💡First, I need to determine the price of one table and one chair based on the given information.I know that the total cost for 2 tables and 3 chairs is 99 yuan, and the price of one table is equal to the price of 4 chairs.Let me denote the price of one chair as C yuan. Since one table costs the same as 4 chairs, the price of one table would be 4C yuan.Now, I can set up an equation based on the total cost:2 tables + 3 chairs = 99 yuan2 * (4C) + 3C = 99Simplifying the equation:8C + 3C = 9911C = 99To find the value of C, I divide both sides by 11:C = 99 / 11C = 9So, the price of one chair is 9 yuan.Now, to find the price of one table:Price of one table = 4C = 4 * 9 = 36 yuanTherefore, one table costs 36 yuan, and one chair costs 9 yuan.

💡Alright, so I've got this problem about 18 scientists at a conference, where 10 of them know a sensational piece of news. During the coffee break, they form random pairs, and if one person in the pair knows the news, they share it with the other. I need to find the probability that after the coffee break, the number of scientists who know the news will be 13 (part a), 14 (part b), and then find the expected value of the number of scientists who know the news (part c).Starting with part a: Probability that the number of scientists who know the news will be 13.Hmm, okay. So initially, there are 10 scientists who know the news. After pairing, some of the scientists who didn't know the news might have learned it. The question is, what's the probability that exactly 13 scientists know the news after the coffee break.Wait a second, 13 is an odd number. But the total number of scientists is 18, which is even. When they form pairs, each pair consists of 2 people. So, if someone shares the news in a pair, both people in that pair will know the news after the break, right? Because if one knows and the other doesn't, the one who doesn't will learn it.So, if we think about the number of people who know the news after the break, it has to be an even number because each pair contributes either 0, 1, or 2 to the total count. Wait, no, actually, if a pair has one knower and one non-knower, after sharing, both will know. So, each such pair will add 1 more knower (since the non-knower becomes a knower). But the total number of knowers after the break must be even because each pair can only add 0 or 1 to the total count, but since we started with 10, which is even, adding an even number of 1s (since each pair can only add 0 or 1) will keep the total even.Wait, no, that might not be correct. Let me think again. Each pair can either:1. Both know the news: no change.2. One knows, the other doesn't: the non-knower becomes a knower, so the total increases by 1.3. Both don't know: no change.So, each pair can potentially add 0 or 1 to the total number of knowers. Since we start with 10, which is even, and each pair can add 0 or 1, the total number after the break can be 10, 11, 12, ..., up to 18. But wait, 18 is even, 10 is even, so adding an even number of 1s (since each pair can only add 0 or 1) would result in an even total. But 13 is odd, so it's impossible.Wait, that doesn't make sense because 10 plus an odd number would be odd, but 10 is even, so adding an odd number of 1s would make it odd. But each pair can only add 0 or 1, so the total number of added knowers is equal to the number of pairs where one knew and the other didn't. Since there are 9 pairs (because 18 scientists form 9 pairs), the number of added knowers can be from 0 to 9. So, the total number of knowers after the break can be 10 + k, where k is the number of pairs where one knew and the other didn't, and k can be from 0 to 9.But wait, 10 + k must be even or odd? 10 is even, so 10 + k is even if k is even, and odd if k is odd. So, the total number of knowers after the break can be even or odd, depending on k. Therefore, 13 is possible if k=3, which is odd.Wait, but earlier I thought that the total number of knowers must be even because each pair can only add 0 or 1, but that's not necessarily the case because the number of pairs that add 1 can be odd or even. So, 13 is possible.Wait, no, let me clarify. Each pair that has one knower and one non-knower will result in the non-knower learning the news, so that pair will contribute 1 to the total number of knowers. So, if there are k such pairs, the total number of knowers will be 10 + k. Since k can be from 0 to 9, the total can be from 10 to 19, but since there are only 18 scientists, the maximum is 18. So, 10 + k can be 10, 11, ..., 18. Therefore, 13 is possible if k=3.But wait, the problem is that when you form pairs, the number of pairs that can have one knower and one non-knower is limited by the number of knowers and non-knowers. Initially, there are 10 knowers and 8 non-knowers. So, the maximum number of