💡Okay, so I have this problem about a tetrahedron ABCD. The faces ABC and BCD meet at a dihedral angle of 45 degrees. The area of face ABC is 150, the area of face BCD is 90, and the length of edge BC is 12 cm. I need to find the volume of this tetrahedron.Hmm, let me recall what a tetrahedron is. It's a three-dimensional shape with four triangular faces, right? So, ABCD has four vertices, and each pair of vertices is connected by an edge. In this case, the two faces ABC and BCD share the common edge BC.The dihedral angle between two faces is the angle between the two planes that make up those faces. So, the angle between face ABC and face BCD is 45 degrees. I think this angle is important because it tells me how "tilted" one face is relative to the other.Given the areas of the two faces and the length of their common edge, I need to find the volume. I remember that the volume of a tetrahedron can be calculated if I know the area of a base and the height perpendicular to that base. The formula is:[ text{Volume} = frac{1}{3} times text{Base Area} times text{Height} ]So, if I can figure out the height from the opposite vertex to the base, I can compute the volume. But how do I find this height?Let me think about the two faces ABC and BCD. Both share the edge BC, which is 12 cm long. The area of ABC is 150, and the area of BCD is 90. Maybe I can find the heights of each triangle with respect to the base BC.For triangle ABC, the area is 150, and the base is BC = 12 cm. The area of a triangle is given by:[ text{Area} = frac{1}{2} times text{base} times text{height} ]So, for triangle ABC:[ 150 = frac{1}{2} times 12 times h_{ABC} ][ 150 = 6 times h_{ABC} ][ h_{ABC} = frac{150}{6} = 25 text{ cm} ]So, the height from A to BC is 25 cm.Similarly, for triangle BCD, the area is 90, and the base is BC = 12 cm:[ 90 = frac{1}{2} times 12 times h_{BCD} ][ 90 = 6 times h_{BCD} ][ h_{BCD} = frac{90}{6} = 15 text{ cm} ]So, the height from D to BC is 15 cm.Now, I need to relate these heights to the dihedral angle between the two faces. The dihedral angle is 45 degrees. I think this angle is between the two planes, so it might relate to the angle between the normals of the two faces or something like that.Wait, maybe I can use the dihedral angle to find the height from D to the plane ABC. Because if I can find that height, I can use it in the volume formula.Let me visualize this. The two faces ABC and BCD meet along BC, and the dihedral angle between them is 45 degrees. So, if I imagine the two triangles hinged along BC, the angle between them is 45 degrees.I think the height from D to BC (which is 15 cm) can be related to the height from D to the plane ABC. If I can find that, then I can use it as the height in the volume formula.So, if I consider the height from D to BC as 15 cm, and the dihedral angle between the two faces as 45 degrees, then the height from D to the plane ABC would be the component of this height perpendicular to the plane ABC.In other words, if I have a right triangle where one leg is the height from D to BC (15 cm), and the angle between the two faces is 45 degrees, then the height from D to the plane ABC is 15 cm multiplied by sin(45°).Because in the right triangle, the height from D to the plane ABC is opposite the 45-degree angle, and the height from D to BC is the hypotenuse.So, let me write that:[ h_{text{perpendicular}} = h_{BCD} times sin(45^circ) ][ h_{text{perpendicular}} = 15 times frac{sqrt{2}}{2} ][ h_{text{perpendicular}} = frac{15sqrt{2}}{2} text{ cm} ]Okay, so the perpendicular height from D to the plane ABC is (15√2)/2 cm.Now, I can use this height in the volume formula. The base area is the area of triangle ABC, which is 150 cm².So, plugging into the volume formula:[ text{Volume} = frac{1}{3} times 150 times frac{15sqrt{2}}{2} ]Let me compute this step by step.First, multiply 150 by 15:[ 150 times 15 = 2250 ]Then, multiply by √2:[ 2250 times sqrt{2} = 2250sqrt{2} ]Now, divide by 3 and then by 2:Dividing by 3 first:[ frac{2250sqrt{2}}{3} = 750sqrt{2} ]Then dividing by 2:[ frac{750sqrt{2}}{2} = 375sqrt{2} ]So, the volume is 375√2 cubic centimeters.