answer：First, I need to find the vector (vec{a} - 2vec{b}) by subtracting twice vector (vec{b}) from vector (vec{a}).Given:[vec{a} = (sqrt{3}, 1)][vec{b} = (0, -1)][vec{c} = (k, sqrt{3})]Calculating (vec{a} - 2vec{b}):[vec{a} - 2vec{b} = (sqrt{3} - 2 times 0, 1 - 2 times (-1)) = (sqrt{3}, 3)]Next, since (vec{a} - 2vec{b}) is perpendicular to (vec{c}), their dot product must be zero:[(vec{a} - 2vec{b}) cdot vec{c} = 0][(sqrt{3}, 3) cdot (k, sqrt{3}) = 0][sqrt{3} times k + 3 times sqrt{3} = 0]Solving for (k):[sqrt{3}k + 3sqrt{3} = 0][sqrt{3}k = -3sqrt{3}][k = -3]

answer：Okay, so I have this problem here about a curve y = f(x) that has a tangent line at x = 5, and the equation of that tangent line is y = -x + 8. I need to find f(5) and f'(5), and then choose the correct option from A to D. First, let me recall what a tangent line means. A tangent line to a curve at a particular point is a straight line that just touches the curve at that point and has the same slope as the curve at that point. So, the slope of the tangent line at x = 5 should be equal to the derivative of f at x = 5, which is f'(5). Given the equation of the tangent line is y = -x + 8. That's a straight line, and I know that the general form of a straight line is y = mx + b, where m is the slope and b is the y-intercept. So, in this case, the slope m is -1. Therefore, I can say that f'(5) should be equal to -1 because the derivative at that point gives the slope of the tangent line.Now, I need to find f(5). Since the tangent line touches the curve at x = 5, the point (5, f(5)) lies on both the curve and the tangent line. So, if I plug x = 5 into the equation of the tangent line, I should get the y-coordinate, which is f(5).Let me calculate that. Plugging x = 5 into y = -x + 8 gives y = -5 + 8. That simplifies to y = 3. So, f(5) is 3.Wait, let me double-check that. If x is 5, then y = -5 + 8 is indeed 3. So, f(5) = 3 and f'(5) = -1. Looking back at the options given:A: 3, 3 B: 3, -1 C: -1, 3 D: -1, -1So, f(5) is 3 and f'(5) is -1, which corresponds to option B. Hmm, is there anything I might have missed? Let me think again. The tangent line at x = 5 has the equation y = -x + 8. So, the slope is -1, which is the derivative at that point. Plugging x = 5 into the tangent line equation gives the y-value, which is f(5). That seems right.I don't think I made any mistakes here. The calculations are straightforward. So, I'm confident that the correct answer is option B: 3, -1.

answer：First, determine the number of sections needed for boys and girls separately, ensuring each section has at most 24 students.For boys:- Number of boys = 408- Maximum students per section = 24- Number of sections for boys = 408 / 24 = 17For girls:- Number of girls = 240- Maximum students per section = 24- Number of sections for girls = 240 / 24 = 10Next, adjust these numbers to maintain the ratio of boys' sections to girls' sections as 3:2.Let the number of boys' sections be 3x and the number of girls' sections be 2x. Given that the actual number of boys' sections is 17:- 3x = 17- x = 17 / 3 ≈ 5.6667Since we can't have a fraction of a section, we round up to the nearest whole number, x = 6.Therefore:- Number of boys' sections = 3x = 3 * 6 = 18- Number of girls' sections = 2x = 2 * 6 = 12Verify if these sections can accommodate all students:- Boys: 18 sections * 24 students/section = 432 spots (sufficient for 408 boys)- Girls: 12 sections * 24 students/section = 288 spots (sufficient for 240 girls)Finally, calculate the total number of sections:- Total sections = 18 (boys) + 12 (girls) = 30

