answer：First, I recognize that the base 9 number 726₉ needs to be converted to base 3. Since 9 is equal to 3 squared (3²), each digit in base 9 can be directly converted into two digits in base 3.Starting with the leftmost digit, 7 in base 9. To convert 7 to base 3, I divide 7 by 3, which gives a quotient of 2 and a remainder of 1. This means 7 in base 9 is equivalent to 21 in base 3.Next, I convert the middle digit, 2 in base 9. Since 2 is less than 3, it remains the same in base 3. However, to maintain the two-digit format required for base 3 conversion from base 9, I write it as 02.Finally, I convert the rightmost digit, 6 in base 9. Dividing 6 by 3 gives a quotient of 2 and a remainder of 0, so 6 in base 9 is equivalent to 20 in base 3.By concatenating these base 3 equivalents, I obtain the final result of 210220₃.

answer：First, I need to determine the typing rate of one typist. Given that some typists can type 46 letters in 20 minutes, I'll assume there are ( n ) typists in the initial group. The total typing rate for ( n ) typists is:[frac{46 text{ letters}}{20 text{ minutes}} = frac{46}{20} text{ letters per minute per typist}]Next, I'll calculate how many letters one typist can type in 1 hour (60 minutes). Since 60 minutes is three times 20 minutes, the rate for one typist in 60 minutes is:[frac{46}{20} times 3 = frac{138}{20} = 6.9 text{ letters per minute per typist}]Now, with 30 typists working at this rate for 1 hour, the total number of letters they can type is:[30 text{ typists} times 6.9 text{ letters per minute per typist} times 60 text{ minutes} = 207 text{ letters}]Setting up the equation to find ( n ):[frac{46}{n} times 3 = frac{207}{30}]Simplifying:[frac{138}{n} = 6.9]Solving for ( n ):[n = frac{138}{6.9} = 20]Therefore, there were 20 typists in the initial group.

answer：First, I need to determine how to handle the family of three who must stay together. To simplify the problem, I'll treat the family as a single unit or "block" when considering the lineup.By grouping the family together, I effectively reduce the number of entities to arrange. Originally, there are seven individuals, but treating the family as one unit means I now have five units to arrange: the family block and the four other people.Next, I'll calculate the number of ways to arrange these five units. This is a straightforward permutation problem, and the number of ways to arrange five distinct units is 5 factorial (5!).However, within the family block, the three family members can be arranged among themselves in different ways. The number of ways to arrange three distinct individuals is 3 factorial (3!).To find the total number of possible lineups, I'll multiply the number of ways to arrange the five units by the number of ways to arrange the family members within their block. This gives me the total number of valid lineups where the family stays together.Finally, I'll compute the numerical value of 5! multiplied by 3! to get the final answer.

answer：Alright, so I have this question here asking which function represents y being directly proportional to x. The options are A: y = -0.1x, B: y = 2x², C: y² = 4x, and D: y = 2x + 1. Hmm, okay, let me think about what direct proportionality means.From what I remember, direct proportionality means that y is equal to some constant multiplied by x. So, mathematically, it's like y = kx, where k is a constant. That means the relationship between y and x should be linear, right? The graph of y versus x should be a straight line passing through the origin because there's no constant term added or subtracted.Let me go through each option one by one.Option A: y = -0.1x. Okay, this looks like y equals some constant (-0.1) times x. So, that fits the form y = kx. The negative sign just means that y and x are inversely related in direction, but they're still directly proportional. So, this seems like a possible answer.Option B: y = 2x². Hmm, this is a quadratic function because of the x squared term. Direct proportionality requires the exponent of x to be 1, not 2. So, this doesn't fit the definition. Plus, the graph of this would be a parabola, not a straight line through the origin. So, I think B is out.Option C: y² = 4x. This one is a bit trickier. If I solve for y, I get y = ±2√x. That means y is proportional to the square root of x, not x itself. So, this isn't a direct proportionality between y and x. Instead, it's a different kind of relationship, maybe a square root relationship. So, C doesn't fit either.Option D: y = 2x + 1. This is a linear function, but it has a constant term added, which is the "+1". In direct proportionality, there shouldn't be any constant term; the relationship should pass through the origin. If it had just y = 2x, that would be direct proportionality, but the "+1" means it's not. So, D is also not the right answer.So, going back to the options, A is the only one that fits the definition of direct proportionality because it's in the form y = kx with k being a constant (-0.1 in this case). The other options either have a quadratic relationship, a square root relationship, or a linear relationship with an extra constant term.I think I've got it. Option A is the correct answer because it's the only function where y is directly proportional to x without any additional terms or different exponents.

