answer：Okay, let's try to figure out this frog jump problem. So, there are three points on the ground: A, B, and C. The frog starts at point P, which is 0.27 meters away from point C. The frog does these symmetric jumps over points A, B, and C repeatedly. After 2009 jumps, we need to find how far the frog is from the starting point P.First, I need to understand what a "symmetric jump" means. I think it means reflecting the current position over a point. So, if the frog is at point P and it does a symmetric jump over point A, it will end up at a new point P1 such that A is the midpoint between P and P1. Similarly, the next jump over B will make B the midpoint between P1 and P2, and so on.Let me try to visualize this. Suppose I have points A, B, and C on a line. The frog starts at P, which is 0.27 meters from C. Let's assume all points are on a straight line for simplicity, maybe on the x-axis. So, let's assign coordinates to these points.Let me set point C at the origin (0,0) for simplicity. Then, point P is 0.27 meters away from C, so P is at (0.27, 0). Now, I need to figure out where points A and B are. Hmm, the problem doesn't specify their positions, so maybe I need to figure out a general solution or perhaps the positions of A and B don't matter because of the periodicity of the jumps.Wait, maybe the positions of A, B, and C are arbitrary, but the frog's movement is periodic. So, after a certain number of jumps, the frog returns to its original position. If that's the case, then the distance after 2009 jumps would depend on where 2009 falls in the cycle.Let me think about the reflection process. Each reflection over a point is like a translation. If I reflect over A, then over B, then over C, and so on, the frog's position changes each time. Maybe after a certain number of reflections, the frog comes back to where it started.Let me try to model this with coordinates. Suppose point A is at (a, 0), point B is at (b, 0), and point C is at (c, 0). The frog starts at P = (0.27, 0).First jump: Reflect over A. The reflection of P over A is P1 = (2a - 0.27, 0).Second jump: Reflect P1 over B. So, P2 = (2b - (2a - 0.27), 0) = (2b - 2a + 0.27, 0).Third jump: Reflect P2 over C. So, P3 = (2c - (2b - 2a + 0.27), 0) = (2c - 2b + 2a - 0.27, 0).Fourth jump: Reflect P3 over A. So, P4 = (2a - (2c - 2b + 2a - 0.27), 0) = (2a - 2c + 2b - 2a + 0.27, 0) = (-2c + 2b + 0.27, 0).Fifth jump: Reflect P4 over B. So, P5 = (2b - (-2c + 2b + 0.27), 0) = (2b + 2c - 2b - 0.27, 0) = (2c - 0.27, 0).Sixth jump: Reflect P5 over C. So, P6 = (2c - (2c - 0.27), 0) = (2c - 2c + 0.27, 0) = (0.27, 0).Oh! So after six jumps, the frog is back at the starting point P. That means the frog's movement is periodic with a period of 6 jumps.So, every 6 jumps, the frog returns to P. Therefore, to find where the frog is after 2009 jumps, I can divide 2009 by 6 and find the remainder. The remainder will tell me how many jumps beyond the last complete cycle of 6 jumps the frog has made.Let me calculate 2009 divided by 6. 2009 ÷ 6 = 334 with a remainder of 5 because 334 × 6 = 2004, and 2009 - 2004 = 5.So, the frog completes 334 full cycles (each of 6 jumps) and then makes 5 more jumps. Therefore, the position after 2009 jumps is the same as after 5 jumps.From my earlier calculations, after 5 jumps, the frog is at P5 = (2c - 0.27, 0). But wait, in my coordinate system, I set C at (0,0). So, c = 0. Therefore, P5 = (2×0 - 0.27, 0) = (-0.27, 0).Wait, that can't be right because in my earlier step-by-step, when I set C at (0,0), then P is at (0.27, 0). After 5 jumps, the frog is at (-0.27, 0). So, the distance from P is the distance between (0.27, 0) and (-0.27, 0), which is 0.54 meters or 54 centimeters.But hold on, in my initial setup, I assumed all points are on the x-axis. What if points A, B, and C are not colinear? Would this affect the result? Hmm, the problem doesn't specify, so maybe it's safe to assume they are colinear for simplicity, especially since the frog's starting position is given relative to C.Alternatively, if A, B, and C form a triangle, the reflections would be more complex, but since the problem doesn't specify their positions, perhaps the key insight is that after 6 jumps, the frog returns to the starting point, regardless of the positions of A, B, and C. Therefore, the distance after 5 jumps would be twice the distance from P to C, which is 0.27 meters, so 0.54 meters or 54 centimeters.Wait, let me verify that. If the frog is at P5, which is the reflection of P over C, then the distance from P to P5 would be twice the distance from P to C. Since P is 0.27 meters from C, P5 would be 0.27 meters on the other side of C, making the total distance between P and P5 equal to 0.54 meters.Yes, that makes sense. So, regardless of the positions of A and B, after 5 jumps, the frog is diametrically opposite to P with respect to C, hence twice the distance.Therefore, the distance between P2009 and P is 54 centimeters.**Final Answer**The distance between ( P_{2009} ) and the starting point ( P ) is boxed{54} centimeters.