answer：Okay, so I have this problem where I need to find the magnitude of the vector (overrightarrow{a} + 2overrightarrow{b}). The given information is that the angle between vectors (overrightarrow{a}) and (overrightarrow{b}) is (60^circ), vector (overrightarrow{a}) is ((1, sqrt{3})), and the magnitude of vector (overrightarrow{b}) is 1. First, I remember that the magnitude of a vector (overrightarrow{v} = (x, y)) is calculated using the formula (sqrt{x^2 + y^2}). So, maybe I can use that somewhere here. I also recall that when adding vectors, the resultant vector's magnitude can be found using the law of cosines if I know the magnitudes of the individual vectors and the angle between them. The formula for the magnitude of the sum of two vectors (overrightarrow{a}) and (overrightarrow{b}) is:[|overrightarrow{a} + overrightarrow{b}| = sqrt{|overrightarrow{a}|^2 + |overrightarrow{b}|^2 + 2|overrightarrow{a}||overrightarrow{b}| cos theta}]But in this problem, it's not just (overrightarrow{a} + overrightarrow{b}), it's (overrightarrow{a} + 2overrightarrow{b}). Hmm, so maybe I can adjust the formula accordingly. Let me think about how to handle the scalar multiplication.If I have (2overrightarrow{b}), then the magnitude of that vector would be (2|overrightarrow{b}|). Since (|overrightarrow{b}| = 1), then (|2overrightarrow{b}| = 2 times 1 = 2). So, now I have two vectors: (overrightarrow{a}) and (2overrightarrow{b}). The angle between them is still (60^circ) because scaling a vector doesn't change the angle between it and another vector. So, I can use the law of cosines formula with these two vectors.First, let me find the magnitude of (overrightarrow{a}). Given (overrightarrow{a} = (1, sqrt{3})), the magnitude is:[|overrightarrow{a}| = sqrt{1^2 + (sqrt{3})^2} = sqrt{1 + 3} = sqrt{4} = 2]Okay, so (|overrightarrow{a}| = 2) and (|2overrightarrow{b}| = 2). The angle between them is (60^circ). Plugging these into the law of cosines formula:[|overrightarrow{a} + 2overrightarrow{b}| = sqrt{|overrightarrow{a}|^2 + |2overrightarrow{b}|^2 + 2|overrightarrow{a}||2overrightarrow{b}| cos 60^circ}]Let me compute each part step by step. First, (|overrightarrow{a}|^2 = 2^2 = 4).Second, (|2overrightarrow{b}|^2 = 2^2 = 4).Third, the cross term: (2|overrightarrow{a}||2overrightarrow{b}| cos 60^circ). Calculating this:(2 times 2 times 2 times cos 60^circ). Wait, hold on. Let me make sure I get this right. The formula is (2|overrightarrow{a}||2overrightarrow{b}| cos theta). So, that's (2 times 2 times 2 times cos 60^circ). Wait, that seems too much. Let me double-check. The formula is:[|overrightarrow{a} + overrightarrow{b}| = sqrt{|overrightarrow{a}|^2 + |overrightarrow{b}|^2 + 2|overrightarrow{a}||overrightarrow{b}| cos theta}]So, in this case, it's (|overrightarrow{a} + 2overrightarrow{b}|), so the cross term should be (2 times |overrightarrow{a}| times |2overrightarrow{b}| times cos 60^circ). So, (2 times 2 times 2 times cos 60^circ). But (cos 60^circ = 0.5), so this becomes (2 times 2 times 2 times 0.5). Let me compute that:First, 2 times 2 is 4, times 2 is 8, times 0.5 is 4. So, the cross term is 4.Putting it all together:[|overrightarrow{a} + 2overrightarrow{b}| = sqrt{4 + 4 + 4} = sqrt{12}]Simplifying (sqrt{12}), which is equal to (2sqrt{3}).Wait, that seems straightforward, but let me verify if I did everything correctly. Maybe I should try another approach to confirm.Another way to find the magnitude is to express both vectors in component form and then add them, and then compute the magnitude of the resulting vector.Given that (overrightarrow{a} = (1, sqrt{3})), and (overrightarrow{b}) has a magnitude of 1 and makes a (60^circ) angle with (overrightarrow{a}). I can represent (overrightarrow{b}) in terms of its components. Since the angle between (overrightarrow{a}) and (overrightarrow{b}) is (60^circ), and (overrightarrow{a}) is in the direction of ((1, sqrt{3})), which makes a (60^circ) angle with the positive x-axis because:The magnitude of (overrightarrow{a}) is 2, as we found earlier, and its direction is given by (tan theta = frac{sqrt{3}}{1}), so (theta = 60^circ). Therefore, if (overrightarrow{a}) is at (60^circ) from the x-axis, and (overrightarrow{b}) makes a (60^circ) angle with (overrightarrow{a}), then (overrightarrow{b}) could be at (60^circ + 60^circ = 120^circ) from the x-axis, or (60^circ - 60^circ = 0^circ) from the x-axis. Wait, that might not necessarily be the case. The angle between (overrightarrow{a}) and (overrightarrow{b}) is (60^circ), but (overrightarrow{a}) itself is at (60^circ) from