answer：Okay, so I have this problem about a high school with 800 first-year students who took a math test. The question is asking what work needs to be done to find out the proportions of students who scored in different ranges: no less than 120 points, between 90 and 120 points, between 75 and 90 points, between 60 and 75 points, and below 60 points. The options are A, B, C, and D, which involve drawing a sample, calculating the average score, conducting a frequency distribution, or calculating the variance.Alright, let me break this down. First, I need to understand what each of these options entails. Option A is about drawing a sample and estimating the population based on that sample. That sounds like inferential statistics, where you take a subset of the data to make generalizations about the whole population. But in this case, we have data from all 800 students, right? So if we have the complete data, maybe we don't need to sample. Unless the data isn't readily available, but the problem doesn't specify that. Hmm.Option B is calculating the average score. The average, or mean, gives us a central tendency of the scores. But the question isn't asking for the average; it's asking for the proportions in different score ranges. So while the average might be useful to know, it's not directly answering the question. Maybe it's not necessary here.Option C is conducting a frequency distribution. Frequency distribution involves organizing and summarizing data by showing how often each value or range of values occurs. That sounds exactly like what we need because we're interested in how many students fall into each score range. So if we can create a frequency distribution of the scores, we can easily determine the proportions for each category.Option D is calculating the variance. Variance measures how spread out the scores are from the mean. While variance gives us information about the dispersion of scores, the question is specifically about proportions in different score ranges, not about the spread or variability of the scores. So variance might not be directly necessary here.So, to recap: We need to find out the proportions of students in various score ranges. If we have all the scores, the most straightforward way is to count how many students are in each range and then express those counts as proportions. That's essentially creating a frequency distribution. Option A, drawing a sample, might not be necessary if we already have data from all 800 students. Option B, calculating the average, while useful, doesn't directly give us the proportions we need. Option D, calculating variance, also doesn't address the specific question about proportions in different ranges.Therefore, the necessary work to be done is to conduct a frequency distribution, which is option C.

answer：Okay, so I've got this problem here: If (2(z + overline{z}) + 3(z - overline{z}) = 4 + 6i), then (z = left( right))And the options are:A: (1 - 2i)B: (1 + 2i)C: (1 + i)D: (1 - i)Alright, let's try to figure this out step by step. I'm a bit rusty on complex numbers, but I remember that (z) is a complex number, and (overline{z}) is its conjugate. So if (z = a + bi), then (overline{z} = a - bi), right?So, maybe I should start by expressing (z) and (overline{z}) in terms of (a) and (b). Let me write that down:Let (z = a + bi), so (overline{z} = a - bi).Now, let's substitute these into the equation given:(2(z + overline{z}) + 3(z - overline{z}) = 4 + 6i)Substituting (z) and (overline{z}):(2((a + bi) + (a - bi)) + 3((a + bi) - (a - bi)) = 4 + 6i)Hmm, let's simplify the expressions inside the parentheses first.For (z + overline{z}):((a + bi) + (a - bi) = a + a + bi - bi = 2a)And for (z - overline{z}):((a + bi) - (a - bi) = a - a + bi + bi = 2bi)So now, substitute these back into the equation:(2(2a) + 3(2bi) = 4 + 6i)Simplify the left side:(4a + 6bi = 4 + 6i)Okay, so now we have an equation where the left side is (4a + 6bi) and the right side is (4 + 6i). Since two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal, we can set up a system of equations.So, equating the real parts:(4a = 4)And equating the imaginary parts:(6b = 6)Now, solve for (a) and (b):From the first equation:(4a = 4)Divide both sides by 4:(a = 1)From the second equation:(6b = 6)Divide both sides by 6:(b = 1)So, (a = 1) and (b = 1). Therefore, the complex number (z) is:(z = a + bi = 1 + 1i = 1 + i)Looking back at the options, that's option C: (1 + i).Wait, let me double-check my steps to make sure I didn't make a mistake.1. Expressed (z) as (a + bi) and (overline{z}) as (a - bi). That seems correct.2. Substituted into the equation: (2(z + overline{z}) + 3(z - overline{z})) Which became: (2(2a) + 