answer：Okay, so I have this problem about 15 planets, each with an astronomer observing the nearest planet. I need to prove that there's at least one planet that no astronomer observes. Hmm, interesting. Let me try to break this down step by step.First, I know that all the distances between the planets are pairwise distinct. That means no two pairs of planets are the same distance apart. So, for any planet, there's a unique nearest neighbor. That makes sense because if distances were the same, there could be ambiguity about which planet is the nearest.Now, each astronomer is observing the nearest planet to them. So, for each planet, there's exactly one other planet that it's pointing to as the nearest. But the problem is asking to show that there's at least one planet that isn't being observed by anyone. That is, there's a planet that isn't the nearest to any other planet.Let me think about how to approach this. Maybe I can use some kind of pairing or graph theory concept here. If I imagine each planet as a node in a graph, and draw an arrow from each planet to its nearest neighbor, then each node has exactly one outgoing arrow. But the question is about the incoming arrows—there must be at least one node with no incoming arrows.Wait, that sounds like a directed graph where each node has out-degree 1. So, we have 15 nodes, each with one outgoing edge. Now, in such a graph, what can we say about the in-degrees? Since there are 15 edges, the sum of all in-degrees is 15. But there are 15 nodes, so if every node had at least one incoming edge, the sum would be exactly 15, meaning each node has exactly one incoming edge. But that would imply that the graph is a collection of cycles, right? Because each node has one incoming and one outgoing edge.But wait, in our case, the edges are determined by the nearest neighbor, which might not necessarily form cycles. Hmm, maybe I'm overcomplicating it. Let me think differently.Suppose I pick any planet, say Planet A. Planet A is observing its nearest neighbor, Planet B. Now, Planet B is observing its nearest neighbor, which could be Planet A or another planet. If Planet B's nearest neighbor is Planet A, then they form a mutual pair. But if Planet B's nearest neighbor is another planet, say Planet C, then Planet C is observing its nearest neighbor, and so on.This seems like it could form a chain or a cycle. But since there are 15 planets, which is an odd number, if we try to pair them up, there must be at least one planet that isn't paired, right? Wait, but in this case, each planet is pointing to exactly one other planet, so it's more like a functional graph, where each node has one outgoing edge.In functional graphs, the components are either cycles or trees leading into cycles. So, in our case, the graph would consist of cycles and trees directed towards the cycles. But since each node has exactly one outgoing edge, the graph is made up of cycles and trees directed towards those cycles.But how does this help me? I need to show that there's at least one node with no incoming edges. If all nodes had at least one incoming edge, then the graph would consist solely of cycles, right? Because every node would have both an incoming and an outgoing edge, forming cycles.But in reality, not all nodes can have an incoming edge because the number of edges is equal to the number of nodes. If every node had at least one incoming edge, then each node would have exactly one incoming edge, making it a permutation, which is a collection of cycles. But in our case, the edges are determined by the nearest neighbor, which might not form a permutation.Wait, maybe I'm on the right track. If the graph were a permutation, meaning every node has exactly one incoming and one outgoing edge, then it would be a collection of cycles. But in reality, the edges are determined by the nearest neighbor, which might not form such a structure.Let me think about the closest pair of planets. Suppose Planet A and Planet B are the closest pair. Then, Planet A's nearest neighbor is Planet B, and Planet B's nearest neighbor is Planet A. So, they form a mutual pair. Now, consider the next closest pair. Let's say Planet C and Planet D are the next closest. Similarly, they form a mutual pair.Continuing this way, since there are 15 planets, which is an odd number, we can't pair them all up. There will be one planet left without a pair. This leftover planet would not have its nearest neighbor pointing to it because all the other planets are already paired up, each pointing to their mutual nearest neighbor.Wait, is that correct? Let me see. If I pair up the