answer：Okay, so I've got this problem here where I need to figure out which function best fits some given data points. The options are A: ( y = 3 times 2^{x-1} ), B: ( y = log_{2}x ), C: ( y = 3x ), and D: ( y = x^{2} ). Hmm, let me think about how to approach this.First, I should probably figure out what data points we're dealing with. The solution mentioned substituting the points (1,3), (2,5.99), and (3,12.01) into each function. I guess these are the data points we need to fit. So, let me write those down:- When ( x = 1 ), ( y = 3 )- When ( x = 2 ), ( y approx 6 )- When ( x = 3 ), ( y approx 12 )Wait, actually, the y-values are 3, 5.99, and 12.01. So, they're very close to 3, 6, and 12. That seems like they might be following a specific pattern. Let me check each function one by one.Starting with option A: ( y = 3 times 2^{x-1} ). Let's plug in the x-values and see what we get.For ( x = 1 ):( y = 3 times 2^{1-1} = 3 times 2^{0} = 3 times 1 = 3 ). That matches the data point (1,3).For ( x = 2 ):( y = 3 times 2^{2-1} = 3 times 2^{1} = 3 times 2 = 6 ). The data point is (2,5.99), which is almost 6. That's really close.For ( x = 3 ):( y = 3 times 2^{3-1} = 3 times 2^{2} = 3 times 4 = 12 ). The data point is (3,12.01), which is almost 12. Again, very close.Okay, so option A seems to fit the data points almost perfectly. Let me check the other options just to be thorough.Option B: ( y = log_{2}x ). Let's plug in the x-values.For ( x = 1 ):( y = log_{2}1 = 0 ). But the data point is (1,3). That's way off.For ( x = 2 ):( y = log_{2}2 = 1 ). The data point is (2,5.99). Still not close.For ( x = 3 ):( y = log_{2}3 approx 1.585 ). The data point is (3,12.01). Nope, doesn't fit at all.So, option B is definitely not the right fit.Moving on to option C: ( y = 3x ). Let's test this.For ( x = 1 ):( y = 3 times 1 = 3 ). That matches the first data point.For ( x = 2 ):( y = 3 times 2 = 6 ). The data point is (2,5.99). That's pretty close, just a tiny bit less.For ( x = 3 ):( y = 3 times 3 = 9 ). The data point is (3,12.01). Hmm, that's quite a bit off. It's 9 versus 12.01. So, the third point doesn't fit well.So, option C is a linear function, but it doesn't fit the third data point as well as option A does.Now, option D: ( y = x^{2} ). Let's see.For ( x = 1 ):( y = 1^{2} = 1 ). The data point is (1,3). Not matching.For ( x = 2 ):( y = 2^{2} = 4 ). The data point is (2,5.99). Still not matching.For ( x = 3 ):( y = 3^{2} = 9 ). The data point is (3,12.01). Again, not matching.So, option D is a quadratic function, but it doesn't fit any of the data points well except maybe the first one a little, but even that's off.Wait, let me double-check my calculations for each function to make sure I didn't make any mistakes.For option A:- ( x = 1 ): 3*2^(0) = 3*1 = 3 ✔️- ( x = 2 ): 3*2^(1) = 3*2 = 6 ✔️ (5.99 is very close)- ( x = 3 ): 3*2^(2) = 3*4 = 12 ✔️ (12.01 is very close)Looks solid.Option B:- ( x = 1 ): log2(1) = 0 ❌- ( x = 2 ): log2(2) = 1 ❌- ( x = 3 ): log2(3) ≈1.585 ❌Nope, doesn't fit.Option C:- ( x = 1 ): 3*1 = 3 ✔️- ( x = 2 ): 3*2 = 6 (5.99 is close) ✔️- ( x = 3 ): 3*3 = 9 vs 12.01 ❌So, only the first two points fit, but the third is way off.Option D:- ( x = 1 ): 1^2 = 1 vs 3 ❌- ( x = 2 ): 4 vs 5.99 ❌- ( x = 3 ): 9 vs 12.01 ❌No fit at all.So, after checking all options, option A is the only one that fits all three data points very closely. The other options either don't fit the first point or the third point, or they don't fit at all.Wait, just to make sure, is there any chance that another function could fit better? For example, maybe a different exponential function or something else? But the options given are only A to D, so I don't need to consider others. Among these four, A is clearly the best fit.I think that's thorough enough. I don't see any mistakes in my reasoning, so I'm confident that option A is the correct answer.**Final Answer**The best fitting function is boxed{A}.

