💡Alright, so I've got this problem about coloring the edges of a cube either red or green. The condition is that every face of the cube must have at least one red edge. I need to figure out the smallest number of red edges required to satisfy this condition. Let me think through this step by step.First, let's recall that a cube has 12 edges and 6 faces. Each face is a square, and each square has 4 edges. So, if I were to color some edges red, I need to make sure that each of these 6 faces has at least one red edge.My initial thought is to minimize the number of red edges. So, I want to use as few red edges as possible while still covering all 6 faces. If I can find a way where each red edge covers multiple faces, that would be efficient.Let me visualize the cube. Each edge is shared between two faces. So, if I color an edge red, it contributes to two different faces. That means, in theory, each red edge can cover two faces. If that's the case, then perhaps I can cover all 6 faces with just 3 red edges, since 3 edges times 2 faces each would cover 6 faces.But wait, is that actually possible? Let me try to imagine placing red edges in such a way that each red edge covers two faces, and all six faces are covered without overlap.If I place a red edge on one of the vertical edges of the cube, it covers the front and right face, for example. Then, if I place another red edge on the back-left vertical edge, it covers the back and left faces. Now, I've covered four faces with two red edges. The top and bottom faces still need to be covered. If I place a red edge on one of the top edges, say the front-top edge, it covers the top and front face. But the front face is already covered by the first red edge. So, that might not be the most efficient.Alternatively, if I place a red edge on the top edge that's not shared with the front or back face, like the top edge connecting the left and right faces, then it would cover the top and one of the side faces. But I'm not sure if this is the best approach.Maybe I need to think about the cube's structure more carefully. Each edge is part of two faces, but each face has four edges. So, if I color one edge per face, that would be six red edges, but that's probably more than necessary since edges are shared.Wait, if I can find a set of edges where each red edge covers two faces, and all six faces are covered, then three red edges should suffice. Let me try to find such a set.Imagine the cube with edges labeled from 1 to 12. Let's say I color edge 1 (front-top edge) red. This covers the front face and the top face. Then, I color edge 7 (back-bottom edge) red. This covers the back face and the bottom face. Now, I've covered four faces with two red edges. The remaining two faces are the left and right faces. If I color edge 4 (left-middle edge) red, it covers the left face and the front face. But the front face is already covered. Alternatively, if I color edge 5 (right-middle edge) red, it covers the right face and the back face, which is already covered.Hmm, this approach is leaving some faces uncovered or overlapping on already covered faces. Maybe I need a different strategy.Perhaps I should consider the cube's space diagonals or something like that. Wait, no, space diagonals are not edges; they're diagonals inside the cube.Let me think about the cube's edges in terms of their positions. There are 12 edges: 4 on the top face, 4 on the bottom face, and 4 vertical edges connecting the top and bottom.If I color all four vertical edges red, that would cover all six faces because each vertical edge is part of two adjacent faces. But that's four red edges, which might not be the minimum.Alternatively, if I color three vertical edges, each covering two faces, that would cover six faces. But wait, three vertical edges would cover six faces, but each vertical edge is shared between two faces, so three vertical edges would cover six faces. But is that possible?Let me try to visualize this. If I color the front-left, front-right, and back-left vertical edges red, then:- Front-left vertical edge covers the front and left faces.- Front-right vertical edge covers the front and right faces.- Back-left vertical edge covers the back and left faces.This way, the front, left, right, and back faces are covered. But the top and bottom faces are still uncovered. So, I need to cover those as well.If I color one of the top edges red, say the front-top edge, it covers the top and front faces. But the front face is already covered. Alternatively, if I color the top edge that's not on the front or back, like the top edge connecting the left and right faces, it would cover the top face and one of the side faces, which are already covered.This seems tricky. Maybe I need to use a combination of vertical and horizontal edges.What if I color three edges, each from different axes? For example, one vertical edge, one top edge, and one bottom edge. Let's see:- Color the front-left vertical edge red: covers front and left faces.- Color the top edge connecting the front and back on the right side red: covers top and right faces.- Color the bottom edge connecting the front and back on the left side red: covers bottom and left faces.Now, let's check which faces are covered:- Front: covered by the vertical edge.- Left: covered by the vertical edge and the bottom edge.- Right: covered by the top edge.- Back: not covered yet.- Top: covered by the top edge.- Bottom: covered by the bottom edge.Oh, the back face is still uncovered. So, I need to cover the back face as well. Maybe I need to color another edge.If I color the back-right vertical edge red, it covers the back and right faces. Now, all faces are covered:- Front: front-left vertical edge.