the x-axis. So, (overrightarrow{b}) can be in two possible directions: one on one side of (overrightarrow{a}) and the other on the opposite side. But since the problem doesn't specify the direction, maybe I can assume a particular orientation for simplicity. Let me choose (overrightarrow{b}) to be at (60^circ) from (overrightarrow{a}) in the counterclockwise direction. So, if (overrightarrow{a}) is at (60^circ), then (overrightarrow{b}) would be at (60^circ + 60^circ = 120^circ) from the x-axis.Alternatively, if I take (overrightarrow{b}) to be at (60^circ - 60^circ = 0^circ) from the x-axis, which is along the x-axis. Wait, but in the previous approach, I considered (overrightarrow{b}) as ((1, 0)), but that might not necessarily be the case because the angle between (overrightarrow{a}) and (overrightarrow{b}) is (60^circ), not the angle of (overrightarrow{b}) with the x-axis.Hmm, maybe I confused the angle of (overrightarrow{b}) with the angle between (overrightarrow{a}) and (overrightarrow{b}). Let me clarify.Since (overrightarrow{a}) is at (60^circ) from the x-axis, and the angle between (overrightarrow{a}) and (overrightarrow{b}) is (60^circ), then (overrightarrow{b}) can be at (60^circ + 60^circ = 120^circ) or (60^circ - 60^circ = 0^circ) from the x-axis.So, let's consider both possibilities.Case 1: (overrightarrow{b}) is at (120^circ) from the x-axis.Then, the components of (overrightarrow{b}) would be:[overrightarrow{b} = (cos 120^circ, sin 120^circ) = (-frac{1}{2}, frac{sqrt{3}}{2})]Case 2: (overrightarrow{b}) is at (0^circ) from the x-axis.Then, the components of (overrightarrow{b}) would be:[overrightarrow{b} = (cos 0^circ, sin 0^circ) = (1, 0)]So, depending on the direction, (overrightarrow{b}) can be either ((1, 0)) or ((- frac{1}{2}, frac{sqrt{3}}{2})).Wait, but in the initial approach, I assumed (overrightarrow{b} = (1, 0)), which is one possibility. But actually, there are two possible vectors for (overrightarrow{b}) given the angle between them is (60^circ). So, I need to consider both cases to see if the magnitude of (overrightarrow{a} + 2overrightarrow{b}) is the same in both cases or different.Let me compute both cases.Case 1: (overrightarrow{b} = (1, 0))Then, (2overrightarrow{b} = (2, 0))Adding to (overrightarrow{a} = (1, sqrt{3})):[overrightarrow{a} + 2overrightarrow{b} = (1 + 2, sqrt{3} + 0) = (3, sqrt{3})]Magnitude:[|overrightarrow{a} + 2overrightarrow{b}| = sqrt{3^2 + (sqrt{3})^2} = sqrt{9 + 3} = sqrt{12} = 2sqrt{3}]Case 2: (overrightarrow{b} = (-frac{1}{2}, frac{sqrt{3}}{2}))Then, (2overrightarrow{b} = (-1, sqrt{3}))Adding to (overrightarrow{a} = (1, sqrt{3})):[overrightarrow{a} + 2overrightarrow{b} = (1 - 1, sqrt{3} + sqrt{3}) = (0, 2sqrt{3})]Magnitude:[|overrightarrow{a} + 2overrightarrow{b}| = sqrt{0^2 + (2sqrt{3})^2} = sqrt{0 + 12} = sqrt{12} = 2sqrt{3}]So, in both cases, the magnitude is the same, (2sqrt{3}). That's reassuring. So, regardless of the direction of (overrightarrow{b}), the magnitude of (overrightarrow{a} + 2overrightarrow{b}) is (2sqrt{3}).Alternatively, I could have used the dot product formula to find the magnitude. The formula for the magnitude squared of the sum is:[|overrightarrow{a} + 2overrightarrow{b}|^2 = |overrightarrow{a}|^2 + |2overrightarrow{b}|^2 + 2 times overrightarrow{a} cdot (2overrightarrow{b})]But wait, that might complicate things. Alternatively, using the formula:[|overrightarrow{a} + 2overrightarrow{b}|^2 = |overrightarrow{a}|^2 + |2overrightarrow{b}|^2 + 4 (overrightarrow{a} cdot overrightarrow{b})]But since (overrightarrow{a} cdot overrightarrow{b} = |overrightarrow{a}||overrightarrow{b}| cos theta), which is (2 times 1 times cos 60^circ = 2 times 0.5 = 1).So, plugging in:[|overrightarrow{a} + 2overrightarrow{b}|^2 = 4 + 4 + 4 times 1 = 4 + 4 + 4 = 12]Therefore, (|overrightarrow{a} + 2overrightarrow{b}| = sqrt{12} = 2sqrt{3}).This method also gives the same result, which is consistent with the previous approaches.I think I've verified it enough times. All methods lead to the same answer, so I'm confident that the magnitude is (2sqrt{3}).

