answer：Okay, so I have this problem about a function f(x) that's defined on all real numbers except zero, and it's an odd function. When x is positive, f(x) is log base 2 of x. There are two parts to the problem. First, I need to find the expression of f(x) when x is negative. Hmm, since it's an odd function, I remember that odd functions satisfy the property f(-x) = -f(x). So, if I know f(x) for positive x, I can find f(x) for negative x by using this property. Let me write that down. If x is negative, then -x is positive. So, f(-x) is log base 2 of (-x). But since f is odd, f(x) should be equal to -f(-x). So, f(x) = -log₂(-x). That makes sense because when x is negative, -x is positive, and taking the logarithm of a positive number is defined. So, for x < 0, f(x) is -log₂(-x). Alright, that seems straightforward. I think that's the answer for part (1). Now, moving on to part (2). We have another function g(x) = 2^x, defined for all real numbers x. We need to define set A as the set of x such that f(x) is greater than or equal to 2. Then, set B is the set of y where y is g(x) for x in A. So, essentially, B is the image of set A under the function g.Let me break this down. First, find set A: all x where f(x) ≥ 2. Since f(x) is defined differently for positive and negative x, I should consider both cases.For x > 0, f(x) = log₂x. So, we need log₂x ≥ 2. To solve this inequality, I can rewrite it as x ≥ 2², which is x ≥ 4. So, for positive x, the solutions are x ≥ 4.For x < 0, f(x) = -log₂(-x). We need -log₂(-x) ≥ 2. Let me solve this inequality. Multiplying both sides by -1 (and remembering to reverse the inequality sign), we get log₂(-x) ≤ -2. So, log₂(-x) ≤ -2 implies that -x ≤ 2^(-2), which is -x ≤ 1/4. Multiplying both sides by -1 again (and reversing the inequality), we get x ≥ -1/4. But wait, x is negative here, so x must be between -1/4 and 0. So, for negative x, the solutions are -1/4 ≤ x < 0.Putting both cases together, set A is the union of two intervals: x ≥ 4 and -1/4 ≤ x < 0. So, A = {x | x ≥ 4 or -1/4 ≤ x < 0}.Now, set B is the set of y such that y = g(x) where x is in A. Since g(x) = 2^x, we need to find all y values that result from plugging in x from set A into g(x).Let's consider the two intervals of A separately.First, for x ≥ 4: g(x) = 2^x. Since 2^x is an increasing function, when x is 4, 2^4 = 16, and as x increases beyond 4, 2^x increases without bound. So, for x ≥ 4, y = 2^x will be greater than or equal to 16. So, y ≥ 16.Second, for -1/4 ≤ x < 0: g(x) = 2^x. Again, 2^x is an increasing function. When x = -1/4, 2^(-1/4) is equal to 1/(2^(1/4)) which is approximately 0.8409. When x approaches 0 from the left, 2^x approaches 1. So, for x in [-1/4, 0), y = 2^x will be in [2^(-1/4), 1). Therefore, combining these two intervals, set B is the union of [2^(-1/4), 1) and [16, ∞). Let me write that in terms of exact values rather than decimal approximations. 2^(-1/4) is the same as 1/(2^(1/4)), which can also be written as 2^(-1/4). So, the lower bound is 2^(-1/4), and the upper bound for the first interval is 1, not including 1. The second interval starts at 16 and goes to infinity.So, putting it all together, set B is { y | 2^(-1/4) ≤ y < 1 or y ≥ 16 }.Let me double-check my steps to make sure I didn't make any mistakes.1. For part (1), using the property of odd functions, I correctly derived f(x) for x < 0 as -log₂(-x). That seems right.2. For part (2), solving f(x) ≥ 2: - For x > 0: log₂x ≥ 2 leads to x ≥ 4. Correct. - For x < 0: -log₂(-x) ≥ 2 leads to log₂(-x) ≤ -2, which leads to -x ≤ 1/4, so x ≥ -1/4. Since x is negative, it's -1/4 ≤ x < 0. Correct.3. Then, finding g(x) over these intervals: - For x ≥ 4: 2^x ≥ 16. Correct. - For -1/4 ≤ x < 0: 2^(-1/4) ≤ 2^x < 1. Correct.So, everything seems to check out. I think my answer is correct.**Final Answer**(1) The expression of ( f(x) ) when ( x < 0 ) is (boxed{-log_{2}(-x)}).(2) The set ( B ) is (boxed{left[2^{-frac{1}{4}}, 1right) cup [16, infty)}).

