💡Alright, let's tackle this problem step by step. It's about an odd function and its derivative, and we need to compare three expressions involving the function evaluated at different points. Hmm, okay, let's break it down.First, the function ( f(x) ) is given as an odd function with domain ( mathbb{R} ). That means ( f(-x) = -f(x) ) for all ( x ). That's a key property we might need later.Next, we're told that for ( x neq 0 ), the derivative ( f'(x) + frac{f(x)}{x} > 0 ). Hmm, that's interesting. It looks like a differential inequality. Maybe we can manipulate this to find something useful about ( f(x) ).Let me think. If I have ( f'(x) + frac{f(x)}{x} > 0 ), perhaps I can write this as ( x f'(x) + f(x) > 0 ) when ( x neq 0 ). Wait, that expression ( x f'(x) + f(x) ) looks familiar. Isn't that the derivative of ( x f(x) )? Let me check:If ( h(x) = x f(x) ), then ( h'(x) = f(x) + x f'(x) ). Yes, exactly! So ( h'(x) > 0 ) for all ( x neq 0 ). That tells us that ( h(x) ) is increasing on intervals where ( x neq 0 ).But wait, ( h(x) = x f(x) ). Since ( f(x) ) is odd, let's see what ( h(-x) ) is:( h(-x) = (-x) f(-x) = (-x)(-f(x)) = x f(x) = h(x) ). So ( h(x) ) is an even function. That means it's symmetric about the y-axis. So the behavior of ( h(x) ) for positive ( x ) will mirror its behavior for negative ( x ).Given that ( h'(x) > 0 ) for all ( x neq 0 ), this tells us that ( h(x) ) is increasing on ( (0, infty) ) and decreasing on ( (-infty, 0) ). But since ( h(x) ) is even, its increasing on the positive side and decreasing on the negative side.Wait, actually, since ( h'(x) > 0 ) for ( x neq 0 ), that means for ( x > 0 ), ( h(x) ) is increasing, and for ( x < 0 ), ( h(x) ) is also increasing because ( h'(x) > 0 ) there as well. But since ( h(x) ) is even, the function's behavior on the negative side is a mirror image of the positive side. So, in terms of increasing or decreasing, it's increasing as ( x ) moves away from zero in both directions.But let's focus on ( x > 0 ) since ( h(x) ) is increasing there. So, if ( x_1 < x_2 ) for ( x_1, x_2 > 0 ), then ( h(x_1) < h(x_2) ).Now, let's look at the definitions of ( a ), ( b ), and ( c ):- ( a = frac{1}{2} fleft( frac{1}{2} right) )- ( b = -2 f(-2) )- ( c = left( ln frac{1}{2} right) fleft( ln frac{1}{2} right) )Let's express each of these in terms of ( h(x) ):- ( a = frac{1}{2} fleft( frac{1}{2} right) = hleft( frac{1}{2} right) )- ( b = -2 f(-2) = -2 (-f(2)) = 2 f(2) = h(2) )- ( c = left( ln frac{1}{2} right) fleft( ln frac{1}{2} right) ). Since ( ln frac{1}{2} = -ln 2 ), and ( f ) is odd, ( f(-ln 2) = -f(ln 2) ). So, ( c = (-ln 2) (-f(ln 2)) = (ln 2) f(ln 2) = h(ln 2) ).So now, ( a = hleft( frac{1}{2} right) ), ( b = h(2) ), and ( c = h(ln 2) ).We need to compare ( a ), ( b ), and ( c ). Since ( h(x) ) is increasing for ( x > 0 ), we can compare the arguments:- ( frac{1}{2} approx 0.5 )- ( ln 2 approx 0.693 )- ( 2 ) is obviously larger.So, ordering the arguments from smallest to largest: ( frac{1}{2} < ln 2 < 2 ).Since ( h(x) ) is increasing, the function values will follow the same order:( hleft( frac{1}{2} right) < h(ln 2) < h(2) ).Therefore, ( a < c < b ).Wait, but let me double-check. Is ( h(x) ) increasing for all ( x > 0 )? Yes, because ( h'(x) > 0 ) there. So, as ( x ) increases, ( h(x) ) increases. So, the larger the ( x ), the larger the ( h(x) ).So, since ( frac{1}{2} < ln 2 < 2 ), then ( hleft( frac{1}{2} right) < h(ln 2) < h(2) ), which means ( a < c < b ).Looking back at the options:A: ( a < c < b )B: ( b < c < a )C: ( a < b < c )D: ( c < a < b )So, the correct answer should be A.But just to make sure, let's recap:1. Defined ( h(x) = x f(x) ).2. Showed ( h(x) ) is even because ( f(x) ) is odd.3. Found ( h'(x) = f(x) + x f'(x) > 0 ) for ( x neq 0 ), so ( h(x) ) is increasing for ( x > 0 ).4. Expressed ( a ), ( b ), ( c ) in terms of ( h(x) ).5. Compared the arguments of ( h(x) ) and ordered them.6. Concluded the order of ( a ), ( c ), ( b ).Everything seems to check out. So, yeah, I think A is the correct answer.**Final Answer**boxed{A}

