💡Okay, so I have this problem about a hyperbola, and I need to find its eccentricity. Let me try to understand what's given and what I need to do.First, the hyperbola is given by the equation (frac{x^2}{a^2} - frac{y^2}{b^2} = 1), where (a > 0) and (b > 0). I know that for hyperbolas, the standard form is similar to this, and the foci are located at ((pm c, 0)), where (c^2 = a^2 + b^2). The eccentricity (e) is given by (e = frac{c}{a}), so ultimately, I need to find (c) in terms of (a) or find the ratio (b/a) to compute the eccentricity.The problem mentions that point F is the right focus of the hyperbola. So, F is at ((c, 0)).Next, the line (y = sqrt{3}x) intersects the hyperbola at points A and B. So, I need to find the points of intersection between the line and the hyperbola. Let me set up the equations to solve for these points.Substituting (y = sqrt{3}x) into the hyperbola equation:[frac{x^2}{a^2} - frac{(sqrt{3}x)^2}{b^2} = 1]Simplify the equation:[frac{x^2}{a^2} - frac{3x^2}{b^2} = 1]Combine the terms:[x^2 left( frac{1}{a^2} - frac{3}{b^2} right) = 1]Let me denote this as:[x^2 left( frac{b^2 - 3a^2}{a^2b^2} right) = 1]So,[x^2 = frac{a^2b^2}{b^2 - 3a^2}]Hmm, okay. So, (x^2) is equal to that fraction. Since (x^2) must be positive, the denominator (b^2 - 3a^2) must also be positive. So, (b^2 > 3a^2). That's an important condition.So, the points A and B are symmetric with respect to the origin because if ((x, y)) is a solution, then ((-x, -y)) is also a solution. So, let me denote point A as ((x, sqrt{3}x)) and point B as ((-x, -sqrt{3}x)).Now, the problem states that AF is perpendicular to BF. So, the vectors AF and BF are perpendicular. Let me write the vectors AF and BF.Vector AF is from F to A: ((x - c, sqrt{3}x - 0) = (x - c, sqrt{3}x))Vector BF is from F to B: ((-x - c, -sqrt{3}x - 0) = (-x - c, -sqrt{3}x))Since AF is perpendicular to BF, their dot product should be zero.So, let's compute the dot product:[(x - c)(-x - c) + (sqrt{3}x)(-sqrt{3}x) = 0]Let me expand this:First term: ((x - c)(-x - c))Multiply them out:[x(-x) + x(-c) - c(-x) - c(-c) = -x^2 - cx + cx + c^2]Simplify:[-x^2 + c^2]Second term: ((sqrt{3}x)(-sqrt{3}x))Multiply them:[-3x^2]So, putting it all together:[(-x^2 + c^2) + (-3x^2) = 0]Combine like terms:[-4x^2 + c^2 = 0]So,[c^2 = 4x^2]Therefore,[c = 2x quad text{or} quad c = -2x]But since (c) is a positive value (as it's the distance from the center to the focus), and (x) is a coordinate which could be positive or negative, but in our case, since point A is ((x, sqrt{3}x)), (x) could be positive or negative. However, since (c) is positive, and (c = 2x), so (x) must be positive. So, we can write (c = 2x).But wait, actually, in the equation (c^2 = 4x^2), (c) is positive, so (c = 2|x|). Since (x) is a coordinate, but in our case, since the hyperbola is symmetric, we can take (x) positive without loss of generality. So, (c = 2x).So, from