- Left: front-left vertical edge and bottom edge.- Right: top edge and back-right vertical edge.- Back: back-right vertical edge.- Top: top edge.- Bottom: bottom edge.So, that's four red edges: front-left vertical, top-right horizontal, bottom-left horizontal, and back-right vertical. This covers all six faces.But is four the minimum? Maybe I can do it with three.Let me try again. Suppose I color three edges such that each red edge covers two faces, and all six faces are covered without overlap.If I color the front-top edge red: covers front and top.Color the back-bottom edge red: covers back and bottom.Color the left-middle edge red: covers left and front.Wait, the front face is covered twice, and the left face is covered once. The right face is still uncovered. So, that doesn't work.Alternatively, color the front-top edge: covers front and top.Color the back-bottom edge: covers back and bottom.Color the right-middle edge: covers right and back.Now, front, top, back, bottom, and right are covered. Left is still uncovered.So, I need to cover the left face as well. Maybe color the left-middle edge: covers left and front. But front is already covered.Alternatively, color the left-top edge: covers left and top. Top is already covered.Hmm, seems like with three red edges, one face always gets left out. Maybe three red edges aren't enough.Wait, what if I choose edges that are not adjacent? For example, color the front-top edge, the back-bottom edge, and the left-right edge on the top face.But the left-right edge on the top face is the same as the front-top edge or the back-top edge. Maybe I'm getting confused.Alternatively, think about the cube's edges in terms of their positions:- There are 4 vertical edges.- 4 edges on the top face.- 4 edges on the bottom face.If I color three edges, each from different axes, maybe I can cover all six faces.But I'm not sure. It seems like three red edges might not be enough because each red edge can only cover two faces, and with three red edges, that's six face coverages, but since each face needs at least one, it's possible in theory, but in practice, due to the cube's structure, it might not be possible without overlapping.Wait, let's think about it differently. If I can find three edges such that each edge covers two different faces, and all six faces are covered without any face being left out, then three red edges would suffice.Let me try to find such a set:1. Color the front-top edge: covers front and top.2. Color the back-bottom edge: covers back and bottom.3. Color the left-right edge on the top face: covers left and right.Wait, the left-right edge on the top face is actually the same as the front-top edge or the back-top edge. No, actually, the left-right edge on the top face is a separate edge. So, if I color that, it covers the left and right faces.So, with these three red edges:- Front-top edge: front and top.- Back-bottom edge: back and bottom.- Left-right edge on top: left and right.Now, all six faces are covered:- Front: front-top edge.- Top: front-top edge and left-right edge.- Back: back-bottom edge.- Bottom: back-bottom edge.- Left: left-right edge.- Right: left-right edge.Yes, that works! So, with three red edges, I can cover all six faces. Each red edge covers two faces, and there's no overlap in the faces they cover.Wait, but is this actually possible? Let me double-check.Front-top edge: covers front and top.Back-bottom edge: covers back and bottom.Left-right edge on top: covers left and right.Yes, that seems to cover all six faces without any face being left out. So, three red edges should be sufficient.But wait, in my earlier attempt, I thought that three red edges might not be enough because of overlapping, but it seems like with careful selection, it is possible.So, maybe the answer is three red edges.But let me think again. Is there a way to do it with two red edges? If I color two edges, each covering two faces, that would cover four faces. But we have six faces, so two red edges would only cover four faces, leaving two faces uncovered. Therefore, two red edges are insufficient.Therefore, the minimum number of red edges required is three.Wait, but earlier I thought that three red edges might not cover all faces, but with careful selection, it seems possible. So, maybe three is the answer.But let me try to visualize this again. If I color the front-top edge, back-bottom edge, and left-right edge on the top face, do they actually cover all six faces?- Front-top edge: front and top.- Back-bottom edge: back and bottom.- Left-right edge on top: left and right.Yes, that's six faces covered. So, three red edges are sufficient.But I'm still a bit unsure because when I tried earlier, I thought that three red edges might leave a face uncovered, but maybe I was making a mistake in the selection.Alternatively, maybe I need to consider that some edges are shared between more than two faces, but in a cube, each edge is only shared between two faces.So, each red edge can only cover two faces. Therefore, to cover six faces, we need at least three red edges, since 3 edges x 2 faces per edge = 6 faces.Therefore, three red edges should be sufficient, and it's the minimum because two red edges would only cover four faces.So, the answer should be three red edges.But wait, let me think about the cube's structure again. If I color three edges, each covering two faces, and all six faces are covered, then three is indeed the minimum.Yes, I think that's correct.

