💡Alright, so I've got this probability question here, and I need to figure out which of the given propositions are true. Let me go through each one step by step.First, let's look at proposition ①: "The probability of a certain event equals 1." Hmm, I remember that in probability, a certain event is something that will definitely happen. So, for example, if I roll a six-sided die, the probability of getting a number between 1 and 6 is 1 because it's certain. Yeah, that makes sense. So, I think ① is true.Next, proposition ②: "The probability of some event equals 1.1." Wait a minute, probabilities are always between 0 and 1, right? An event can't have a probability greater than 1 because that would mean it's more likely than certain, which doesn't make sense. So, 1.1 is outside the possible range of probabilities. Therefore, ② must be false.Moving on to proposition ③: "Mutually exclusive events are necessarily complementary events." Okay, mutually exclusive means that two events can't happen at the same time. For example, if I roll a die, getting a 1 and getting a 2 are mutually exclusive because they can't both occur on a single roll. But complementary events are a specific type of mutually exclusive events where one event is the exact opposite of the other. So, the complement of getting a 1 is getting anything other than 1. So, while complementary events are mutually exclusive, not all mutually exclusive events are complementary. For instance, getting a 1 and getting a 2 are mutually exclusive but not complementary. Therefore, ③ is false.Proposition ④: "Complementary events are necessarily mutually exclusive events." This seems related to the previous one. Complementary events are indeed mutually exclusive because they can't both happen at the same time. For example, if one event is getting a head on a coin flip, the complementary event is getting a tail, and they can't both occur. So, yes, complementary events are mutually exclusive. Therefore, ④ is true.Lastly, proposition ⑤: "Planting a seed under appropriate conditions and observing whether it germinates is a classical probability model." I think the classical probability model refers to situations where all outcomes are equally likely. In this case, planting a seed and observing germination might not necessarily have equally likely outcomes because some seeds might have a higher chance of germinating than others based on factors like soil quality or seed type. However, if we assume that under appropriate conditions, each seed has an equal chance of germinating or not, then it could be considered a classical probability model. But I'm not entirely sure. Maybe it's more of an empirical probability model because it's based on observation rather than equally likely outcomes. Hmm, this one is a bit tricky. I'll have to think about this more.Wait, the classical probability model is typically used when all outcomes are equally likely, like flipping a fair coin or rolling a fair die. In the case of planting a seed, unless we have specific information that each outcome (germinating or not) is equally likely, we can't assume it's a classical model. So, maybe ⑤ is false.But hold on, the question says "planting a seed under appropriate conditions." If the conditions are appropriate, does that imply that the probability is 1? No, because even under appropriate conditions, some seeds might not germinate due to other factors. So, it's not certain. But is it a classical model? I think not, because we don't know if the two outcomes (germinates or doesn't germinate) are equally likely. It might be more of a real-world probability scenario rather than a classical one. So, I think ⑤ is false.Wait, but maybe I'm overcomplicating it. If we consider that under appropriate conditions, each seed has an equal chance of germinating or not, then it could be a classical model. But in reality, that's not always the case. So, perhaps the question is implying that it is a classical model, so ⑤ is true. I'm a bit confused here.Let me recap:①: True②: False③: False④: True⑤: Maybe True or False, depending on interpretation.But given that the options are A: ①③, B: ③⑤, C: ①③⑤, D: ①④⑤If I consider ⑤ as true, then D would be the answer. If ⑤ is false, then only ① and ④ are true, but that's not one of the options. Wait, option D includes ①④⑤, so if ⑤ is true, then D is correct. If ⑤ is false, then none of the options match, but since ⑤ is part of D, I think the intended answer is D.So, I'll go with D: ①④⑤.