3(2bi)) Simplified to: (4a + 6bi) That looks right.3. Set equal to (4 + 6i): (4a + 6bi = 4 + 6i) So, real parts: (4a = 4) => (a = 1) Imaginary parts: (6b = 6) => (b = 1) So, (z = 1 + i), which is option C.Hmm, seems solid. Maybe I can plug the value back into the original equation to verify.Let me compute (2(z + overline{z}) + 3(z - overline{z})) with (z = 1 + i).First, compute (z + overline{z}):(z + overline{z} = (1 + i) + (1 - i) = 2)Then, compute (z - overline{z}):(z - overline{z} = (1 + i) - (1 - i) = 2i)Now, plug into the expression:(2(z + overline{z}) + 3(z - overline{z}) = 2(2) + 3(2i) = 4 + 6i)Which matches the right side of the equation. So, yes, (z = 1 + i) is correct.Therefore, the answer is option C: (1 + i).**Final Answer**boxed{C}

answer：Okay, so I have this problem about an isosceles triangle where the altitude drawn to the base is h, and this altitude is twice its projection onto one of the legs. I need to find the area of the triangle. Hmm, let me try to visualize this.First, let me draw an isosceles triangle ABC where AB equals BC. Wait, no, actually, in an isosceles triangle, two sides are equal. So, maybe AB equals AC? Or BC equals AC? Hmm, actually, the base is AC, so the two equal sides would be AB and BC. So, triangle ABC with base AC and equal sides AB and BC. Then, the altitude from B to the base AC is h. Okay, so BK is the altitude, where K is the foot of the altitude on AC.Now, the problem says that this altitude h is twice its projection onto one of the legs. So, the projection of BK onto, say, BC is half of h. Let me denote the projection as BP, where P is the foot of the perpendicular from K onto BC. So, BP is the projection of BK onto BC, and BP = h/2.Wait, so BK is h, and its projection onto BC is h/2. That means that the angle between BK and BC is such that the projection is half the length of BK. Hmm, maybe I can use trigonometry here. If I consider the right triangle BKP, where angle at B is between BK and BP, then BP is adjacent side, BK is hypotenuse, so cosine of angle at B would be BP/BK, which is (h/2)/h = 1/2. So, the angle at B is 60 degrees because cos(60°) = 0.5.Okay, so angle KBC is 60 degrees. Now, since triangle ABC is isosceles with AB = BC, the base angles at A and C are equal. Wait, no, actually, in triangle ABC, if AB = BC, then the base is AC, and the base angles are at A and C. But wait, angle at B is 60 degrees, so the other two angles must add up to 120 degrees, and since they are equal, each is 60 degrees. Wait, that would make triangle ABC equilateral, but that can't be because then all sides would be equal, but the problem says it's isosceles. Hmm, maybe I made a mistake.Wait, no, if angle at B is 60 degrees, and the triangle is isosceles with AB = BC, then angles at A and C are equal. So, angle at A = angle at C = (180 - 60)/2 = 60 degrees. So, actually, triangle ABC is equilateral. But the problem says it's isosceles, so maybe I misinterpreted the sides.Wait, perhaps AB = AC instead. Let me clarify. If the triangle is isosceles with base AC, then the equal sides are AB and CB. So, AB = CB. Then, the altitude BK is drawn from B to AC, splitting AC into two equal parts, AK and KC. So, AK = KC.Now, the projection of BK onto BC is BP = h/2. So, in triangle BKP, which is a right triangle, angle at B is 60 degrees, as we found earlier. So, in triangle BKP, BK is h, BP is h/2, and angle at B is 60 degrees. So, the side opposite to 60 degrees is KP, which would be h * sin(60°) = h * (√3/2). So, KP = (h√3)/2.But KP is part of BC. Wait, no, KP is the projection on BC? Wait, no, BP is the projection, which is h/2, and KP is the other leg of the right triangle BKP, which is (h√3)/2. So, that's the length from K to P on BC.Wait, but in triangle ABC, BK is the altitude, so BK is perpendicular to AC. So, BK is h, and BP is the projection of BK onto BC, which is h/2. So, using trigonometry, we can find the length of BC.In triangle BKP, which is right-angled at P, we have BP = h/2, BK = h, and angle at B is 60 degrees. So, the length BC can be found using the fact that BP is the projection, so BC is the hypotenuse of triangle BPC, where PC is the other leg.Wait, no, actually, BP is the projection of BK onto BC, so BP = BK * cos(theta), where theta is the angle between BK and BC. So, BP = h * cos(theta) = h/2, so cos(theta) = 1/2, which means theta = 60 degrees.So, angle between BK and BC is 60 degrees. Therefore, in triangle BKC, which is right-angled at K, angle at B is 60 degrees, so angle at C is 30 degrees. Therefore, triangle BKC is a 30-60-90 triangle.In a 30-60-90 triangle, the sides are in the ratio 1 : √3 : 2. So, BK is opposite the 30-degree angle, so BK = h = (1/2) * BC. Therefore, BC = 2h.Wait, no, in triangle BKC, BK is opposite angle C, which is 30 degrees, so BK = (1/2) * BC. So, BC = 2h.But wait, BK is h, so BC = 2h. So, the length of BC is 2h.Now, since triangle ABC is isosceles with AB = BC = 2h, and base AC. The altitude BK is h, so the area of triangle ABC is (AC * BK)/2.But I need to find AC. Since BK is the altitude, it splits AC into two equal parts, AK and KC. So, AK = KC.In triangle BKC, which is right-angled at K, we have BK = h, BC = 2h, so KC can be found using Pythagoras: BC² = BK² + KC².So, (2h)² = h² + KC² => 4h² = h² + KC² => KC² = 3h² => KC = h√3.Therefore, AC = 2 * KC = 2h√3.So, the area of triangle ABC is (AC * BK)/2 = (2h√3 * h)/2 = (2h²√3)/2 = h²√3.Wait, that seems straightforward. Let me just double-check.We have triangle ABC, isosceles with AB = BC = 2h, base AC = 2h√3, altitude BK = h. So, area is (base * height)/2 = (2h√3 * h)/2 = h²√3. Yes, that seems correct.So, the area of the triangle is h squared times the square root of 3.

answer：Okay, so I need to find the maximum value of the function f(x) = -x⁴ + 2x² + 3. Hmm, I remember from my math class that to find maxima or minima of a function, I should look at its critical points. Critical points are where the derivative is zero or undefined. Since this function is a polynomial, its derivative will also be a polynomial, and polynomials are defined everywhere, so I just need to find where the derivative equals zero.First, let me find the derivative of f(x). The function is f(x) = -x⁴ + 2x² + 3. The derivative, f'(x), is found by applying the power rule to each term. The power rule says that the derivative of xⁿ is n*xⁿ⁻¹.So, the derivative of -x⁴ is -4x³, the derivative of 2x² is 4x, and the derivative of the constant term 3 is 0. Putting it all together, f'(x) = -4x³ + 4x.Now, I need to set this derivative equal to zero and solve for x to find the critical points. So, I have:-4x³ + 4x = 0I can factor out a common term from both parts of this equation. Both terms have a factor of 4x, so let's factor that out:4x(-x² + 1) = 0Wait, actually, looking again, it's -4x³ + 4x. So, factoring out 4x gives:4x(-x² + 1) = 0But actually, I think it's better to factor out a -4x to make the equation simpler. Let me try that:-4x³ + 4x = 0Factor out -4x:-4x(x² - 1) = 0So, now we have:-4x(x² - 1) = 0This gives us two factors: -4x and (x² - 1). Setting each factor equal to zero:1. -4x = 0 ⇒ x = 02. x² - 1 = 0 ⇒ x² = 1 ⇒ x = ±1So, the critical points are at x = 0, x = 1, and x = -1.Now, I need to determine whether each critical point is a maximum, minimum, or a saddle point. To do this, I can use the second derivative test. The second derivative test involves finding the second derivative of the function and evaluating it at each critical point.First, let's find the second derivative, f''(x). We already have the first derivative, f'(x) = -4x³ + 4x. Taking the derivative of f'(x):f''(x) = d/dx (-4x³ + 4x) = -12x² + 4So, f''(x) = -12x² + 4.Now, let's evaluate f''(x) at each critical point.1. At x = 0:f''(0) = -12*(0)² + 4 = 0 + 4 = 4Since f''(0) = 4 > 0, this means that the function is concave up at x = 0, so this point is a local minimum.2. At x = 1:f''(1) = -12*(1)² + 4 = -12 + 4 = -8Since f''(1) = -8 < 0, the function is concave down at x = 1, so this is a local maximum.3. At x = -1:f''(-1) = -12*(-1)² + 4 = -12*(1) + 4 = -12 + 4 = -8Similarly, f''(-1) = -8 < 0, so the function is concave down at x = -1, which is also a local maximum.So, we have two local maxima at x = 1 and x = -1, and a local minimum at x = 0.Now, to find the maximum value of the function, we need to evaluate f(x) at these critical points and see which one gives the highest value.Let's compute f(x) at each critical point.1. At x = 0:f(0) = -(0)⁴ + 2*(0)² + 3 = 0 + 0 + 3 = 32. At x = 1:f(1) = -(1)⁴ + 2*(1)² + 3 = -1 + 2 + 3 = 43. At x = -1:f(-1) = -(-1)⁴ + 2*(-1)² + 3 = -1 + 2 + 3 = 4So, f(0) = 3, f(1) = 4, and f(-1) = 4.Comparing these values, the maximum value is 4, which occurs at both x = 1 and x = -1.Wait, but I should also consider the behavior of the function as x approaches infinity or negative infinity to ensure that there isn't a higher value somewhere else. Since the leading term of the function is -x⁴, which is a negative coefficient for an even degree term, the function will tend to negative infinity as x approaches both positive and negative infinity. This means that the function doesn't have any global maxima beyond these critical points because it decreases without bound as x moves away from zero in either direction.Therefore, the highest value the function attains is indeed 4, at x = 1 and x = -1.So, the maximum value of the function f(x) = -x⁴ + 2x² + 3 is 4.