this, (x = frac{c}{2}).But earlier, we found that (x^2 = frac{a^2b^2}{b^2 - 3a^2}). So, substituting (x = frac{c}{2}):[left( frac{c}{2} right)^2 = frac{a^2b^2}{b^2 - 3a^2}]Simplify:[frac{c^2}{4} = frac{a^2b^2}{b^2 - 3a^2}]But we know that (c^2 = a^2 + b^2). So, substitute (c^2) with (a^2 + b^2):[frac{a^2 + b^2}{4} = frac{a^2b^2}{b^2 - 3a^2}]Cross-multiplying:[(a^2 + b^2)(b^2 - 3a^2) = 4a^2b^2]Let me expand the left side:First, multiply (a^2) with each term:(a^2 cdot b^2 = a^2b^2)(a^2 cdot (-3a^2) = -3a^4)Then, multiply (b^2) with each term:(b^2 cdot b^2 = b^4)(b^2 cdot (-3a^2) = -3a^2b^2)So, altogether:[a^2b^2 - 3a^4 + b^4 - 3a^2b^2 = 4a^2b^2]Combine like terms:(a^2b^2 - 3a^2b^2 = -2a^2b^2)So, left side becomes:[-2a^2b^2 - 3a^4 + b^4 = 4a^2b^2]Bring all terms to the left side:[-2a^2b^2 - 3a^4 + b^4 - 4a^2b^2 = 0]Combine like terms:(-2a^2b^2 - 4a^2b^2 = -6a^2b^2)So,[b^4 - 6a^2b^2 - 3a^4 = 0]Hmm, this is a quartic equation in terms of (b^2). Let me set (k = frac{b^2}{a^2}). Then, (b^2 = ka^2). Let's substitute this into the equation.So,[(ka^2)^2 - 6a^2(ka^2) - 3a^4 = 0]Simplify each term:( (k^2a^4) - 6k a^4 - 3a^4 = 0 )Factor out (a^4):[a^4(k^2 - 6k - 3) = 0]Since (a > 0), (a^4 neq 0), so:[k^2 - 6k - 3 = 0]This is a quadratic equation in (k). Let's solve for (k):Using quadratic formula:[k = frac{6 pm sqrt{36 + 12}}{2} = frac{6 pm sqrt{48}}{2} = frac{6 pm 4sqrt{3}}{2} = 3 pm 2sqrt{3}]So, (k = 3 + 2sqrt{3}) or (k = 3 - 2sqrt{3}).But (k = frac{b^2}{a^2}), which must be positive. Let's check both solutions:1. (3 + 2sqrt{3}) is positive, since (2sqrt{3} approx 3.464), so (3 + 3.464 approx 6.464 > 0).2. (3 - 2sqrt{3}) is approximately (3 - 3.464 = -0.464), which is negative. Since (b^2) and (a^2) are positive, (k) must be positive. So, we discard (k = 3 - 2sqrt{3}).Thus, (k = 3 + 2sqrt{3}), so (frac{b^2}{a^2} = 3 + 2sqrt{3}).Now, recall that (c^2 = a^2 + b^2). So,[c^2 = a^2 + (3 + 2sqrt{3})a^2 = (1 + 3 + 2sqrt{3})a^2 = (4 + 2sqrt{3})a^2]Therefore,[c = sqrt{4 + 2sqrt{3}} cdot a]I need to simplify (sqrt{4 + 2sqrt{3}}). Let me see if this can be expressed as (sqrt{a} + sqrt{b}).Let me assume that:[sqrt{4 + 2sqrt{3}} = sqrt{a} + sqrt{b}]Then, squaring both sides:[4 + 2sqrt{3} = a + b + 2sqrt{ab}]Comparing the terms, we have:1. (a + b = 4)2. (2sqrt{ab} = 2sqrt{3}) => (sqrt{ab} = sqrt{3}) => (ab = 3)So, we have a system:(a + b = 4)(ab = 3)This is a quadratic equation in terms of (a) and (b). Let me solve for (a) and (b).Let me denote (a) and (b) as roots of the equation (t^2 - 4t + 3 = 0).Solving this:[t = frac{4 pm sqrt{16 - 12}}{2} = frac{4 pm 2}{2} = 3 text{ or } 1]So, (a = 3) and (b = 1), or vice versa.Therefore,[sqrt{4 + 2sqrt{3}} = sqrt{3} + sqrt{1} = sqrt{3} + 1]Thus,[c = (sqrt{3} + 1)a]Therefore, the eccentricity (e = frac{c}{a} = sqrt{3} + 1).So, the eccentricity of the hyperbola is (sqrt{3} + 1).**Final Answer**The eccentricity of the hyperbola is (boxed{sqrt{3} + 1}).