💡Okay, so I have this problem where I need to choose three numbers, a₁, a₂, a₃, from the numbers 1 to 14. They have to be in increasing order, which means a₁ < a₂ < a₃. But there are some extra conditions: the difference between a₂ and a₁ has to be at least 3, and the difference between a₃ and a₂ also has to be at least 3. So, I need to figure out how many different ways I can choose such numbers.Hmm, let me think about how to approach this. Maybe I can model this problem in a way that makes it easier to count the valid combinations. I remember that sometimes when dealing with differences between numbers, it's helpful to transform the problem into something else where the differences become fixed, and then count the possibilities there.Let me try that. If I have a₁, a₂, a₃ such that a₂ - a₁ ≥ 3 and a₃ - a₂ ≥ 3, maybe I can adjust the numbers so that the differences become at least 1 instead. That might make it easier to count because then I can use combinations.So, if I define new variables to represent the adjusted numbers. Let me set:b₁ = a₁b₂ = a₂ - 2b₃ = a₃ - 4Why did I subtract 2 and 4? Well, because the original differences were at least 3, so subtracting 2 from a₂ and 4 from a₃ should make the differences between b₂ and b₁, and b₃ and b₂ at least 1. Let me check that.If a₂ - a₁ ≥ 3, then (a₂ - 2) - a₁ ≥ 1, which is b₂ - b₁ ≥ 1.Similarly, a₃ - a₂ ≥ 3 implies (a₃ - 4) - (a₂ - 2) ≥ 1, which is b₃ - b₂ ≥ 1.So, the new variables b₁, b₂, b₃ must satisfy b₁ < b₂ < b₃, with each consecutive pair differing by at least 1. But since they are in increasing order, they automatically differ by at least 1. So, the problem reduces to choosing three numbers from a smaller set where the differences are at least 1.But wait, what is the range of these new variables? Let's see.Since a₁ is at least 1, b₁ is also at least 1.a₂ is at least a₁ + 3, so b₂ = a₂ - 2 is at least (a₁ + 3) - 2 = a₁ + 1. Since a₁ is at least 1, b₂ is at least 2.Similarly, a₃ is at least a₂ + 3, so b₃ = a₃ - 4 is at least (a₂ + 3) - 4 = a₂ - 1. Since a₂ is at least a₁ + 3, which is at least 4, so b₃ is at least 3.But what's the maximum value of b₃? Since a₃ is at most 14, b₃ = a₃ - 4 is at most 10.So, the new variables b₁, b₂, b₃ are numbers from 1 to 10, with b₁ < b₂ < b₃. So, how many ways are there to choose three numbers from 1 to 10 without any restrictions except that they are in increasing order?That's just the combination of 10 things taken 3 at a time, which is denoted as C(10,3). The formula for combinations is C(n,k) = n! / (k! (n - k)!).Calculating that, C(10,3) = 10! / (3! * 7!) = (10 * 9 * 8) / (3 * 2 * 1) = 120.So, does that mean the number of valid triplets (a₁, a₂, a₃) is 120? Let me verify.If I map each valid triplet (a₁, a₂, a₃) to (b₁, b₂, b₃) where b₁ = a₁, b₂ = a₂ - 2, b₃ = a₃ - 4, then each triplet in the original problem corresponds to a unique triplet in the transformed problem, and vice versa. So, the number of triplets should indeed be the same as the number of ways to choose 3 numbers from 1 to 10, which is 120.Wait, but let me make sure I didn't make a mistake in the transformation. Let's take an example. Suppose I choose a₁ = 1, a₂ = 4, a₃ = 7. Then, b₁ = 1, b₂ = 4 - 2 = 2, b₃ = 7 - 4 = 3. So, (1, 2, 3) is a valid triplet in the transformed set.Another example: a₁ = 2, a₂ = 5, a₃ = 8. Then, b₁ = 2, b₂ = 5 - 2 = 3, b₃ = 8 - 4 = 4. So, (2, 3, 4) is another valid triplet.What if I choose a₁ = 1, a₂ = 4, a₃ = 8? Then, b₁ = 1, b₂ = 2, b₃ = 4. That's also valid.Wait, but what if a₁ is larger? For example, a₁ = 10, a₂ = 13, a₃ = 14. Then, b₁ = 10, b₂ = 13 - 2 = 11, b₃ = 14 - 4 = 10. Wait, that's a problem because b₃ = 10, which is less than b₂ = 11. That would mean b₁ < b₂ < b₃ is not satisfied because 10 < 11 < 10 is not true. Hmm, so did I make a mistake here?Wait, no. Because a₃ is 14, so b₃ = 14 - 4 = 10. But a₂ is 13, so b₂ = 13 - 2 = 11. So, b₁ = 10, b₂ = 11, b₃ = 10. Wait, that's not possible because b₃ can't be less than b₂. So, that triplet (10, 11, 10) is invalid. But in reality, a₃ is 14, which is greater than a₂ = 13, so b₃ should be greater than b₂. Wait, 14 - 4 = 10, which is less than 