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Calculated the heights of both triangles ABC and BCD with respect to base BC: 25 cm and 15 cm respectively. That seems correct.2. Then, considered the dihedral angle of 45 degrees between the two faces. So, the height from D to the plane ABC is the height from D to BC multiplied by sin(45°). That makes sense because the dihedral angle is the angle between the two planes, so the perpendicular height would be the component of the height from D to BC in the direction perpendicular to the plane ABC.3. Calculated h_perpendicular as 15 * sin(45°) = (15√2)/2 cm. That seems right.4. Then, used the volume formula: (1/3)*base area*height. Plugged in 150 for the base area and (15√2)/2 for the height.5. Computed 150 * (15√2)/2 = (2250√2)/2 = 1125√2. Then, multiplied by 1/3: (1125√2)/3 = 375√2. Wait, hold on, in my initial calculation, I think I messed up the order of division.Wait, let me redo that step:Volume = (1/3) * 150 * (15√2)/2First, multiply 150 and (15√2)/2:150 * (15√2)/2 = (150 * 15√2)/2 = (2250√2)/2 = 1125√2Then, multiply by 1/3:1125√2 * (1/3) = 375√2Yes, so that's correct. So, the volume is 375√2 cubic centimeters.I think that's the right answer. Let me just make sure I didn't confuse the dihedral angle with something else.Another way to think about it is: the dihedral angle is the angle between the two planes. So, if I have two planes intersecting at a line (BC), the dihedral angle is the angle between them. So, if I have a point D in one plane, the distance from D to the other plane can be found by multiplying the distance from D to the line BC by sin(theta), where theta is the dihedral angle.Yes, that seems consistent with what I did earlier. So, I think my reasoning is correct.Therefore, the volume of the tetrahedron is 375√2 cubic centimeters.**Final Answer**The volume of the tetrahedron is boxed{375sqrt{2}} cubic centimeters.

💡First, I need to calculate the total amount Sarah spent on the gifts. She bought two toy cars at 11 each, which amounts to 22. She also purchased a scarf for 10, a beanie for 14, a necklace for 20, a pair of gloves for 12, and a book for 15.Adding up all these amounts:22 (toy cars) + 10 (scarf) + 14 (beanie) + 20 (necklace) + 12 (gloves) + 15 (book) = 93.After buying all the gifts, Sarah has 7 remaining. To find out how much money she started with, I add the total amount she spent to the remaining money:93 + 7 = 100.Therefore, Sarah started with 100.

💡Alright, so I have this problem where I need to find the locus of points D such that the areas of triangles ABD and ACD are equal. Hmm, okay. Let me try to visualize this. I have three points, A, B, and C, somewhere on a plane. I need to find all possible points D where the area of triangle ABD is the same as the area of triangle ACD.First, I remember that the area of a triangle can be calculated using the formula: (1/2)*base*height. So, for triangles ABD and ACD, they both share the base AD. That means if their areas are equal, their heights from B and C to the base AD must be equal as well. Wait, so if the heights from B and C to AD are equal, that means that the perpendicular distances from B and C to the line AD must be the same. That makes sense. So, point D must be such that when you draw line AD, both B and C are at the same distance from this line.Now, how can I find such points D? Well, if the perpendicular distances from B and C to AD are equal, then AD must be a line that is equidistant from both B and C. I think that means AD has to be either the perpendicular bisector of BC or a line parallel to BC.Let me think about that. If AD is the perpendicular bisector of BC, then it would indeed be equidistant from both B and C. But also, if AD is parallel to BC, then the distance from B to AD would be the same as the distance from C to AD, right? Because if two lines are parallel, the distance between them is constant.So, does that mean that the locus of D is the set of all points lying on either the perpendicular bisector of BC or a line parallel to BC passing through A? Hmm, that seems plausible.Wait, but if I take a line parallel to BC through A, then any point D on that line would make AD parallel to BC. In that case, the heights from B and C to AD would indeed be equal because the distance between two parallel lines is constant. So, that would satisfy the condition that the areas of ABD and ACD are equal.Similarly, if D is on the perpendicular bisector of BC, then AD would be the perpendicular bisector, and the distances from