answer：First, I need to simplify the given equation to make it easier to work with. I'll divide every term by 3 to reduce the coefficients.Next, I'll rearrange the equation to group the (x) terms and (y) terms together, which will help in completing the square.For the (x) terms, I'll complete the square by adding and subtracting the square of half the coefficient of (x). Similarly, I'll do the same for the (y) terms.After completing the square for both (x) and (y), I'll rewrite the equation in the standard form of a circle's equation, which is ((x - h)^2 + (y - k)^2 = r^2).From the standard form, I'll identify the radius of the circle. Finally, I'll calculate the area of the circle using the formula (A = pi r^2) and present the answer in terms of (pi).

answer：Alright, so I need to find the inverse of the function ( f(x) = frac{3x - 2}{x + 4} ) and express it in the form ( f^{-1}(x) = frac{ax + b}{cx + d} ). Then, I have to find the ratio ( frac{a}{c} ). Hmm, okay, let's break this down step by step.First, I remember that to find the inverse of a function, I need to switch the roles of ( x ) and ( y ) and then solve for ( y ). So, let me start by writing the function as an equation:( y = frac{3x - 2}{x + 4} )Now, to find the inverse, I'll swap ( x ) and ( y ):( x = frac{3y - 2}{y + 4} )Okay, now I need to solve this equation for ( y ). Let's do that step by step.First, I'll multiply both sides of the equation by ( y + 4 ) to eliminate the denominator:( x(y + 4) = 3y - 2 )Expanding the left side:( xy + 4x = 3y - 2 )Now, I want to collect all the terms containing ( y ) on one side and the rest on the other side. Let me subtract ( 3y ) from both sides and also subtract ( xy ) from both sides:Wait, actually, let me think. If I subtract ( 3y ) and subtract ( xy ), that might complicate things. Maybe a better approach is to get all the ( y ) terms on the left and the constants on the right.So, starting from ( xy + 4x = 3y - 2 ), I can subtract ( 3y ) from both sides:( xy + 4x - 3y = -2 )Now, factor out ( y ) from the terms on the left:( y(x - 3) + 4x = -2 )Wait, actually, let me check that. If I have ( xy - 3y + 4x = -2 ), then factoring ( y ) from the first two terms gives:( y(x - 3) + 4x = -2 )Hmm, that seems correct. Now, I can subtract ( 4x ) from both sides to isolate the term with ( y ):( y(x - 3) = -2 - 4x )Now, to solve for ( y ), I'll divide both sides by ( x - 3 ):( y = frac{-2 - 4x}{x - 3} )Hmm, that looks like the inverse function. Let me write that as:( f^{-1}(x) = frac{-4x - 2}{x - 3} )Wait, but the problem asks for the inverse in the form ( frac{ax + b}{cx + d} ). Let me see if I can rearrange the numerator and denominator to match that form.Looking at the numerator: ( -4x - 2 ) can be written as ( -4x + (-2) ), so ( a = -4 ) and ( b = -2 ).The denominator is ( x - 3 ), which can be written as ( 1x + (-3) ), so ( c = 1 ) and ( d = -3 ).Wait, but in the problem statement, the inverse is written as ( frac{ax + b}{cx + d} ). So, comparing:( f^{-1}(x) = frac{-4x - 2}{1x - 3} )So, ( a = -4 ), ( b = -2 ), ( c = 1 ), and ( d = -3 ).But the question asks for ( frac{a}{c} ). So, ( a = -4 ) and ( c = 1 ), so ( frac{a}{c} = frac{-4}{1} = -4 ).Wait, but let me double-check my steps to make sure I didn't make a mistake.Starting from ( y = frac{3x - 2}{x + 4} ), swapping ( x ) and ( y ):( x = frac{3y - 2}{y + 4} )Multiply both sides by ( y + 4 ):( x(y + 4) = 3y - 2 )Expanding:( xy + 4x = 3y - 2 )Bringing all ( y ) terms to one side:( xy - 3y = -4x - 2 )Factoring ( y ):( y(x - 3) = -4x - 2 )Dividing both sides by ( x - 3 ):( y = frac{-4x - 2}{x - 3} )Yes, that seems correct. So, ( a = -4 ), ( c = 1 ), so ( frac{a}{c} = -4 ).Wait, but in the initial problem, the function is ( f(x) = frac{3x - 2}{x + 4} ). Let me verify if my inverse function is correct by composing ( f ) and ( f^{-1} ) to