rectangle can't extend beyond the triangle, so x can't be larger than b, and y can't be larger than a. So, maybe the maximum area occurs at some specific x and y.Alternatively, maybe I should consider the area as a function of x and find its maximum within the constraints of the triangle. So, the area S = x*y = x*(a/b)x = (a/b)x². To find the maximum, I can take the derivative dS/dx = 2*(a/b)x. Setting this equal to zero gives x = 0, which is the minimum, not the maximum. Hmm, that suggests that the area increases as x increases, which contradicts the physical constraint of the triangle.Wait, perhaps I need to consider that as x increases, y decreases, but in a way that their product might have a maximum. Let me think differently. Maybe instead of expressing y in terms of x, I can express both x and y in terms of a single variable and then find the maximum.Alternatively, maybe I should use calculus to maximize the area function. Let's define the area S = x*y. From the earlier relationship, y = (a/b)(b - x). Wait, no, that's not correct. Wait, from (b - x)/b = (a - y)/a, we have y = a - (a/b)(b - x) = a - a + (a/b)x = (a/b)x. So, y = (a/b)x.So, S = x*(a/b)x = (a/b)x². To find the maximum, I can take the derivative dS/dx = 2*(a/b)x. Setting this equal to zero gives x = 0, which is the minimum. But that doesn't make sense because the maximum area should occur somewhere inside the triangle.Wait, maybe I need to consider that the rectangle can't extend beyond the triangle, so x can't be more than b, and y can't be more than a. So, the maximum area occurs when x = b/2 and y = a/2, giving an area of (b/2)*(a/2) = ab/4. But the area of the triangle is (1/2)*a*b, so ab/4 is half of that, which is (1/2)*(1/2)*a*b = ab/4. Wait, that would mean the maximum area of the rectangle is half the area of the triangle.But wait, that can't be right because if I set x = b/2, then y = (a/b)*(b/2) = a/2, and the area is (b/2)*(a/2) = ab/4, which is indeed half of the triangle's area, since the triangle's area is (1/2)*a*b. So, ab/4 is half of (1/2)*a*b, which is correct.Wait, but earlier I thought that the area function was increasing, but now I see that the maximum occurs at x = b/2, y = a/2, giving the maximum area as ab/4, which is half of the triangle's area. So, that seems to make sense.But let me double-check. If I take x = b/2, then y = (a/b)*(b/2) = a/2, so the rectangle has dimensions b/2 and a/2, area ab/4. The triangle's area is (1/2)*a*b, so ab/4 is indeed half of that. So, that seems correct.But wait, is this the only case? What if the rectangle is placed differently? For example, not with one vertex at the right angle. Maybe the maximum area occurs when the rectangle is placed such that its base is along the base of the triangle, but not necessarily starting at the right angle.Hmm, maybe I should generalize this. Let's consider any triangle, not necessarily right-angled. Let's say we have a triangle with base length b and height h. The area of the triangle is (1/2)*b*h.Now, inscribe a rectangle inside this triangle. The rectangle will have its base along the base of the triangle and its top side parallel to the base, touching the two other sides of the triangle. Let's denote the height of the rectangle as y, so the remaining height above the rectangle is h - y.At height y from the base, the width of the triangle is reduced proportionally. Since the triangle's sides are straight lines, the width at height y is b*(h - y)/h. So, the width of the rectangle is b*(h - y)/h.Therefore, the area of the rectangle is width times height, which is [b*(h - y)/h] * y = (b/h)*y*(h - y).To find the maximum area, we can take the derivative of this function with respect to y and set it equal to zero. Let's denote S(y) = (b/h)*y*(h - y).Taking the derivative, dS/dy = (b/h)*(h - y) + (b/h)*y*(-1) = (b/h)*(h - y - y) = (b/h)*(h - 2y).Setting dS/dy = 0, we get (b/h)*(h - 2y) = 0. Since b/h is not zero, we have h - 2y = 0, so y = h/2.So, the maximum area occurs when y = h/2. Plugging this back into the area formula, S = (b/h)*(h/2)*(h - h/2) = (b/h)*(h/2)*(h/2) = (b/h)*(h²/4) = b*h/4.The area of the triangle is (1/2)*b*h, so the maximum area of the rectangle is (b*h)/4, which is half of the triangle's area, since (1/2)*b*h / 2 = b*h/4.Wait, that's the same result as before. So, regardless of the triangle being right-angled or not, the maximum area of an inscribed rectangle is half the area of the triangle.But wait, in the right-angled triangle case, we had the same result, but in that case, the rectangle was placed with one vertex at the right angle. In the general triangle case, the rectangle is placed with its base along the base of the triangle. So, both cases lead to the same