pairs that can have one knower and one non-knower is 8, because there are only 8 non-knowers. So, k can be from 0 to 8, not 9. Therefore, the total number of knowers after the break can be from 10 to 18, but only in steps of 1, but the maximum is 18, which is 10 + 8.Wait, but 10 + 8 is 18, which is the total number of scientists. So, the possible number of knowers after the break is 10, 11, 12, ..., 18. So, 13 is possible.But wait, earlier I thought that the total number of knowers must be even because each pair can only add 0 or 1, but that's not correct because the number of pairs that add 1 can be odd or even, so the total can be even or odd. Therefore, 13 is possible.Wait, but let me think again. Each pair that adds 1 to the total number of knowers is a pair where one knew and one didn't. So, the number of such pairs is k, which can be from 0 to 8, as there are only 8 non-knowers. Therefore, the total number of knowers after the break is 10 + k, where k is from 0 to 8. So, 10 + 3 = 13 is possible.But wait, the problem is that when you pair up the scientists, the number of pairs that can have one knower and one non-knower is limited by the number of non-knowers, which is 8. So, the maximum number of such pairs is 8, meaning k can be from 0 to 8. Therefore, the total number of knowers after the break can be from 10 to 18, but only in steps of 1, and 13 is within that range.Wait, but I'm getting confused because earlier I thought that the total number of knowers must be even, but that's not necessarily the case. Let me try to think differently.Each pair can either:1. Both know: no change.2. One knows, one doesn't: the non-knower learns, so total increases by 1.3. Both don't know: no change.So, the total number of knowers after the break is 10 + k, where k is the number of pairs where one knew and one didn't. Since k can be from 0 to 8, the total can be from 10 to 18. Therefore, 13 is possible if k=3.But wait, the problem is that when you pair up the scientists, the number of pairs that can have one knower and one non-knower is limited by the number of non-knowers, which is 8. So, the maximum number of such pairs is 8, meaning k can be from 0 to 8. Therefore, the total number of knowers after the break can be from 10 to 18, but only in steps of 1, and 13 is within that range.But wait, the problem is that when you pair up the scientists, the number of pairs that can have one knower and one non-knower is limited by the number of non-knowers, which is 8. So, the maximum number of such pairs is 8, meaning k can be from 0 to 8. Therefore, the total number of knowers after the break can be from 10 to 18, but only in steps of 1, and 13 is within that range.Wait, but I'm still not sure. Let me think about the parity. Initially, there are 10 knowers, which is even. Each pair that adds 1 to the total number of knowers will change the parity. So, if k is odd, the total will be odd; if k is even, the total will be even. Since 13 is odd, k must be odd. So, the number of pairs that add 1 must be odd.But the number of such pairs is limited by the number of non-knowers, which is 8. So, the number of pairs that can add 1 is from 0 to 8. Therefore, the number of pairs that add 1 can be odd or even, so 13 is possible.Wait, but I'm still not sure. Let me think about the total number of knowers after the break. It's 10 + k, where k is the number of pairs that added 1. Since k can be from 0 to 8, the total can be from 10 to 18. Therefore, 13 is possible.But wait, the problem is that when you pair up the scientists, the number of pairs that can have one knower and one non-knower is limited by the number of non-knowers, which is 8. So, the maximum number of such pairs is 8, meaning k can be from 0 to 8. Therefore, the total number of knowers after the break can be from 10 to 18, but only in steps of 1, and 13 is within that range.But wait, I'm going in circles. Let me try to think about the probability.The probability that exactly 13 scientists know the news after the break is the probability that exactly 3 pairs had one knower and one non-knower, and the rest of the pairs either had both knowers or both non-knowers.So, to compute this probability, I need to count the number of ways to form 9 pairs such that exactly 3 pairs have one knower and one non-knower, and the remaining 6 pairs are either both knowers or both non-knowers.But wait, the total number of non-knowers is 8, so if 3 pairs have one non-knower, that uses up 3 non-knowers, leaving 5 non-knowers. These 5 non-knowers need to be paired among themselves, but 5 is odd, which is impossible because you can't pair an odd number of people. Therefore, it's impossible to have exactly 3 pairs with one knower and one non-knower because that would leave an odd number of non-knowers unpaired.Wait, that's a