see if I get ( x ).Let me compute ( f(f^{-1}(x)) ):( fleft( frac{-4x - 2}{x - 3} right) = frac{3left( frac{-4x - 2}{x - 3} right) - 2}{left( frac{-4x - 2}{x - 3} right) + 4} )Simplify numerator:( 3left( frac{-4x - 2}{x - 3} right) = frac{-12x - 6}{x - 3} )Subtract 2:( frac{-12x - 6}{x - 3} - 2 = frac{-12x - 6 - 2(x - 3)}{x - 3} = frac{-12x - 6 - 2x + 6}{x - 3} = frac{-14x}{x - 3} )Denominator:( frac{-4x - 2}{x - 3} + 4 = frac{-4x - 2 + 4(x - 3)}{x - 3} = frac{-4x - 2 + 4x - 12}{x - 3} = frac{-14}{x - 3} )So, ( f(f^{-1}(x)) = frac{-14x}{x - 3} div frac{-14}{x - 3} = frac{-14x}{x - 3} times frac{x - 3}{-14} = x )Yes, that checks out. So, my inverse function is correct.Therefore, ( a = -4 ), ( c = 1 ), so ( frac{a}{c} = -4 ).Wait, but in the initial problem, the function is given as ( f(x) = frac{3x - 2}{x + 4} ). Let me make sure I didn't make a mistake in the signs when solving for ( y ).Starting again from:( x = frac{3y - 2}{y + 4} )Multiply both sides by ( y + 4 ):( x(y + 4) = 3y - 2 )Expanding:( xy + 4x = 3y - 2 )Bring all ( y ) terms to the left:( xy - 3y = -4x - 2 )Factor ( y ):( y(x - 3) = -4x - 2 )Divide by ( x - 3 ):( y = frac{-4x - 2}{x - 3} )Yes, that's correct. So, ( a = -4 ), ( c = 1 ), so ( frac{a}{c} = -4 ).I think that's the correct answer. I don't see any mistakes in my steps, and the verification by composition worked out.

answer：Okay, so I have this function f(x) = ln(√(1 + x²) - x) + 1, and I know that f(a) = 4. I need to find f(-a). Hmm, let's break this down step by step.First, let me understand the function f(x). It's composed of a natural logarithm and some algebraic expressions. The natural logarithm function, ln, is only defined for positive arguments, so I should check if √(1 + x²) - x is always positive. Well, since √(1 + x²) is always greater than or equal to |x|, subtracting x will still keep the argument positive, right? So, the function is defined for all real numbers x.Now, the function f(x) can be split into two parts: the logarithm part and the constant 1. Maybe if I consider the logarithm part separately, I can find some properties that might help me. Let me define g(x) = ln(√(1 + x²) - x). Then, f(x) = g(x) + 1.Given that f(a) = 4, that means g(a) + 1 = 4, so g(a) = 3. Therefore, ln(√(1 + a²) - a) = 3. If I exponentiate both sides to eliminate the natural logarithm, I get √(1 + a²) - a = e³.Wait, that's interesting. So, √(1 + a²) - a = e³. Maybe I can solve for a here, but I'm not sure if that's necessary. The problem is asking for f(-a), so perhaps I can find a relationship between g(a) and g(-a) without knowing the exact value of a.Let me compute g(-a). That would be ln(√(1 + (-a)²) - (-a)) = ln(√(1 + a²) + a). Hmm, that looks similar to the expression I had earlier, but with a plus a instead of minus a.Is there a relationship between g(a) and g(-a)? Let's see. If I multiply √(1 + a²) - a and √(1 + a²) + a, what do I get? That's (√(1 + a²))² - (a)² = (1 + a²) - a² = 1. So, (√(1 + a²) - a)(√(1 + a²) + a) = 1. Therefore, √(1 + a²) + a = 1 / (√(1 + a²) - a).So, g(-a) = ln(√(1 + a²) + a) = ln(1 / (√(1 + a²) - a)) = -ln(√(1 + a²) - a) = -g(a). Ah, so g(-a) = -g(a). That means g is an odd function. That's a useful property. Therefore, since g(a) = 3, then g(-a) = -3.Now, going back to f(-a). Since f(x) = g(x) + 1, then f(-a) = g(-a) + 1 = -3 + 1 = -2.Wait, that seems straightforward. Let me verify my steps to make sure I didn't make a mistake.1. Defined g(x) = ln(√(1 + x²) - x), so f(x) = g(x) + 1.2. Given f(a) = 4, so g(a) = 3.3. Calculated g(-a) by recognizing that √(1 + a²) + a is the reciprocal of √(1 + a²) - a, so ln of the reciprocal is the negative of the original ln.4. Therefore, g(-a) = -g(a) = -3.5. Thus, f(-a) = g(-a) + 1 = -3 + 1 = -2.Everything seems to check out. I don't see any errors in my reasoning. So, I think the value of f(-a) is -2.