B and C to AD would be equal because D is equidistant from B and C. So, that also satisfies the area condition.But wait, is that all? Are there any other points D that could satisfy the condition? Let me see. Suppose D is not on either of these lines. Then, the distances from B and C to AD would not be equal, so the areas of ABD and ACD would not be equal. Therefore, only points on these two lines would satisfy the condition.So, putting it all together, the locus of points D is the union of the perpendicular bisector of BC and the line through A parallel to BC.Wait, but I should make sure that these are the only possibilities. Let me consider some specific cases. Suppose points B and C are on the same side of AD. Then, for the areas to be equal, the heights must be equal, which would require AD to be parallel to BC. On the other hand, if B and C are on opposite sides of AD, then the distances from B and C to AD must still be equal, which would imply that AD is the perpendicular bisector of BC.Therefore, these two cases cover all possibilities, and thus the locus is indeed the combination of these two lines.I think I've got it. The locus of points D consists of two lines: one is the perpendicular bisector of BC, and the other is the line through A parallel to BC. So, any point D lying on either of these lines will satisfy the condition that the areas of triangles ABD and ACD are equal.Just to double-check, let me consider a simple example. Let’s say A is at (0,0), B is at (1,0), and C is at (-1,0). Then, the perpendicular bisector of BC would be the y-axis, and the line through A parallel to BC would be the x-axis itself. So, any point D on the y-axis or the x-axis would satisfy the area condition. For instance, if D is at (0,1), then triangles ABD and ACD both have base AD of length 1 and height 1, so their areas are equal. Similarly, if D is at (1,0), then AD is along the x-axis, and since B and C are symmetric with respect to the y-axis, the areas of ABD and ACD are equal. This seems to confirm my reasoning. Therefore, I'm confident that the locus is indeed the union of the perpendicular bisector of BC and the line through A parallel to BC.**Final Answer**The locus of points ( D ) is the union of the line through ( A ) parallel to ( BC ) and the perpendicular bisector of ( BC ). Thus, the final answer is boxed{e_1 text{ and } e_2}, where ( e_1 ) is the line through ( A ) parallel to ( BC ) and ( e_2 ) is the perpendicular bisector of ( BC ).

💡Alright, so I'm trying to figure out this problem about connecting computers in an organization. Let me read it again to make sure I understand all the details.There are 40 employees total. Out of these, 25 have brand A computers, and 15 have brand B computers. The cables can only connect brand A to brand B, not A to A or B to B. Also, each brand B computer must be connected to at least two distinct brand A computers. Initially, no computers are connected. The technician is connecting them following these rules until everyone can communicate either directly or indirectly. We need to find the maximum possible number of cables that could be used.Hmm, okay. So, this seems like a graph theory problem, specifically dealing with bipartite graphs since we're only connecting A to B. Each B must have at least two connections to A. And we want the maximum number of cables, which would mean maximizing the number of edges in this bipartite graph while satisfying the constraints.First, let's recall that in a bipartite graph with partitions A and B, the maximum number of edges without any constraints would be the product of the sizes of A and B. So, in this case, that would be 25 * 15 = 375. But we have an additional constraint: each B must be connected to at least two A's. So, does this constraint affect the maximum number of edges? Or is it just ensuring that the graph is connected in a certain way?Wait, actually, the constraint might not reduce the maximum number of edges because even with the constraint, we can still have all possible connections. The constraint just ensures that each B is connected to at least two A's, but it doesn't prevent any A from being connected to multiple B's. So, in theory, we could still have all 375 connections.But hold on, the problem also mentions that the technician connects computers until every