answer：First, determine the number of adult and children tickets. There are 10 people in total, with 4 children, so there are 6 adults.Calculate the cost for adult tickets:6 adults * 30/adult = 180Calculate the cost for children tickets (half price):4 children * (30/2) = 4 * 15 = 60Add the costs to find the total admission price before discounts:180 (adults) + 60 (children) = 240Apply the 20% discount on the total admission price:20% of 240 = 0.20 * 240 = 48Subtract the discount from the total admission price:240 - 48 = 192Add the cost of the soda:192 + 5 = 197Janet paid a total of 197 for everything.

answer：Alright, so I've got this problem here where I need to find the sum of all possible distinct values of (a^2 + b^2 + c^2) given that (a + b + c = 23) and (gcd(a, b) + gcd(b, c) + gcd(c, a) = 9). Hmm, okay, let's break this down step by step.First, I know that (a), (b), and (c) are positive integers. Their sum is 23, which is an odd number. That makes me think about the parity of (a), (b), and (c). Since the sum is odd, they can't all be even because the sum of three even numbers is even. So, either one of them is odd and the other two are even, or all three are odd.Let me consider the two cases separately.**Case 1: One odd and two even numbers.**Without loss of generality, let's assume (a) is odd, and (b) and (c) are even. Now, let's think about the gcds. The gcd of an odd and an even number is 1 because they don't share any common divisors other than 1. So, (gcd(a, b)) and (gcd(a, c)) would both be 1. On the other hand, (gcd(b, c)) would be at least 2 because both (b) and (c) are even. So, the sum of the gcds would be (1 + text{something} + 1), which is (2 + text{something}). But we are given that the sum is 9, which is odd. However, (2 + text{something}) would be even if (text{something}) is even, or odd if (text{something}) is odd. Wait, but (gcd(b, c)) is at least 2, so it's even or odd? Since both (b) and (c) are even, their gcd is also even. So, (gcd(b, c)) is even, meaning the sum (1 + text{even} + 1 = text{even} + 2), which is even. But 9 is odd, so this case is impossible. Therefore, Case 1 can't happen.**Case 2: All three numbers are odd.**Since all three are odd, their sum is odd, which matches the given condition. Now, all the gcds (gcd(a, b)), (gcd(b, c)), and (gcd(c, a)) must be odd because the gcd of two odd numbers is odd. So, the sum of three odd numbers is odd, which is consistent with the given sum of 9.Now, I need to find triples ((a, b, c)) such that (a + b + c = 23) and the sum of their pairwise gcds is 9. Let me think about possible combinations of gcds that add up to 9.Possible sets of gcds (since they are positive integers and odd):1. (1, 1, 7)2. (1, 3, 5)3. (3, 3, 3)Let me check each possibility.**Subcase 2.1: (gcd(a, b) = 1), (gcd(b, c) = 1), (gcd(c, a) = 7)**Here, (a) and (c) share a gcd of 7, so both (a) and (c) must be multiples of 7. Let me denote (a = 7k) and (c = 7m), where (k) and (m) are positive integers. Since (gcd(a, c) = 7), it means that (gcd(k, m) = 1).Now, (a + b + c = 7k + b + 7m = 23). So, (b = 23 - 7(k + m)). Since (b) must be a positive integer, (7(k + m) < 23), so (k + m leq 3) because (7*4 = 28 > 23).Possible values for (k + m):- (k + m = 1): Not possible because (k) and (m) are at least 1, so (k + m geq 2).- (k + m = 2): Then (b = 23 - 14 = 9).