answer：Okay, so I have this problem about lines on the coordinate plane. It's asking if there's a family of infinitely many lines ( l_1, l_2, ldots, l_n, ldots ) that satisfy three conditions. Let me try to unpack each condition and see what they mean.First, condition (I) says that each line ( l_n ) must pass through the point (1,1). That seems straightforward. So every line in this family has to go through (1,1). That probably means that each line can be written in a form that includes (1,1). Maybe using the point-slope form?Condition (II) is a bit more involved. It says that the slope of the next line, ( k_{n+1} ), is equal to ( a_n - b_n ), where ( a_n ) is the x-intercept and ( b_n ) is the y-intercept of the current line ( l_n ). Hmm, so each subsequent line's slope is determined by the difference between the x-intercept and y-intercept of the previous line. That seems like a recursive relationship.Condition (III) requires that the product of consecutive slopes ( k_n ) and ( k_{n+1} ) is non-negative. So ( k_n times k_{n+1} geq 0 ). This means that either both slopes are positive or both are negative. They can't have opposite signs. So the slopes can't alternate between positive and negative; they have to maintain the same sign throughout.Alright, so let's try to formalize this. Let's start by writing the equation of each line ( l_n ). Since each line passes through (1,1), we can write it in point-slope form:( y - 1 = k_n(x - 1) )Simplifying this, we get:( y = k_n x - k_n + 1 )So that's the equation of each line. Now, let's find the x-intercept ( a_n ) and y-intercept ( b_n ) for this line.To find the x-intercept, set ( y = 0 ):( 0 = k_n x - k_n + 1 )Solving for x:( k_n x = k_n - 1 )( x = frac{k_n - 1}{k_n} )So, ( a_n = frac{k_n - 1}{k_n} )To find the y-intercept, set ( x = 0 ):( y = 0 - k_n + 1 )So, ( b_n = 1 - k_n )Alright, so now we have expressions for ( a_n ) and ( b_n ) in terms of ( k_n ). According to condition (II), ( k_{n+1} = a_n - b_n ). Let's substitute the expressions we found:( k_{n+1} = frac{k_n - 1}{k_n} - (1 - k_n) )Simplify this:First, distribute the negative sign:( k_{n+1} = frac{k_n - 1}{k_n} - 1 + k_n )Now, let's combine the terms:( k_{n+1} = frac{k_n - 1}{k_n} + k_n - 1 )Let me get a common denominator for the first two terms:( frac{k_n - 1}{k_n} + frac{k_n^2 - k_n}{k_n} )Combine the numerators:( frac{k_n - 1 + k_n^2 - k_n}{k_n} )Simplify the numerator:( k_n - 1 + k_n^2 - k_n = k_n^2 - 1 )So, ( k_{n+1} = frac{k_n^2 - 1}{k_n} )Simplify this:( k_{n+1} = k_n - frac{1}{k_n} )Okay, so we have a recursive formula for the slopes:( k_{n+1} = k_n - frac{1}{k_n} )Now, condition (III) says that ( k_n times k_{n+1} geq 0 ). So, the product of consecutive slopes must be non-negative. Let's see what this implies.First, let's consider the sign of ( k_n ). Suppose ( k_n ) is positive. Then, for ( k_{n+1} ) to satisfy ( k_n times k_{n+1} geq 0 ), ( k_{n+1} ) must also be positive. Similarly, if ( k_n ) is negative, ( k_{n+1} ) must also be negative.So, the sequence of slopes must maintain the same sign throughout. They can't switch from positive to negative or vice versa.Let's analyze the recursive formula ( k_{n+1} = k_n - frac{1}{k_n} ).Case 1: ( k_n > 0 )Then, ( k_{n+1} = k_n - frac{1}{k_n} )Since ( k_n > 0 ), ( frac{1}{k_n} > 0 ), so ( k_{n+1} = k_n - text{positive} ). Therefore, ( k_{n+1} ) is less than ( k_n ). So, the sequence is decreasing.But we also have that ( k_{n+1} > 0 ) because of condition (III). So, as long as ( k_n > 0 ), ( k_{n+1} ) is positive but less than ( k_n ). So, the sequence is decreasing and bounded below by 0.What happens as ( n ) approaches infinity? If the sequence is decreasing and bounded below, it must converge to some limit ( L geq 0 ).Let's assume that ( lim_{n to infty} k_n = L ). Then, taking the limit on both sides of the recursive formula:( L = L - frac{1}{L} )Subtract ( L ) from both sides:( 0 = -frac{1}{L} )Multiply both sides by ( L ):( 0 = -1 )Wait, that's a contradiction. So, this suggests that our assumption that the sequence converges to a finite limit ( L ) is wrong. Therefore, the sequence can't converge to a finite limit. But since it's decreasing and bounded below by 0, it must converge to 0.But if ( L = 0 ), then as ( n ) approaches infinity, ( k_n ) approaches 0. Let's see if that's possible.If ( k_n ) approaches 0, then ( frac{1}{k_n} ) approaches infinity. So, ( k_{n+1} = k_n - frac{1}{k_n} ) would approach negative infinity. But that would mean ( k_{n+1} ) becomes negative, which contradicts condition (III) because ( k_n ) is approaching 0 from the positive side, and ( k_{n+1} ) would become negative, making their product negative.Therefore, if we start with a positive ( k_1 ), the sequence ( k_n ) would eventually become negative, violating condition (III). So, that's a problem.Case 2: ( k_n < 0 )Similarly, let's analyze if ( k_n ) is negative.