💡Okay, so I need to figure out how much I have to pay in the tenth month for this household appliance I bought on an installment plan. The total price is 1150 yuan, and I paid 150 yuan upfront. That leaves me with a debt of 1000 yuan. Each month, I have to pay 50 yuan plus the interest on the remaining debt. The interest rate is 1% per month. The first payment starts after the initial 150 yuan payment, so the first month is the first installment.Alright, let me break this down. After paying 150 yuan, I owe 1000 yuan. Each month, I pay 50 yuan towards the principal and 1% interest on the remaining debt. So, the interest part changes every month because the remaining debt decreases as I pay more principal.Let me think about the first few months to see if I can find a pattern.In the first month, the remaining debt is 1000 yuan. The interest is 1% of 1000, which is 10 yuan. So, the total payment for the first month is 50 + 10 = 60 yuan.After paying 60 yuan, the principal paid is 50 yuan, so the remaining debt is 1000 - 50 = 950 yuan.In the second month, the interest is 1% of 950 yuan, which is 9.5 yuan. So, the total payment is 50 + 9.5 = 59.5 yuan.After paying 59.5 yuan, the principal paid is another 50 yuan, so the remaining debt is 950 - 50 = 900 yuan.In the third month, the interest is 1% of 900 yuan, which is 9 yuan. So, the total payment is 50 + 9 = 59 yuan.Hmm, I see a pattern here. Each month, the interest decreases by 0.5 yuan because the principal decreases by 50 yuan each month, and 1% of 50 yuan is 0.5 yuan. So, the interest part decreases by 0.5 yuan each month.Let me verify that. If each month the principal decreases by 50 yuan, then the interest, which is 1% of the remaining principal, decreases by 1% of 50 yuan, which is 0.5 yuan. Yes, that makes sense.So, the interest part starts at 10 yuan in the first month and decreases by 0.5 yuan each subsequent month. Therefore, the total payment each month is 50 yuan plus an interest that decreases by 0.5 yuan each month.So, the payment in the nth month can be expressed as:Payment_n = 50 + (10 - 0.5*(n-1)) yuanLet me check this formula with the first few months.For n=1: Payment_1 = 50 + (10 - 0) = 60 yuan. Correct.For n=2: Payment_2 = 50 + (10 - 0.5) = 59.5 yuan. Correct.For n=3: Payment_3 = 50 + (10 - 1.0) = 59 yuan. Correct.Good, the formula seems to work.Now, the question is asking for the payment in the tenth month. So, let's plug n=10 into the formula.Payment_10 = 50 + (10 - 0.5*(10-1)) = 50 + (10 - 4.5) = 50 + 5.5 = 55.5 yuan.So, in the tenth month, I should pay 55.5 yuan.Now, the second part of the question is asking how much money has actually been spent after the entire loan is paid off.I need to calculate the total amount paid, which includes the initial 150 yuan plus all the monthly payments.First, let's figure out how many months it takes to pay off the loan. Since I'm paying 50 yuan each month towards the principal, and the initial principal is 1000 yuan, it should take 