11. That's a problem.So, does that mean that some triplets in the original problem don't map correctly? Or maybe my transformation is flawed?Wait, let's check the original triplet: a₁ = 10, a₂ = 13, a₃ = 14. The differences are a₂ - a₁ = 3, which is acceptable, and a₃ - a₂ = 1, which is less than 3. So, actually, this triplet doesn't satisfy the original condition because a₃ - a₂ is only 1, which is less than 3. So, it's not a valid triplet in the first place. Therefore, in the transformed problem, it's correctly not a valid triplet because b₃ is less than b₂, which is invalid.So, actually, the transformation correctly excludes such invalid triplets. So, the mapping is correct.Another example: a₁ = 12, a₂ = 15, a₃ = 18. Wait, but our original set only goes up to 14, so a₂ can't be 15. So, in our problem, the maximum a₃ is 14. So, let me pick a valid triplet: a₁ = 1, a₂ = 4, a₃ = 7. Then, b₁ = 1, b₂ = 2, b₃ = 3. That works.Another valid triplet: a₁ = 3, a₂ = 6, a₃ = 9. Then, b₁ = 3, b₂ = 4, b₃ = 5. That works.What about a₁ = 5, a₂ = 8, a₃ = 11. Then, b₁ = 5, b₂ = 6, b₃ = 7. That works.Wait, but what if a₁ is 10? Then, a₂ has to be at least 13, and a₃ has to be at least 16, but 16 is beyond our original set of 1 to 14. So, the maximum a₁ can be is such that a₂ is at least a₁ + 3, and a₃ is at least a₂ + 3. So, a₁ can be at most 14 - 6 = 8, because a₁ + 3 + 3 = a₁ + 6 ≤ 14, so a₁ ≤ 8.Wait, so a₁ can be from 1 to 8, a₂ from a₁ + 3 to 11, and a₃ from a₂ + 3 to 14.But in the transformed problem, b₁ can be from 1 to 10, b₂ from b₁ + 1 to 9, and b₃ from b₂ + 1 to 10. So, the number of triplets is C(10,3) = 120.But let me think again. If a₁ can only go up to 8, then in the transformed problem, b₁ = a₁ can go up to 8, but in the transformed problem, b₁ can go up to 10. So, does that mean that some triplets in the transformed problem correspond to invalid original triplets?Wait, no, because in the transformed problem, b₁ can be up to 10, but in the original problem, a₁ can only be up to 8. So, does that mean that some triplets in the transformed problem are invalid?Wait, no, because when we map back, if b₁ is 9, then a₁ = 9, but then a₂ = b₂ + 2. If b₂ is at least b₁ + 1 = 10, then a₂ = 10 + 2 = 12. Then, a₃ = b₃ + 4. If b₃ is at least b₂ + 1 = 11, then a₃ = 11 + 4 = 15, which is beyond 14. So, that triplet would be invalid in the original problem.Wait, so does that mean that the transformed problem includes some invalid triplets? But earlier, I thought the mapping was one-to-one. Hmm, maybe I need to adjust the transformed set.Wait, perhaps I need to adjust the transformed set so that b₃ doesn't exceed 10, but in the original problem, a₃ can't exceed 14. So, when I subtract 4 from a₃, I get b₃ = a₃ - 4, which can be at most 10, since a₃ is at most 14.But in the transformed problem, b₃ can be up to 10, which is fine because a₃ is up to 14. So, maybe the mapping is correct.Wait, let me take an example where b₁ is 9. So, a₁ = 9. Then, b₂ must be at least 10, so a₂ = b₂ + 2 must be at least 12. Then, b₃ must be at least 11, so a₃ = b₃ + 4 must be at least 15. But 15 is beyond 14, so that triplet is invalid. So, in the transformed problem, the triplet (9,10,11) would map to (9,12,15), which is invalid because 15 >14.So, does that mean that the transformed problem includes some invalid triplets? But in the transformed problem, we're only choosing b₁, b₂, b₃ from 1 to 10, so b₃ can't be more than 10. So, in the transformed problem, (9,10,11) is invalid because b₃ can't be 11. So, actually, in the transformed problem, b₃ is at most 10, so a₃ is at most 14, which is valid.Wait, so in the transformed problem, b₃ is at most 10, so a₃ is at most 14. So, the triplet (9,10,10) would map to (9,12,14). Let's check: a₁=9, a₂=12, a₃=14. Then, a₂ - a₁ = 3, which is acceptable, and a₃ - a₂ = 2, which is less than 3. So, that triplet is invalid in the original problem. So, in the transformed problem, (9,10,10) is invalid because b₃ must be greater than b₂, but in this case, b₃=10 is equal to b₂=10, which is not allowed because they must be in increasing order.Wait, so in the transformed problem, b₁ < b₂ < b₃, so b₃ must be at least b₂ +1. So, in the example above, b₂=10, so b₃ must be at least 11, but since b₃ is at most 10, that triplet is invalid. So, in the transformed problem, such triplets are excluded because b₃ can't be greater than 10. So, the mapping is correct.Therefore, the number of valid triplets in the original problem is equal to the number of triplets in the transformed problem, which is C(10,3) = 120.Wait, but let me think again. If a₁ can be up to 8, then in the transformed problem, b₁ can be up to 8, but in the transformed problem, b₁ can be up to 10. So, does that mean that some triplets in the transformed problem correspond to a₁ beyond 8, which would make a₂ and a₃ beyond 14? But no, because in the transformed problem, b₃ is at most 10, so a₃ is at most 14. So, even if b₁ is 9 or 10, a₁ is 9 or 10, but then a₂ and a₃ are adjusted accordingly, but a₃ can't exceed 14.Wait, let me take b₁=9, b₂=10, b₃=10. Then, a₁=9, a₂=12, a₃=14. But a₃ - a₂=2, which is less than 3, so this triplet is invalid. So, in the transformed problem, (9,10,10) is invalid because b₃ must be greater than b₂, but in this case, b₃=10 is equal to b₂=10, which is not allowed. So, such triplets are excluded.Therefore, the mapping is correct, and the number of valid triplets is indeed C(10,3)=120.Wait, but let me think of another way to approach this problem to confirm. Maybe using stars and bars or something similar.So, the problem is to choose three numbers a₁ < a₂ < a₃ from 1 to 14, with a₂ - a₁ ≥3 and a₃ - a₂ ≥3. So, we can think of the gaps between the numbers.Let me define the gaps as follows:Let x₁ = a₁ - 1x₂ = a₂ - a₁ - 2 (since a₂ - a₁ ≥3, so x₂ ≥1)x₃ = a₃ - a₂ - 2 (since a₃ - a₂ ≥3, so x₃ ≥1)x₄ = 14 - a₃So, the total sum of these gaps is:x₁ + x₂ + x₃ + x₄ = (a₁ -1) + (a₂ - a₁ -2) + (a₃ - a₂ -2) + (14 - a₃) = 14 -1 -2 -2 = 9So, x₁ + x₂ + x₃ + x₄ =9, with x₁ ≥0, x₂ ≥1, x₃ ≥1, x₄ ≥0.But since x₂ and x₃ are at least 1, let me set y₂ = x₂ -1 and y₃ = x₃ -1, so y₂ ≥0 and y₃ ≥0.Then, the equation becomes:x₁ + (y₂ +1) + (y₃ +1) + x₄ =9Which simplifies to:x₁ + y₂ + y₃ + x₄ =7Now, we have four non-negative integers x₁, y₂, y₃, x₄ that sum to 7.The number of solutions is C(7 +4 -1,4 -1)=C(10,3)=120.Wait, that's the same answer as before. So, this confirms that the number of valid triplets is 120.So, both methods give the same result, which is reassuring.Therefore, the number of different ways to choose such numbers is 120.

💡Okay, so I have this geometry problem here about a quadrilateral (ABCD). The midpoints of its sides are (K, L, M, N) on (AB, BC, CD, DA) respectively. Then, lines (AL) and (CK) intersect at point (P), and lines (AM) and (CN) intersect at point (Q). It's given that (APCQ) is a parallelogram, and I need to prove that (ABCD) is also a parallelogram.Hmm, let me start by recalling some properties of midpoints and parallelograms. I remember that the line segment connecting the midpoints of two sides of a triangle is parallel to the third side and half its length. That's the Midline Theorem. Maybe that can help here.Since (K, L, M, N) are midpoints, connecting them should form another quadrilateral, which I think is called the Varignon parallelogram. So, (KLMN) should be a parallelogram because the midlines are parallel to the sides of the original quadrilateral and half their lengths. That makes sense.Now, (APCQ) is given as a parallelogram. So, (AP) is parallel to (CQ) and (PC) is parallel to (AQ). Also, in a parallelogram, the diagonals bisect each other. So, if I can show that the diagonals of (ABCD) bisect each other, then (ABCD) would be a parallelogram.Wait, but how does (APCQ) being a parallelogram relate to (ABCD)? Maybe I need to find some relationships between the points (P) and (Q) and the midpoints (K, L, M, N).Let me try to visualize this. Points (K) and (M) are midpoints of (AB) and (CD), respectively. Similarly, (L) and (N) are midpoints of (BC) and (DA). So, lines (AL) and (CK) intersect at (P), and lines (AM) and (CN) intersect at (Q).Since (KLMN) is a parallelogram, its diagonals (KM) and (LN) bisect each other at some point, say (O). So, (O) is the midpoint of both (KM) and (LN).If I can show that (O) is also the midpoint of (PQ), then maybe that can help me relate it back to (ABCD). Since (APCQ) is a parallelogram, its diagonals (AC) and (PQ) should bisect each other. So, the midpoint of (PQ) should be the same as the midpoint of (AC).But wait, (O) is the midpoint of (KM) and (LN). How does that relate to (AC)? Maybe I need to use coordinate geometry to make this more precise.Let me assign coordinates to the quadrilateral (ABCD). Let's place point (A) at ((0, 0)), (B) at ((2a, 0)), (C) at ((2b, 2c)), and (D) at ((2d, 2e)). I'm doubling the coordinates to make the midpoints have integer coordinates, which might simplify calculations.So, midpoints:- (K) is midpoint of (AB): ((a, 0))- (L) is midpoint of (BC): ((a + b, c))- (M) is midpoint of (CD): ((b + d, c + e))- (N) is midpoint of (DA): ((d, e))Now, let's find equations of lines (AL) and (CK) to find point (P).Line (AL) goes from (A(0,0)) to (L(a + b, c)). The parametric equations can be written as:(x = t(a + b)), (y = t c), where (t) ranges from 0 to 1.Line (CK) goes from (C(2b, 2c)) to (K(a, 0)). The parametric equations can be written as:(x = 2b + s(a - 2b)), (y = 2c + s(-2c)), where (s) ranges from 0 to 1.To find the intersection (P), set the coordinates equal:(t(a + b) = 2b + s(a - 2b))(t c = 2c - 2c s)From the second equation:(t c = 2c(1 - s))Assuming (c neq 0), divide both sides by (c):(t = 2(1 - s))Substitute (t = 2(1 - s)) into the first equation:(2(1 - s)(a + b) = 2b + s(a - 2b))Expand:(2(a + b) - 2s(a + b) = 2b + s(a - 2b))Bring all terms to left:(2(a + b) - 2s(a + b) - 2b - s(a - 2b) = 0)Simplify:(2a + 2b - 2b - 2s(a + b) - s(a - 2b) = 0)Which simplifies to:(2a - 2s(a + b) - s(a - 2b) = 0)Factor out (s):(2a - s[2(a + b) + (a - 2b)] = 0)Simplify inside the brackets:(2a - s[2a + 2b + a - 2b] = 0)Which is:(2a - s(3a) = 0)So:(2a = 3a s)Assuming (a neq 0), divide both sides by (a):(2 = 3s)Thus, (s = frac{2}{3})Then, (t = 2(1 - frac{2}{3}) = 2(frac{1}{3}) = frac{2}{3})So, point (P) has coordinates:From line (AL): (x = frac{2}{3}(a + b)), (y = frac{2}{3}c)Similarly, let's find point (Q) as the intersection of (AM) and (CN).Line (AM) goes from (A(0,0)) to (M(b + d, c + e)). Parametric equations:(x = t(b + d)), (y = t(c + e)), (t) from 0 to 1.Line (CN) goes from (C(2b, 2c)) to (N(d, e)). Parametric equations:(x = 2b + s(d - 2b)), (y = 2c + s(e - 2c)), (s) from 0 to 1.Set coordinates equal:(t(b + d) = 2b + s(d - 2b))(t(c + e) = 2c + s(e - 2c))From the second equation:(t(c + e) = 2c + s(e - 2c))Let me solve for (t) and (s).This seems a bit complicated. Maybe I can express (t) from the second equation and substitute into the first.From the second equation:(t = frac{2c + s(e - 2c)}{c + e})Substitute into the first equation:(frac{2c + s(e - 2c)}{c + e}(b + d) = 2b + s(d - 2b))Multiply both sides by (c + e):((2c + s(e - 2c))(b + d) = (2b + s(d - 2b))(c + e))Let me expand both sides:Left side:(2c(b + d) + s(e - 2c)(b + d))Right side:(2b(c + e) + s(d - 2b)(c + e))Bring all terms to left:(2c(b + d) + s(e - 2c)(b + d) - 2b(c + e) - s(d - 2b)(c + e) = 0)Factor out (s):(2c(b + d) - 2b(c + e) + s[(e - 2c)(b + d) - (d - 2b)(c + e)] = 0)Simplify the constant terms:(2c b + 2c d - 2b c - 2b e = 2c d - 2b e)Now, the coefficient of (s):((e - 2c)(b + d) - (d - 2b)(c + e))Expand both products:(e b + e d - 2c b - 2c d - [d c + d e - 2b c - 2b e])Simplify:(e b + e d - 2c b - 2c d - d c - d e + 2b c + 2b e)Combine like terms:(e b + e d - 2c b - 2c d - c d - d e + 2b c + 2b e)Simplify:(e b + (e d - d e) + (-2c b + 2b c) + (-2c d - c d) + 2b e)Which becomes:(e b + 0 + 0 - 3c d + 2b e)So, total:(3b e - 3c d)Putting it all together:(2c d - 2b e + s(3b e - 3c d) = 0)Factor out 3:(2c d - 2b e + 3s(b e - c d) = 0)Let me write it as:(2(c d - b e) + 3s(b e - c d) = 0)Factor out ((c d - b e)):((c d - b e)(2 - 3s) = 0)So, either (c d - b e = 0) or (2 - 3s = 0).If (c d - b e = 0), then (c d = b e). But we don't know that yet. Alternatively, (s = frac{2}{3}).Assuming (s = frac{2}{3}), then from earlier:(t = frac{2c + frac{2}{3}(e - 2c)}{c + e} = frac{2c + frac{2}{3}e - frac{4}{3}c}{c + e} = frac{frac{2}{3}c + frac{2}{3}e}{c + e} = frac{2}{3})So, (t = frac{2}{3}), and point (Q) has coordinates:From line (AM): (x = frac{2}{3}(b + d)), (y = frac{2}{3}(c + e))So, points (P) and (Q) have coordinates:(Pleft(frac{2}{3}(a + b), frac{2}{3}cright))(Qleft(frac{2}{3}(b + d), frac{2}{3}(c + e)right))Now, since (APCQ) is a parallelogram, the midpoint of (AC) should be the same as the midpoint of (PQ).Midpoint of (AC): (left(frac{0 + 2b}{2}, frac{0 + 2c}{2}right) = (b, c))Midpoint of (PQ): (left(frac{frac{2}{3}(a + b) + frac{2}{3}(b + d)}{2}, frac{frac{2}{3}c + frac{2}{3}(c + e)}{2}right))Simplify:(x)-coordinate: (frac{frac{2}{3}(a + 2b + d)}{2} = frac{2}{3} cdot frac{a + 2b + d}{2} = frac{a + 2b + d}{3})(y)-coordinate: (frac{frac{2}{3}(2c + e)}{2} = frac{2}{3} cdot frac{2c + e}{2} = frac{2c + e}{3})Since midpoints must be equal:(frac{a + 2b + d}{3} = b) and (frac{2c + e}{3} = c)Solving the (x)-coordinate equation:(frac{a + 2b + d}{3} = b)Multiply both sides by 3:(a + 2b + d = 3b)So, (a + d = b)Similarly, for the (y)-coordinate:(frac{2c + e}{3} = c)Multiply both sides by 3:(2c + e = 3c)So, (e = c)So, from these, we have (a + d = b) and (e = c).Now, let's recall the coordinates of points (A, B, C, D):(A(0, 0)), (B(2a, 0)), (C(2b, 2c)), (D(2d, 2e))But since (e = c), point (D) is ((2d, 2c))Also, (a + d = b), so (d = b - a)So, point (D) becomes ((2(b - a), 2c))Now, let's see the coordinates:- (A(0, 0))- (B(2a, 0))- (C(2b, 2c))- (D(2(b - a), 2c))Now, let's check if (ABCD) is a parallelogram. For that, we need to check if opposite sides are equal and parallel.Compute vectors:- Vector (AB = B - A = (2a, 0))- Vector (DC = C - D = (2b - 2(b - a), 2c - 2c) = (2a, 0))So, (AB = DC), which is good.- Vector (AD = D - A = (2(b - a), 2c))- Vector (BC = C - B = (2b - 2a, 2c - 0) = (2(b - a), 2c))So, (AD = BC), which is also good.Since both pairs of opposite sides are equal and parallel, (ABCD) is a parallelogram.So, the conclusion is that (ABCD) must be a parallelogram given the conditions.**Final Answer**boxed{ABCD text{ is a parallelogram}}

💡Okay, so I have this problem here where two positive numbers are in the ratio 1/(6x) to 1/(5y²), and the sum of the numbers is such that 2x + 3y = 42. I need to find by what percent the second number is more than the first. Hmm, let me try to break this down step by step.First, let's understand what the ratio means. If two numbers are in the ratio 1/(6x) : 1/(5y²), that means if I call the first number A and the second number B, then A/B = (1/(6x)) / (1/(5y²)). Simplifying that, A/B = (5y²)/(6x). So, A = (5y²)/(6x) * B. Okay, that makes sense.Now, the sum of the numbers is given by 2x + 3y = 42. Wait, that seems a bit confusing. Is the sum of the numbers equal to 42, or is it 2x + 3y = 42? The wording says "the sum of the numbers is such that 2x + 3y = 42." So, I think it means that 2x + 3y equals 42, not necessarily that the sum of the numbers themselves is 42. That’s a bit tricky.So, we have two equations:1. A/B = (5y²)/(6x)2. 2x + 3y = 42But we need to find by what percent B is more than A. To find that, I need to express B in terms of A or vice versa and then compute the percentage increase.From the first equation, A = (5y²)/(6x) * B. So, if I rearrange that, B = (6x)/(5y²) * A. That means B is (6x)/(5y²) times A. To find the percentage increase, I need to calculate (B - A)/A * 100%.Substituting B from above, we get:(B - A)/A * 100% = [(6x/(5y²) * A - A)/A] * 100% = [(6x/(5y²) - 1)] * 100%So, the percentage increase is [(6x/(5y²) - 1)] * 100%. Now, to find this, I need to find the relationship between x and y from the second equation: 2x + 3y = 42.Let me solve for x in terms of y:2x = 42 - 3yx = (42 - 3y)/2Now, substitute this x back into the percentage increase expression:Percentage increase = [(6*(42 - 3y)/2)/(5y²) - 1] * 100%= [(6*(42 - 3y))/(2*5y²) - 1] * 100%= [(6*(42 - 3y))/(10y²) - 1] * 100%= [(252 - 18y)/(10y²) - 1] * 100%= [(252 - 18y - 10y²)/(10y²)] * 100%Hmm, this seems complicated. I have a quadratic in the numerator. Maybe I can simplify it further or find specific values for y that satisfy 2x + 3y = 42. But without additional information, it's hard to find exact values for x and y. Maybe I