planets based on their closest distances, each pair consists of two planets observing each other. Since 15 is odd, after pairing as much as possible, there will be one planet left. This planet's nearest neighbor must be someone in one of the pairs, but since the pairs are already mutual, that planet's nearest neighbor is already accounted for. Therefore, the leftover planet's nearest neighbor is someone in a pair, but that someone is already observing their mutual pair. So, the leftover planet is not being observed by anyone because all the other planets are either in mutual pairs or observing someone else.Therefore, there must be at least one planet that no astronomer observes. That makes sense.Wait, but I need to make sure that the leftover planet's nearest neighbor isn't also the leftover planet. No, because the distances are pairwise distinct, so the leftover planet must have a unique nearest neighbor, which is someone in a pair. Since the pairs are mutual, the nearest neighbor of the leftover planet is already observing their mutual pair, so the leftover planet isn't being observed by anyone.Yes, that seems to work. So, by considering the closest pairs and realizing that with an odd number of planets, one planet must be left out, we can conclude that there's at least one planet that no astronomer observes.I think that's a solid argument. Let me try to summarize it step by step to make sure I haven't missed anything.1. Consider all pairs of planets and their distances. Since all distances are distinct, each planet has a unique nearest neighbor.2. Start by identifying the closest pair of planets, say A and B. They observe each other.3. Next, identify the next closest pair, say C and D. They also observe each other.4. Continue this process, pairing up planets based on their closest distances.5. Since there are 15 planets, which is odd, after pairing as much as possible, there will be one planet left without a pair.6. This leftover planet must have a nearest neighbor, which is someone in one of the already paired planets.7. However, each paired planet is already observing their mutual nearest neighbor, so the leftover planet isn't being observed by anyone.8. Therefore, there must be at least one planet that no astronomer observes.Yes, that seems to cover it. I think I've got it!

answer：Okay, so I need to find out how many moles of ethane (C2H6) are formed when I combine 3 moles of hydrogen (H2) and 3 moles of ethylene (C2H4). Hmm, let's see. I remember that chemical reactions involve balancing equations to make sure the number of atoms of each element is the same on both sides. So, first, I should probably write down the chemical equation for this reaction.Alright, ethylene is C2H4, and hydrogen is H2. When they react, they form ethane, which is C2H6. So, the reaction should look something like this:C2H4 + H2 → C2H6Now, I need to balance this equation. Let's count the number of each type of atom on both sides. On the left side, I have 2 carbon atoms, 4 hydrogen atoms from C2H4, and 2 hydrogen atoms from H2, making a total of 6 hydrogen atoms. On the right side, I have 2 carbon atoms and 6 hydrogen atoms in C2H6. Wait, that seems balanced already! So, the balanced equation is:C2H4 + H2 → C2H6Okay, so from this balanced equation, I can see that 1 mole of C2H4 reacts with 1 mole of H2 to produce 1 mole of C2H6. That means the mole ratio is 1:1:1.Now, I have 3 moles of H2 and 3 moles of C2H4. Since the ratio is 1:1, I can use either reactant to determine how much product is formed. Let's see, if I have 3 moles of C2H4, and each mole reacts with 1 mole of H2, then theoretically, I should be able to produce 3 moles of C2H6. But wait, I also have 3 moles of H2, which is exactly what's needed for 3 moles of C2H4. So, neither reactant is in excess; they are both completely consumed.But just to double-check, what if one of them was in excess? Let's say I had more H2 than C2H4. Then, C2H4 would be the limiting reactant, and the amount of C2H6 produced would be based on the amount of C2H4 available. Similarly, if I had more C2H4 than H2, H2 would be the limiting reactant. But in this case, since both are exactly 3 moles, they perfectly match the stoichiometric ratio, so both are completely used up, and 3 moles of C2H6 are formed.Wait a second, is there any possibility of side reactions or something else happening? I don't think so in this simple reaction. It's a straightforward addition of hydrogen to ethylene to form ethane. So, I think my initial conclusion is correct.Therefore, combining 3 moles of H2 and 3 moles of C2H4 will produce 3 moles of C2H6.