( k_{n+1} = k_n - frac{1}{k_n} )Since ( k_n < 0 ), ( frac{1}{k_n} < 0 ), so ( -frac{1}{k_n} > 0 ). Therefore, ( k_{n+1} = k_n + text{positive} ). So, ( k_{n+1} ) is greater than ( k_n ). So, the sequence is increasing.But since ( k_n < 0 ), and the sequence is increasing, it's moving towards zero. Again, let's assume that ( lim_{n to infty} k_n = L ). Then, taking the limit:( L = L - frac{1}{L} )Again, subtract ( L ):( 0 = -frac{1}{L} )Multiply by ( L ):( 0 = -1 )Contradiction again. So, similar to the positive case, the sequence can't converge to a finite limit. Therefore, it must approach zero.But if ( k_n ) approaches zero from the negative side, then ( frac{1}{k_n} ) approaches negative infinity, so ( k_{n+1} = k_n - frac{1}{k_n} ) would approach positive infinity. That would mean ( k_{n+1} ) becomes positive, which again violates condition (III) because ( k_n ) is negative and ( k_{n+1} ) becomes positive, making their product negative.Therefore, whether we start with a positive or negative slope, the sequence of slopes ( k_n ) will eventually switch sign, violating condition (III). Hence, it's impossible to have such an infinite family of lines that satisfy all three conditions.Wait, but what if we choose ( k_n ) such that the sequence doesn't approach zero? Is that possible?Well, if we start with ( k_1 ) such that ( k_{n} ) doesn't approach zero, but from the recursive formula ( k_{n+1} = k_n - frac{1}{k_n} ), it seems that regardless of the starting value, the sequence will either decrease (if positive) or increase (if negative) towards zero, but crossing zero would cause the slope to switch sign, which is not allowed.Alternatively, maybe there's a fixed point where ( k_{n+1} = k_n ). Let's check:Set ( k_{n+1} = k_n ):( k_n = k_n - frac{1}{k_n} )Subtract ( k_n ):( 0 = -frac{1}{k_n} )Which implies ( frac{1}{k_n} = 0 ), which is impossible because ( k_n ) is finite. So, there's no fixed point except at infinity, which isn't practical.Therefore, it seems that no matter what starting slope we choose, the sequence will eventually lead to a contradiction with condition (III). Hence, such an infinite family of lines cannot exist.I think that's the conclusion. There's no way to have infinitely many lines passing through (1,1) with slopes following the given recursive relationship without eventually violating the sign condition on the product of consecutive slopes.

answer：Okay, so I have this problem where I need to find the ordered pair (A, B), where A is the sum of the x-coordinates of the intersection points of two graphs, and B is the sum of the y-coordinates of those points. The graphs given are a cubic function and a linear equation.First, let me write down the equations to make sure I have them correctly:1. The cubic function: ( y = x^3 - 2x^2 + x + 1 )2. The linear equation: ( x + 5y = 5 )I need to find the points where these two graphs intersect. That means I need to solve these equations simultaneously. To do that, I can substitute the expression for y from the linear equation into the cubic equation.Let me rearrange the linear equation to express y in terms of x. Starting with ( x + 5y = 5 ), if I subtract x from both sides, I get ( 5y = -x + 5 ). Then, dividing both sides by 5, I have ( y = -frac{x}{5} + 1 ). Okay, so that's the expression for y in terms of x.Now, I can substitute this into the cubic equation. So, replacing y in the cubic equation with ( -frac{x}{5} + 1 ), the equation becomes:( -frac{x}{5} + 1 = x^3 - 2x^2 + x + 1 )Hmm, let me write that out step by step:( x^3 - 2x^2 + x + 1 = -frac{x}{5} + 1 )Now, I need to solve for x. Let me bring all the terms to one side so that I can have a polynomial equation equal to zero. Subtract ( -frac{x}{5} + 1 ) from both sides:( x^3 - 2x^2 + x + 1 - (-frac{x}{5} + 1) = 0 )Simplify that:( x^3 - 2x^2 + x + 1 + frac{x}{5} - 1 = 0 )Combine like terms. The constants 1 and -1 cancel out. For the x terms, I have x and ( frac{x}{5} ). Let's combine those:( x + frac{x}{5} = frac{5x}{5} + frac{x}{5} = frac{6x}{5} )So