answer：Okay, so I have this problem where I need to find the smallest possible value of ( n ) such that there exist nonnegative real numbers ( x_1, x_2, dots, x_n ) satisfying two conditions: their sum is 1, and the sum of their squares is at most ( frac{1}{50} ). Hmm, that sounds interesting. Let me try to break it down step by step.First, I remember that there are some inequalities that relate the sum of numbers and the sum of their squares. Maybe the Cauchy-Schwarz inequality or the QM-AM inequality could be useful here. Let me recall what those are.The QM-AM inequality states that the quadratic mean is greater than or equal to the arithmetic mean. In mathematical terms, for nonnegative real numbers ( x_1, x_2, dots, x_n ), it says:[sqrt{frac{x_1^2 + x_2^2 + dots + x_n^2}{n}} ge frac{x_1 + x_2 + dots + x_n}{n}]So, if I apply this to our problem, the left side is the quadratic mean, and the right side is the arithmetic mean. Since we know the sum of the ( x_i )s is 1, the arithmetic mean is ( frac{1}{n} ). Therefore, plugging that into the inequality, we get:[sqrt{frac{x_1^2 + x_2^2 + dots + x_n^2}{n}} ge frac{1}{n}]Now, squaring both sides to eliminate the square root gives:[frac{x_1^2 + x_2^2 + dots + x_n^2}{n} ge frac{1}{n^2}]Multiplying both sides by ( n ) simplifies this to:[x_1^2 + x_2^2 + dots + x_n^2 ge frac{1}{n}]But wait, the problem states that the sum of the squares is at most ( frac{1}{50} ). So, putting these together:[frac{1}{n} le x_1^2 + x_2^2 + dots + x_n^2 le frac{1}{50}]This implies that:[frac{1}{n} le frac{1}{50}]Which simplifies to:[n ge 50]So, ( n ) has to be at least 50. But is 50 actually possible? Let me check.If ( n = 50 ), can we have all ( x_i ) equal? Let's try setting each ( x_i = frac{1}{50} ). Then, the sum would be:[x_1 + x_2 + dots + x_{50} = 50 times frac{1}{50} = 1]Which satisfies the first condition. Now, the sum of the squares would be:[x_1^2 + x_2^2 + dots + x_{50}^2 = 50 times left(frac{1}{50}right)^2 = 50 times frac{1}{2500} = frac{50}{2500} = frac{1}{50}]So, that exactly meets the second condition. Therefore, ( n = 50 ) is achievable.But just to be thorough, what if ( n ) were less than 50? Let's say ( n = 49 ). Then, using the same logic, the sum of squares would have to be at least ( frac{1}{49} ), which is approximately 0.0204. However, ( frac{1}{50} ) is 0.02, which is less than 0.0204. So, it's impossible for ( n = 49 ) because the sum of squares can't be both at least ( frac{1}{49} ) and at most ( frac{1}{50} ). Therefore, 50 is indeed the smallest possible value.I think that covers it. The key was applying the QM-AM inequality to relate the sum and the sum of squares, then checking if the minimal ( n ) found actually works. Since it does, we can confidently say that 50 is the smallest ( n ) needed.**Final Answer**The smallest possible value of ( n ) is boxed{50}.