💡Okay, so I have this geometry problem here with points A, B, C, D, E, F, and G on a diagram. The problem gives me some lengths and ratios, and I need to find the length of FC. Let me try to figure this out step by step.First, let me parse the given information:- DC = 10- CB = 12- AB = (1/3)AD- ED = (2/3)ADAnd I need to find FC.Looking at the Asymptote code, it seems like the figure is a triangle with points A, D, and E, with various lines drawn inside. Points B, C, F, and G are along the base AD, and F is connected to E, forming another triangle. It might be similar triangles we're dealing with here.Let me try to visualize the figure. Point A is at (0,0), D is at (-30,0), and E is at (-30,24). So, AD is a horizontal line from A to D, and E is directly above D at ( -30,24). Then, there are points B and C on AD, with B at (-6,0) and C at (-18,0). Points F and G are above C and B, respectively, with F at (-18,16) and G at (-6,6). So, from the figure, it seems like triangles ABG, ACF, and ADE are similar.The user mentioned that these triangles are similar, so I can use the properties of similar triangles to find FC.Let me note down the given lengths:- DC = 10- CB = 12So, the distance from D to C is 10 units, and from C to B is 12 units. That means from D to B is DC + CB = 10 + 12 = 22 units.Now, AB is given as (1/3)AD. So, if I let AD be some length, say x, then AB = x/3. Therefore, BD, which is the distance from B to D, would be AD - AB = x - x/3 = (2/3)x.But from earlier, BD is 22 units. So, (2/3)x = 22. Solving for x, we get x = 22 * (3/2) = 33. So, AD is 33 units.Therefore, AB = (1/3) * 33 = 11 units. That makes sense because AB is from A (0,0) to B (-6,0), which is 6 units in the Asymptote code, but wait, in the actual problem, AB is 11 units. Hmm, maybe the Asymptote code is just a scaled drawing.Anyway, moving on. ED is given as (2/3)AD. Since AD is 33, ED = (2/3)*33 = 22 units. So, ED is 22 units.Now, I need to find FC. Looking at the figure, FC is the vertical segment from C (-18,0) up to F (-18,16). So, in the Asymptote code, FC is 16 units, but in the actual problem, it's a different scale.Since the triangles ABG, ACF, and ADE are similar, their corresponding sides are proportional. Let me write down the ratios.First, triangle ADE is the largest triangle. Its base is AD = 33, and its height is ED = 22.Triangle ACF is similar to ADE. The base of ACF is AC. AC is from A to C, which is AB + BC = 11 + 12 = 23 units. So, AC = 23.Let me denote FC as h. Then, the ratio of similarity between ACF and ADE would be AC/AD = 23/33. Therefore, the height FC should be proportional to ED by the same ratio. So, FC = ED * (AC/AD) = 22 * (23/33).Calculating that: 22 * 23 = 506, and 506 divided by 33 is approximately 15.333...Wait, let me double-check that. 33 goes into 506 how many times? 33*15 = 495, so 506 - 495 = 11, so it's 15 and 11/33, which simplifies to 15 and 1/3, which is 15.333...So, FC is approximately 15.333 units.Let me make sure I didn't make any mistakes here. I found AD by using the given ratios and the lengths of DC and CB. Then, I used the similarity of triangles to find FC. The key was recognizing that triangle ACF is similar to triangle ADE, so their corresponding sides are proportional.Another way to think about it is using coordinates. If I assign coordinates to the points, maybe I can calculate FC using coordinate geometry.Let me try that approach.Let me place point A at (0,0). Then, since AD is 33 units long, point D is at (-33,0). Wait, in the Asymptote code, D is at (-30,0), but in the problem, AD is 33. So, maybe the Asymptote code is just a specific case, but in the actual problem, AD is 33.So, point A is (0,0), D is (-33,0), and E is (-33,22) because ED is 22 units.Point B is at AB = 11 units from A, so since AB is along AD, which is the x-axis, point B is at (-11,0). Wait, but in the Asymptote code, B is at (-6,0). Hmm, maybe the Asymptote code is not to scale.But in the problem, AB is 11, CB is 12, so AC is AB + BC = 11 + 12 = 23. So, point C is 23 units from A along AD, which would be at (-23,0). Wait, but in the Asymptote code, C is at (-18,0). Hmm, maybe the Asymptote code is just a different coordinate system.But