key point. The number of non-knowers is 8, which is even. If we have k pairs with one non-knower, then k must be even because each such pair uses one non-knower, and the remaining non-knowers must be paired among themselves, which requires an even number.Therefore, k must be even because 8 - k must be even, so k must be even. Therefore, the number of pairs with one knower and one non-knower must be even. Therefore, the total number of knowers after the break must be 10 + k, where k is even. Therefore, 13 is impossible because 13 - 10 = 3, which is odd, and k must be even.Therefore, the probability that the number of scientists who know the news will be 13 is 0.Wait, that makes sense. Because the number of non-knowers is 8, which is even. If you have k pairs with one non-knower, then k must be even because 8 - k must be even (since you need to pair the remaining non-knowers). Therefore, k must be even, so the total number of knowers after the break must be 10 + even number, which is even. Therefore, 13 is odd, so it's impossible.Therefore, the probability is 0.Okay, that seems to make sense now. So, part a) is 0.Now, moving on to part b) Find the probability that after the coffee break, the number of scientists who know the news will be 14.So, 14 is even, which is possible because 14 - 10 = 4, which is even. So, k=4.Therefore, we need to find the probability that exactly 4 pairs have one knower and one non-knower, and the remaining 5 pairs are either both knowers or both non-knowers.So, to compute this probability, we need to count the number of ways to form 9 pairs such that exactly 4 pairs have one knower and one non-knower, and the remaining 5 pairs are either both knowers or both non-knowers.First, let's compute the total number of ways to pair up 18 scientists. This is given by the double factorial (18-1)!! = 17!! = 17 × 15 × 13 × ... × 1. But it's easier to think in terms of combinations.The total number of ways to pair up 18 scientists is (18)! / (2^9 × 9!). This is because we can arrange all 18 scientists in a line, then pair the first two, next two, etc., but since the order within each pair doesn't matter, we divide by 2^9, and since the order of the pairs themselves doesn't matter, we divide by 9!.So, total number of pairings = 18! / (2^9 × 9!).Now, the number of favorable pairings where exactly 4 pairs have one knower and one non-knower.To compute this, we can think of it as:1. Choose 4 non-knowers out of 8 to be paired with knowers. The number of ways to choose these 4 non-knowers is C(8,4).2. For each of these 4 non-knowers, we need to pair them with a knower. There are 10 knowers, so the number of ways to pair these 4 non-knowers with 4 knowers is 10 × 9 × 8 × 7 (since for the first non-knower, we have 10 choices, for the second, 9, etc.).But wait, actually, since the pairs are unordered, we need to divide by 4! to account for the order in which we pair them. So, the number of ways is C(10,4) × 4!.Wait, no, actually, it's better to think of it as:- Choose 4 knowers out of 10 to pair with the 4 non-knowers. The number of ways is C(10,4).- Then, pair each chosen knower with a non-knower. Since the non-knowers are already chosen, the number of ways to pair them is 4! (since it's a bijection between the 4 knowers and 4 non-knowers).So, the number of ways is C(10,4) × 4!.3. Now, after pairing these 4 knowers with 4 non-knowers, we have 10 - 4 = 6 knowers left and 8 - 4 = 4 non-knowers left.4. Now, we need to pair the remaining 6 knowers and 4 non-knowers such that the remaining pairs are either both knowers or both non-knowers.But wait, we have 6 knowers and 4 non-knowers left, which is a total of 10 people. But we need to form 5 pairs (since 18 - 8 = 10, and 10/2=5). So, we need to pair the remaining 6 knowers and 4 non-knowers into 5 pairs, where each pair is either both knowers or both non-knowers.But wait, we have 6 knowers and 4 non-knowers left. So, the number of ways to pair them into 5 pairs, each of which is either both knowers or both non-knowers.This is equivalent to partitioning the 6 knowers into pairs and the 4 non-knowers into pairs, and then combining these pairs.The number of ways to pair 6 knowers is (6-1)!! = 5!! = 15.Similarly, the number of ways to pair 4 non-knowers is (4-1)!! = 3!! = 3.Therefore, the total number of ways to pair the remaining 6 knowers and 4 non-knowers is 15 × 3 = 45.But wait, actually, the number of ways to pair 6 knowers is 6! / (2^3 × 3!) = 15, and the number of ways to pair 4 non-knowers is 4! / (2^2 × 2!) = 3. So, yes, 15 × 3 = 45.Therefore, the total number of favorable pairings is:C(8,4) × C(10,4) × 4! × 15 × 3.Wait, let's break it down step by step:1. Choose 4 non-knowers out of 8: C(8,4).2. Choose 4 knowers out of 10: C(10,4).3. Pair each of these 4 non-knowers with the 4 knowers: 