answer：Okay, so I've got this problem about electricity prices and revenue. Let me try to understand what's going on here. Last year, the price was 0.80 yuan per kWh, and they used 'a' kWh in total. This year, they want to lower the price to somewhere between 0.55 and 0.75 yuan per kWh. But users are hoping for even lower, like 0.40 yuan per kWh. First, I need to write a function that shows how the revenue (y) of the power department changes with the actual price (x) after the reduction. Revenue is basically the amount of money they make, which would be the price per kWh multiplied by the number of kWh sold. But here, it seems like the amount of electricity used might change based on the price. If the price goes down, people might use more electricity, right? So, the revenue isn't just price times a fixed amount; it's price times a variable amount that depends on the price itself.The problem mentions that the increased consumption is proportional to the price difference, with a proportionality constant 'k'. So, if the price is lowered, the consumption increases by some factor related to how much the price was lowered. Specifically, it says "k = 0.2a", so that's a given value. Let me try to write this out. If the new price is x yuan per kWh, then the increase in consumption would be k divided by (x - 0.40). Wait, why (x - 0.40)? Because 0.40 is the user's expected price, so the difference from that expected price affects how much more they consume. So, the total consumption would be the original 'a' plus this increase. So, the total consumption is a + (k / (x - 0.40)). Then, the revenue y would be this total consumption multiplied by the new price x, but I think we also have to consider the cost, which is 0.30 yuan per kWh. So, actually, the revenue is the total consumption times (x - 0.30), because 0.30 is the cost, and they make a profit on each kWh sold.Putting that together, the function would be y = (a + (k / (x - 0.40))) * (x - 0.30). That seems right. Now, for the second part, they want to know the minimum price x where the revenue increases by at least 20% compared to last year. Last year's revenue was 0.80 yuan per kWh times 'a' kWh, which is 0.80a. But actually, since they have a cost of 0.30 yuan per kWh, their profit last year was (0.80 - 0.30)a = 0.50a. So, a 20% increase on that would be 0.50a * 1.20 = 0.60a.So, we need to find the minimum x such that y >= 0.60a. Plugging in k = 0.2a into our function, we get y = (a + (0.2a / (x - 0.40))) * (x - 0.30). We need this to be greater than or equal to 0.60a.Let me write that out:(a + (0.2a / (x - 0.40))) * (x - 0.30) >= 0.60aI can factor out 'a' since it's positive:[1 + (0.2 / (x - 0.40))] * (x - 0.30) >= 0.60Now, let's simplify this inequality. Let's denote t = x - 0.40 to make it easier:[1 + (0.2 / t)] * (t + 0.10) >= 0.60Expanding this:(1 * (t + 0.10)) + (0.2 / t * (t + 0.10)) >= 0.60Simplify each term:(t + 0.10) + (0.2 + 0.02 / t) >= 0.60Combine like terms:t + 0.10 + 0.2 + 0.02 / t >= 0.60So, t + 0.30 + 0.02 / t >= 0.60Subtract 0.60 from both sides:t + 0.02 / t - 0.30 >= 0Multiply both sides by t to eliminate the denominator (assuming t > 0, which it is since x > 0.40):t^2 + 0.02 - 0.30t >= 0Rearrange:t^2 - 0.30t + 0.02 >= 0This is a quadratic inequality. Let's find the roots:t = [0.30 ± sqrt(0.09 - 0.08)] / 2t = [0.30 ± sqrt(0.01)] / 2t = [0.30 ± 0.10] / 2So, t = (0.40)/2 = 0.20 or t = (0.20)/2 = 0.10So, the quadratic is positive outside the roots, meaning t <= 0.10 or t >= 0.20. But t = x - 0.40, and x is between 0.55 and 0.75, so t is between 0.15 and 0.35. Therefore, the relevant interval is t >= 0.20, which means x - 0.40 >= 0.20, so x >= 0.60.Therefore, the minimum price x is 0.60 yuan per kWh.