maximum area, which is half the area of the triangle.Therefore, it seems that the maximum area of a rectangle inscribed in a triangle is indeed half the area of the triangle, regardless of the triangle's type.But let me think again. Is this always true? What if the triangle is very tall and skinny? Would the maximum area of the inscribed rectangle still be half the triangle's area?Let's consider a very tall and skinny triangle, say with base b and height h, where h is much larger than b. The area of the triangle is (1/2)*b*h. The maximum area of the inscribed rectangle would be (b*h)/4, which is half of the triangle's area. So, even in this case, it holds.Alternatively, if the triangle is very wide and short, with h much smaller than b, the same result applies. The maximum area of the rectangle is still half the triangle's area.Therefore, it seems that regardless of the triangle's dimensions, the maximum area of an inscribed rectangle is half the area of the triangle.But wait, let me consider another configuration. What if the rectangle is not placed with its base along the base of the triangle? For example, what if it's placed such that one side is along one of the other sides?Hmm, in that case, the analysis might be different. Let's say we have a triangle ABC, and we inscribe a rectangle such that one side is along side AB, and the opposite side touches side AC and BC. Would the maximum area still be half the triangle's area?I think it would, because the problem is symmetric with respect to the sides. So, regardless of which side the rectangle is placed along, the maximum area should still be half the triangle's area.Alternatively, maybe the maximum area occurs when the rectangle is placed such that its sides are parallel to the base and height of the triangle, regardless of the triangle's orientation.In any case, from the calculations above, it seems that the maximum area of an inscribed rectangle is indeed half the area of the triangle.But let me try to formalize this proof more rigorously.Consider a triangle with base length b and height h. The area of the triangle is (1/2)*b*h.Now, inscribe a rectangle inside this triangle such that its base is along the base of the triangle and its top side is parallel to the base, touching the two other sides of the triangle.Let the height of the rectangle be y. Then, the width of the rectangle at height y is reduced proportionally. Since the triangle's sides are straight lines, the width at height y is b*(h - y)/h.Therefore, the area of the rectangle is:S = width * height = [b*(h - y)/h] * y = (b/h)*y*(h - y)To find the maximum area, we can take the derivative of S with respect to y and set it equal to zero.dS/dy = (b/h)*(h - y) + (b/h)*y*(-1) = (b/h)*(h - y - y) = (b/h)*(h - 2y)Setting dS/dy = 0:(b/h)*(h - 2y) = 0Since b/h ≠ 0, we have:h - 2y = 0 ⇒ y = h/2So, the maximum area occurs when y = h/2. Plugging this back into the area formula:S = (b/h)*(h/2)*(h - h/2) = (b/h)*(h/2)*(h/2) = (b/h)*(h²/4) = b*h/4The area of the triangle is (1/2)*b*h, so the maximum area of the rectangle is (b*h)/4, which is half of the triangle's area.Therefore, the area of a rectangle inscribed in a triangle does not exceed half the area of the triangle.But wait, in the right-angled triangle case, we had the same result, but the rectangle was placed with one vertex at the right angle. In the general triangle case, the rectangle is placed with its base along the base of the triangle. So, both cases lead to the same maximum area, which is half the area of the triangle.This suggests that regardless of how the rectangle is placed inside the triangle, as long as it's inscribed, its maximum area cannot exceed half the area of the triangle.But to be thorough, let's consider another configuration. Suppose the rectangle is placed such that one of its sides is along one of the other sides of the triangle, not the base. For example, in a right-angled triangle, placing the rectangle along the vertical side instead of the base.Let's say we have a right-angled triangle with legs of length a and b, and hypotenuse c. If we place the rectangle along the vertical leg of length a, then the width of the rectangle would be x, and the height would be y.Using similar triangles, the relationship between x and y would be similar to the previous case. The area would again be maximized when x = a/2 and y = b/2, giving an area of (a/2)*(b/2) = ab/4, which is half the area of the triangle, since the triangle's area is (1/2)*a*b.So, regardless of the orientation, the maximum area of the inscribed rectangle is half the area of the triangle.Therefore, we can conclude that the area of a rectangle inscribed in a triangle does not exceed half the area of the triangle.