1000 / 50 = 20 months.So, there are 20 monthly payments. Each payment is 50 yuan plus the decreasing interest.We already have the formula for the payment in the nth month: Payment_n = 50 + (10 - 0.5*(n-1)).This is an arithmetic sequence where the first term a1 = 60 yuan, and the common difference d = -0.5 yuan.The total sum of the monthly payments can be calculated using the formula for the sum of an arithmetic series:S_n = n/2 * (2a1 + (n-1)d)Where n is the number of terms, a1 is the first term, and d is the common difference.Plugging in the values:S_20 = 20/2 * (2*60 + (20-1)*(-0.5)) = 10 * (120 + 19*(-0.5)) = 10 * (120 - 9.5) = 10 * 110.5 = 1105 yuan.So, the total of the monthly payments is 1105 yuan.Adding the initial payment of 150 yuan, the total expenditure is 150 + 1105 = 1255 yuan.Let me double-check the arithmetic:Sum of monthly payments:First term a1 = 60, last term a20 = 50 + (10 - 0.5*(20-1)) = 50 + (10 - 9.5) = 50 + 0.5 = 50.5 yuan.Sum S_20 = 20/2 * (60 + 50.5) = 10 * 110.5 = 1105 yuan. Correct.Total expenditure: 150 + 1105 = 1255 yuan. Correct.So, the total amount spent after paying off the loan is 1255 yuan.

💡Okay, so I have this equation here: 9 over x squared equals x over 81. Hmm, let me write that down to make it clearer. So, it's 9 divided by x squared equals x divided by 81. I need to find the value of x that makes this equation true.Alright, first, I remember that when you have a proportion like this, you can cross-multiply to solve for the unknown. That means I can multiply the numerator of one fraction by the denominator of the other fraction. So, if I cross-multiply, I'll get 9 times 81 equals x squared times x. Let me write that out: 9 * 81 = x^3. Wait, let me make sure I did that right. Yeah, because 9 is in the numerator on the left, and 81 is in the denominator on the right, so 9 times 81 should equal x squared times x, which is x cubed. Okay, that makes sense.Now, I need to calculate 9 multiplied by 81. Let me do that step by step. 9 times 80 is 720, and 9 times 1 is 9, so adding those together gives me 729. So, 9 * 81 equals 729. That means x cubed equals 729. To find x, I need to take the cube root of both sides of the equation. The cube root of x cubed is just x, and the cube root of 729 is... Hmm, what number multiplied by itself three times gives 729? Let me think. I know that 9 times 9 is 81, and 81 times 9 is 729. So, 9 cubed is 729. That means the cube root of 729 is 9. So, x equals 9. Let me double-check that to make sure I didn't make a mistake. If I plug x = 9 back into the original equation, does it hold true? Let's see. On the left side, it's 9 divided by 9 squared, which is 9 over 81, which simplifies to 1/9. On the right side, it's 9 divided by 81, which is also 1/9. Yep, both sides are equal, so x = 9 is definitely the correct solution.I think that's it. I didn't run into any problems, and my answer checks out when I substitute it back into the original equation. So, I'm confident that x is 9.