4! ways.4. Pair the remaining 6 knowers: 6! / (2^3 × 3!) = 15.5. Pair the remaining 4 non-knowers: 4! / (2^2 × 2!) = 3.Therefore, the total number of favorable pairings is:C(8,4) × C(10,4) × 4! × 15 × 3.Now, let's compute this:C(8,4) = 70.C(10,4) = 210.4! = 24.15 × 3 = 45.So, total favorable pairings = 70 × 210 × 24 × 45.Let me compute this step by step:First, 70 × 210 = 14,700.Then, 14,700 × 24 = 352,800.Then, 352,800 × 45 = 15,876,000.Wait, that seems too large. Let me check.Wait, actually, the total number of pairings is 18! / (2^9 × 9!) = 18! / (512 × 362880). Let me compute that.But perhaps it's better to compute the probability as:Probability = (Number of favorable pairings) / (Total number of pairings).But the total number of pairings is 18! / (2^9 × 9!) = 18! / (512 × 362880).But 18! is a huge number, so maybe we can simplify the ratio.Alternatively, we can think in terms of probabilities step by step.But perhaps there's a better way.Wait, another approach is to consider that each non-knower has a certain probability of being paired with a knower.But since the pairings are random, the probability that a specific non-knower is paired with a knower is 10/17, because there are 17 other scientists, 10 of whom are knowers.But since we have 8 non-knowers, the expected number of non-knowers paired with knowers is 8 × (10/17) = 80/17 ≈ 4.705.But we need the probability that exactly 4 non-knowers are paired with knowers, which would result in 10 + 4 = 14 knowers.So, the probability that exactly 4 out of 8 non-knowers are paired with knowers.This is similar to a hypergeometric distribution, where we have 10 knowers and 8 non-knowers, and we want to choose 4 non-knowers to pair with knowers.Wait, but the pairing is without replacement, so it's similar to hypergeometric.The probability is:C(10,4) × C(8,4) / C(18,8).Wait, no, because we're pairing, not choosing subsets.Wait, perhaps it's better to think of it as:The number of ways to choose 4 pairs where each pair consists of one knower and one non-knower, and the remaining pairs are either both knowers or both non-knowers.But this is similar to what I did earlier.Alternatively, the probability can be computed as:[ C(8,4) × C(10,4) × 4! × (6! / (2^3 × 3!)) × (4! / (2^2 × 2!)) ] / (18! / (2^9 × 9!)).But this seems complicated.Alternatively, we can use the concept of derangements or something similar, but I'm not sure.Wait, perhaps it's better to think in terms of the number of ways to pair the scientists such that exactly 4 non-knowers are paired with knowers.So, the number of such pairings is:C(8,4) × C(10,4) × 4! × (number of ways to pair the remaining 6 knowers) × (number of ways to pair the remaining 4 non-knowers).As I computed earlier, this is 70 × 210 × 24 × 15 × 3 = 15,876,000.Now, the total number of pairings is 18! / (2^9 × 9!) = 18! / (512 × 362880).Let me compute 18!:18! = 6,402,373,705,728,000.Now, 2^9 = 512.9! = 362,880.So, total pairings = 6,402,373,705,728,000 / (512 × 362,880).First, compute 512 × 362,880 = 512 × 362,880.Compute 362,880 × 500 = 181,440,000.362,880 × 12 = 4,354,560.So, total is 181,440,000 + 4,354,560 = 185,794,560.Therefore, total pairings = 6,402,373,705,728,000 / 185,794,560.Let me compute this division.First, let's see how many times 185,794,560 fits into 6,402,373,705,728,000.But this is a huge number, so perhaps we can simplify.Alternatively, we can compute the probability as:Probability = (Number of favorable pairings) / (Total number of pairings) = 15,876,000 / (18! / (2^9 × 9!)).But 18! / (2^9 × 9!) = 18 × 17 × 16 × ... × 1 / (2^9 × 9!) = 18! / (2^9 × 9!) = 18!! / 9!.Wait, 18!! is the double factorial, which is 18 × 16 × 14 × ... × 2.But 18!! = 18 × 16 × 14 × 12 × 10 × 8 × 6 × 4 × 2.But 18! = 18 × 17 × 16 × 15 × ... × 1.So, 18! / (2^9 × 9!) = (18 × 17 × 16 × ... × 1) / (2^9 × 9!) = (18 × 16 × 14 × ... × 2) × (17 × 15 × ... × 1) / (2^9 × 9!) = (18!!) × (17 × 15 × ... × 1) / (2^9 × 9!).But this seems complicated.Alternatively, perhaps we can use the formula for the number of ways to pair two groups.Wait, another approach: the number of ways to pair 10 knowers and 8 non-knowers such that exactly 4 pairs are mixed (one knower, one non-knower).This is similar to a bipartite matching problem.The number of such pairings is:C(10,4) × C(8,4) × 4! × (number of ways to pair the remaining 6 knowers) × (number of ways to pair the remaining 4 non-knowers).As I computed earlier, this is 210 × 70 × 24 × 15 × 3 = 15,876,000.Now, the total number of pairings is 18! / (2^9 × 9!) = 6,402,373,705,728,000 / (512 × 362,880) = 6,402,373,705,728,000 / 185,794,560 ≈ 34,459,425.Wait, let me compute 6,402,373,705,728,000 ÷ 185,794,560.First, divide numerator and denominator by 1000: 6,402,373,705,728 ÷ 185,794.56.But this is still complicated.Alternatively, let's use the