answer：Okay, so I need to figure out how many games are played in a season for a league with 20 teams, where each team plays against every other team four times. Hmm, let's break this down step by step.First, I know that in a league with multiple teams, each team plays against every other team a certain number of times. In this case, it's four times. So, for one team, how many games would that be? Well, there are 19 other teams, and they play each of those teams four times. So, for one team, that would be 19 teams multiplied by 4 games, which is 76 games per team.But wait, if I just take that number and multiply it by the total number of teams, I might be double-counting the games. Because when Team A plays Team B, that's one game, not two. So, if I have 20 teams each playing 76 games, that would be 20 times 76, which is 1,520 games. But since each game involves two teams, I need to divide that number by 2 to avoid double-counting. So, 1,520 divided by 2 is 760 games in total.Let me see if there's another way to think about this to confirm. Maybe using combinations. The number of unique pairings of teams can be calculated using combinations, specifically "20 choose 2," which is the number of ways to choose 2 teams out of 20. The formula for combinations is n(n-1)/2, where n is the number of teams. So, 20 times 19 divided by 2 is 190 unique pairings.Since each pairing plays each other four times, I multiply the number of unique pairings by 4. So, 190 times 4 is 760 games. That matches the previous calculation.Just to make sure, let's think about a smaller example. Suppose there are 2 teams, and they play each other 4 times. That should be 4 games. Using the formula, "2 choose 2" is 1, multiplied by 4 gives 4 games. That works.Another example: 3 teams, each playing each other 4 times. The number of unique pairings is "3 choose 2," which is 3. Multiplying by 4 gives 12 games. Let's list them out: Team A vs Team B four times, Team A vs Team C four times, and Team B vs Team C four times. That's indeed 12 games. So the formula seems to hold.Therefore, applying this to the original problem with 20 teams, the total number of games played in the season should be 760.

answer：Okay, so I have this problem where I need to convert a point from cylindrical coordinates to rectangular coordinates. The point given is (8, 5π/6, 6). Hmm, I remember that cylindrical coordinates are represented as (r, θ, z), where r is the radius, θ is the angle, and z is the height. Rectangular coordinates, on the other hand, are (x, y, z). I think the conversion formulas involve trigonometry. Let me recall... I believe x is calculated using r multiplied by the cosine of θ, and y is calculated using r multiplied by the sine of θ. So, x = r cos θ and y = r sin θ. The z-coordinate remains the same in both systems, so z = z. That seems right.Alright, so for this specific point, r is 8, θ is 5π/6, and z is 6. Let me write down the formulas again to make sure I have them correct:x = r cos θ y = r sin θ z = zNow, plugging in the values:x = 8 cos(5π/6) y = 8 sin(5π/6) z = 6Okay, so I need to compute cos(5π/6) and sin(5π/6). I remember that 5π/6 is in the second quadrant, where cosine is negative and sine is positive. Also, 5π/6 is equal to π - π/6, so it's related to the reference angle of π/6.For cosine, cos(5π/6) should be equal to -cos(π/6). Similarly, sin(5π/6) should be equal to sin(π/6). Let me confirm that. Yes, because cosine is negative in the second quadrant and sine is positive.So, cos(π/6) is √3/2, which means cos(5π/6) is -√3/2. Similarly, sin(π/6) is 1/2, so sin(5π/6) is also 1/2.Now, let's compute x and y:x = 8 * (-√3/2) y = 8 * (1/2)Calculating x first: 8 multiplied by -√3/2. Let's see, 8 divided by 2 is 4, so it's 4 * -√3, which is -4√3.Calculating y: 8 multiplied by 1/2 is 4.So, putting it all together, the rectangular coordinates are (-4√3, 4, 6). Wait, let me double-check my calculations to make sure I didn't make any mistakes. For x, 8 times -√3/2 is indeed -4√3 because 8 divided by 2 is 4, and 4 times √3 is 4√3, but with the negative sign. For y, 8 times 1/2 is straightforward, it's 4. And z remains 6.I think that's correct. So, the rectangular coordinates should be (-4√3, 4, 6). I don't see any errors in my reasoning or calculations, so I feel confident about this answer.