💡Alright, so I've got this math problem here, and I need to figure out which grades have 30 fewer students than grades 1 and 2 combined. Let me try to break it down step by step.First, the problem says, "The total number of students in grades 1 and 2 is 30 more than the total number of students in some grades." Okay, so if I let G1 be the number of students in grade 1 and G2 be the number in grade 2, then G1 + G2 is 30 more than some other set of grades. Let's call that set of grades Set X. So, G1 + G2 = Set X + 30.Next, it says, "The number of students in grade 5 is 30 less than the number of students in grade 1." So, if G5 is the number of students in grade 5, then G5 = G1 - 30.Now, the question is asking which grades have 30 fewer students than grades 1 and 2 combined. From the first part, we know that G1 + G2 = Set X + 30. So, Set X must be the grades that have 30 fewer students than grades 1 and 2 combined.But wait, the problem also mentions grade 5. If G5 = G1 - 30, then grade 5 has 30 fewer students than grade 1. But the question is about grades that have 30 fewer students than grades 1 and 2 combined, not just grade 1.So, maybe Set X includes grade 5 and possibly other grades? Or is it just grade 5?Let me think. If G1 + G2 = Set X + 30, and Set X is the grades that have 30 fewer students than G1 + G2, then Set X could be grade 5 alone or a combination of grades.But since G5 = G1 - 30, and G1 + G2 = Set X + 30, maybe Set X is grade 5 plus grade 2? Let's see:If Set X = G5 + G2, then:G1 + G2 = (G5 + G2) + 30But G5 = G1 - 30, so:G1 + G2 = ((G1 - 30) + G2) + 30Simplify:G1 + G2 = G1 - 30 + G2 + 30G1 + G2 = G1 + G2That checks out, but it doesn't tell me anything new.Maybe Set X is just grade 5? Let's test that:If Set X = G5, then:G1 + G2 = G5 + 30But G5 = G1 - 30, so:G1 + G2 = (G1 - 30) + 30Simplify:G1 + G2 = G1That doesn't make sense because G2 would have to be zero, which isn't possible.So, Set X can't be just grade 5. Maybe it's grade 5 and another grade?Suppose Set X includes grade 5 and grade 3. Let's say Set X = G5 + G3.Then:G1 + G2 = (G5 + G3) + 30Again, G5 = G1 - 30, so:G1 + G2 = (G1 - 30 + G3) + 30Simplify:G1 + G2 = G1 + G3So, G2 = G3That means grade 3 has the same number of students as grade 2. Is that possible? Maybe, but the problem doesn't specify that.Alternatively, maybe Set X is grade 5 and grade 4.Set X = G5 + G4Then:G1 + G2 = (G5 + G4) + 30Again, G5 = G1 - 30, so:G1 + G2 = (G1 - 30 + G4) + 30Simplify:G1 + G2 = G1 + G4So, G2 = G4Again, grade 4 would have the same number of students as grade 2. Not necessarily impossible, but the problem doesn't provide that information.Wait, maybe Set X is grade 5 and grade 6?Set X = G5 + G6Then:G1 + G2 = (G5 + G6) + 30G5 = G1 - 30, so:G1 + G2 = (G1 - 30 + G6) + 30Simplify:G1 + G2 = G1 + G6So, G2 = G6Again, grade 6 would have the same number of students as grade 2.This pattern suggests that Set X could be grade 5 plus any other grade, and that other grade would have the same number of students as grade 2. But since the problem doesn't specify, I can't determine exactly which grades are in Set X.Alternatively, maybe Set X is just grade 5 and grade 2?Set X = G5 + G2Then:G1 + G2 = (G5 + G2) + 30G5 = G1 - 30, so:G1 + G2 = (G1 - 30 + G2) + 30Simplify:G1 + G2 = G1 + G2That's true, but it doesn't give me new information.Hmm, this is confusing. Maybe I need to approach it differently.Let me assign variables:Let G1 = number of students in grade 1G2 = number of students in grade 2G5 = number of students in grade 5Given:G1 + G2 = Set X + 30G5 = G1 - 30We need to find Set X such that Set X = G1 + G2 - 30But Set X is the total number of students in some grades. So, Set X could be a single grade or multiple grades.If Set X is a single grade, say grade 5, then:G5 = G1 + G2 - 30But we also have G5 = G1 - 30So:G1 - 30 = G1 + G2 - 30Simplify:G1 - 30 = G1 + G2 - 30Subtract G1 from both sides:-30 = G2 - 30Add 30 to both sides:0 = G2That can't be right because grade 2 can't have zero students.So, Set X can't be just grade 5.Maybe Set X is grade 5 and grade 4.Set X = G5 + G4Then:G1 + G2 = G5 + G4 + 30But G5 = G1 - 30, so:G1 + G2 = (G1 - 30) + G4 + 30Simplify:G1 + G2 = G1 + G4So, G2 = G4Again, grade 4 has the same number of students as grade 2.Alternatively, Set X could be grade 5 and grade 3.Set X = G5 + G3Then:G1 + G2 = G5 + G3 + 30G5 = G1 - 30, so:G1 + G2 = (G1 - 30) + G3 + 30Simplify:G1 + G2 = G1 + G3So, G2 = G3Grade 3 has the same number of students as grade 2.This pattern suggests that Set X is grade 5 plus any other grade, and that other grade has the same number of students as grade 2.But since the problem doesn't specify which other grade, I can't determine exactly which grades are in Set X.Alternatively, maybe Set X is grade 5 and grade 6.Set X = G5 + G6Then:G1 + G2 = G5 + G6 + 30G5 = G1 - 30, so:G1 + G2 = (G1 - 30) + G6 + 30Simplify:G1 + G2 = G1 + G6So, G2 = G6Grade 6 has the same number of students as grade 2.Again, the problem doesn't specify, so I can't determine exactly.Wait, maybe Set X is just grade 5 and grade 2.Set X = G5 + G2Then:G1 + G2 = G5 + G2 + 30G5 = G1 - 30, so:G1 + G2 = (G1 - 30) + G2 + 30Simplify:G1 + G2 = G1 + G2That's true, but it doesn't give me new information.So, it seems like Set X could be grade 5 plus any other grade, and that other grade would have the same number of students as grade 2. But without more information, I can't pinpoint exactly which grades are in Set X.Maybe the problem expects a general answer, like "grades 5 and another grade with the same number of students as grade 2." But I'm not sure.Alternatively, perhaps the problem is simpler. Maybe Set X is just grade 5, but that doesn't work because G5 = G1 - 30, and G1 + G2 = G5 + 30 would imply G2 = 0, which isn't possible.So, the only way for Set X to have 30 fewer students than G1 + G2 is if Set X includes grade 5 and another grade with the same number of students as grade 2.Therefore, the grades that have 30 fewer students than grades 1 and 2 combined are grade 5 and another grade with the same number of students as grade 2.But since the problem doesn't specify which other grade, I can't give a specific answer. Maybe it's grade 3, 4, or 6. Without more information, it's impossible to determine exactly.Wait, maybe the problem is asking for which grades in general, not specific grades. So, the answer would be that the grades with 30 fewer students than grades 1 and 2 combined are grade 5 and another grade with the same number of students as grade 2.But I'm not sure if that's what the problem is asking.Alternatively, maybe the problem is asking for the total number of students in Set X, which is G1 + G2 - 30. But the question is asking which grades, not the number.Hmm, this is tricky. I think I need to go back to the problem statement."The total number of students in grades 1 and 2 is 30 more than the total number of students in some grades. The number of students in grade 5 is 30 less than the number of students in grade 1. Which grades have 30 fewer students than grades 1 and 2 combined?"So, rephrasing: G1 + G2 = Set X + 30G5 = G1 - 30Find Set X.From the first equation, Set X = G1 + G2 - 30But Set X is the total number of students in some grades. So, Set X could be a single grade or multiple grades.If Set X is a single grade, say grade 5, then G5 = G1 + G2 - 30But we also have G5 = G1 - 30So:G1 - 30 = G1 + G2 - 30Which simplifies to G2 = 0, which is impossible.Therefore, Set X must be multiple grades.If Set X is grade 5 and another grade, say grade k, then:G5 + Gk = G1 + G2 - 30But G5 = G1 - 30, so:(G1 - 30) + Gk = G1 + G2 - 30Simplify:G1 - 30 + Gk = G1 + G2 - 30Cancel out G1 and -30:Gk = G2So, grade k has the same number of students as grade 2.Therefore, Set X is grade 5 and any grade with the same number of students as grade 2.But since the problem doesn't specify which grade that is, I can't name it.Alternatively, maybe Set X is grade 5 and grade 2.Set X = G5 + G2Then:G5 + G2 = G1 + G2 - 30G5 = G1 - 30So:(G1 - 30) + G2 = G1 + G2 - 30Simplify:G1 - 30 + G2 = G1 + G2 - 30Which is true, but it doesn't give me new information.So, in conclusion, the grades that have 30 fewer students than grades 1 and 2 combined are grade 5 and another grade with the same number of students as grade 2. However, without specific information about which grade that is, I can't provide a more precise answer.