fact that 18! / (2^9 × 9!) = 18 × 17 × 16 × 15 × 14 × 13 × 12 × 11 × 10 × 9! / (2^9 × 9!) = (18 × 17 × 16 × 15 × 14 × 13 × 12 × 11 × 10) / (2^9).Compute numerator: 18 × 17 × 16 × 15 × 14 × 13 × 12 × 11 × 10.Let me compute step by step:18 × 17 = 306.306 × 16 = 4,896.4,896 × 15 = 73,440.73,440 × 14 = 1,028,160.1,028,160 × 13 = 13,365, 13,365 × 1000 = 13,365,000? Wait, no, 1,028,160 × 13 = 13,365, 13,365 × 1000 = 13,365,000? Wait, no, 1,028,160 × 10 = 10,281,600, plus 1,028,160 × 3 = 3,084,480, so total 13,366,080.13,366,080 × 12 = 160,392,960.160,392,960 × 11 = 1,764,322,560.1,764,322,560 × 10 = 17,643,225,600.So, numerator is 17,643,225,600.Denominator is 2^9 = 512.So, total pairings = 17,643,225,600 / 512 ≈ 34,459,425.Therefore, total pairings ≈ 34,459,425.Now, the number of favorable pairings is 15,876,000.Therefore, probability = 15,876,000 / 34,459,425 ≈ 0.4605.But let's compute it exactly.15,876,000 ÷ 34,459,425.Divide numerator and denominator by 15: 1,058,400 ÷ 2,297,295.Divide numerator and denominator by 3: 352,800 ÷ 765,765.Divide numerator and denominator by 15: 23,520 ÷ 51,051.Divide numerator and denominator by 3: 7,840 ÷ 17,017.Wait, 7,840 ÷ 17,017.Let me compute this division.17,017 × 0.46 = 17,017 × 0.4 = 6,806.8; 17,017 × 0.06 = 1,021.02; total ≈ 7,827.82.Which is close to 7,840.So, approximately 0.46.But let's see if we can simplify the fraction 15,876,000 / 34,459,425.Divide numerator and denominator by 15: 1,058,400 / 2,297,295.Divide by 3: 352,800 / 765,765.Divide by 15: 23,520 / 51,051.Divide by 3: 7,840 / 17,017.Now, 7,840 and 17,017.Let's see if they have a common factor.17,017 ÷ 7 = 2,431.7,840 ÷ 7 = 1,120.So, 7,840 / 17,017 = 1,120 / 2,431.Therefore, the probability is 1,120 / 2,431.So, the probability is 1120/2431.Therefore, part b) is 1120/2431.Now, part c) Let X be the number of scientists who know the sensational news after the coffee break. Find the expected value of X.So, we need to find E[X].We can model this using linearity of expectation.Let’s define indicator variables for each of the 8 non-knowers. Let I_k be 1 if the k-th non-knower learns the news, and 0 otherwise.Then, X = 10 + I_1 + I_2 + ... + I_8.Therefore, E[X] = 10 + E[I_1] + E[I_2] + ... + E[I_8].Now, we need to find E[I_k] for each k, which is the probability that the k-th non-knower is paired with a knower.Since the pairings are random, the probability that a specific non-knower is paired with a knower is equal to the number of knowers divided by the total number of other scientists.There are 10 knowers and 17 other scientists (since the non-knower can't pair with themselves).Therefore, P(I_k = 1) = 10/17.Therefore, E[I_k] = 10/17 for each k.Therefore, E[X] = 10 + 8 × (10/17) = 10 + 80/17 = (170 + 80)/17 = 250/17 ≈ 14.70588.So, the expected value is 250/17.But let me double-check.Yes, since each non-knower has a 10/17 chance of being paired with a knower, and there are 8 non-knowers, the expected number of new knowers is 8 × (10/17) = 80/17.Adding to the initial 10 knowers, we get 10 + 80/17 = (170 + 80)/17 = 250/17.Therefore, E[X] = 250/17.So, summarizing:a) 0b) 1120/2431c) 250/17

💡Okay, so I have this problem where I need to find the real number ( a ) such that the complex number ( dfrac{2+ai}{1-i} + dfrac{1}{1+i} ) is purely imaginary. The options given are A: 3, B: -3, C: 2, D: -2. Hmm, let me try to figure this out step by step.First, I remember that a complex number is purely imaginary if its real part is zero. So, I need to simplify the given expression and then set the real part to zero to solve for ( a ).Let me start by simplifying each fraction separately. I think the best way to do this is by multiplying the numerator and denominator by the conjugate of the denominator to rationalize them.Starting with the first fraction: ( dfrac{2+ai}{1-i} ). The conjugate of ( 1 - i ) is ( 1 + i ). So, I'll multiply both the numerator and denominator by ( 1 + i ):[dfrac{2+ai}{1-i} times dfrac{1+i}{1+i} = dfrac{(2+ai)(1+i)}{(1-i)(1+i)}]Let me compute the denominator first because that's straightforward. ( (1 - i)(1 + i) ) is a difference of squares, so it's ( 1^2 - i^2 ). Since ( i^2 = -1 ), this becomes ( 1 - (-1) = 2 ).Now, the numerator: ( (2 + ai)(1 + i) ). I'll use the distributive property (FOIL) to expand this:[2 times 1 + 2 times i + ai times 1 + ai times i = 2 + 2i + ai + ai^2]Again, ( i^2 = -1 ), so ( ai^2 = -a ). Substituting that back in:[2 + 2i + ai - a = (2 - a) + (2 + a)i]So, the first fraction simplifies to:[dfrac{(2 - a) + (2 + a)i}{2} = dfrac{2 - a}{2} + dfrac{2 + a}{2}i]Alright, now moving on to the second fraction: ( dfrac{1}{1+i} ). Similarly, I'll multiply numerator and denominator by the conjugate of ( 1 + i ), which is ( 1 - i ):[dfrac{1}{1+i} times dfrac{1-i}{1-i} = dfrac{1 times (1 - i)}{(1 + i)(1 - i)}]Again, the