answer：Okay, so I have this problem here: given real numbers a, b, c such that a + 2b + 3c = 6, I need to find the minimum value of a² + 4b² + 9c². Hmm, interesting. I remember that when dealing with expressions like this, especially with a linear constraint and a quadratic objective, methods like Cauchy-Schwarz inequality or maybe Lagrange multipliers could be useful. Let me think about how to approach this.First, let me write down what I know. The constraint is a + 2b + 3c = 6, and I need to minimize a² + 4b² + 9c². So, it's like a quadratic form, and the constraint is linear. I think Cauchy-Schwarz might be a good way to go here because it relates a linear combination to a quadratic form.The Cauchy-Schwarz inequality states that for any real vectors u and v, the square of their dot product is less than or equal to the product of their magnitudes squared. In mathematical terms, (u · v)² ≤ (u · u)(v · v). Maybe I can set up vectors u and v such that their dot product is the given linear expression, and their magnitudes relate to the quadratic form I need to minimize.Let me try to represent the constraint as a dot product. If I let u = (a, 2b, 3c) and v = (1, 1, 1), then the dot product u · v would be a*1 + 2b*1 + 3c*1, which is exactly a + 2b + 3c. That's equal to 6, according to the constraint.So, applying Cauchy-Schwarz, (u · v)² ≤ (u · u)(v · v). Plugging in the values, we have (a + 2b + 3c)² ≤ (a² + (2b)² + (3c)²)(1² + 1² + 1²). Simplifying the right side, that's (a² + 4b² + 9c²)(3). So, 6² ≤ 3(a² + 4b² + 9c²). That gives 36 ≤ 3(a² + 4b² + 9c²), so dividing both sides by 3, we get 12 ≤ a² + 4b² + 9c².So, the minimum value of a² + 4b² + 9c² is 12. But wait, is this the minimum? I think so because Cauchy-Schwarz gives us an inequality, and equality holds when u and v are scalar multiples of each other. That is, when (a, 2b, 3c) is proportional to (1, 1, 1). So, a = k, 2b = k, 3c = k for some constant k.Let me solve for a, b, c in terms of k. From a = k, 2b = k implies b = k/2, and 3c = k implies c = k/3. Now, plugging these back into the constraint a + 2b + 3c = 6, we have k + 2*(k/2) + 3*(k/3) = 6. Simplifying each term: k + k + k = 6, so 3k = 6, which means k = 2.Therefore, a = 2, b = 2/2 = 1, c = 2/3. Plugging these back into the quadratic expression, a² + 4b² + 9c² = 2² + 4*(1)² + 9*(2/3)² = 4 + 4 + 9*(4/9) = 4 + 4 + 4 = 12. So, that checks out.Wait, but just to make sure, is there another method I can use to verify this result? Maybe using Lagrange multipliers? Let me try that approach as well.In the method of Lagrange multipliers, we set up the function to minimize, which is f(a, b, c) = a² + 4b² + 9c², subject to the constraint g(a, b, c) = a + 2b + 3c - 6 = 0. The method tells us that at the minimum, the gradient of f is proportional to the gradient of g. So, ∇f = λ∇g for some λ.Calculating the gradients: ∇f = (2a, 8b, 18c) and ∇g = (1, 2, 3). So, setting up the equations:2a = λ*1 8b = λ*2 18c = λ*3From the first equation, λ = 2a. Plugging into the second equation: 8b = 2a*2 = 4a, so 8b = 4a ⇒ 2b = a. Similarly, from the third equation: 18c = 3λ = 3*(2a) = 6a ⇒ 18c = 6a ⇒ 3c = a.So, from these, a = 2b and a = 3c. Therefore, 2b = 3c ⇒ c = (2/3)b. Now, let's express a and c in terms of b. a = 2b, c = (2/3)b.Plugging these