need to assume some values or find a relationship that allows me to express the percentage in terms of y.Alternatively, perhaps I made a mistake in interpreting the problem. Let me double-check. The sum of the numbers is such that 2x + 3y = 42. Does that mean 2x + 3y equals the sum of the numbers? If so, then A + B = 2x + 3y = 42. That would make more sense because then I have two equations:1. A/B = (5y²)/(6x)2. A + B = 42That seems more manageable. Let's try that approach.From equation 1: A = (5y²)/(6x) * BSubstitute A into equation 2:(5y²)/(6x) * B + B = 42B*(5y²/(6x) + 1) = 42B = 42 / (5y²/(6x) + 1)But I still have x and y in the equation. Maybe I can express x in terms of y from 2x + 3y = 42:x = (42 - 3y)/2Substitute this into the expression for B:B = 42 / (5y²/(6*(42 - 3y)/2) + 1)= 42 / (5y²/(3*(42 - 3y)) + 1)= 42 / (5y²/(126 - 9y) + 1)This is getting quite involved. Maybe I need to find specific values for y that make the denominator a whole number or something. Alternatively, perhaps I can assume a value for y and see if it fits.Wait, maybe there's a better way. Let's consider that A and B are in the ratio 1/(6x) : 1/(5y²). Let's denote the ratio as k:1, where k = (1/(6x))/(1/(5y²)) = (5y²)/(6x). So, A = k*B.Then, A + B = 42 => k*B + B = 42 => B(k + 1) = 42 => B = 42/(k + 1)Similarly, A = k*B = k*42/(k + 1)Now, we need to find k in terms of y. Since k = (5y²)/(6x) and x = (42 - 3y)/2, substitute x:k = (5y²)/(6*(42 - 3y)/2) = (5y²)/(3*(42 - 3y)) = (5y²)/(126 - 9y)So, k = (5y²)/(126 - 9y)Now, B = 42/(k + 1) = 42/[(5y²)/(126 - 9y) + 1] = 42/[(5y² + 126 - 9y)/(126 - 9y)] = 42*(126 - 9y)/(5y² + 126 - 9y)Similarly, A = k*B = (5y²)/(126 - 9y) * [42*(126 - 9y)/(5y² + 126 - 9y)] = 42*5y²/(5y² + 126 - 9y)Now, to find the percentage increase from A to B, we calculate (B - A)/A * 100%.Let's compute B - A:B - A = [42*(126 - 9y)/(5y² + 126 - 9y)] - [42*5y²/(5y² + 126 - 9y)] = 42*(126 - 9y - 5y²)/(5y² + 126 - 9y)So, (B - A)/A = [42*(126 - 9y - 5y²)/(5y² + 126 - 9y)] / [42*5y²/(5y² + 126 - 9y)] = (126 - 9y - 5y²)/(5y²)Therefore, the percentage increase is [(126 - 9y - 5y²)/(5y²)] * 100%This expression still depends on y, and without additional information, I can't find a numerical value. Maybe I need to find y such that 2x + 3y = 42 and the ratio makes sense. Let's try to find possible integer values for y that make x positive.From 2x + 3y = 42, x = (42 - 3y)/2. For x to be positive, 42 - 3y > 0 => y < 14.Let's try y = 6:x = (42 - 18)/2 = 24/2 = 12Now, k = (5*(6)^2)/(126 - 9*6) = (5*36)/(126 - 54) = 180/72 = 2.5So, A = 42*5y²/(5y² + 126 - 9y) = 42*180/(180 + 126 - 54) = 42*180/252 = 42*(5/7) = 30B = 42 - 30 = 12Wait, that can't be right because if y = 6, then B should be larger than A, but here B = 12 and A = 30, which contradicts. Maybe I made a mistake in calculations.Wait, let's recalculate:A = 42*5y²/(5y² + 126 - 9y) with y=6:A = 42*180/(180 + 126 - 54) = 42*180/252 = 42*(5/7) = 30B = 42 - 30 = 12But according to the ratio, A should be (5y²)/(6x) * B. With y=6, x=12:A = (5*36)/(6*12) * B = 180/72 * B = 2.5 * BIf B=12, then A=30, which matches. So, the percentage increase from A to B is (12 - 30)/30 * 100% = (-18)/30 * 100% = -60%. That doesn't make sense because B should be more than A.Wait, maybe I got the ratio reversed. The ratio is A:B = 1/(6x) : 1/(5y²). So, A/B = (1/(6x))/(1/(5y²)) = 5y²/(6x). So, A = (5y²)/(6x) * B. If y=6 and x=12, then A = (5*36)/(6*12) * B = 180/72 * B = 2.5 * B. So, A is 2.5 times B, meaning A is larger than B. But the problem asks by what percent the second number is more than the first, which would be negative in this case, which doesn't make sense. So, maybe y=6 is not the right value.Let's try y=4:x = (42 - 12)/2 = 30/2 = 15k = (5*16)/(126 - 36) = 80/90 = 8/9So, A = 42*80/(80 + 126 - 36) = 42*80/170 = 42*(8/17) ≈ 19.76B = 42 - 19.76 ≈ 22.24Percentage increase = (22.24 - 19.76)/19.76 * 100% ≈ (2.48)/19.76 * 100% ≈ 12.55%That seems more reasonable. So, with y=4, the second number is approximately 12.55% more than the first.But I need to check if this is consistent with the ratio. A = (5y²)/(6x) * B = (5*16)/(6*15) * B = 80/90 * B ≈ 0.8889 * B. So, A ≈ 0.8889B, which