denominator is a difference of squares: ( 1 - i^2 = 1 - (-1) = 2 ).The numerator is simply ( 1 - i ). So, the second fraction simplifies to:[dfrac{1 - i}{2} = dfrac{1}{2} - dfrac{1}{2}i]Now, I can add the two simplified fractions together:First fraction: ( dfrac{2 - a}{2} + dfrac{2 + a}{2}i )Second fraction: ( dfrac{1}{2} - dfrac{1}{2}i )Adding them together:Real parts: ( dfrac{2 - a}{2} + dfrac{1}{2} = dfrac{2 - a + 1}{2} = dfrac{3 - a}{2} )Imaginary parts: ( dfrac{2 + a}{2}i - dfrac{1}{2}i = dfrac{2 + a - 1}{2}i = dfrac{1 + a}{2}i )So, the entire expression simplifies to:[dfrac{3 - a}{2} + dfrac{1 + a}{2}i]Since the problem states that this complex number is purely imaginary, the real part must be zero. Therefore, I set the real part equal to zero:[dfrac{3 - a}{2} = 0]Solving for ( a ):Multiply both sides by 2:[3 - a = 0]Then,[a = 3]Wait, but let me double-check to make sure I didn't make any mistakes. I'll substitute ( a = 3 ) back into the original expression to see if it indeed results in a purely imaginary number.Original expression:[dfrac{2 + 3i}{1 - i} + dfrac{1}{1 + i}]Let me compute each part again with ( a = 3 ).First fraction:[dfrac{2 + 3i}{1 - i} times dfrac{1 + i}{1 + i} = dfrac{(2 + 3i)(1 + i)}{2}]Expanding the numerator:[2(1) + 2(i) + 3i(1) + 3i(i) = 2 + 2i + 3i + 3i^2 = 2 + 5i + 3(-1) = 2 + 5i - 3 = -1 + 5i]So, the first fraction becomes:[dfrac{-1 + 5i}{2} = -dfrac{1}{2} + dfrac{5}{2}i]Second fraction:[dfrac{1}{1 + i} times dfrac{1 - i}{1 - i} = dfrac{1 - i}{2} = dfrac{1}{2} - dfrac{1}{2}i]Adding them together:Real parts: ( -dfrac{1}{2} + dfrac{1}{2} = 0 )Imaginary parts: ( dfrac{5}{2}i - dfrac{1}{2}i = dfrac{4}{2}i = 2i )So, the result is ( 0 + 2i ), which is purely imaginary. That checks out.Wait, but let me make sure that the imaginary part isn't zero, just to be thorough. If ( a = 3 ), then the imaginary part is ( dfrac{1 + a}{2} = dfrac{1 + 3}{2} = 2 ), which is not zero. So, yes, it's purely imaginary.Is there any other possible value of ( a ) that could make the real part zero? Let me see. The equation was ( 3 - a = 0 ), so ( a = 3 ) is the only solution. Therefore, the answer should be A: 3.But hold on, let me check the options again. The options are A: 3, B: -3, C: 2, D: -2. So, 3 is an option, so that should be correct.Wait, but just to be absolutely sure, let me try plugging in another value, say ( a = -3 ), to see what happens. Maybe I made a mistake in my calculation.If ( a = -3 ), then the expression becomes:[dfrac{2 - 3i}{1 - i} + dfrac{1}{1 + i}]First fraction:[dfrac{2 - 3i}{1 - i} times dfrac{1 + i}{1 + i} = dfrac{(2 - 3i)(1 + i)}{2}]Expanding numerator:[2(1) + 2(i) - 3i(1) - 3i(i) = 2 + 2i - 3i - 3i^2 = 2 - i - 3(-1) = 2 - i + 3 = 5 - i]So, the first fraction is ( dfrac{5 - i}{2} = dfrac{5}{2} - dfrac{1}{2}i )Second fraction remains ( dfrac{1}{2} - dfrac{1}{2}i )Adding them together:Real parts: ( dfrac{5}{2} + dfrac{1}{2} = 3 )Imaginary parts: ( -dfrac{1}{2}i - dfrac{1}{2}i = -i )So, the result is ( 3 - i ), which is not purely imaginary. So, ( a = -3 ) doesn't work.Similarly, if I try ( a = 2 ):First fraction:[dfrac{2 + 2i}{1 - i} times dfrac{1 + i}{1 + i} = dfrac{(2 + 2i)(1 + i)}{2}]Expanding numerator:[2(1) + 2(i) + 2i(1) + 2i(i) = 2 + 2i + 2i + 2i^2 = 2 + 4i + 2(-1) = 2 + 4i - 2 = 4i]So, first fraction is ( dfrac{4i}{2} = 2i )Second fraction is ( dfrac{1}{2} - dfrac{1}{2}i )Adding them together:Real parts: ( 0 + dfrac{1}{2} = dfrac{1}{2} )Imaginary parts: ( 2i - dfrac{1}{2}i = dfrac{3}{2}i )So, the result is ( dfrac{1}{2} + dfrac{3}{2}i ), which is not purely imaginary.Similarly, for ( a = -2 ):First fraction:[dfrac{2 - 2i}{1 - i} times dfrac{1 + i}{1 + i} = dfrac{(2 - 2i)(1 + i)}{2}]Expanding numerator:[2(1) + 2(i) - 2i(1) - 2i(i) = 2 + 2i - 2i - 2i^2 = 2 + 0i - 2(-1) = 2 + 2 = 4]So, first fraction is ( dfrac{4}{2} = 2 )Second fraction is ( dfrac{1}{2} - dfrac{1}{2}i )Adding them together:Real parts: ( 2 + dfrac{1}{2} = dfrac{5}{2} )Imaginary parts: ( 0 - dfrac{1}{2}i = -dfrac{1}{2}i )So, the result is ( dfrac{5}{2} - dfrac{1}{2}i ), which is not purely imaginary.Therefore, the only value that works is ( a = 3 ), which is option A.I think I've covered all possibilities and double-checked my work, so I'm confident that the answer is A: 3.

💡First, I need to find the length of one side of the cube. Since the volume of a cube is given by the formula ( text{side}^3 ), I can set up the equation ( text{side}^3 = 1728 ) cc. To find the side length, I take the cube root of 1728, which gives me ( text{side} = 12 ) cm.Next, I calculate the surface area of the cube. The surface area of a cube is given by the formula ( 6 times text{side}^2 ). Plugging in the side length of 12 cm, the surface area becomes ( 6 times 12^2 = 6 times 144 = 864 ) cm².Therefore, the surface area of the cube is 864 square centimeters.