now, the equation simplifies to:( x^3 - 2x^2 + frac{6x}{5} = 0 )Hmm, that's a cubic equation. To make it easier, maybe I can eliminate the fraction by multiplying the entire equation by 5. Let's do that:( 5x^3 - 10x^2 + 6x = 0 )Okay, so now I have a cubic equation: ( 5x^3 - 10x^2 + 6x = 0 ). I can factor out an x:( x(5x^2 - 10x + 6) = 0 )So, this gives me two factors: x and ( 5x^2 - 10x + 6 ). Setting each factor equal to zero:1. ( x = 0 )2. ( 5x^2 - 10x + 6 = 0 )Let's solve the quadratic equation ( 5x^2 - 10x + 6 = 0 ) using the quadratic formula. The quadratic formula is ( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ), where a = 5, b = -10, and c = 6.Plugging in the values:( x = frac{-(-10) pm sqrt{(-10)^2 - 4*5*6}}{2*5} )Simplify:( x = frac{10 pm sqrt{100 - 120}}{10} )( x = frac{10 pm sqrt{-20}}{10} )Oh, wait, the discriminant is negative (( sqrt{-20} )), which means there are no real solutions from this quadratic equation. So, the only real solution is x = 0.Therefore, the cubic equation has only one real root at x = 0, and two complex roots. But since we're dealing with the intersection points of two graphs, which are real points, we only consider the real solution. So, there is only one intersection point at x = 0.But the problem statement mentions three points of intersection: ( (x_1, y_1) ), ( (x_2, y_2) ), and ( (x_3, y_3) ). Hmm, that's confusing because according to my calculation, there's only one real intersection point. Maybe I made a mistake somewhere.Let me double-check my steps.Starting from the substitution:( y = -frac{x}{5} + 1 ) into ( y = x^3 - 2x^2 + x + 1 )So:( -frac{x}{5} + 1 = x^3 - 2x^2 + x + 1 )Subtracting ( -frac{x}{5} + 1 ) from both sides:( 0 = x^3 - 2x^2 + x + 1 + frac{x}{5} - 1 )Simplify:( 0 = x^3 - 2x^2 + frac{6x}{5} )Multiply by 5:( 0 = 5x^3 - 10x^2 + 6x )Factor:( 0 = x(5x^2 - 10x + 6) )Quadratic formula on ( 5x^2 - 10x + 6 ):Discriminant: ( (-10)^2 - 4*5*6 = 100 - 120 = -20 )So, yes, discriminant is negative, so only one real solution at x=0.But the problem says three points of intersection. That suggests that maybe I made a mistake in the substitution or in simplifying.Wait, let me check the substitution again.Original cubic: ( y = x^3 - 2x^2 + x + 1 )Linear equation: ( x + 5y = 5 ) => ( y = (5 - x)/5 = 1 - x/5 )So, substituting into the cubic:( 1 - x/5 = x^3 - 2x^2 + x + 1 )Subtract 1 from both sides:( -x/5 = x^3 - 2x^2 + x )Bring all terms to left:( x^3 - 2x^2 + x + x/5 = 0 )Combine x terms:( x + x/5 = (5x + x)/5 = 6x/5 )So, equation becomes:( x^3 - 2x^2 + (6x)/5 = 0 )Multiply by 5:( 5x^3 - 10x^2 + 6x = 0 )Factor:( x(5x^2 - 10x + 6) = 0 )So, same result as before. So, perhaps the problem is expecting complex roots as well? But in reality, the graphs only intersect once. So, maybe the problem is considering all roots, real and complex, but in the context of the problem, it's talking about points of intersection, which are real. So, perhaps the problem is incorrect in stating three points? Or maybe I made a mistake.Wait, no, actually, the cubic equation is degree three, so it can have up to three real roots, but in this case, it only has one real root and two complex roots. So, perhaps the problem is considering all roots, even the complex ones, for the purpose of computing A and B.But in the context of the problem, it says "points" of intersection, which are (x, y) points. Since complex x and y wouldn't correspond to real points on the graph, perhaps the problem is only considering the real intersection point, but then A and B would just be the coordinates of that single point. But the problem says three points, so maybe it's expecting to include the complex roots as well for the sum.Alternatively, perhaps I made a mistake in the substitution.Wait, let me try another approach. Maybe I can express x in terms of y from the linear equation and substitute into the cubic.From ( x + 5y = 5 ), we have ( x = 5 - 5y ).Substitute this into the cubic equation:( y = (5 - 5y)^3 - 2(5 - 5y)^2 + (5 - 5y) + 1 )This might be more complicated, but let's try.First, compute ( (5 - 5y)^3 ):( (5 - 5y)^3 = 125 - 375y + 375y^2 - 125y^3 )Then, compute ( -2(5 - 5y)^2 ):( (5 - 5y)^2 = 25 - 50y + 25y^2 ), so multiplying by -2: ( -50 + 100y - 50y^2 )Then, ( (5 - 5y) = 5 - 5y )Adding all together:( y = [125 - 375y + 375y^2 - 125y^3] + [-50 + 100y - 50y^2] + [5 - 5y] + 1 )Now, let's combine like terms:Constants: 125 - 50 + 5 + 1 = 81y terms: -375y + 100y -5y = (-375 + 100 -5)y = (-280)yy^2 terms: 375y^2 -50y^2 = 325y^2y^3 term: -125y^3So, the equation becomes:( y = -125y^3 + 325y^2 - 280y + 81 )Bring all terms to the left:( 125y^3 - 325y^2 + 280y - 81 + y = 0 )Wait, that's because I subtracted y from both sides:( 0 = -125y^3 + 325y^2 - 280y + 81 - y )Which simplifies to:( 0 = -125y^3 + 325y^2 - 281y + 81 )Multiply both sides by -1:( 125y^3 - 325y^2 + 281y - 81 = 0 )Hmm, that's a cubic in y. Maybe this approach is more complicated. Perhaps I should stick with the previous substitution.Wait, but in the previous substitution, I only found one real x. Maybe the problem is expecting to consider all roots, real and complex, for the sum A and B.In that case, for the cubic equation ( 5x^3 - 10x^2 + 6x = 0 ), which factors as ( x(5x^2 - 10x + 6) = 0 ), the roots are x=0, and the roots of ( 5x^2 - 10x + 6 = 0 ), which are complex.So, for the sum of the roots, Vieta's formula tells us that the sum of the roots of the cubic equation ( ax^3 + bx^2 + cx + d = 0 ) is -b/a.In our case, the cubic equation is ( 5x^3 - 10x^2 + 6x = 0 ). So, a=5, b=-10.Thus, the sum of the roots is -b/a = -(-10)/5 = 10/5 = 2. So, A = 2.Similarly, for the sum of the y-coordinates, since each y is determined by x, we can express y in terms of x as ( y = -x/5 + 1 ). So, the sum of the y's is ( y_1 + y_2 + y_3 = (-x_1/5 + 1) + (-x_2/5 + 1) + (-x_3/5 + 1) ).Simplify:( (-1/5)(x_1 + x_2 + x_3) + 3*1 )We already know that ( x_1 + x_2 + x_3 = 2 ), so:( (-1/5)(2) + 3 = -2/5 + 3 = (-2/5) + (15/5) = 13/5 )Wait, but in the initial problem, the user's solution said that y1 + y2 + y3 = 1, but according to this, it should be 13/5. Hmm, that's conflicting.Wait, let me check again.From the cubic equation, the sum of x's is 2. Each y is ( y = -x/5 + 1 ). So, sum of y's is:( sum y_i = sum (-x_i/5 + 1) = -frac{1}{5}sum x_i + 3*1 )Which is:( -frac{1}{5}(2) + 3 = -frac{2}{5} + 3 = frac{-2 + 15}{5} = frac{13}{5} )So, B should be 13/5, not 1. But in the initial solution, the user said that y1 + y2 + y3 = 1, considering only the real root. But if we consider all roots, including complex ones, the sum is 13/5.But in the problem statement, it says "the graphs... intersect in points (x1,y1), (x2,y2), and (x3,y3)". Since the graphs are real, the intersection points are real, so only x=0 is a real solution, meaning only one point of intersection. So, why does the problem mention three points? Maybe it's a misstatement, or perhaps I'm misunderstanding.Alternatively, perhaps the user made a mistake in their initial solution by considering the complex roots for the sum, but in reality, only the real roots contribute to the sum of the coordinates of the intersection points.Wait, but the problem says "the graphs... intersect in points (x1,y1), (x2,y2), and (x3,y3)", which implies three real points. But according to my calculations, there's only one real intersection point. So, perhaps I made a mistake in the substitution.Wait, let me try solving the cubic equation again.From the substitution:( x^3 - 2x^2 + x + 1 = -x/5 + 1 )Subtracting -x/5 +1 from both sides:( x^3 - 2x^2 + x + 1 + x/5 -1 = 0 )Simplify:( x^3 - 2x^2 + (6x)/5 = 0 )Multiply by 5:( 5x^3 -10x^2 +6x=0 )Factor:( x(5x^2 -10x +6)=0 )So, x=0 or solving quadratic: 5x² -10x +6=0Discriminant: 100 -120= -20 <0, so only x=0 is real.So, only one real intersection point.But the problem says three points. Maybe the problem is considering multiplicity? Or perhaps I made a mistake in the substitution.Wait, let me check the substitution again.Original cubic: y =x³ -2x² +x +1Linear equation: x +5y=5 => y=(5 -x)/5=1 -x/5Substitute into cubic:1 -x/5 =x³ -2x² +x +1Subtract 1 from both sides:-x/5 =x³ -2x² +xBring all terms to left:x³ -2x² +x +x/5=0Combine x terms:x +x/5=6x/5So, equation is:x³ -2x² +6x/5=0Multiply by5:5x³ -10x² +6x=0Factor:x(5x² -10x +6)=0Same as before. So, only x=0 is real.So, the problem might have a typo, or perhaps I'm misunderstanding the question.Alternatively, maybe the user's initial solution is incorrect because they considered the complex roots for the sum, but in reality, since the problem is about intersection points, which are real, only x=0 is relevant, so A=0 and B=1.But the problem says three points, so maybe it's expecting to consider all roots, including complex ones, for the sum. So, in that case, A=2 and B=13/5.But in the initial solution, the user said B=1, which is incorrect because they only considered the real root.Wait, let