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question:Let ((1+sqrt{2}+sqrt{3})^{n}=q_{n}+r_{n} sqrt{2}+s_{n} sqrt{3}+t_{n} sqrt{6}), where (q_{n}, r_{n}, s_{n}, t_{n}) are natural numbers. Compute the limits (lim _{n rightarrow infty} frac{r_{n}}{q_{n}}, lim _{n rightarrow infty} frac{s_{n}}{q_{n}}), and (lim _{n rightarrow infty} frac{t_{n}}{q_{n}}).
answer:# Problem: Let (1 + sqrt{2} + sqrt{3})^n = q_n + r_n sqrt{2} + s_n sqrt{3} + t_n sqrt{6}, where (q_n, r_n, s_n, t_n) are natural numbers. Compute the limits: [ lim_{n to infty} frac{r_n}{q_n}, quad lim_{n to infty} frac{s_n}{q_n}, quad text{and} quad lim_{n to infty} frac{t_n}{q_n}. ] : 1. **Define roots and expressions:** Define the following four roots for simplicity: [ lambda_1 = 1 + sqrt{2} + sqrt{3}, quad lambda_2 = 1 - sqrt{2} + sqrt{3}, quad lambda_3 = 1 + sqrt{2} - sqrt{3}, quad lambda_4 = 1 - sqrt{2} - sqrt{3} ] 2. **Express roots raised to the power (n):** From the problem's information, [ lambda_1^n = q_n + r_n sqrt{2} + s_n sqrt{3} + t_n sqrt{6} ] Similarly, consider: [ lambda_2^n = q_n - r_n sqrt{2} + s_n sqrt{3} - t_n sqrt{6} ] [ lambda_3^n = q_n + r_n sqrt{2} - s_n sqrt{3} - t_n sqrt{6} ] [ lambda_4^n = q_n - r_n sqrt{2} - s_n sqrt{3} + t_n sqrt{6} ] 3. **Combine expressions:** By adding and subtracting the above expressions suitably: For the first sum: [ lambda_1^n + lambda_2^n + lambda_3^n + lambda_4^n = 4q_n ] For the second sum: [ lambda_1^n + lambda_2^n - lambda_3^n - lambda_4^n = 4s_n sqrt{3} ] For the third sum: [ lambda_1^n - lambda_2^n + lambda_3^n - lambda_4^n = 4r_n sqrt{2} ] For the fourth sum: [ lambda_1^n - lambda_2^n - lambda_3^n + lambda_4^n = 4t_n sqrt{6} ] 4. **Limit behavior of other roots:** Notice that: [ lim_{n to infty} frac{lambda_2^n}{lambda_1^n} = lim_{n to infty} frac{lambda_3^n}{lambda_1^n} = lim_{n to infty} frac{lambda_4^n}{lambda_1^n} = 0 ] because (|lambda_2|, |lambda_3|, |lambda_4| < |lambda_1|). 5. **Limits:** From the above limit behavior, [ lim_{n to infty} frac{lambda_1^n}{q_n} = 4 ] and similarly, [ lim_{n to infty} frac{r_n}{lambda_1^n} = frac{1}{4 sqrt{2}}, quad lim_{n to infty} frac{s_n}{lambda_1^n} = frac{1}{4 sqrt{3}}, quad lim_{n to infty} frac{t_n}{lambda_1^n} = frac{1}{4 sqrt{6}} ] 6. **Final calculations:** Therefore, [ lim_{n to infty} frac{r_n}{q_n} = frac{frac{1}{4 sqrt{2}}}{frac{1}{4}} = frac{1}{sqrt{2}}, quad lim_{n to infty} frac{s_n}{q_n} = frac{frac{1}{4 sqrt{3}}}{frac{1}{4}} = frac{1}{sqrt{3}}, quad lim_{n to infty} frac{t_n}{q_n} = frac{frac{1}{4 sqrt{6}}}{frac{1}{4}} = frac{1}{sqrt{6}} ] # Conclusion: [ boxed{frac{1}{sqrt{2}}, quad frac{1}{sqrt{3}}, quad frac{1}{sqrt{6}}} ]
question:Longteng Primary School's fifth grade has a total of four classes. Class 1 has 42 students, Class 2 has frac{6}{7} of the number of students in Class 1, Class 3 has frac{5}{6} of the number of students in Class 2, and Class 4 has 1.2 times the number of students in Class 3. How many students are there in total in the fifth grade?