answer：Okay, so I need to find the sum of all the non-trivial factors of 2015. Hmm, non-trivial factors are the ones other than 1 and the number itself, right? So, first, I guess I should figure out what all the factors of 2015 are.Let me start by recalling that factors of a number are the numbers that divide it without leaving a remainder. So, I need to find all numbers that can multiply together in pairs to give 2015. First, I know that 1 and 2015 are factors, but those are the trivial ones. I need to find the others. Maybe I can start by checking small prime numbers to see if they divide 2015.Let me try dividing 2015 by some primes. Starting with 2: 2015 divided by 2 is 1007.5, which isn't an integer, so 2 isn't a factor. Next, 3: 2015 divided by 3 is approximately 671.666..., which also isn't an integer. How about 5? 2015 divided by 5 is 403, which is an integer. So, 5 is a factor.Now, since 5 is a factor, I can divide 2015 by 5 to get 403. So, 5 and 403 are factors. Now, I need to factorize 403 further. Let's see if 403 can be divided by any primes. Trying 3 again: 403 divided by 3 is about 134.333..., not an integer. Next, 7: 403 divided by 7 is approximately 57.571..., not an integer. How about 11? 403 divided by 11 is around 36.636..., still not an integer. Next prime is 13: 403 divided by 13 is 31, which is an integer. So, 13 is a factor, and 31 is another factor.So, putting it all together, the prime factors of 2015 are 5, 13, and 31. Therefore, the factors of 2015 are all the products of these primes. Let me list them out:1 (trivial), 5, 13, 31, 5×13=65, 5×31=155, 13×31=403, and 2015 (trivial). So, the non-trivial factors are 5, 13, 31, 65, 155, and 403.Now, to find the sum of these non-trivial factors, I'll add them up:5 + 13 + 31 + 65 + 155 + 403.Let me do the addition step by step:5 + 13 = 1818 + 31 = 4949 + 65 = 114114 + 155 = 269269 + 403 = 672So, the sum of all the non-trivial factors of 2015 is 672.Wait, just to make sure I didn't miss any factors, let me double-check. The prime factors are 5, 13, and 31. The factors are all combinations of these primes:- 1 (trivial)- 5- 13- 31- 5×13=65- 5×31=155- 13×31=403- 5×13×31=2015 (trivial)Yes, that's all of them. So, the non-trivial ones are indeed 5, 13, 31, 65, 155, and 403. Adding them gives 672. I think that's correct.Alternatively, I remember that the sum of all factors of a number can be found using the formula related to its prime factorization. For a number n = p^a * q^b * r^c..., the sum of its factors is (1 + p + p^2 + ... + p^a)(1 + q + q^2 + ... + q^b)(1 + r + r^2 + ... + r^c)... So, for 2015, which is 5^1 * 13^1 * 31^1, the sum of all factors would be (1 + 5)(1 + 13)(1 + 31). Let me calculate that:(1 + 5) = 6(1 + 13) = 14(1 + 31) = 32Now, multiply these together: 6 * 14 = 84, then 84 * 32. Let me compute 84 * 32:84 * 30 = 252084 * 2 = 168So, 2520 + 168 = 2688So, the sum of all factors is 2688. Now, subtract the trivial factors, which are 1 and 2015. So, 2688 - 1 - 2015 = 2688 - 2016 = 672. That matches the earlier result. So, both methods give me 672. Therefore, I'm confident that the sum of all non-trivial factors of 2015 is 672.