💡Alright, so I have this problem here about negating a proposition. The original proposition is: "There exists an irrational number whose square is rational." And I need to find its negation from the given options. Let me try to break this down step by step.First, let's understand what the original statement is saying. It's saying that we can find at least one irrational number (let's call it 'x') such that when we square it, the result is a rational number. In logical terms, this can be written as: ∃x (x is irrational ∧ x² is rational). So, it's an existential statement—it claims that such an x exists.Now, to find the negation of this statement, I remember that the negation of an existential quantifier (∃) is a universal quantifier (∀), and we also need to negate the inner statement. So, the negation of ∃x (P(x)) is ∀x (¬P(x)). Applying this to our problem, the negation would be: For all irrational numbers x, x² is not rational. In symbols, that's ∀x (x is irrational → x² is not rational).Looking at the options provided:A: Every rational number has a rational square.- Hmm, this is talking about rational numbers, but our original statement was about irrational numbers. So, this doesn't seem directly related to the negation.B: Every irrational number has a square that is not rational.- This matches exactly with what I derived earlier. It's saying that for every irrational number, its square isn't rational, which is the negation of the original statement.C: There exists a rational number whose square is rational.- Again, this is about rational numbers, not irrational ones. Plus, it's an existential statement, whereas the negation should be universal.D: There exists an irrational number whose square is not rational.- This is still an existential statement, just like the original. The negation should be universal, so this isn't quite right.So, based on this reasoning, option B seems to be the correct negation of the original proposition. It correctly changes the existential quantifier to a universal one and negates the inner statement appropriately.Wait, let me double-check to make sure I'm not making a mistake here. The original statement is about the existence of an irrational number with a rational square. If I negate that, it should mean that no irrational number has a rational square, which is the same as saying every irrational number does not have a rational square. Yes, that aligns with option B.I think I'm confident now that option B is the right answer. It's important to remember how negations work with quantifiers—switching ∃ to ∀ and negating the inner condition. That helps avoid confusion between similar-sounding options.Another way to think about it is through examples. For instance, we know that √2 is irrational, and (√2)² = 2, which is rational. So, the original statement is actually true because √2 serves as an example. Therefore, the negation should be false. Looking at the options, option B would be false because we have at least one irrational number (√2) whose square is rational. So, if B were true, that would mean no irrational number has a rational square, which contradicts the existence of √2. Hence, B is indeed the correct negation.I guess it's also helpful to recall that in logic, the structure of the statement matters a lot. The original is ∃x (P(x) ∧ Q(x)), so the negation would be ∀x (P(x) → ¬Q(x)). Translating back into English, that's exactly what option B is saying.Just to cover all bases, let me consider why the other options don't fit:- Option A talks about rational numbers, which isn't relevant because the original statement was about irrational numbers. It's like changing the subject entirely.- Option C is similar but again focuses on rational numbers and doesn't negate the original statement.- Option D is still about irrational numbers but keeps the existential quantifier, which doesn't negate the original statement properly.So, yeah, after considering all these points, I'm pretty sure option B is the correct negation.

