answer：Okay, so I have this problem where I need to find the equation of a line that's symmetric to the line x - y + 1 = 0 with respect to the y-axis. The options are A: x + y - 1 = 0, B: x - y + 1 = 0, C: x + y + 1 = 0, and D: x - y - 1 = 0. Hmm, let me think about how to approach this.First, I remember that reflecting a line over the y-axis involves changing the sign of the x-coordinate. So, if I have a point (x, y) on the original line, its reflection over the y-axis would be (-x, y). Maybe I can use this idea to find the equation of the symmetric line.Let me write down the original equation: x - y + 1 = 0. I can rearrange this to get y in terms of x, which might make it easier to see what's happening. So, subtracting x and 1 from both sides gives me y = x + 1. Okay, so the original line has a slope of 1 and a y-intercept at (0, 1).Now, reflecting this line over the y-axis. If I take a general point (x, y) on the original line, its reflection would be (-x, y). So, if I substitute x with -x in the original equation, I should get the equation of the reflected line. Let me try that.Starting with y = x + 1, replacing x with -x gives y = (-x) + 1, which simplifies to y = -x + 1. Hmm, that looks like a different line. Let me write this in standard form. Subtracting y from both sides, I get x + y - 1 = 0. Wait, that's option A. So, is that the answer?But let me double-check to make sure I didn't make a mistake. Another way to think about reflection over the y-axis is that it changes the direction of the slope. The original line has a slope of 1, so the reflected line should have a slope of -1. That makes sense because reflecting over the y-axis flips the direction of the line. So, the slope changes sign.Also, the y-intercept should remain the same because reflecting over the y-axis doesn't change the y-value. So, the original line crosses the y-axis at (0, 1), and the reflected line should also cross the y-axis at (0, 1). That matches with the equation y = -x + 1, which has a y-intercept at 1.Let me visualize this. The original line y = x + 1 is a diagonal line going up to the right, passing through (0,1). The reflected line y = -x + 1 should be a diagonal line going down to the right, also passing through (0,1). So, they should be mirror images across the y-axis, which seems correct.Wait, let me check the options again. Option A is x + y - 1 = 0, which is the same as y = -x + 1. That's exactly what I got. Option B is the original equation, so that's not it. Option C is x + y + 1 = 0, which would have a y-intercept at -1, which is different. Option D is x - y - 1 = 0, which simplifies to y = x - 1, which has a different y-intercept and the same slope as the original line, so that's not the reflection.Just to be thorough, maybe I can test a point. Let's take a point on the original line, say when x = 1, y = 1 + 1 = 2. So, the point (1, 2) is on the original line. Reflecting this over the y-axis should give (-1, 2). Let me plug (-1, 2) into the equation of option A: (-1) + 2 - 1 = 0. That simplifies to 0, so it works. What about option D? Plugging (-1, 2) into x - y - 1 = 0: (-1) - 2 - 1 = -4, which is not 0. So, that doesn't work.Another point: when x = 0, y = 1 on both lines, which makes sense because that's the y-intercept and it's on the y-axis, so it doesn't change when reflected. So, (0,1) is on both lines, which is consistent.I think I'm confident now that option A is the correct answer. The reflection over the y-axis changes the slope's sign and keeps the y-intercept the same, leading to the equation x + y - 1 = 0.**Final Answer**boxed{A}

answer：To determine the rate of the current, let's denote the speed of the current as ( c ) km/hr.1. **Downstream Speed**: When rowing downstream, the effective speed is the sum of the man's rowing speed and the current's speed, which is ( 4.5 + c ) km/hr.2. **Upstream Speed**: When rowing upstream, the effective speed is the difference between the man's rowing speed and the current's speed, which is ( 4.5 - c ) km/hr.3. **Time Relationship**: It takes twice as long to row upstream as it does to row downstream. If we let ( d ) be the distance rowed in each direction, the time taken to row downstream is ( frac{d}{4.5 + c} ) hours, and the time taken to row upstream is ( frac{d}{4.5 - c} ) hours.4. **Setting Up the Equation**: According to the problem, ( frac{d}{4.5 - c} = 2 times frac{d}{4.5 + c} ). By canceling ( d ) from both sides, we get ( frac{1}{4.5 - c} = 2 times frac{1}{4.5 + c} ).5. **Solving for ( c )**: [ 1 = 2 times frac{4.5 - c}{4.5 + c} ] [ 4.5 + c = 2 times (4.5 - c) ] [ 4.5 + c = 9 - 2c ] [ 3c = 4.5 ] [ c = frac{4.5}{3} = 1.5 ]Therefore, the rate of the current is ( 1.5 ) km/hr.