- (k + m = 3): Then (b = 23 - 21 = 2), but (b) must be odd because all numbers are odd. So, (b = 2) is invalid.So, only (k + m = 2) is possible. This gives (k = 1) and (m = 1) because (gcd(k, m) = 1). Therefore, (a = 7), (c = 7), and (b = 9).Let me check the gcds:- (gcd(7, 9) = 1)- (gcd(9, 7) = 1)- (gcd(7, 7) = 7)Yes, that adds up to 1 + 1 + 7 = 9. So, this is a valid solution. Now, calculate (a^2 + b^2 + c^2 = 7^2 + 9^2 + 7^2 = 49 + 81 + 49 = 179).**Subcase 2.2: (gcd(a, b) = 1), (gcd(b, c) = 3), (gcd(c, a) = 5)**Here, (a) and (c) share a gcd of 5, so (a = 5k) and (c = 5m) with (gcd(k, m) = 1). Also, (b) and (c) share a gcd of 3, so (b = 3n) and (c = 3p) with (gcd(n, p) = 1). Wait, but (c) is already expressed as (5m), so (c) must be a multiple of both 5 and 3, meaning (c) is a multiple of 15. Let me denote (c = 15q), where (q) is a positive integer.So, (a = 5k), (c = 15q), and (b = 3n). Now, let's write the sum:(a + b + c = 5k + 3n + 15q = 23).We need to find positive integers (k), (n), (q) such that this equation holds, and also satisfy the gcd conditions.First, let's note that (c = 15q). Since (c) is positive, (q) must be at least 1. Let's try (q = 1):Then, (c = 15), so the equation becomes (5k + 3n + 15 = 23), which simplifies to (5k + 3n = 8).Now, we need to solve (5k + 3n = 8) for positive integers (k) and (n).Let me try possible values for (k):- (k = 1): (5 + 3n = 8) → (3n = 3) → (n = 1). So, (k = 1), (n = 1).- (k = 2): (10 + 3n = 8) → Negative (n), invalid. So, the only solution is (k = 1), (n = 1), (q = 1). Therefore, (a = 5), (b = 3), (c = 15).Let me check the gcds:- (gcd(5, 3) = 1)- (gcd(3, 15) = 3)- (gcd(15, 5) = 5)Yes, that adds up to 1 + 3 + 5 = 9. So, this is another valid solution. Now, calculate (a^2 + b^2 + c^2 = 5^2 + 3^2 + 15^2 = 25 + 9 + 225 = 259).**Subcase 2.3: (gcd(a, b) = 3), (gcd(b, c) = 3), (gcd(c, a) = 3)**Here, all pairwise gcds are 3. So, (a), (b), and (c) are all multiples of 3. Let me denote (a = 3k), (b = 3m), (c = 3n), where (k), (m), (n) are positive integers with (gcd(k, m) = gcd(m, n) = gcd(n, k) = 1).Now, the sum becomes (3k + 3m + 3n = 23), which simplifies to (k + m + n = frac{23}{3}). But 23 is not divisible by 3, so (k + m + n) is not an integer, which is impossible because (k), (m), (n) are integers. Therefore, this case is impossible.So, the only valid solutions are from Subcases 2.1 and 2.2, giving (a^2 + b^2 + c^2) as 179 and 259, respectively.Now, the question asks for the sum of all possible distinct values of (a^2 + b^2 + c^2). So, I need to add 179 and 259.179 + 259 = 438.Wait, but let me double-check if there are any other possible combinations of gcds that add up to 9. For example, could we have (gcd(a, b) = 1), (gcd(b, c) = 5), (gcd(c, a) = 3)? That would still sum to 9, but is that a different case?Actually, in Subcase 2.2, I considered (gcd(a, b) = 1), (gcd(b, c) = 3), (gcd(c, a) = 5). If I permute the gcds, it might lead to different triples, but in this case, since (a), (b), and (c) are just labels, the resulting (a^2 + b^2 + c^2) would be the same. So, it's not a distinct value.Similarly, other permutations of the same gcds would result in the same sum of squares. Therefore, I don't need to consider them separately.So, I think I've covered all possible cases, and the sum of the distinct values is indeed 438.