💡Okay, so I have this problem where I need to find the sum of ( C_{100j+19, 17} ) for all ( j ) from 0 to infinity. ( C_{k, n} ) represents the number of paths from (0,0) to (k, n) with certain constraints. Let me try to understand what each part means and how to approach this.First, the problem says that each path can only move directly upward or directly rightward, and you can only change direction at lattice points. So, basically, it's like moving on a grid where each step is either to the right or up, and you can't move diagonally or in any other direction. Also, each horizontal segment (a sequence of rightward moves) can be at most 99 units long. That means between any two upward moves, you can't move right more than 99 times.So, ( C_{k, n} ) is the number of such paths from (0,0) to (k, n). In this case, we're looking at paths to points where the x-coordinate is ( 100j + 19 ) and the y-coordinate is 17. So, we're summing over all possible j, which means we're considering all points that are 19 more than a multiple of 100 in the x-direction and 17 in the y-direction.Let me think about how to count these paths. Normally, without the horizontal segment constraint, the number of paths from (0,0) to (k, n) would be the binomial coefficient ( binom{k + n}{k} ) because you have to make a total of ( k + n ) moves, choosing k of them to be rightward moves (or equivalently, n upward moves). But here, we have the added constraint that each horizontal segment can't be longer than 99.So, how does this constraint affect the count? It means that between any two upward moves, the number of rightward moves can't exceed 99. So, if we have 17 upward moves, there are 18 horizontal segments (including the one before the first upward move and the one after the last upward move). Each of these horizontal segments must be between 0 and 99 rightward moves.Therefore, the problem reduces to finding the number of ways to distribute ( 100j + 19 ) rightward moves into 18 horizontal segments, each of which can be at most 99. This is a classic stars and bars problem with an upper limit on each bin.The formula for the number of non-negative integer solutions to ( x_1 + x_2 + dots + x_m = n ) with each ( x_i leq k ) is given by inclusion-exclusion. But in this case, since each ( x_i ) can be up to 99, and we have 18 variables, the number of solutions is ( binom{100j + 19 + 18 - 1}{18 - 1} ) minus the cases where one or more ( x_i ) exceed 99. But this seems complicated because we have to consider all possible overlaps where multiple ( x_i ) exceed 99.However, since we're summing over all j, maybe there's a generating function approach that can simplify this. The generating function for each horizontal segment is ( 1 + x + x^2 + dots + x^{99} ), which is a geometric series. So, the generating function for 18 such segments would be ( (1 + x + x^2 + dots + x^{99})^{18} ).We need the coefficient of ( x^{100j + 19} ) in this generating function for each j, and then sum all those coefficients. But summing over all j would essentially be evaluating the generating function at x = 1, but only considering the terms where the exponent is congruent to 19 modulo 100. Hmm, that might not be straightforward.Wait, another thought: since each horizontal segment can be at most 99, and we have 18 segments, the maximum total rightward moves would be ( 18 times 99 = 1782 ). But we're considering ( 100j + 19 ), which can be larger than 1782 for large j. But since we're summing over all j, including those where ( 100j + 19 ) exceeds 1782, those terms would be zero because you can't have more rightward moves than the maximum allowed by the horizontal segments.But wait, actually, for each j, ( C_{100j+19,17} ) counts the number of ways to distribute ( 100j + 19 ) rightward moves into 18 segments, each at most 99. So, for each j, if ( 100j + 19 leq 1782 ), then ( C_{100j+19,17} ) is non-zero, otherwise zero. But the problem is asking for the sum over all j from 0 to infinity, so we need to find the total number of such paths for all possible valid j.But how does this relate to the generating function? The generating function ( (1 + x + x^2 + dots + x^{99})^{18} ) encodes all possible numbers of rightward moves, with coefficients giving the number of ways. So, the sum we're asked for is the sum of coefficients of ( x^{100j + 19} ) for all j. This is equivalent to evaluating the generating function at the 100th roots of unity and using the roots of unity filter to extract the coefficients where the exponent is congruent to 19 modulo 100.This is getting a bit complex, but I think it's the right direction. The formula for extracting such coefficients