answer：First, I need to identify the two smallest elements in the set S, which are -4 and -1. To minimize the increase in the mean, I should replace these with the smallest possible prime numbers, which are 2 and 3.Next, I'll calculate the original mean of the set S. The sum of the elements is -4 - 1 + 0 + 6 + 9, which equals 10. Dividing by the number of elements (5), the original mean is 2.After replacing -4 and -1 with 2 and 3, the new set becomes {2, 3, 0, 6, 9}. The sum of the new set is 2 + 3 + 0 + 6 + 9, totaling 20. Dividing by 5, the new mean is 4.Finally, to find the percentage increase in the mean, I'll use the formula: ((New Mean - Original Mean) / Original Mean) * 100. Plugging in the values, ((4 - 2) / 2) * 100 equals 100%.

answer：Okay, so I have this problem where I'm given a polynomial ( f(x) ) of degree ( n ) that takes integer values at the points ( x = 0, 1, ldots, n ). I need to prove that this polynomial can be expressed in terms of binomial coefficients with integer coefficients. Specifically, I need to show that:[f(x) = d_0 binom{x}{0} + d_1 binom{x}{1} + ldots + d_n binom{x}{n}]where ( d_0, d_1, ldots, d_n ) are integers.Alright, let's start by recalling some basics. A polynomial of degree ( n ) is determined uniquely by its values at ( n+1 ) distinct points. In this case, we know the values at ( x = 0, 1, ldots, n ). So, in principle, we can express ( f(x) ) using interpolation. But the problem is asking for a specific form involving binomial coefficients, so I need to think about how to connect these two ideas.I remember that binomial coefficients ( binom{x}{k} ) are polynomials in ( x ) of degree ( k ). For example, ( binom{x}{0} = 1 ), ( binom{x}{1} = x ), ( binom{x}{2} = frac{x(x-1)}{2} ), and so on. These polynomials form a basis for the space of polynomials. That means any polynomial can be expressed as a linear combination of these binomial coefficients. So, in theory, ( f(x) ) can be written as:[f(x) = d_0 binom{x}{0} + d_1 binom{x}{1} + ldots + d_n binom{x}{n}]for some coefficients ( d_0, d_1, ldots, d_n ). The question is, are these coefficients necessarily integers?Given that ( f(x) ) takes integer values at ( x = 0, 1, ldots, n ), and the binomial coefficients ( binom{x}{k} ) evaluated at integer points ( x ) are integers, it seems plausible that the coefficients ( d_k ) must also be integers. But I need to make this precise.Maybe I can use the method of finite differences. I recall that the ( k )-th finite difference of a polynomial of degree ( n ) is a polynomial of degree ( n - k ). Moreover, the ( n )-th finite difference of a degree ( n ) polynomial is a constant. This might be useful.Let me define the finite difference operator ( Delta ) as:[Delta f(x) = f(x + 1) - f(x)]Then, the ( k )-th finite difference ( Delta^k f(x) ) can be defined recursively:[Delta^k f(x) = Delta(Delta^{k-1} f(x))]I also remember that the ( k )-th finite difference of ( binom{x}{k} ) is 1. Let me verify that:[Delta binom{x}{k} = binom{x + 1}{k} - binom{x}{k} = binom{x}{k - 1}]Yes, that's correct. So, applying the finite difference operator ( k ) times to ( binom{x}{k} ) gives 1. This suggests that the coefficients ( d_k ) can be related to the finite differences of ( f(x) ) at ( x = 0 ).Specifically, if I take the ( k )-th finite difference of ( f(x) ) at ( x = 0 ), I should get ( d_k ). Let me see:[Delta^k f(0) = d_k]Since ( f(x) ) takes integer values at ( x = 0, 1, ldots, n ), the finite differences ( Delta^k f(0) ) must also be integers. Therefore, ( d_k ) are integers.Putting this all together, since ( f(x) ) can be expressed as a linear combination of binomial coefficients with coefficients equal to the finite differences of ( f ) at 0, and these coefficients are integers, we have:[f(x) = d_0 binom{x}{0} + d_1 binom{x}{1} + ldots + d_n binom{x}{n}]where ( d_0, d_1, ldots, d_n ) are integers.I think this makes sense. The key steps were recognizing that binomial coefficients form a basis for polynomials, understanding how finite differences relate to the coefficients, and using the fact that the finite differences must be integers because ( f(x) ) takes integer values at integer points.Just to double-check, let me consider a simple example. Suppose ( n = 2 ), so ( f(x) ) is a quadratic polynomial taking integer values at ( x = 0, 1, 2 ). Let's say ( f(0) = a ), ( f(1) = b ), ( f(2) = c ), where ( a, b, c ) are integers.Expressing ( f(x) ) in terms of binomial coefficients:[f(x) = d_0 binom{x}{0} + d_1 binom{x}{1} + d_2 binom{x}{2}]Evaluating at ( x = 0 ):[f(0) = d_0 = a]At ( x = 1 ):[f(1) = d_0 + d_1 = b implies d_1 = b - a]At ( x = 2 ):[f(2) = d_0 + 2d_1 + d_2 = c implies d_2 = c - 2d_1 - d_0 = c - 2(b - a) - a = c - 2b + 2a - a = c - 2b + a]So, ( d_0 = a ), ( d_1 = b - a ), ( d_2 = c - 2b + a ). Since ( a, b, c ) are integers, ( d_0, d_1, d_2 ) are integers as well. This confirms that the coefficients are integers in this case.Therefore, the general proof should hold by induction or by using the properties of finite differences as I outlined earlier.