💡Alright, so I have this problem about vectors in a triangle. Let me try to visualize it first. There's triangle ABC, and AD is the median to side BC. That means D is the midpoint of BC. Then, E is the midpoint of AD. I need to find the vector EB in terms of vectors AB and AC.Okay, let's start by recalling some vector basics. In a triangle, a median connects a vertex to the midpoint of the opposite side. So, since AD is the median, D is the midpoint of BC. That means BD equals DC.Now, E is the midpoint of AD. So, AE equals ED. I need to express vector EB in terms of AB and AC. Hmm, maybe I can express EB as a combination of AB and AC by breaking it down into smaller vectors.Let me think about how to express EB. Vector EB goes from point E to point B. Maybe I can express this vector in terms of vectors from other points that I know.Since E is the midpoint of AD, perhaps I can express vector AE first and then relate it to vector EB. Vector AE is half of vector AD because E is the midpoint. So, AE = (1/2)AD.But what is vector AD? Since D is the midpoint of BC, vector AD can be expressed in terms of vectors AB and AC. Let me recall that the midpoint of BC can be expressed as (B + C)/2 in vector terms. So, vector AD would be from A to D, which is (B + C)/2 - A.Wait, in vector terms, if we consider point A as the origin, then vector AB is B - A, and vector AC is C - A. But maybe it's easier to express everything in terms of vectors AB and AC without setting A as the origin.Let me try another approach. Let's express vector AD. Since D is the midpoint of BC, vector AD is equal to (AB + AC)/2. Is that right? Wait, no, that's not quite accurate. Vector AD is from A to D, and D is the midpoint of BC. So, vector AD is equal to vector AB plus half of vector BC.But vector BC is AC - AB, right? Because from B to C is AC - AB. So, vector BC = AC - AB. Therefore, half of vector BC is (AC - AB)/2.So, vector AD = vector AB + (AC - AB)/2. Let's compute that:AD = AB + (AC - AB)/2= (2AB + AC - AB)/2= (AB + AC)/2Okay, so vector AD is indeed (AB + AC)/2. That makes sense because D is the midpoint of BC, so AD is the average of AB and AC.Now, since E is the midpoint of AD, vector AE is half of vector AD. So, vector AE = (1/2)AD = (1/2)(AB + AC)/2 = (AB + AC)/4.So, vector AE = (AB + AC)/4.Now, I need vector EB. Vector EB is from E to B. To find this, I can express it as vector AB minus vector AE.Wait, let me think. If I have point E, and I want to go from E to B, that's the same as going from A to B minus going from A to E. So, vector EB = vector AB - vector AE.Yes, that seems right. So, vector EB = AB - AE.We already have vector AE as (AB + AC)/4. So, substituting that in:EB = AB - (AB + AC)/4= (4AB - AB - AC)/4= (3AB - AC)/4So, vector EB is (3/4)AB - (1/4)AC.Looking at the options, that's option A: (3/4)AB - (1/4)AC.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Vector AD is (AB + AC)/2 because D is the midpoint of BC.2. Vector AE is half of AD, so (AB + AC)/4.3. Vector EB is AB - AE, which gives (3/4)AB - (1/4)AC.Yes, that seems correct. So, the answer should be option A.