answer:We begin by defining the number of students in each class. We start by noting the number of students in Class 1 and then calculating for the other classes step-by-step according to the given ratios. 1. **Determine the number of students in Class 1**: [ text{Number of students in Class 1} = 42 ] 2. **Calculate the number of students in Class 2**: [ text{Number of students in Class 2} = 42 times frac{6}{7} ] Performing the multiplication: [ 42 times frac{6}{7} = 42 div 7 times 6 = 6 times 6 = 36 ] So, the number of students in Class 2 is: [ 36 ] 3. **Calculate the number of students in Class 3**: [ text{Number of students in Class 3} = 36 times frac{5}{6} ] Performing the multiplication: [ 36 times frac{5}{6} = 36 div 6 times 5 = 6 times 5 = 30 ] So, the number of students in Class 3 is: [ 30 ] 4. **Calculate the number of students in Class 4**: [ text{Number of students in Class 4} = 30 times 1.2 ] Performing the multiplication: [ 30 times 1.2 = 30 times frac{12}{10} = 30 times 1.2 = 36 ] So, the number of students in Class 4 is: [ 36 ] 5. **Sum the total number of students in all four classes**: Adding the number of students from each class: [ 42 + 36 + 30 + 36 ] [ 42 + 36 = 78 ] [ 78 + 30 = 108 ] [ 108 + 36 = 144 ] Therefore, the total number of students in the fifth grade is: [ 144 ] # Conclusion: [ boxed{144} ]
question:The graph of [y^4 16x^4 = 8y^2 4] is the union of the graphs of two different conic sections. Which two types of conic sections are they?
answer:Rewrite the given equation as: [ y^4 - 8y^2 + 4 = 16x^4 ] The equation can be rewritten by factoring the left-hand side as a perfect square of a binomial: [ (y^2 - 4)^2 = 16x^4 ] By taking the square root of both sides, we have: [ y^2 - 4 = pm 4x^2 ] Rearranging these gives two separate equations: 1. ( y^2 - 4x^2 = 4 ) 2. ( y^2 + 4x^2 = 4 ) Observing these equations: - The first equation ( y^2 - 4x^2 = 4 ) can be rearranged as ((y^2 - 4x^2) = 4) which represents a scaled hyperbola. - The second equation ( y^2 + 4x^2 = 4 ) is equivalent to (frac{y^2}{4} + frac{x^2}{1} = 1) which represents an ellipse. Thus, the conic sections are a hyperbola and an ellipse. Therefore, the answer is boxed{text{H, E}}.
question:To further improve the urban quality of our city and enhance the functional layout of historical and cultural blocks, the municipal government has decided to implement a project to renovate and upgrade the old urban areas. Zhenxing Construction Company has contracted the earthwork transportation task for a certain section, planning to dispatch two types of earthwork transport vehicles, large and small, to transport earthwork. It is known that 3 large earthwork transport vehicles and 4 small earthwork transport vehicles can transport a total of 44 tons of earthwork at once, and 4 large earthwork transport vehicles and 6 small earthwork transport vehicles can transport a total of 62 tons of earthwork at once. (1) How many tons of earthwork does one large earthwork transport vehicle and one small earthwork transport vehicle transport at a time? (2) The earthwork transport company decides to dispatch a total of 12 vehicles, including both large and small earthwork transport vehicles, to participate in the transportation work. If the total amount of earthwork transported each time is not less than 78 tons, and at least 4 small earthwork transport vehicles are dispatched, what are the possible dispatch plans? Please list all the dispatch plans after calculation.
answer:# Solution: Part 1: Calculating the Earthwork Capacity of Each Vehicle Type Let's denote: - x = the amount of earthwork transported by one large earthwork transport vehicle at a time (in tons), - y = the amount of earthwork transported by one small earthwork transport vehicle at a time (in tons). From the problem, we have two equations based on the given conditions: 1. For 3 large and 4 small vehicles: 3x + 4y = 44, 2. For 4 large and 6 small vehicles: 4x + 6y = 62. To solve these equations, we can use the method of substitution or elimination. Here, we'll use elimination for simplicity: - Multiplying the first equation by 2 gives us 6x + 8y = 88, - The second equation is already in a good form to eliminate y: 4x + 6y = 62. Now, multiplying the second equation by -frac{4}{3} to align it with the first equation for elimination: - -frac{4}{3} cdot (4x + 6y) = -frac{4}{3} cdot 62 simplifies to -5.33x - 8y = -82.67. Adding this to the first equation (6x + 8y = 88) allows us to cancel y and solve for x: - 6x - 5.33x = 88 - 82.67, - 0.67x = 5.33, - x = 8. Substituting x = 8 back into one of the original equations to solve for y: - 3(8) + 4y = 44, - 24 + 4y = 44, - 4y = 20, - y = 5. Therefore, we find that one large earthwork transport vehicle transports 8 tons at a time, and one small vehicle transports 5 tons at a time. Thus, the answer to part (1) is: - boxed{text{One large vehicle: 8 tons, One small vehicle: 5 tons}}. Part 2: Finding Possible Dispatch Plans Given that m is the number of small vehicles dispatched and the total number of vehicles is 12, the number of large vehicles dispatched is 12 - m. The conditions are: 1. At least 4 small vehicles must be dispatched: m geq 4, 2. The total earthwork transported must be at least 78 tons: 8(12-m) + 5m geq 78. Substituting the values of x and y into the second condition gives us: - 96 - 8m + 5m geq 78, - -3m geq -18, - m leq 6. Combining this with m geq 4, we find 4 leq m leq 6. Therefore, m can be 4, 5, or 6, leading to three possible dispatch plans: 1. 8 large and 4 small vehicles, 2. 7 large and 5 small vehicles, 3. 6 large and 6 small vehicles. Thus, the possible dispatch plans are: - boxed{text{Plan 1: 8 large, 4 small}}, - boxed{text{Plan 2: 7 large, 5 small}}, - boxed{text{Plan 3: 6 large, 6 small}}.