is:[sum_{j=0}^{infty} C_{100j+19,17} = frac{1}{100} sum_{k=0}^{99} omega^{-19k} cdot (1 + omega^k + omega^{2k} + dots + omega^{99k})^{18}]where ( omega = e^{2pi i / 100} ) is a primitive 100th root of unity.But this seems quite involved, and I'm not sure if there's a simpler way. Maybe instead of dealing with roots of unity, there's a combinatorial interpretation or a simplification.Wait, another approach: since each horizontal segment is at most 99, and we have 18 segments, the total number of rightward moves is at most 1782. So, the sum over all j is just the total number of ways to distribute any number of rightward moves (from 0 up to 1782) into 18 segments, each at most 99, and then for each such distribution, we have a corresponding path.But actually, no, because we're specifically looking at the number of paths where the total rightward moves are ( 100j + 19 ). So, it's not the total number of paths, but the sum over all j of the number of paths with exactly ( 100j + 19 ) rightward moves.But maybe there's a way to see that this sum is equal to the total number of paths without the horizontal segment constraint, modulo some periodicity.Alternatively, perhaps we can model this as a recurrence relation. Let me think about the number of paths to (k, n) with the horizontal segment constraint. For each step, if we're moving right, we can do so up to 99 times before needing to move up. So, the recurrence would involve considering how many right steps we've taken since the last up step.But this might get complicated, especially since we're dealing with a sum over all j.Wait, going back to the generating function idea. The generating function for each horizontal segment is ( 1 + x + x^2 + dots + x^{99} ), which is ( frac{1 - x^{100}}{1 - x} ). So, the generating function for 18 segments is ( left( frac{1 - x^{100}}{1 - x} right)^{18} ).We need the sum of coefficients of ( x^{100j + 19} ) in this generating function. This is equivalent to evaluating the generating function at the 100th roots of unity and using the discrete Fourier transform to extract the coefficients.Specifically, the sum we want is:[sum_{j=0}^{infty} C_{100j+19,17} = frac{1}{100} sum_{k=0}^{99} omega^{-19k} cdot left( frac{1 - omega^{100k}}{1 - omega^k} right)^{18}]But since ( omega^{100k} = (omega^{100})^k = 1^k = 1 ), the numerator becomes ( 1 - 1 = 0 ). Wait, that can't be right because then the whole expression would be zero, which doesn't make sense.Wait, no, actually, ( omega^{100k} = (omega^{100})^k = 1^k = 1 ), so ( 1 - omega^{100k} = 0 ). Therefore, the generating function evaluated at ( omega^k ) is ( left( frac{0}{1 - omega^k} right)^{18} ), which is 0 for all k except when ( omega^k = 1 ), i.e., when k=0.But when k=0, ( omega^0 = 1 ), so the generating function becomes ( left( frac{1 - 1}{1 - 1} right)^{18} ), which is indeterminate. However, we can take the limit as ( x ) approaches 1. The generating function ( left( frac{1 - x^{100}}{1 - x} right)^{18} ) approaches ( (100)^{18} ) as ( x ) approaches 1, because each term ( frac{1 - x^{100}}{1 - x} ) approaches 100.Therefore, the sum we want is:[frac{1}{100} left[ 100^{18} + sum_{k=1}^{99} 0 right] = frac{100^{18}}{100} = 100^{17}]So, the sum ( sum_{j=0}^{infty} C_{100j+19,17} ) is equal to ( 100^{17} ).Wait, let me double-check this reasoning. The generating function approach seems to suggest that the sum of coefficients where the exponent is congruent to 19 modulo 100 is equal to ( 100^{17} ). But why is that? It seems that when we use the roots of unity filter, all terms except when k=0 vanish because the generating function becomes zero at those roots. But when k=0, we have to evaluate the generating function at x=1, which gives the total number of paths without any modulo constraint, which is ( 100^{18} ). Then, dividing by 100 gives ( 100^{17} ).But wait, actually, the total number of paths without any horizontal segment constraint would be ( binom{100j + 19 + 17}{17} ), but with the constraint, it's different. However, in our case, we're summing over all j, so it's like considering all possible rightward moves, which is equivalent to the total number of ways to distribute any number of rightward moves into 18 segments, each at most 99. But since each segment can be up to 99, the total number of rightward moves can be up to 1782, but we're summing over all j, which includes all possible rightward moves, so it's actually the total number of such distributions, which is ( 100^{18} ). But since we're only considering the coefficients where the exponent is 19 modulo 100, the sum is ( 100^{17} ).Yes, that makes sense. So, the final answer should be ( 100^{17} ).