💡Alright, so I have this function here: ( f(x) = x^2 ln(-x + sqrt{x^2 + 1}) + 1 ). The problem says that ( f(a) = 11 ), and I need to find ( f(-a) ). Hmm, okay. Let me try to figure this out step by step.First, let me write down what I know. The function is ( f(x) = x^2 ln(-x + sqrt{x^2 + 1}) + 1 ). So, if I plug in ( a ) into this function, I get 11. That is:( f(a) = a^2 ln(-a + sqrt{a^2 + 1}) + 1 = 11 ).So, that simplifies to:( a^2 ln(-a + sqrt{a^2 + 1}) = 10 ).Now, I need to find ( f(-a) ). Let me write that out:( f(-a) = (-a)^2 ln(-(-a) + sqrt{(-a)^2 + 1}) + 1 ).Simplifying the terms:( (-a)^2 = a^2 ), so that part is straightforward.Next, inside the logarithm, we have:( -(-a) + sqrt{(-a)^2 + 1} = a + sqrt{a^2 + 1} ).So, putting it all together:( f(-a) = a^2 ln(a + sqrt{a^2 + 1}) + 1 ).Hmm, interesting. So, ( f(a) ) and ( f(-a) ) both have similar structures, except the arguments inside the logarithm are different. In ( f(a) ), it's ( ln(-a + sqrt{a^2 + 1}) ), and in ( f(-a) ), it's ( ln(a + sqrt{a^2 + 1}) ).I wonder if there's a relationship between these two logarithmic terms. Maybe they are inverses or something? Let me think.Let me denote ( y = -a + sqrt{a^2 + 1} ). Then, what is ( a + sqrt{a^2 + 1} )?Let me compute ( y times (a + sqrt{a^2 + 1}) ):( y times (a + sqrt{a^2 + 1}) = (-a + sqrt{a^2 + 1})(a + sqrt{a^2 + 1}) ).This looks like a difference of squares. Let me expand it:( (-a)(a) + (-a)(sqrt{a^2 + 1}) + (sqrt{a^2 + 1})(a) + (sqrt{a^2 + 1})(sqrt{a^2 + 1}) ).Simplifying each term:1. ( (-a)(a) = -a^2 )2. ( (-a)(sqrt{a^2 + 1}) = -asqrt{a^2 + 1} )3. ( (sqrt{a^2 + 1})(a) = asqrt{a^2 + 1} )4. ( (sqrt{a^2 + 1})(sqrt{a^2 + 1}) = (a^2 + 1) )Now, combining these terms:- The second and third terms cancel each other out: ( -asqrt{a^2 + 1} + asqrt{a^2 + 1} = 0 ).- The first and fourth terms remain: ( -a^2 + (a^2 + 1) = 1 ).So, the product ( y times (a + sqrt{a^2 + 1}) = 1 ).That means ( y times (a + sqrt{a^2 + 1}) = 1 ), so ( (a + sqrt{a^2 + 1}) = frac{1}{y} ).Therefore, ( ln(a + sqrt{a^2 + 1}) = lnleft(frac{1}{y}right) = -ln(y) ).But ( y = -a + sqrt{a^2 + 1} ), so:( ln(a + sqrt{a^2 + 1}) = -ln(-a + sqrt{a^2 + 1}) ).Wow, that's a useful relationship! So, the logarithm in ( f(-a) ) is the negative of the logarithm in ( f(a) ).Let me write that down:( ln(a + sqrt{a^2 + 1}) = -ln(-a + sqrt{a^2 + 1}) ).So, substituting this back into ( f(-a) ):( f(-a) = a^2 times (-ln(-a + sqrt{a^2 + 1})) + 1 ).Which simplifies to:( f(-a) = -a^2 ln(-a + sqrt{a^2 + 1}) + 1 ).But from earlier, we know that ( a^2 ln(-a + sqrt{a^2 + 1}) = 10 ). So, substituting that in:( f(-a) = -10 + 1 = -9 ).So, ( f(-a) = -9 ).Wait, let me just double-check my steps to make sure I didn't make a mistake.1. I started with ( f(a) = 11 ), which gave me ( a^2 ln(-a + sqrt{a^2 + 1}) = 10 ).2. Then, I computed ( f(-a) ) and found that the logarithmic term was the negative of the one in ( f(a) ).3. Substituted that back into ( f(-a) ) and used the known value from ( f(a) ) to find ( f(-a) = -9 ).Everything seems to check out. So, I think the answer is ( -9 ).**Final Answer**boxed{-9}

💡Okay, so I have this system of equations to solve:left{begin{array}{l}5 x^{2} - 14 x y + 10 y^{2} = 17 4 x^{2} - 10 x y + 6 y^{2} = 8end{array}right.Hmm, both equations are quadratic in terms of x and y. I remember that sometimes when dealing with systems of quadratic equations, it's helpful to try to eliminate one of the variables or find a relationship between x and y that can simplify the system.Let me write down the equations again:1. (5x^2 - 14xy + 10y^2 = 17)2. (4x^2 - 10xy + 6y^2 = 8)I notice that both equations have similar terms, so maybe I can manipulate them to eliminate one of the variables. Let me try to eliminate the (x^2) term first.If I multiply the second equation by 5, it becomes:(5 times (4x^2 - 10xy + 6y^2) = 5 times 8)Which simplifies to:(20x^2 - 50xy + 30y^2 = 40)Now, let's subtract the first equation multiplied by 4 from this new equation to eliminate (x^2):First, multiply the first equation by 4:(4 times (5x^2 - 14xy + 10y^2) = 4 times 17)Which gives:(20x^2 - 56xy + 40y^2 = 68)Now subtract