employee can communicate either directly or indirectly. That sounds like we need the graph to be connected. In a bipartite graph, being connected means there's a path between any two nodes, which in this case would mean that all A's and B's are connected through some sequence of cables.But if we have all 375 cables, that's a complete bipartite graph, which is definitely connected. So, in that case, the maximum number of cables is 375. But let me think again because sometimes these problems have a twist.Wait, maybe I'm missing something. The constraint is that each B must have at least two A connections. So, if we have 15 B's, each needing at least two A's, that's a minimum of 30 connections. But since we have 25 A's, each A can be connected to multiple B's. So, the minimum number of connections is 30, and the maximum is 375. But the question is about the maximum number of cables that could be used while ensuring the entire network is connected.But in a complete bipartite graph, it's already connected, so why would the maximum number of cables be less than 375? Maybe I'm overcomplicating it.Alternatively, perhaps the question is about the maximum number of cables before the network becomes connected, but that doesn't seem to be the case because it says the technician connects until everyone can communicate. So, the technician would keep adding cables until the network is connected, and we need to find the maximum number of cables that could be used in this process.But in that case, the maximum number of cables would still be 375 because once you have all possible connections, the network is definitely connected. So, why is 375 one of the options? It is option D.Wait, but let me double-check. Maybe there's a different interpretation. If we have to ensure that each B is connected to at least two A's, but we also need the entire network to be connected. So, perhaps the maximum number of cables is not just the complete bipartite graph because of some constraints on how the connections are made.But no, in a complete bipartite graph, every B is connected to every A, which certainly satisfies the condition of being connected to at least two A's. So, 375 should be the maximum number of cables.But let me think about it differently. Maybe the problem is asking for the maximum number of cables before the network becomes connected, but that doesn't make sense because the technician is connecting until it's connected. So, the technician would stop once it's connected, but we need the maximum number of cables that could be used in the process.Wait, no, the technician is connecting until it's connected, so the number of cables would be the minimum number needed to make it connected, which is different. But the question is about the maximum possible number of cables that could be used, not the minimum.So, perhaps the technician could add cables in such a way that they use as many cables as possible while still ensuring that the network becomes connected. But in that case, the maximum number of cables would still be 375 because you can have all possible connections and still have a connected network.I'm getting confused here. Let me try to break it down step by step.1. We have 25 A's and 15 B's.2. Each B must be connected to at least two A's.3. The network must be connected, meaning there's a path between any two computers.4. We need to find the maximum number of cables that could be used while satisfying these conditions.So, to maximize the number of cables, we want as many connections as possible. The only constraint is that each B has at least two A connections, and the entire network is connected.In a complete bipartite graph, all B's are connected to all A's, which satisfies the condition of each B having at least two A connections. Also, the network is definitely connected because every computer is connected to every other computer through some path.Therefore, the maximum number of cables is 25 * 15 = 375, which is option D.But wait, let me think again. Is there a scenario where adding more cables could disconnect the network? No, because adding more cables can only increase connectivity, not decrease it. So, the more cables you have, the more connected the network is.Therefore, the maximum number of cables is indeed 375.I think I've convinced myself that the answer is D) 375.