💡First, I need to determine the total time Brooke spends on math and science problems.For math problems:15 problems multiplied by 2 minutes per problem equals 30 minutes.For science problems:10 problems multiplied by 1.5 minutes per problem equals 15 minutes.Adding the time spent on math and science problems:30 minutes (math) plus 15 minutes (science) equals 45 minutes.Brooke spends a total of 48 minutes on all his homework, so the remaining time for social studies problems is:48 minutes (total) minus 45 minutes (math and science) equals 3 minutes.Since each social studies problem takes 30 seconds (which is 0.5 minutes) to answer, I can calculate the number of social studies problems by dividing the total time spent on social studies by the time per problem:3 minutes divided by 0.5 minutes per problem equals 6 problems.Therefore, Brooke has 6 social studies problems.

💡First, I need to solve the equation ((x + 3)^2 / (3x + 29) = 2).To eliminate the fraction, I'll multiply both sides by (3x + 29), resulting in ((x + 3)^2 = 2(3x + 29)).Next, I'll expand the left side and distribute the right side:[x^2 + 6x + 9 = 6x + 58]Then, I'll move all terms to one side to set the equation to zero:[x^2 + 6x + 9 - 6x - 58 = 0]Simplifying the equation gives:[x^2 - 49 = 0]This is a difference of squares, which can be factored as:[(x + 7)(x - 7) = 0]Setting each factor equal to zero provides the two possible values of (x):[x + 7 = 0 quad text{or} quad x - 7 = 0][x = -7 quad text{or} quad x = 7]Finally, the difference between the two possible values of (x) is:[7 - (-7) = 14]

💡Okay, so I've got this problem about a ship and its cargo. Let me try to understand what's going on here. The ship has a maximum capacity of 20,000 tons. Initially, it's loaded with 5,973 tons of cargo, which is broken down into three categories: steel, timber, and electronics. Then, it stops in the Bahamas, and they load an additional 8,723 tons of cargo, which includes steel, textiles, timber, and electronics. First, I need to figure out if the ship exceeds its maximum capacity after loading the additional cargo in the Bahamas. If it does, I have to calculate how much cargo needs to be offloaded. Then, assuming we need to offload some cargo to maximize profit, I need to determine which types of cargo should be prioritized for offloading based on their total weight and import tax. Finally, I need to calculate the total import tax cost for the final cargo load.Alright, let's start by calculating the total cargo after the ship stops in the Bahamas. The initial cargo is 5,973 tons, and they add 8,723 tons. So, 5,973 plus 8,723 equals... let me do the math... 5,973 plus 8,723 is 14,696 tons. Now, the ship's maximum capacity is 20,000 tons. So, 20,000 minus 14,696 equals... 