💡Okay, so I have this problem about a hyperbola, and I need to find the intersection point of two mutually perpendicular lines that pass through the foci of the hyperbola. The hyperbola is given by the equation (frac{x^2}{16} - frac{y^2}{9} = 1), and there's a point (A(6, -2)) that lies on the line passing through the right focus. Hmm, let me try to break this down step by step.First, I remember that the standard form of a hyperbola is (frac{x^2}{a^2} - frac{y^2}{b^2} = 1), which means it opens left and right. In this case, (a^2 = 16) and (b^2 = 9), so (a = 4) and (b = 3). Now, the foci of a hyperbola are located at ((pm c, 0)), where (c) is the distance from the center to each focus. I recall that (c^2 = a^2 + b^2), so let me calculate that:(c^2 = 16 + 9 = 25), so (c = 5).Therefore, the foci are at ((-5, 0)) and ((5, 0)). Let me note these as (F_1(-5, 0)) and (F_2(5, 0)).The problem mentions two mutually perpendicular lines passing through these foci. One of these lines passes through the right focus, which is (F_2(5, 0)), and the point (A(6, -2)) lies on this line. So, I need to find the equation of the line passing through (F_2) and (A), and then find a line perpendicular to this that also passes through the other focus (F_1). The intersection of these two lines will be the point we're looking for.Let me start by finding the equation of the line passing through (F_2(5, 0)) and (A(6, -2)). To find the equation of a line, I need the slope first. The slope (m) between two points ((x_1, y_1)) and ((x_2, y_2)) is given by:(m = frac{y_2 - y_1}{x_2 - x_1})Plugging in the coordinates of (F_2) and (A):(m = frac{-2 - 0}{6 - 5} = frac{-2}{1} = -2)So, the slope of the line (AF_2) is (-2). Now, using the point-slope form of the equation of a line, which is:(y - y_1 = m(x - x_1))Using point (F_2(5, 0)):(y - 0 = -2(x - 5))Simplifying:(y = -2x + 10)So, the equation of line (AF_2) is (y = -2x + 10). Now, since the two lines are mutually perpendicular, the other line passing through (F_1(-5, 0)) must have a slope that is the negative reciprocal of (-2). The negative reciprocal of (-2) is (frac{1}{2}). So, the slope of the line perpendicular to (AF_2) is (frac{1}{2}).Using the point-slope form again for the line passing through (F_1(-5, 0)) with slope (frac{1}{2}):(y - 0 = frac{1}{2}(x - (-5)))Simplifying:(y = frac{1}{2}(x + 5))So, the equation of the perpendicular line is (y = frac{1}{2}x + frac{5}{2}).Now, I need to find the intersection point of these two lines: (y = -2x + 10) and (y = frac{1}{2}x + frac{5}{2}). To find the intersection, I can set the two equations equal to each other:(-2x + 10 = frac{1}{2}x + frac{5}{2})Let me solve for (x). First, multiply both sides by 2 to eliminate the fraction:(-4x + 20 = x + 5)Now, bring all terms to one side:(-4x - x + 20 - 5 = 0)Simplify:(-5x + 15 = 0)So,(-5x = -15)Divide both sides by (-5):(x = 3)Now, plug (x = 3) back into one of the equations to find (y). Let me use (y = -2x + 10):(y = -2(3) + 10 = -6 + 10 = 4)So, the intersection point is ((3, 4)).Wait, let me double-check by plugging (x = 3) into the other equation (y = frac{1}{2}x + frac{5}{2}):(y = frac{1}{2}(3) + frac{5}{2} = frac{3}{2} + frac{5}{2} = frac{8}{2} = 4)Yep, that checks out. So, the intersection point is indeed ((3, 4)).Just to make sure I didn't make any mistakes, let me recap:1. Found the foci of the hyperbola: ((-5, 0)) and ((5, 0)).2. Calculated the slope of the line through (F_2(5, 0)) and (A(6, -2)) as (-2).3. Derived the equation of that line: (y = -2x + 10).4. Found the slope of the perpendicular line as (frac{1}{2}).5. Derived the equation of the perpendicular line through (F_1(-5, 0)): (y = frac{1}{2}x + frac{5}{2}).6. Solved the system of equations to find the intersection at ((3, 4)).Everything seems to add up. I think I'm confident with this answer.**Final Answer**The coordinates of the intersection point are boxed{(3, 4)}.