💡Okay, so I'm trying to figure out this problem about the man who buys beer in two different countries, Anchuria and Gvaiasuela, and ends up with the same amount of money he started with. That seems a bit confusing at first, but I'll try to break it down step by step.First, both countries initially had their dollars valued equally. That means one Anchurian dollar was worth the same as one Gvaiasuelan dollar. But then, the government of Gvaiasuela decided to change the exchange rate so that one Anchurian dollar was worth ninety Gvaiasuelan cents. The next day, Anchuria did the same, setting the exchange rate for Gvaiasuelan dollars to ninety Anchurian cents.So, now, one Anchurian dollar is worth ninety Gvaiasuelan cents, and one Gvaiasuelan dollar is worth ninety Anchurian cents. That seems like a reciprocal exchange rate, but I need to be careful with how that affects the value of each currency.The man goes to a tavern in Anchuria and buys a beer that costs ten cents. He pays with an Anchurian dollar and receives change in Gvaiasuelan currency. Then he crosses the border to Gvaiasuela, buys another beer, pays with a Gvaiasuelan dollar, and gets change in Anchurian currency. After all that, he has the same amount of money he started with. So, who actually paid for the beers?Let me try to model this mathematically. Let's denote:- ( A ) as the value of an Anchurian dollar.- ( G ) as the value of a Gvaiasuelan dollar.Initially, ( A = G ).After the exchange rates are set:- ( 1A = 0.90G )- ( 1G = 0.90A )So, each dollar is worth ninety cents in the other country's currency.Now, the man buys a beer in Anchuria for ten cents. Since he's paying with an Anchurian dollar, which is worth 0.90G, he should receive change in Gvaiasuelan currency. The beer costs ten cents, which in Gvaiasuelan currency would be ( 10 ) cents. But since ( 1A = 0.90G ), the value of ten cents in Gvaiasuela is ( frac{10}{0.90}A approx 11.11A ). Wait, that doesn't make sense because he's paying with an Anchurian dollar, which is worth more than ten cents in Gvaiasuela.Hold on, maybe I need to think differently. If the beer costs ten cents in Anchuria, and he pays with an Anchurian dollar, the change he should receive is ( 1A - 0.10A = 0.90A ). But he receives this change in Gvaiasuelan currency. Since ( 1A = 0.90G ), the change in Gvaiasuela would be ( 0.90A times frac{1G}{0.90A} = 1G ). So, he gets one Gvaiasuelan dollar as change.Now, he crosses the border to Gvaiasuela with this one Gvaiasuelan dollar. He buys another beer, which should also cost ten cents. But in Gvaiasuela, the beer costs ten cents, so he pays with his Gvaiasuelan dollar and should receive change. The beer costs ten cents, so the change would be ( 1G - 0.10G = 0.90G ). But he receives this change in Anchurian currency. Since ( 1G = 0.90A ), the change in Anchuria would be ( 0.90G times frac{1A}{0.90G} = 1A ).So, after these two transactions, he started with one Anchurian dollar, paid for a beer, got one Gvaiasuelan dollar as change, then used that to buy another beer and got one Anchurian dollar back. So, he ends up with the same amount of money he started with.But who actually paid for the beers? The man didn't spend any of his own money because he ended up with the same amount. It seems like the exchange rates allowed him to effectively get free beers. But that doesn't make sense in a real-world scenario. Maybe the tavern keepers are the ones who are losing money because of the exchange rate discrepancies.Let me think again. When he buys the first beer in Anchuria, he pays one Anchurian dollar, which is worth ninety Gvaiasuelan cents. The beer costs ten cents, so the tavern keeper in Anchuria is only receiving ten cents worth of value but giving back ninety cents in change. That means the tavern keeper is effectively losing eighty cents on the first transaction.Then, when the man goes to Gvaiasuela, he pays one Gvaiasuelan dollar, which is worth ninety Anchurian cents, for a beer that costs ten cents. The tavern keeper in Gvaiasuela is only receiving ten cents worth of value but giving back ninety cents in change. So, the tavern keeper in Gvaiasuela is also losing eighty cents.Therefore, both tavern keepers are losing money on each transaction because of the exchange rate manipulation. The man is essentially exploiting the exchange rate to get free beers, and the cost is being borne by the tavern keepers in both countries.So, in conclusion, the tavern keepers are the ones paying for the beers because they're the ones losing money due to the exchange rate discrepancies.

💡First, I recognize that (tan alpha = 3) implies that the ratio of (sin alpha) to (cos alpha) is 3. This means I can express (sin alpha) as (3 cos alpha).Next, I substitute (sin alpha = 3 cos alpha) into the given expression:[frac{sin alpha - cos alpha}{2 sin alpha + cos alpha} = frac{3 cos alpha - cos alpha}{2 cdot 3 cos alpha + cos alpha}]Simplifying the numerator and the denominator:[frac{2 cos alpha}{7 cos alpha}]Since (cos alpha) is not zero, it cancels out:[frac{2}{7}]Therefore, the value of the expression is (boxed{frac{2}{7}}).