me think again. If we consider all roots, real and complex, then for the cubic equation, the sum of x's is 2, and the sum of y's is 13/5.But in the problem, it says "the graphs... intersect in points (x1,y1), (x2,y2), and (x3,y3)", which are real points. So, if there's only one real intersection point, then A and B should be based on that single point, meaning A=0 and B=1.But the problem mentions three points, so perhaps it's expecting to include the complex roots as well, even though they don't correspond to real intersection points.Alternatively, maybe I made a mistake in the substitution.Wait, let me try solving the system again.From the linear equation: x +5y=5 => x=5-5ySubstitute into cubic:y=(5-5y)^3 -2(5-5y)^2 + (5-5y) +1Compute (5-5y)^3:=125 - 375y + 375y² -125y³Compute -2(5-5y)^2:= -2*(25 -50y +25y²)= -50 +100y -50y²Compute (5-5y):=5 -5yAdd all together:y= [125 -375y +375y² -125y³] + [-50 +100y -50y²] + [5 -5y] +1Combine constants: 125-50+5+1=81Combine y terms: -375y +100y -5y= -280yCombine y² terms: 375y² -50y²=325y²Combine y³ terms: -125y³So, equation becomes:y= -125y³ +325y² -280y +81Bring all terms to left:125y³ -325y² +280y -81 +y=0Wait, that's because I subtracted y from both sides:0= -125y³ +325y² -280y +81 -yWhich simplifies to:0= -125y³ +325y² -281y +81Multiply by -1:125y³ -325y² +281y -81=0This is a cubic in y. Let's try to find rational roots using Rational Root Theorem. Possible roots are factors of 81 over factors of 125, so ±1, ±3, ±9, ±27, ±81, ±1/5, etc.Let me test y=1:125(1)^3 -325(1)^2 +281(1) -81=125-325+281-81= (125+281)-(325+81)=406-406=0Oh, y=1 is a root!So, we can factor out (y-1):Using polynomial division or synthetic division.Divide 125y³ -325y² +281y -81 by (y-1).Using synthetic division:Coefficients: 125 | -325 | 281 | -81Bring down 125Multiply by 1: 125Add to next coefficient: -325 +125= -200Multiply by1: -200Add to next coefficient:281 + (-200)=81Multiply by1:81Add to last coefficient: -81 +81=0So, the cubic factors as (y-1)(125y² -200y +81)=0Now, solve 125y² -200y +81=0Using quadratic formula:y=(200±√(40000 -4*125*81))/250Compute discriminant:40000 -4*125*81=40000 -40500= -500Negative discriminant, so only real solution is y=1.Thus, the only real intersection point is when y=1. Then, from x=5-5y=5-5(1)=0. So, x=0, y=1.So, indeed, only one real intersection point at (0,1). So, the problem's mention of three points is confusing. Maybe it's a misstatement, or perhaps it's considering complex intersections, but in reality, only one real point.But the problem says "the graphs... intersect in points (x1,y1), (x2,y2), and (x3,y3)", which suggests three real points. So, perhaps I made a mistake in the substitution.Wait, but both substitution methods led to only one real solution. So, perhaps the problem is incorrect in stating three points, or perhaps I'm misunderstanding.Alternatively, maybe the cubic and the line intersect at three points, but two of them are complex. So, in that case, the sum A would be the sum of all x's, including complex ones, and similarly for B.In that case, for the cubic equation (5x^3 -10x^2 +6x=0), the sum of roots is 2, as per Vieta's formula. So, A=2.For the sum of y's, since y= -x/5 +1, then sum of y's is:( y1 + y2 + y3 = (-x1/5 +1) + (-x2/5 +1) + (-x3/5 +1) = -frac{1}{5}(x1 +x2 +x3) +3 = -frac{2}{5} +3 = frac{13}{5} )So, B=13/5.But in the initial solution, the user said B=1, which is incorrect because they only considered the real root. So, the correct answer should be A=2 and B=13/5.But wait, in the problem statement, it says "the graphs... intersect in points (x1,y1), (x2,y2), and (x3,y3)", which are real points. So, if there's only one real intersection point, then A and B should be based on that single point, meaning A=0 and B=1.But the problem mentions three points, so perhaps it's expecting to include the complex roots as well, even though they don't correspond to real intersection points. In that case, A=2 and B=13/5.Alternatively, maybe the problem is correct, and I made a mistake in the substitution.Wait, let me check the substitution again.From the linear equation: x +5y=5 => y=1 -x/5Substitute into cubic:1 -x/5 =x³ -2x² +x +1Subtract 1 from both sides:-x/5 =x³ -2x² +xBring all terms to left:x³ -2x² +x +x/5=0Combine x terms:x +x/5=6x/5So, equation is:x³ -2x² +6x/5=0Multiply by5:5x³ -10x² +6x=0Factor:x(5x² -10x +6)=0So, x=0 or solving quadratic:5x² -10x +6=0Discriminant:100 -120= -20 <0, so only x=0 is real.Thus, only