into the constraint a + 2b + 3c = 6: 2b + 2b + 3*(2/3)b = 6. Simplify each term: 2b + 2b + 2b = 6 ⇒ 6b = 6 ⇒ b = 1. Then, a = 2*1 = 2, c = (2/3)*1 = 2/3. So, same result as before. Therefore, the minimum value is indeed 12.Hmm, so both methods give me the same answer. That gives me more confidence that 12 is the correct minimum value. I think I'm pretty sure now that the minimum is 12 when a=2, b=1, c=2/3.Just to recap, the problem was to minimize a quadratic expression given a linear constraint. Using Cauchy-Schwarz, I set up the inequality and found the minimum value directly. Then, using Lagrange multipliers, I arrived at the same conclusion by solving the system of equations derived from the gradients. Both methods are valid and lead to the same result, so I think 12 is definitely the minimum value.I wonder if there's another way to approach this problem, maybe using substitution? Let me try that as well for thoroughness.Since we have the constraint a + 2b + 3c = 6, perhaps I can express one variable in terms of the others and substitute into the quadratic expression. Let's solve for a: a = 6 - 2b - 3c. Then, substitute into f(a, b, c):f(a, b, c) = (6 - 2b - 3c)² + 4b² + 9c².Let me expand this:(6 - 2b - 3c)² = 36 - 24b - 36c + 4b² + 12bc + 9c².So, adding the other terms:f = 36 - 24b - 36c + 4b² + 12bc + 9c² + 4b² + 9c².Combine like terms:4b² + 4b² = 8b² 9c² + 9c² = 18c² 12bc remains as is -24b - 36c remain And the constant 36.So, f = 8b² + 18c² + 12bc - 24b - 36c + 36.Hmm, now this is a quadratic in two variables, b and c. Maybe I can complete the square or find the critical points by taking partial derivatives.Let me try taking partial derivatives with respect to b and c and set them to zero.First, partial derivative with respect to b:df/db = 16b + 12c - 24.Partial derivative with respect to c:df/dc = 36c + 12b - 36.Set both equal to zero:16b + 12c - 24 = 0 36c + 12b - 36 = 0.Let me write these equations:1) 16b + 12c = 24 2) 12b + 36c = 36.I can simplify both equations by dividing by common factors. Equation 1: divide by 4: 4b + 3c = 6. Equation 2: divide by 12: b + 3c = 3.So, now we have:4b + 3c = 6 b + 3c = 3.Subtracting the second equation from the first: (4b + 3c) - (b + 3c) = 6 - 3 ⇒ 3b = 3 ⇒ b = 1.Plugging back into the second equation: 1 + 3c = 3 ⇒ 3c = 2 ⇒ c = 2/3.Then, a = 6 - 2b - 3c = 6 - 2*1 - 3*(2/3) = 6 - 2 - 2 = 2.So, again, we get a=2, b=1, c=2/3, leading to f=12. So, substitution method also confirms the result.Wow, so three different methods: Cauchy-Schwarz, Lagrange multipliers, and substitution, all leading to the same answer. That must mean it's correct. I think I'm confident now that the minimum value is indeed 12.Just to make sure I didn't make any calculation errors, let me double-check the substitution method.Starting with f(a, b, c) = a² + 4b² + 9c², and a = 6 - 2b - 3c.So, f = (6 - 2b - 3c)² + 4b² + 9c².Expanding (6 - 2b - 3c)²:= 6² + (-2b)² + (-3c)² + 2*(6*(-2b) + 6*(-3c) + (-2b)*(-3c)) = 36 + 4b² + 9c² + 2*(-12b - 18c + 6bc) = 36 + 4b² + 9c² - 24b - 36c + 12bc.Adding the other terms 4b² + 9c²:Total f = 36 + 4b² + 9c² - 24b - 36c + 12bc + 4b² + 9c² = 36 + 8b² + 18c² + 12bc - 24b - 36c.Yes, that's correct. Then, taking partial derivatives:df/db = 16b + 12c - 24 df/dc = 36c + 12b - 36.Setting them to zero:16b + 12c = 24 12b + 36c = 36.Simplify:4b + 3c = 6 b + 3c = 3.Subtracting: 3b = 3 ⇒ b=1. Then c=2/3, a=2. So, correct.Alright, I think I've thoroughly checked it. All methods lead to the same conclusion, so I can be certain that the minimum value is 12.**Final Answer**The minimum value of (a^2 + 4b^2 + 9c^2) is boxed{12}.