regardless, in the problem, AC is 23, AD is 33, so the ratio is 23/33.Since triangle ACF is similar to triangle ADE, the height FC corresponds to ED. So, FC = ED * (AC/AD) = 22 * (23/33) ≈ 15.333.Alternatively, using coordinates, if I can find the coordinates of F, then I can find FC.Since triangle ACF is similar to triangle ADE, the coordinates of F can be found by scaling.Point E is at (-33,22). So, scaling down by the ratio AC/AD = 23/33, the coordinates of F would be:x-coordinate: (-33)*(23/33) = -23y-coordinate: 22*(23/33) ≈ 15.333So, point F is at (-23,15.333). Therefore, FC is the vertical distance from C (-23,0) to F (-23,15.333), which is 15.333 units.Wait, but in the Asymptote code, point C is at (-18,0) and F is at (-18,16). So, in the Asymptote code, FC is 16 units, but in the actual problem, FC is approximately 15.333 units.So, that seems consistent. The Asymptote code is just a specific drawing, but the actual problem has different lengths.Therefore, I think my answer is correct. FC is approximately 15.333 units.Just to recap:1. Found AD using the given ratios and lengths.2. Calculated AB and ED.3. Recognized the similarity of triangles ACF and ADE.4. Used the ratio of similarity to find FC.I think that's solid. I don't see any mistakes in my reasoning.**Final Answer**boxed{15.333}

💡First, identify the initial number of hotdogs, which is 99.Next, determine the number of hotdogs remaining after some were sold, which is 97.To find out how many hotdogs were sold, subtract the remaining hotdogs from the initial number: 99 - 97 = 2.Therefore, the restaurant sold 2 hotdogs.

💡Okay, so I need to find the smallest four-digit palindrome that's divisible by 5. Hmm, let's break this down step by step. First, what exactly is a palindrome? A palindrome is a number that reads the same forwards and backwards. So, for a four-digit number, it would have the form ABBA, where A and B are digits. That means the first digit is the same as the last digit, and the second digit is the same as the third digit.Alright, so the number looks like this: A B B A. Now, since it's a four-digit number, A can't be zero because that would make it a three-digit number. So, A has to be from 1 to 9. The smallest four-digit palindrome would then start with the smallest possible A, which is 1. So, the smallest four-digit palindrome is 1001. But wait, is 1001 divisible by 5? Let me check.Divisibility by 5 is straightforward: a number is divisible by 5 if its last digit is either 0 or 5. In this case, the last digit is 1, which isn't 0 or 5. So, 1001 isn't divisible by 5. I need to find the next smallest four-digit palindrome that ends with 0 or 5.Let's think about the structure again: A B B A. The last digit is A, so for the number to be divisible by 5, A must be either 0 or 5. But A can't be 0 because that would make it a three-digit number. So, A has to be 5. That means the number will be 5 B B 5.Now, to find the smallest such number, I need the smallest possible B. Since B is a digit, the smallest value is 0. So, plugging that in, the number would be 5 0 0 5, which is 5005. Let me check if this is divisible by 5. The last digit is 5, so yes, it is divisible by 5. Also, 5005 divided by 5 is 1001, which is an integer, so that works.Wait, but is there a smaller four-digit palindrome that ends with 5? Let me think. If I try A=5, the smallest B is 0, so 5005 is the smallest. If I try A=5 and B=0, that's the smallest possible. If I try A=5 and B=1, that would be 5115, which is larger than 5005. So, 5005 is indeed the smallest.Just to make sure I'm not missing anything, let's consider if there's a four-digit palindrome starting with a digit less than 5 that ends with 0 or 5. But since A can't be 0, the next possible A is 1, 2, 3, or 4. If A is 1, the number would be 1 B B 1. The last digit is 1, which isn't 0 or 5, so it's not divisible by 5. Similarly, for A=2, 3, or 4, the last digit would be 2, 3, or 4, none of which are 0 or 5. So, those numbers won't be divisible by 5.Therefore, the smallest four-digit palindrome that's divisible by 5 must start and end with 5, and have the smallest possible middle digits, which are 0s. So, the number is 5005. I think that's the answer.