💡Okay, so I have this problem here: I need to find the equation of the tangent line that passes through the point M(-√3, 1) on the circle x² + y² = 4. The options are given as A, B, C, D. Hmm, let me think about how to approach this.First, I remember that the equation of a circle is x² + y² = r², where r is the radius. In this case, the equation is x² + y² = 4, so the radius r is 2. The center of the circle is at the origin (0,0).Now, I need to find the tangent line at point M(-√3, 1). I recall that a tangent line to a circle is perpendicular to the radius at the point of tangency. So, the radius here is the line from the center (0,0) to the point M(-√3, 1). Let me find the slope of this radius first. The slope formula is (y2 - y1)/(x2 - x1). So, plugging in the points (0,0) and (-√3, 1), the slope would be (1 - 0)/(-√3 - 0) = 1/(-√3) = -1/√3. Since the tangent line is perpendicular to this radius, its slope should be the negative reciprocal of -1/√3. The negative reciprocal of -1/√3 is √3. So, the slope of the tangent line is √3.Now that I have the slope of the tangent line and a point it passes through, which is M(-√3, 1), I can use the point-slope form of the equation of a line. The point-slope form is y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point.Plugging in the values, we get:y - 1 = √3(x - (-√3))Simplifying the equation:y - 1 = √3(x + √3)Expanding the right side:y - 1 = √3*x + (√3)*(√3)Since √3*√3 is 3, the equation becomes:y - 1 = √3*x + 3Now, let's bring all terms to one side to get the standard form:y - 1 - √3*x - 3 = 0Simplifying further:-√3*x + y - 4 = 0Multiplying both sides by -1 to make the coefficients positive (optional, but sometimes preferred):√3*x - y + 4 = 0Looking at the options provided:A: x - √3 y + 2√3 = 0B: √3 x - y + 4 = 0C: √3 x + y + 2 = 0D: √3 x - y - 4 = 0The equation I derived is √3 x - y + 4 = 0, which matches option B. So, I think the correct answer is B.Wait, let me double-check my calculations to make sure I didn't make any mistakes. Starting from the slope of the radius: from (0,0) to (-√3,1), the slope is (1 - 0)/(-√3 - 0) = 1/(-√3) = -1/√3. That seems correct.Then, the slope of the tangent line is the negative reciprocal, which would be √3. That makes sense because the product of the slopes of two perpendicular lines is -1, and (-1/√3)*(√3) = -1. So that's correct.Using point-slope form: y - 1 = √3(x + √3). Expanding gives y - 1 = √3 x + 3. Then, moving all terms to one side: -√3 x + y - 4 = 0, which is the same as √3 x - y + 4 = 0 when multiplied by -1. So, yes, that's correct.Therefore, I'm confident that option B is the right answer.**Final Answer**boxed{B}

💡Okay, so I need to find the minimum value of the function y = 5x² + 10x + 20. Hmm, I remember that quadratic functions have either a minimum or maximum value, depending on the coefficient of the x² term. Since the coefficient here is 5, which is positive, the parabola opens upwards, meaning it has a minimum point. That makes sense, so there should be a lowest point on this graph.I think there are a couple of ways to find the minimum value. One way is by completing the square, which I've heard about before but I'm not entirely sure how to do it. Another way is using calculus, finding the derivative and setting it to zero to find critical points. But since this is a quadratic function, maybe there's a formula or something simpler.Wait, I think there's a formula for the vertex of a parabola, which gives the minimum or maximum point. The vertex form of a quadratic function is y = a(x - h)² + k, where (h, k) is the vertex. So if I can rewrite the given function in this form, I can easily identify the minimum value.Let me try completing the square then. The function is y = 5x² + 10x + 20. First, I should factor out the coefficient of x² from the first two terms. That would give me y = 5(x² + 2x) + 20. Okay, so I factored out 5 from 5x² and 10x, which leaves me with x² + 2x inside the parentheses.Now, to complete the square, I need to add and subtract the square of half the coefficient of x inside the parentheses. The coefficient of x is 2, so half of that is 1, and squaring it gives 1. So I add and subtract 1 inside the parentheses: y = 5(x² + 2x + 1 - 1) + 20. This simplifies to y = 5[(x + 1)² - 1] + 20.Now, distributing the 5, I get y = 5(x + 1)² - 5 + 20. Combining the constants, -5 + 20 is 15, so the equation becomes y = 5(x + 1)² + 15. In this form, it's clear that the vertex is at (-1, 15). Since the coefficient of the squared term is positive, this vertex is the minimum point. Therefore, the minimum value of the function is 15.Alternatively, I could use the formula for the vertex of a parabola. The x-coordinate of the vertex is given by -b/(2a), where a is the coefficient of x² and b is the coefficient of x. In this case, a = 5 and b = 10, so x = -10/(2*5) = -10/10 = -1. Plugging this back into the original equation, y = 5(-1)² + 10(-1) + 20 = 5(1) - 10 + 20 = 5 - 10 + 20 = 15. So that confirms the minimum value is indeed 15.I think both methods work, but completing the square gives a clearer picture of how the function is structured, while the vertex formula is quicker once you remember it. Either way, the minimum value is 15.