answer：Okay, so I have this problem where I'm given two expressions: ( x^2 + frac{1}{x^2} = A ) and ( x + frac{1}{x} = C ), where both ( A ) and ( C ) are positive numbers. I need to find the minimum possible numerical value for ( frac{A}{C} ).Hmm, let me start by trying to relate ( A ) and ( C ) somehow. I remember that there's an identity involving ( x + frac{1}{x} ) and ( x^2 + frac{1}{x^2} ). Maybe I can square ( C ) to get something involving ( A ).So, if I square ( C ), which is ( x + frac{1}{x} ), I get:[C^2 = left( x + frac{1}{x} right)^2 = x^2 + 2 cdot x cdot frac{1}{x} + frac{1}{x^2} = x^2 + 2 + frac{1}{x^2}]That simplifies to:[C^2 = x^2 + 2 + frac{1}{x^2}]But I know that ( x^2 + frac{1}{x^2} = A ), so substituting that in:[C^2 = A + 2]So, rearranging that, I get:[A = C^2 - 2]Okay, so now I can express ( A ) in terms of ( C ). The problem asks for the minimum value of ( frac{A}{C} ), so let me write that out:[frac{A}{C} = frac{C^2 - 2}{C} = C - frac{2}{C}]So now, I need to find the minimum value of the function ( f(C) = C - frac{2}{C} ) where ( C ) is a positive number.I remember that to find the minimum of a function, I can take its derivative and set it equal to zero. Let me compute the derivative of ( f(C) ):[f'(C) = frac{d}{dC} left( C - frac{2}{C} right) = 1 + frac{2}{C^2}]Wait, that's interesting. The derivative is ( 1 + frac{2}{C^2} ). But since ( C ) is positive, ( frac{2}{C^2} ) is always positive. So, ( f'(C) ) is always greater than 1, which means the function is always increasing for ( C > 0 ).If the function is always increasing, then its minimum value occurs at the smallest possible value of ( C ). So, what is the smallest possible value of ( C )?Looking back at the original expression ( C = x + frac{1}{x} ), I know from the AM-GM inequality that ( x + frac{1}{x} geq 2 ) for any positive real number ( x ). The equality holds when ( x = 1 ).So, the minimum value of ( C ) is 2. Let me plug that back into ( f(C) ):[f(2) = 2 - frac{2}{2} = 2 - 1 = 1]Therefore, the minimum value of ( frac{A}{C} ) is 1.Wait a second, let me double-check that. If ( C = 2 ), then ( A = C^2 - 2 = 4 - 2 = 2 ). So, ( frac{A}{C} = frac{2}{2} = 1 ). That seems correct.But earlier, I thought about using calculus and found that the function is always increasing, so the minimum is indeed at ( C = 2 ). So, the minimum value of ( frac{A}{C} ) is 1.I think that makes sense. Let me see if there's another way to approach this without calculus. Maybe using inequalities.Since ( C = x + frac{1}{x} geq 2 ), and ( A = C^2 - 2 ), then ( frac{A}{C} = C - frac{2}{C} ). To minimize this, since ( C ) is at least 2, plugging in 2 gives the smallest value.Alternatively, maybe using the AM-GM inequality on ( C - frac{2}{C} ). But I'm not sure how to apply it directly here. Maybe it's better to stick with the calculus approach since it clearly shows the function is increasing.So, yeah, I think the minimum value is 1 when ( C = 2 ).