💡Alright, so I have this sequence problem here. It says that in the sequence {a_n}, the first term a_1 is 1, and each subsequent term is defined by the recurrence relation a_{n+1} = 2a_n + 3 for n ≥ 1. I need to find the general term a_n.Hmm, okay. Let me think about how to approach this. I remember that recurrence relations can sometimes be solved by finding a pattern or by using techniques like characteristic equations or generating functions. But since this is a linear recurrence relation, maybe there's a simpler way.First, let's write down the first few terms to see if I can spot a pattern.Given a_1 = 1.Then, a_2 = 2a_1 + 3 = 2*1 + 3 = 5.Next, a_3 = 2a_2 + 3 = 2*5 + 3 = 13.Then, a_4 = 2a_3 + 3 = 2*13 + 3 = 29.And a_5 = 2a_4 + 3 = 2*29 + 3 = 61.Hmm, so the terms are 1, 5, 13, 29, 61, and so on. I notice that each term seems to be roughly doubling and then adding 3. But I need a general formula for a_n.I recall that for linear recurrence relations of the form a_{n+1} = k*a_n + c, where k and c are constants, we can solve them by finding the homogeneous solution and a particular solution.Let me try that approach. The recurrence relation is a_{n+1} = 2a_n + 3. So, k is 2 and c is 3.First, let's solve the homogeneous equation: a_{n+1} = 2a_n. The solution to this is straightforward; it's a geometric sequence where each term is twice the previous one. So, the homogeneous solution is a_n^{(h)} = A*2^n, where A is a constant.Next, we need a particular solution. Since the nonhomogeneous term is a constant (3), we can assume that the particular solution is a constant, say a_n^{(p)} = B.Plugging this into the recurrence relation:a_{n+1}^{(p)} = 2a_n^{(p)} + 3But since a_n^{(p)} = B, we have:B = 2B + 3Solving for B:B - 2B = 3- B = 3B = -3So, the particular solution is a_n^{(p)} = -3.Therefore, the general solution is the sum of the homogeneous and particular solutions:a_n = a_n^{(h)} + a_n^{(p)} = A*2^n - 3Now, we need to determine the constant A using the initial condition. We know that a_1 = 1.Plugging n = 1 into the general solution:a_1 = A*2^1 - 3 = 2A - 3 = 1Solving for A:2A - 3 = 12A = 4A = 2So, the general term is:a_n = 2*2^n - 3 = 2^{n+1} - 3Wait, let me double-check this. If I plug n = 1 into this formula, I should get a_1 = 2^{2} - 3 = 4 - 3 = 1, which is correct.For n = 2: 2^{3} - 3 = 8 - 3 = 5, which matches a_2.n = 3: 2^{4} - 3 = 16 - 3 = 13, which is a_3.n = 4: 2^{5} - 3 = 32 - 3 = 29, which is a_4.n = 5: 2^{6} - 3 = 64 - 3 = 61, which is a_5.So, it seems to work for the first few terms. Therefore, I think the general term is a_n = 2^{n+1} - 3.Alternatively, another way to approach this is by transforming the recurrence relation into a geometric sequence. Let me try that method as well to confirm.Given a_{n+1} = 2a_n + 3.If I add 3 to both sides, I get:a_{n+1} + 3 = 2a_n + 3 + 3 = 2a_n + 6Wait, that doesn't seem helpful. Maybe I should subtract something instead.Wait, let me think. If I can write the recurrence in terms of a shifted sequence, maybe I can make it geometric.Let me define b_n = a_n + c, where c is a constant to be determined. Then, the recurrence becomes:b_{n+1} - c = 2(b_n - c) + 3Expanding this:b_{n+1} - c = 2b_n - 2c + 3Bring the -c to the other side:b_{n+1} = 2b_n - 2c + 3 + cSimplify:b_{n+1} = 2b_n - c + 3Now, if I choose c such that the constant term cancels out, i.e., -c + 3 = 0, then c = 3.So, if I set b_n = a_n + 3, then the recurrence becomes:b_{n+1} = 2b_nWhich is a simple geometric sequence with common ratio 2.Now, we can find b_n. Since b_n is a geometric sequence, b_n = b_1 * 2^{n-1}.We know that b_1 = a_1 + 3 = 1 + 3 = 4.Therefore, b_n = 4 * 2^{n-1} = 2^2 * 2^{n-1} = 2^{n+1}.But b_n = a_n + 3, so:a_n + 3 = 2^{n+1}Therefore, a_n = 2^{n+1} - 3This confirms the earlier result. So, the general term is indeed a_n = 2^{n+1} - 3.I think I've approached this problem from two different angles: solving the linear recurrence using homogeneous and particular solutions, and transforming the sequence into a geometric sequence by shifting it. Both methods led me to the same conclusion, so I'm confident that the general term is correct.Just to be thorough, let me test it for n = 6.Using the recurrence relation:a_6 = 2a_5 + 3 = 2*61 + 3 = 122 + 3 = 125Using the general formula:a_6 = 2^{7} - 3 = 128 - 3 = 125Perfect, it matches. So, the formula works for n = 6 as well.I think I've covered all the necessary steps and verified the solution multiple times. I don't see any mistakes in my reasoning, so I'm ready to present the final answer.