means B ≈ A / 0.8889 ≈ 1.125A, which is a 12.5% increase. That matches our earlier calculation.So, it seems that when y=4, the percentage increase is approximately 12.5%. But is this the only solution? Let's try y=3:x = (42 - 9)/2 = 33/2 = 16.5k = (5*9)/(126 - 27) = 45/99 = 5/11 ≈ 0.4545A = 42*45/(45 + 126 - 27) = 42*45/144 = 42*(5/16) ≈ 13.125B = 42 - 13.125 ≈ 28.875Percentage increase = (28.875 - 13.125)/13.125 * 100% ≈ 15.75/13.125 * 100% ≈ 120%That's a huge increase, which seems unlikely. Also, checking the ratio:A = (5y²)/(6x) * B = (5*9)/(6*16.5) * B = 45/99 * B ≈ 0.4545B, so B ≈ 2.2B, which is a 120% increase. That matches.But this seems too high. Maybe y=4 is the correct value. Alternatively, perhaps there's a unique solution. Let's see if we can solve for y.From earlier, we have:Percentage increase = [(126 - 9y - 5y²)/(5y²)] * 100%Let me set this equal to P% and try to solve for y:(126 - 9y - 5y²)/(5y²) = P/100But without knowing P, I can't solve for y. Alternatively, maybe I can find y such that the percentage is a whole number.From y=4, we got approximately 12.5%, which is 1/8. y=6 gave a negative percentage, which doesn't make sense. y=3 gave 120%, which seems too high. Maybe y=5:x = (42 - 15)/2 = 27/2 = 13.5k = (5*25)/(126 - 45) = 125/81 ≈ 1.543A = 42*125/(125 + 126 - 45) = 42*125/196 ≈ 42*(1.25)/2.45 ≈ 21.43B = 42 - 21.43 ≈ 20.57Percentage increase = (20.57 - 21.43)/21.43 * 100% ≈ (-0.86)/21.43 * 100% ≈ -4.01%Negative again, which doesn't make sense. So, y=5 is not good.y=2:x = (42 - 6)/2 = 36/2 = 18k = (5*4)/(126 - 18) = 20/108 ≈ 0.1852A = 42*20/(20 + 126 - 18) = 42*20/128 ≈ 42*(0.15625) ≈ 6.5625B = 42 - 6.5625 ≈ 35.4375Percentage increase = (35.4375 - 6.5625)/6.5625 * 100% ≈ 28.875/6.5625 * 100% ≈ 440%That's extremely high. So, y=2 is not suitable.y=7:x = (42 - 21)/2 = 21/2 = 10.5k = (5*49)/(126 - 63) = 245/63 ≈ 3.8889A = 42*245/(245 + 126 - 63) = 42*245/308 ≈ 42*(0.7955) ≈ 33.33B = 42 - 33.33 ≈ 8.67Percentage increase = (8.67 - 33.33)/33.33 * 100% ≈ (-24.66)/33.33 * 100% ≈ -74%Negative again.So, from these trials, only y=4 gives a reasonable positive percentage increase of approximately 12.5%. Therefore, I think y=4 and x=15 is the correct pair, leading to the second number being approximately 12.5% more than the first.But to be precise, let's calculate it exactly with y=4:A = 42*5*(4)^2 / (5*(4)^2 + 126 - 9*4) = 42*80 / (80 + 126 - 36) = 3360 / 170 ≈ 19.7647B = 42 - 19.7647 ≈ 22.2353Percentage increase = (22.2353 - 19.7647)/19.7647 * 100% ≈ 2.4706/19.7647 * 100% ≈ 12.5%So, exactly 12.5%.Therefore, the second number is 12.5% more than the first.

💡First, I need to determine the amount of ethanol in fuel A. Since fuel A contains 12% ethanol by volume and 98 gallons of fuel A were added, I can calculate the ethanol in fuel A by multiplying 98 gallons by 12%.Next, I know that the total ethanol in the full tank is 30 gallons. By subtracting the ethanol from fuel A, I can find out how much ethanol comes from fuel B.After finding the ethanol in fuel B, I need to determine the volume of fuel B added to the tank. Since the tank's total capacity is 212 gallons and 98 gallons were filled with fuel A, the remaining volume must be filled with fuel B.Finally, to find the percentage of ethanol in fuel B, I divide the ethanol in fuel B by the volume of fuel B and multiply by 100 to get the percentage.

💡First, I need to determine the initial amount of potatoes shipped by the agricultural cooperative. I know that 150 kg of potatoes were damaged during transport and could not be sold. Therefore, the sellable amount of potatoes is the initial amount minus 150 kg.The potatoes are sold in 50 kg bags, and each bag is sold for 72. The total revenue from the sale of the potatoes is 9,144.To find out how many bags were sold, I can divide the total revenue by the price per bag:9144 ÷ 72 = 127 bags.Since each bag contains 50 kg of potatoes, the total sellable weight is:127 bags × 50 kg/bag = 6,350 kg.Finally, to find the initial amount of potatoes shipped, I add back the 150 kg that were damaged:6,350 kg + 150 kg = 6,500 kg.Therefore, the initial amount of potatoes shipped was 6,500 kg.