answer：Okay, so I'm trying to figure out how many birds there are if there are 20 bird wings in the sky. Hmm, let's see. First, I know that birds have wings, and most birds have two wings. So, if each bird has two wings, then the total number of wings would be twice the number of birds. Wait, but is that always true? Are there any birds that have more than two wings? I don't think so. I mean, I know that some birds might have different numbers of wings due to injuries or mutations, but generally, birds have two wings. So, for the sake of this problem, I'll assume that each bird has two wings.So, if there are 20 wings in total, and each bird contributes two wings, then the number of birds should be 20 divided by 2, which is 10. That seems straightforward. But let me double-check to make sure I'm not missing anything.Is there any other factor I need to consider? Maybe the possibility of some wings being hidden or not visible? But the problem says there are 20 bird wings in the sky, so I think it's safe to assume that all the wings are accounted for. Also, are there any other creatures in the sky that have wings? Like, maybe bats or insects? But the problem specifically mentions bird wings, so I don't need to worry about other creatures.Another thing to consider is whether the wings are from the same type of bird or different types. But since the problem doesn't specify, I'll stick with the general assumption that each bird has two wings.So, going back to the calculation: 20 wings divided by 2 wings per bird equals 10 birds. That seems to make sense. I don't see any complications or additional factors that I need to account for in this problem.Just to visualize, if I imagine 10 birds flying in the sky, each with two wings, that would indeed give me 20 wings in total. So, the answer should be 10 birds.I think I've covered all the bases here. I've considered the number of wings per bird, whether there are any exceptions, and whether there are other creatures involved. Everything points to the number of birds being 10.

answer：First, I need to determine the work rates of one man and one woman. I'll denote the work rate of one man as ( M ) and the work rate of one woman as ( W ). The total work is considered as 1 unit.From the problem, 10 men and 15 women can complete the work in 5 days. This gives the equation:[(10M + 15W) times 5 = 1]It's also given that one man alone takes 100 days to complete the work, which provides:[M times 100 = 1 quad Rightarrow quad M = frac{1}{100}]Substituting ( M = frac{1}{100} ) into the first equation:[left(10 times frac{1}{100} + 15Wright) times 5 = 1][left(frac{1}{10} + 15Wright) times 5 = 1][frac{1}{10} + 15W = frac{1}{5}][15W = frac{1}{5} - frac{1}{10} = frac{1}{10}][W = frac{1}{150}]Finally, to find how many days one woman alone will take to complete the work:[W times text{days} = 1 quad Rightarrow quad frac{1}{150} times text{days} = 1 quad Rightarrow quad text{days} = 150]