💡Okay, so Xiaoming wants to buy flowers for his mom on Mother's Day. He's looking at carnations and lilies. Carnations cost 2 yuan each, and lilies cost 3 yuan each. He has 30 yuan to spend and wants to buy both types of flowers. The question is asking how many different purchasing plans he has. The options are 3, 4, 5, or 6 plans.Alright, let's break this down. First, I need to figure out how many combinations of carnations and lilies Xiaoming can buy without exceeding his 30 yuan budget. Since he wants to buy both types, he can't spend all his money on just one type.Let me define some variables to make this clearer. Let's say the number of carnations he buys is x, and the number of lilies is y. Each carnation is 2 yuan, so the total cost for carnations would be 2x yuan. Similarly, each lily is 3 yuan, so the total cost for lilies would be 3y yuan. The sum of these should equal 30 yuan because that's his total budget.So, the equation would be:2x + 3y = 30Now, I need to find all the possible positive integer solutions for x and y that satisfy this equation. That means both x and y have to be whole numbers greater than zero because you can't buy a fraction of a flower or a negative number of flowers.Let me try to solve for one variable in terms of the other to make it easier. Let's solve for y in terms of x:3y = 30 - 2xy = (30 - 2x)/3For y to be an integer, (30 - 2x) must be divisible by 3. So, 30 is already divisible by 3, which is good. So, 2x must also leave a remainder that makes the whole numerator divisible by 3.Let me think about the possible values of x. Since each carnation is 2 yuan, the maximum number of carnations he can buy is 15 (because 15*2=30). But since he wants to buy both carnations and lilies, x has to be less than 15.Let me list possible values of x and see if y comes out as an integer.Starting with x=1:y = (30 - 2*1)/3 = (30 - 2)/3 = 28/3 ≈9.333Not an integer, so this doesn't work.x=2:y = (30 - 4)/3 = 26/3 ≈8.666Still not an integer.x=3:y = (30 - 6)/3 = 24/3 = 8Okay, y=8 is an integer. So, x=3 and y=8 is a valid plan.x=4:y = (30 - 8)/3 = 22/3 ≈7.333Not an integer.x=5:y = (30 - 10)/3 = 20/3 ≈6.666Not an integer.x=6:y = (30 - 12)/3 = 18/3 = 6y=6 is an integer. So, x=6 and y=6 is another plan.x=7:y = (30 - 14)/3 = 16/3 ≈5.333Not an integer.x=8:y = (30 - 16)/3 = 14/3 ≈4.666Not an integer.x=9:y = (30 - 18)/3 = 12/3 = 4y=4 is an integer. So, x=9 and y=4 is another plan.x=10:y = (30 - 20)/3 = 10/3 ≈3.333Not an integer.x=11:y = (30 - 22)/3 = 8/3 ≈2.666Not an integer.x=12:y = (30 - 24)/3 = 6/3 = 2y=2 is an integer. So, x=12 and y=2 is another plan.x=13:y = (30 - 26)/3 = 4/3 ≈1.333Not an integer.x=14:y = (30 - 28)/3 = 2/3 ≈0.666Not an integer.x=15:y = (30 - 30)/3 = 0/3 = 0But y=0 means he's not buying any lilies, which contradicts the requirement of buying both types. So, this doesn't count.So, the valid combinations are:- x=3, y=8- x=6, y=6- x=9, y=4- x=12, y=2That's four different purchasing plans. So, the answer should be 4 plans, which is option B.