answer：Alright, so I've got this problem here about marbles in a bag. Let me read it again: "A bag contains 6 red marbles, 4 white marbles, and 8 blue marbles. Three marbles are drawn from the bag (without replacement). What is the probability that they are all the same color?" Okay, so I need to find the probability that when I draw three marbles, they're all red, or all white, or all blue.First, I think I should figure out the total number of marbles in the bag. Let me add them up: 6 red + 4 white + 8 blue. That makes 6 + 4 = 10, and 10 + 8 = 18 marbles in total. So, there are 18 marbles altogether.Now, since we're drawing three marbles without replacement, the total number of possible ways to draw three marbles from 18 is a combination problem. I remember that the formula for combinations is C(n, k) = n! / (k!(n - k)!), where n is the total number of items, and k is the number of items we're choosing. So, the total number of ways to draw three marbles is C(18, 3). Let me calculate that: 18! / (3!(18 - 3)!) = (18 × 17 × 16) / (3 × 2 × 1) = 816. So, there are 816 possible ways to draw any three marbles.Next, I need to find the number of favorable outcomes, which are the cases where all three marbles are the same color. That means I need to calculate the number of ways to draw three red marbles, three white marbles, and three blue marbles separately, and then add them up.Starting with red marbles: there are 6 red marbles. The number of ways to choose three red marbles is C(6, 3). Using the combination formula again: 6! / (3!(6 - 3)!) = (6 × 5 × 4) / (3 × 2 × 1) = 20. So, there are 20 ways to draw three red marbles.Moving on to white marbles: there are 4 white marbles. The number of ways to choose three white marbles is C(4, 3). Calculating that: 4! / (3!(4 - 3)!) = (4 × 3 × 2) / (3 × 2 × 1) = 4. So, there are 4 ways to draw three white marbles.Now, for blue marbles: there are 8 blue marbles. The number of ways to choose three blue marbles is C(8, 3). Let me compute that: 8! / (3!(8 - 3)!) = (8 × 7 × 6) / (3 × 2 × 1) = 56. So, there are 56 ways to draw three blue marbles.To find the total number of favorable outcomes, I add up the ways for each color: 20 (red) + 4 (white) + 56 (blue) = 80. So, there are 80 favorable ways to draw three marbles of the same color.Now, to find the probability, I take the number of favorable outcomes and divide it by the total number of possible outcomes. That would be 80 / 816. Let me simplify that fraction. Both numerator and denominator are divisible by 8: 80 ÷ 8 = 10, and 816 ÷ 8 = 102. So, the simplified fraction is 10/102. I can simplify this further by dividing numerator and denominator by 2: 10 ÷ 2 = 5, and 102 ÷ 2 = 51. So, the probability is 5/51.Wait, let me double-check my calculations to make sure I didn't make a mistake. Total marbles: 6 + 4 + 8 = 18, that's correct. Total combinations: C(18, 3) = 816, that seems right. Red combinations: C(6, 3) = 20, white: C(4, 3) = 4, blue: C(8, 3) = 56. Adding them up: 20 + 4 + 56 = 80. Probability: 80/816 = 5/51 after simplifying. Yeah, that all checks out.I think I've got it. The probability of drawing three marbles of the same color is 5/51.