one real intersection point at x=0, y=1.Therefore, the problem's mention of three points is incorrect, or perhaps it's considering complex intersections, which don't correspond to real points.In that case, if we consider all roots, including complex ones, then A=2 and B=13/5.But if we consider only real intersection points, then A=0 and B=1.But the problem says "the graphs... intersect in points (x1,y1), (x2,y2), and (x3,y3)", which are real points, so only one point exists. Therefore, the problem might have a mistake.Alternatively, perhaps I made a mistake in the substitution.Wait, let me try solving the system again using another method.From the linear equation: x +5y=5 => x=5-5ySubstitute into cubic:y=(5-5y)^3 -2(5-5y)^2 + (5-5y) +1Compute (5-5y)^3:=125 - 375y + 375y² -125y³Compute -2(5-5y)^2:= -2*(25 -50y +25y²)= -50 +100y -50y²Compute (5-5y):=5 -5yAdd all together:y= [125 -375y +375y² -125y³] + [-50 +100y -50y²] + [5 -5y] +1Combine constants:125-50+5+1=81Combine y terms:-375y +100y -5y= -280yCombine y² terms:375y² -50y²=325y²Combine y³ terms:-125y³So, equation becomes:y= -125y³ +325y² -280y +81Bring all terms to left:125y³ -325y² +280y -81 +y=0Wait, that's because I subtracted y from both sides:0= -125y³ +325y² -280y +81 -yWhich simplifies to:0= -125y³ +325y² -281y +81Multiply by -1:125y³ -325y² +281y -81=0As before, we found y=1 is a root, so factor out (y-1):125y³ -325y² +281y -81=(y-1)(125y² -200y +81)=0Solve quadratic:125y² -200y +81=0Discriminant:40000 -4*125*81=40000 -40500= -500 <0So, only real solution is y=1, leading to x=0.Thus, only one real intersection point.Therefore, the problem's mention of three points is incorrect, or perhaps it's considering complex intersections, which don't correspond to real points.In conclusion, if we consider all roots (real and complex), then A=2 and B=13/5. If we consider only real intersection points, then A=0 and B=1. But since the problem mentions three points, it's likely expecting the sum of all roots, including complex ones, so A=2 and B=13/5.However, in the initial solution, the user incorrectly concluded B=1 by only considering the real root. The correct approach is to use Vieta's formula for the sum of roots, considering all roots, leading to A=2 and B=13/5.But wait, in the initial problem, the user's solution said B=1, which is incorrect. The correct B should be 13/5.Wait, but let me double-check the sum of y's.Given that y= -x/5 +1, and sum of x's is 2, then sum of y's is:Sum y = (-1/5)(sum x) +3 = (-1/5)(2) +3= -2/5 +15/5=13/5.Yes, that's correct.Therefore, the ordered pair (A,B) is (2,13/5).But in the initial solution, the user said B=1, which is incorrect. So, the correct answer should be (2,13/5).But perhaps the problem expects the answer in fractions, so 13/5 is 2.6, but as a fraction, it's 13/5.Alternatively, maybe the user made a mistake in their initial solution, and the correct answer is (2,13/5).But wait, let me check again.From the cubic equation, sum of x's is 2.Each y is determined by y= -x/5 +1.So, sum of y's is sum(-x/5 +1)= -1/5 sum x +3= -2/5 +3=13/5.Yes, that's correct.Therefore, the ordered pair is (2,13/5).But the initial solution said (2,1), which is incorrect.So, the correct answer is (2,13/5).But let me check if 13/5 is correct.Yes, because:Sum x=2Sum y= -2/5 +3=13/5.Yes.Therefore, the ordered pair is (2,13/5).But in the initial solution, the user said B=1, which is incorrect.So, the correct answer is (2,13/5).But perhaps the problem expects the answer as fractions, so 13/5 is 2 and 3/5.Alternatively, maybe the problem expects the answer in a different form.But in any case, the correct ordered pair is (2,13/5).

answer：First, I need to determine the relative speed between the two trains since they are moving in opposite directions. The man's train is traveling at 36 km/h, and the goods train is moving at 50.4 km/h. To find the relative speed, I add these two speeds together.Next, I'll convert the relative speed from kilometers per hour to meters per second to match the units of the train's length. This conversion is necessary for accurate time calculation.After obtaining the relative speed in meters per second, I can use the formula for time, which is the length of the goods train divided by the relative speed. This will give me the time it takes for the entire goods train to pass the man.Finally, I'll perform the calculations step by step to ensure accuracy and determine the exact time in seconds.