answer：First, I need to convert the number 1,268,000,000 into scientific notation. Scientific notation is expressed in the form ( a times 10^{n} ), where ( a ) is a number between 1 and 10, and ( n ) is an integer.To do this, I'll start by placing the decimal point after the first non-zero digit of the number. In this case, the first digit is 1, so I'll place the decimal point after the 1, resulting in 1.268.Next, I need to determine how many places I moved the decimal point to get from 1,268,000,000 to 1.268. Counting the number of places, I moved the decimal point 9 places to the right.Since I moved the decimal point 9 places to the right, the exponent ( n ) will be 9. Therefore, the number 1,268,000,000 in scientific notation is ( 1.268 times 10^{9} ).Finally, I'll compare this result with the given options:- A: ( 1.268 times 10^{9} )- B: ( 1.268 times 10^{8} )- C: ( 1.268 times 10^{7} )- D: ( 1.268 times 10^{6} )The correct choice is A.

answer：Okay, so I have this multiple-choice question about planes and lines, and I need to figure out which statement is correct. Let me go through each option one by one and think it through step by step.First, let's look at option A: "Two planes can have only one intersection point." Hmm, I remember that in geometry, two planes can either be parallel or intersect. If they're parallel, they don't intersect at all. But if they do intersect, they form a line, right? So, if two planes intersect, they intersect along an entire line, not just at a single point. That means option A is saying something that's not true because they can't have just one intersection point; it's either none or infinitely many points along a line. So, A is incorrect.Moving on to option B: "A line and a plane can have at most one common point." Okay, so if a line intersects a plane, it can do so at one point. But wait, what if the line lies entirely within the plane? In that case, every point on the line is a common point with the plane, which means there are infinitely many common points. So, the statement says "at most one," but in reality, there can be infinitely many. Therefore, B is also incorrect.Now, option C: "If two planes have one common point, they may intersect." Let me think. If two planes have one common point, does that mean they must intersect? Well, if two planes have at least one common point, they must intersect along a line, right? So, having one common point implies that they intersect along a line, which means they definitely intersect. The statement says "they may intersect," which is true because they must intersect if they share a common point. So, C seems correct.Finally, option D: "If two planes have three common points, they must coincide." Hmm, if two planes have three common points, does that mean they are the same plane? Well, if the three points are non-collinear, meaning they don't all lie on a straight line, then yes, the two planes must be the same because three non-collinear points define a unique plane. However, if the three points are collinear, meaning they lie on a straight line, then the two planes could intersect along that line without being the same plane. So, the statement says "they must coincide," but that's only true if the three points are non-collinear. Since the statement doesn't specify, it's not necessarily always true. Therefore, D is incorrect.So, after going through each option, I think only option C is correct.