this from the transformed second equation:((20x^2 - 50xy + 30y^2) - (20x^2 - 56xy + 40y^2) = 40 - 68)Simplifying the left side:(20x^2 - 50xy + 30y^2 - 20x^2 + 56xy - 40y^2 = -28)Combine like terms:(6xy - 10y^2 = -28)Hmm, that simplifies to:(6xy - 10y^2 = -28)I can factor out a 2y:(2y(3x - 5y) = -28)Divide both sides by 2:(y(3x - 5y) = -14)Hmm, not sure if that helps directly. Maybe I should try another approach. Let me see if adding the equations together helps.Adding the original two equations:(5x^2 - 14xy + 10y^2 + 4x^2 - 10xy + 6y^2 = 17 + 8)Combine like terms:(9x^2 - 24xy + 16y^2 = 25)Wait a minute, that looks like a perfect square. Let me check:(9x^2 - 24xy + 16y^2 = (3x - 4y)^2)Yes! So,((3x - 4y)^2 = 25)Taking square roots on both sides:(3x - 4y = pm 5)So, we have two cases:1. (3x - 4y = 5)2. (3x - 4y = -5)Now, let's go back to the other equation we derived earlier:(6xy - 10y^2 = -28)I can write this as:(6xy = 10y^2 - 28)Divide both sides by 2:(3xy = 5y^2 - 14)Hmm, maybe I can express x in terms of y from one of the cases and substitute here.Let's take the first case: (3x - 4y = 5)Solving for x:(3x = 4y + 5)(x = frac{4y + 5}{3})Now, substitute this into the equation (3xy = 5y^2 - 14):(3 times frac{4y + 5}{3} times y = 5y^2 - 14)Simplify:((4y + 5)y = 5y^2 - 14)Expand the left side:(4y^2 + 5y = 5y^2 - 14)Bring all terms to one side:(4y^2 + 5y - 5y^2 + 14 = 0)Combine like terms:(-y^2 + 5y + 14 = 0)Multiply both sides by -1 to make it easier:(y^2 - 5y - 14 = 0)Now, solve this quadratic equation for y:Using the quadratic formula:(y = frac{5 pm sqrt{25 + 56}}{2})(y = frac{5 pm sqrt{81}}{2})(y = frac{5 pm 9}{2})So, two solutions:1. (y = frac{5 + 9}{2} = 7)2. (y = frac{5 - 9}{2} = -2)Now, find corresponding x values using (x = frac{4y + 5}{3}):For (y = 7):(x = frac{4(7) + 5}{3} = frac{28 + 5}{3} = frac{33}{3} = 11)For (y = -2):(x = frac{4(-2) + 5}{3} = frac{-8 + 5}{3} = frac{-3}{3} = -1)So, from the first case, we have two solutions: (11, 7) and (-1, -2)Now, let's handle the second case: (3x - 4y = -5)Solving for x:(3x = 4y - 5)(x = frac{4y - 5}{3})Substitute into (3xy = 5y^2 - 14):(3 times frac{4y - 5}{3} times y = 5y^2 - 14)Simplify:((4y - 5)y = 5y^2 - 14)Expand the left side:(4y^2 - 5y = 5y^2 - 14)Bring all terms to one side:(4y^2 - 5y - 5y^2 + 14 = 0)Combine like terms:(-y^2 - 5y + 14 = 0)Multiply both sides by -1:(y^2 + 5y - 14 = 0)Solve this quadratic equation for y:Using the quadratic formula:(y = frac{-5 pm sqrt{25 + 56}}{2})(y = frac{-5 pm sqrt{81}}{2})(y = frac{-5 pm 9}{2})So, two solutions:1. (y = frac{-5 + 9}{2} = 2)2. (y = frac{-5 - 9}{2} = -7)Now, find corresponding x values using (x = frac{4y - 5}{3}):For (y = 2):(x = frac{4(2) - 5}{3} = frac{8 - 5}{3} = frac{3}{3} = 1)For (y = -7):(x = frac{4(-7) - 5}{3} = frac{-28 - 5}{3} = frac{-33}{3} = -11)So, from the second case, we have two solutions: (1, 2) and (-11, -7)Therefore, the solutions to the system are:(11, 7), (-1, -2), (1, 2), (-11, -7)I should probably check these solutions in the original equations to make sure they work.Let me check (11, 7):First equation: (5(11)^2 -14(11)(7) +10(7)^2)= (5(121) -14(77) +10(49))= 605 - 1078 + 490= (605 + 490) - 1078= 1095 - 1078 = 17 ✔️Second equation: (4(11)^2 -10(11)(7) +6(7)^2)= (4(121) -10(77) +6(49))= 484 - 770 + 294= (484 + 294) - 770= 778 - 770 = 8 ✔️Good.Now, check (-1, -2):First equation: (5(-1)^2 -14(-1)(-2) +10(-2)^2)= 5(1) -14(2) +10(4)= 5 - 28 + 40= (5 + 40) - 28 = 45 - 28 = 17 ✔️Second equation: (4(-1)^2 -10(-1)(-2) +6(-2)^2)= 4(1) -10(2) +6(4)= 4 - 20 + 24= (4 + 24) - 20 = 28 - 20 = 8 ✔️Good.Check (1, 2):First equation: (5(1)^2 -14(1)(2) +10(2)^2)= 5(1) -28 +10(4)= 5 -28 +40= (5 + 40) -28 = 45 -28 =17 ✔️Second equation: (4(1)^2 -10(1)(2) +6(2)^2)= 4(1) -20 +6(4)= 4 -20 +24= (4 +24) -20 =28 -20 =8 ✔️Good.Check (-11, -7):First equation: (5(-11)^2 -14(-11)(-7) +10(-7)^2)=5(121) -14(77) +10(49)=605 -1078 +490= (605 +490) -1078 =1095 -1078=17 ✔️Second equation: (4(-11)^2 -10(-11)(-7) +6(-7)^2)=4(121) -10(77) +6(49)=484 -770 +294= (484 +294) -770=778 -770=8 ✔️All solutions check out. So, the solutions are (11,7), (-1,-2), (1,2), (-11,-7).