5,304 tons. That means the ship still has 5,304 tons of capacity left. So, it doesn't exceed its capacity, and no cargo needs to be offloaded. But just to be thorough, let's break down the cargo by type after loading in the Bahamas. Initially, there were 3,428 tons of steel, 1,244 tons of timber, and 1,301 tons of electronics. After loading in the Bahamas, they added 3,057 tons of steel, 2,364 tons of textiles, 1,517 tons of timber, and 1,785 tons of electronics. So, the total steel cargo is 3,428 plus 3,057, which is 6,485 tons. The total timber cargo is 1,244 plus 1,517, which is 2,761 tons. The total electronics cargo is 1,301 plus 1,785, which is 3,086 tons. And the textiles are just the 2,364 tons loaded in the Bahamas.Now, let's calculate the import taxes for each type of cargo. The import taxes are as follows: steel is 50 per ton, timber is 75 per ton, electronics are 100 per ton, and textiles are 40 per ton.So, for steel, 6,485 tons times 50 per ton equals... 6,485 times 50 is 324,250. For timber, 2,761 tons times 75 per ton equals... 2,761 times 75 is 207,075. For electronics, 3,086 tons times 100 per ton equals... 3,086 times 100 is 308,600. For textiles, 2,364 tons times 40 per ton equals... 2,364 times 40 is 94,560. Now, let's add up all these import taxes to get the total import tax cost. So, 324,250 plus 207,075 is 531,325. Then, 531,325 plus 308,600 is 839,925. Finally, 839,925 plus 94,560 is 934,485. So, the total import tax cost for the final cargo load is 934,485.Since the ship does not exceed its capacity, there's no need to offload any cargo. However, if we were to consider offloading to maximize profit, we would prioritize offloading the cargo with the lowest import tax per ton. In this case, textiles have the lowest import tax at 40 per ton, followed by steel at 50 per ton, then timber at 75 per ton, and finally electronics at 100 per ton. So, if offloading were necessary, textiles would be the first to go to minimize the loss in import taxes.But in this scenario, since the ship is well within its capacity, we don't need to worry about offloading any cargo. The total import tax cost remains at 934,485.

💡To solve for the natural number ( n ) such that ( sqrt{12 - n} ) is an integer, we can follow these steps:1. **Understand the Condition**: The expression ( sqrt{12 - n} ) must be an integer. Let's denote this integer as ( k ). Therefore, we have: [ sqrt{12 - n} = k ] 2. **Square Both Sides**: To eliminate the square root, square both sides of the equation: [ 12 - n = k^2 ] 3. **Solve for ( n )**: Rearrange the equation to solve for ( n ): [ n = 12 - k^2 ] 4. **Determine Possible Values of ( k )**: Since ( n ) is a natural number, ( 12 - k^2 ) must also be a natural number. This means ( k^2 ) must be less than or equal to 12. The possible integer values of ( k ) are 0, 1, 2, and 3 because: [ 0^2 = 0,quad 1^2 = 1,quad 2^2 = 4,quad 3^2 = 9 ] Note that ( k = 4 ) would result in ( 4^2 = 16 ), which is greater than 12 and thus not valid.5. **Calculate Corresponding Values of ( n )**: - For ( k = 0 ): [ n = 12 - 0^2 = 12 ] - For ( k = 1 ): [ n = 12 - 1^2 = 11 ] - For ( k = 2 ): [ n = 12 - 2^2 = 8 ] - For ( k = 3 ): [ n = 12 - 3^2 = 3 ] 6. **Select the Appropriate Value**: The problem asks for one possible value of ( n ). Among the possible values (3, 8, 11, 12), any of these would satisfy the condition. However, if we are to provide a single answer, we can choose the largest value, which is 12.Therefore, one possible value of the natural number ( n ) that satisfies the condition is:

💡Okay, so I have this equation of a hyperbola: (frac{x^2}{4} - frac{y^2}{9} = 1). I need to find the equations of its asymptotes. Hmm, I remember that asymptotes are lines that the hyperbola approaches but never touches. They kind of guide the direction in which the hyperbola opens.First, I should recall the standard form of a hyperbola. I think it's either (frac{x^2}{a^2} - frac{y^2}{b^2} = 1) or (frac{y^2}{b^2} - frac{x^2}{a^2} = 1). Since in my equation, the (x^2) term is positive, it must be the first form, meaning it's a hyperbola that opens left and right along the x-axis.Okay, so in this case, (a^2 = 4) and (b^2 = 9). To find (a) and (b), I just take the square roots. So, (a = sqrt{4} = 2) and (b = sqrt{9} = 3). Got that down.Now, I remember that the asymptotes of a hyperbola are straight lines that pass through the center of the hyperbola. Since the hyperbola is centered at the origin (because there are no shifts in the equation), the asymptotes will also pass through the origin.The formula for the asymptotes of a hyperbola that opens horizontally (which is the case here) is (y = pm frac{b}{a}x). Let me make sure I remember that correctly. Yeah, I think it's plus or minus (b/a) times x. So, plugging in the values I found, (b = 3) and (a = 2), so the slopes are (pm frac{3}{2}).Therefore, the equations of the asymptotes should be (y = frac{3}{2}x) and (y = -frac{3}{2}x). Let me visualize this. If I plot these lines, they should intersect at the origin and have slopes of 1.5 and -1.5. The hyperbola will approach these lines as x and y get larger in magnitude.Wait, let me double-check if I mixed up (a) and (b). Sometimes I get confused between the two. In the standard form, (a) is associated with the transverse axis, which is the one that the hyperbola opens along. Since this hyperbola opens along the x-axis, (a) is under the (x^2) term, so (a = 2). (b) is under the (y^2) term, so (b = 3). So, the asymptotes should indeed be (y = pm frac{b}{a}x = pm frac{3}{2}x).I think that's correct. To be thorough, maybe I can derive the asymptotes from the hyperbola equation itself. If I rearrange the hyperbola equation:[frac{x^2}{4} - frac{y^2}{9} = 1]If I consider the asymptotes, they are the lines that the hyperbola approaches as (x) and (y) become very large. So, for very large (x) and (y), the constant term 1 on the right side becomes negligible compared to the other terms. So, the equation approximately becomes:[frac{x^2}{4} - frac{y^2}{9} approx 0]Which can be rewritten as:[frac{x^2}{4} = frac{y^2}{9}]Taking square roots on both sides (and considering both positive and negative roots):[frac{x}{2} = pm frac{y}{3}]Multiplying both sides by 2:[x = pm frac{2y}{3}]Wait, that seems a bit different. Let me solve for y instead:Starting from (frac{x^2}{4} = frac{y^2}{9}), take square roots:[frac{x}{2} = pm frac{y}{3}]Then, solving for y:[y = pm frac{3}{2}x]Ah, okay, that matches what I had earlier. So, the asymptotes are (y = pm frac{3}{2}x). Good, that confirms my initial answer.Just to make sure, maybe I can plug in a large value of x and see if y approaches the asymptote. Let's say x = 1000. Plugging into the hyperbola equation:[frac{(1000)^2}{4} - frac{y^2}{9} = 1]Simplify:[250000 - frac{y^2}{9} = 1]So,[frac{y^2}{9} = 250000 - 1 = 249999]Multiply both sides by 9:[y^2 = 249999 times 9 = 2249991]Take square root:[y approx sqrt{2249991} approx 1499.997]So, y is approximately 1500 when x is 1000. Now, according to the asymptote (y = frac{3}{2}x), when x = 1000, y should be (frac{3}{2} times 1000 = 1500). That's exactly what we got. So, the hyperbola approaches the asymptote as x increases.Similarly, for negative x, say x = -1000, y would be approximately -1500, which again matches the asymptote (y = -frac{3}{2}x).Therefore, I'm confident that the asymptotes are (y = pm frac{3}{2}x). I think I've got it right.