answer：Okay, so I have this triangle ABC, and the sides opposite angles A, B, and C are a, b, and c respectively. The problem gives me an equation: (2b + √3 c) cos A + √3 a cos C = 0. I need to find the measure of angle A first, and then, if a = 2, find the range of values for b + √3 c.Alright, starting with part (1). I remember that in triangles, the Law of Sines and the Law of Cosines are often useful. The equation given involves cosines of angles and sides, so maybe I can use the Law of Sines to express the sides in terms of the sines of their opposite angles.The Law of Sines says that a/sin A = b/sin B = c/sin C = 2R, where R is the radius of the circumscribed circle. Maybe I can express b and c in terms of a, sin B, and sin C. Let me try that.So, if I let a = 2R sin A, b = 2R sin B, and c = 2R sin C, then I can substitute these into the given equation. Let's see:(2b + √3 c) cos A + √3 a cos C = 0Substituting:[2*(2R sin B) + √3*(2R sin C)] cos A + √3*(2R sin A) cos C = 0Simplify:[4R sin B + 2√3 R sin C] cos A + 2√3 R sin A cos C = 0I can factor out 2R:2R [2 sin B cos A + √3 sin C cos A + √3 sin A cos C] = 0Since R is not zero, the expression inside the brackets must be zero:2 sin B cos A + √3 sin C cos A + √3 sin A cos C = 0Hmm, this seems a bit complicated. Maybe I can rearrange terms or use some trigonometric identities. I notice that the last two terms have √3 and involve sin C cos A and sin A cos C. That looks similar to the sine addition formula: sin(C + A) = sin C cos A + cos C sin A.Wait, but in our case, it's √3 times (sin C cos A + sin A cos C). So, let me write that:√3 (sin C cos A + sin A cos C) = √3 sin(C + A)So, substituting back into the equation:2 sin B cos A + √3 sin(C + A) = 0But in a triangle, A + B + C = π, so C + A = π - B. Therefore, sin(C + A) = sin(π - B) = sin B.So, substituting that in:2 sin B cos A + √3 sin B = 0Factor out sin B:sin B (2 cos A + √3) = 0Now, sin B = 0 would imply that angle B is 0 or π, which is impossible in a triangle. So, sin B ≠ 0, which means that the other factor must be zero:2 cos A + √3 = 0So, cos A = -√3 / 2What angle A has cosine equal to -√3 / 2? I remember that cos(π/6) = √3 / 2, so cos(5π/6) = -√3 / 2. Since in a triangle, angles are between 0 and π, so A must be 5π/6.Alright, so angle A is 5π/6 radians, which is 150 degrees. That seems reasonable because it's an obtuse angle, and the other angles must be acute.Okay, moving on to part (2): If a = 2, find the range of values for b + √3 c.Since we know angle A is 5π/6, and a = 2, we can use the Law of Sines again. Let me write down the Law of Sines:a / sin A = b / sin B = c / sin CGiven that a = 2 and A = 5π/6, let's compute sin A:sin(5π/6) = sin(π - π/6) = sin(π/6) = 1/2So, 2 / (1/2) = 4. Therefore, the circumradius R is 2, since 2R = 4, so R = 2.Wait, actually, in the Law of Sines, a / sin A = 2R. So, 2 / (1/2) = 4 = 2R, so R = 2. Correct.Therefore, b = 2R sin B = 4 sin B, and c = 2R sin C = 4 sin C.But since A = 5π/6, the sum of angles B and C is π - 5π/6 = π/6. So, B + C = π/6. Therefore, C = π/6 - B.So, c = 4 sin(π/6 - B). Let me express that:sin(π/6 - B) = sin π/6 cos B - cos π/6 sin B = (1/2) cos B - (√3 / 2) sin BTherefore, c = 4 [ (1/2) cos B - (√3 / 2) sin B ] = 2 cos B - 2√3 sin BSo, now, let's compute b + √3 c:b + √3 c = 4 sin B + √3 (2 cos B - 2√3 sin B )Let me compute that:= 4 sin B + 2√3 cos B - 2*3 sin B= 4 sin B + 2√3 cos B - 6 sin B= (4 sin B - 6 sin B) + 2√3 cos B= (-2 sin B) + 2√3 cos BFactor out 2:= 2( - sin B + √3 cos B )Hmm, this looks like a form that can be expressed as a single sine or cosine function. Let me recall that expressions like A sin x + B cos x can be written as C sin(x + φ) or C cos(x + φ), where C = √(A² + B²) and φ is some phase shift.In our case, we have - sin B + √3 cos B. Let me write it as √3 cos B - sin B.So, √3 cos B - sin B. Let me factor out a 2:= 2 [ (√3 / 2) cos B - (1/2) sin B ]Notice that √3 / 2 is cos(π/6) and 1/2 is sin(π/6). So, this is equal to:= 2 [ cos(π/6) cos B - sin(π/6) sin B ]Which is equal to:= 2 cos(π/6 + B )Because cos(A + B) = cos A cos B - sin A sin B.So, putting it all together:b + √3 c = 2 * 2 cos(π/6 + B ) = 4 cos(π/6 + B )Wait, hold on, I think I messed up the factoring earlier. Let me go back.We had:b + √3 c = 2( - sin B + √3 cos B ) = 2( √3 cos B - sin B )Then, factor out 2:= 2 [ (√3 / 2) cos B - (1/2) sin B ]Which is:= 2 [ cos(π/6) cos B - sin(π/6) sin B ]= 2 cos(π/6 + B )Yes, that's correct. So, b + √3 c = 2 * 2 cos(π/6 + B )? Wait, no, wait. Wait, 2 times [ cos(π/6 + B ) ] is 2 cos(π/6 + B ). So, b + √3 c = 2 * 2 cos(π/6 + B )? Wait, no, let me re-examine.Wait, starting from:b + √3 c = 4 sin B + √3 (2 cos B - 2√3 sin B )= 4 sin B + 2√3 cos B - 6 sin B= (4 sin B - 6 sin B) + 2√3 cos B= (-2 sin B) + 2√3 cos B= 2( - sin B + √3 cos B )Then, factor out 2:= 2 [ √3 cos B - sin B ]Then, factor out 2 from inside:Wait, no, √3 cos B - sin B can be written as 2 [ (√3 / 2 cos B - 1/2 sin B ) ]Which is 2 cos(π/6 + B )Therefore, b + √3 c = 2 * 2 cos(π/6 + B ) = 4 cos(π/6 + B )Wait, no, that's not correct. Wait, let me do it step by step.We have:√3 cos B - sin B = 2 [ (√3 / 2 cos B - 1/2 sin B ) ]= 2 [ cos(π/6) cos B - sin(π/6) sin B ]= 2 cos(π/6 + B )So, √3 cos B - sin B = 2 cos(π/6 + B )Therefore, b + √3 c = 2 * [ √3 cos B - sin B ] = 2 * 2 cos(π/6 + B ) = 4 cos(π/6 + B )Wait, no, wait. Wait, the expression was:b + √3 c = 2( √3 cos B - sin B ) = 2 * [ 2 cos(π/6 + B ) ] ?Wait, no, the expression inside was:√3 cos B - sin B = 2 cos(π/6 + B )Therefore, b + √3 c = 2 * [ √3 cos B - sin B ] = 2 * 2 cos(π/6 + B ) = 4 cos(π/6 + B )Wait, that seems right.So, b + √3 c = 4 cos(π/6 + B )Now, since B is an angle in the triangle, and we know that A = 5π/6, so B + C = π/6. Therefore, B must be between 0 and π/6.So, B ∈ (0, π/6). Therefore, π/6 + B ∈ (π/6, π/3). So, the argument of the cosine function is between π/6 and π/3.Now, cosine is a decreasing function in [0, π], so as the angle increases, cosine decreases. Therefore, cos(π/6 + B ) decreases as B increases.So, when B approaches 0, π/6 + B approaches π/6, so cos(π/6 + B ) approaches cos(π/6) = √3 / 2.When B approaches π/6, π/6 + B approaches π/3, so cos(π/6 + B ) approaches cos(π/3) = 1/2.Therefore, cos(π/6 + B ) ∈ (1/2, √3 / 2 )Therefore, b + √3 c = 4 cos(π/6 + B ) ∈ (4 * 1/2, 4 * √3 / 2 ) = (2, 2√3 )So, the range of b + √3 c is (2, 2√3 )Wait, let me double-check that.We have:b + √3 c = 4 cos(π/6 + B )Since B ∈ (0, π/6), π/6 + B ∈ (π/6, π/3)cos(π/6) = √3 / 2 ≈ 0.866cos(π/3) = 1/2 = 0.5So, cos(π/6 + B ) decreases from √3 / 2 to 1/2 as B increases from 0 to π/6.Therefore, 4 cos(π/6 + B ) decreases from 4*(√3 / 2 ) = 2√3 to 4*(1/2 ) = 2.Hence, the range is (2, 2√3 )So, the possible values for b + √3 c are between 2 and 2√3.Therefore, the range is (2, 2√3 )Wait, but the problem says "find the range of values for b + √3 c". So, is it open interval or closed? Since B is strictly between 0 and π/6, π/6 + B is strictly between π/6 and π/3, so cos(π/6 + B ) is strictly between 1/2 and √3 / 2. Therefore, 4 cos(π/6 + B ) is strictly between 2 and 2√3. So, the range is (2, 2√3 )Hence, the answer is (2, 2√3 )**Final Answer**(1) The measure of angle ( A ) is (boxed{dfrac{5pi}{6}}).(2) The range of values for ( b + sqrt{3}c ) is (boxed{(2, 2sqrt{3})}).