💡First, I need to determine the total distance the train travels while passing the bridge. This includes both the length of the train and the length of the bridge.Next, I'll calculate the total distance by adding the length of the train (360 meters) to the length of the bridge (140 meters).With the total distance known, I can then find the speed of the train in meters per second by dividing the total distance by the time it takes to pass the bridge (50 seconds).Finally, I'll convert the speed from meters per second to kilometers per hour by multiplying by the conversion factor of 3.6.

💡First, I recognize that the expression (sin 30^{circ}cos 60^{circ} + cos 30^{circ}sin 60^{circ}) resembles the sine addition formula, which is (sin(A + B) = sin A cos B + cos A sin B).By setting (A = 30^{circ}) and (B = 60^{circ}), the expression simplifies to (sin(30^{circ} + 60^{circ})), which is (sin 90^{circ}).Since (sin 90^{circ} = 1), the value of the original expression is (1).

💡Okay, so I have this problem here: for real numbers ( w ) and ( z ), the expression ( frac{frac{1}{w} + frac{1}{z}}{frac{1}{w} - frac{1}{z}} ) equals 2023. I need to find ( frac{w + z}{w - z} ). Hmm, let me try to figure this out step by step.First, let me write down the given equation:[frac{frac{1}{w} + frac{1}{z}}{frac{1}{w} - frac{1}{z}} = 2023]I think I can simplify this by combining the fractions in the numerator and the denominator. Let me find a common denominator for both the numerator and the denominator.For the numerator, ( frac{1}{w} + frac{1}{z} ), the common denominator is ( wz ). So, combining them:[frac{z + w}{wz}]Similarly, for the denominator, ( frac{1}{w} - frac{1}{z} ), the common denominator is also ( wz ):[frac{z - w}{wz}]So now, the original expression becomes:[frac{frac{z + w}{wz}}{frac{z - w}{wz}} = 2023]Since both the numerator and the denominator have the same denominator ( wz ), they cancel out when I divide them. So, this simplifies to:[frac{z + w}{z - w} = 2023]Wait, that's interesting. So, ( frac{z + w}{z - w} = 2023 ). But the problem is asking for ( frac{w + z}{w - z} ). Hmm, let me see. If I look closely, ( frac{w + z}{w - z} ) is just the negative of ( frac{z + w}{z - w} ). Because if I factor out a negative from the denominator:[frac{w + z}{w - z} = frac{w + z}{-(z - w)} = -frac{w + z}{z - w}]So, since ( frac{z + w}{z - w} = 2023 ), then:[frac{w + z}{w - z} = -2023]That seems straightforward. Let me double-check to make sure I didn't make any mistakes.Starting from the given equation:[frac{frac{1}{w} + frac{1}{z}}{frac{1}{w} - frac{1}{z}} = 2023]Combine the fractions:Numerator: ( frac{z + w}{wz} )Denominator: ( frac{z - w}{wz} )So, dividing them:[frac{z + w}{z - w} = 2023]Which implies:[frac{w + z}{w - z} = -2023]Yes, that seems correct. I don't think I made any errors in the algebra here. The key was recognizing that the expression simplifies to ( frac{z + w}{z - w} ) and then realizing that flipping the denominator introduces a negative sign.Alternatively, I could have approached this by letting ( frac{1}{w} = a ) and ( frac{1}{z} = b ). Then the equation becomes:[frac{a + b}{a - b} = 2023]Cross-multiplying gives:[a + b = 2023(a - b)]Expanding the right side:[a + b = 2023a - 2023b]Bringing like terms to one side:[a - 2023a = -2023b - b][-2022a = -2024b][2022a = 2024b][frac{a}{b} = frac{2024}{2022}]Since ( a = frac{1}{w} ) and ( b = frac{1}{z} ), this implies:[frac{frac{1}{w}}{frac{1}{z}} = frac{2024}{2022}][frac{z}{w} = frac{2024}{2022}][frac{w}{z} = frac{2022}{2024}]So, ( w = frac{2022}{2024} z ). Now, let's compute ( frac{w + z}{w - z} ):Substitute ( w = frac{2022}{2024} z ):[frac{frac{2022}{2024} z + z}{frac{2022}{2024} z - z} = frac{left( frac{2022}{2024} + 1 right) z}{left( frac{2022}{2024} - 1 right) z}]The ( z ) terms cancel out:[frac{frac{2022}{2024} + 1}{frac{2022}{2024} - 1}]Convert 1 to ( frac{2024}{2024} ) to have a common denominator:[frac{frac{2022 + 2024}{2024}}{frac{2022 - 2024}{2024}} = frac{frac{4046}{2024}}{frac{-2}{2024}} = frac{4046}{-2} = -2023]So, this method also leads me to the same conclusion: ( frac{w + z}{w - z} = -2023 ).Just to make sure, let me try plugging in some numbers. Suppose ( w = 1 ) and ( z ) is such that the given equation holds. Let's solve for ( z ):Given ( frac{frac{1}{1} + frac{1}{z}}{frac{1}{1} - frac{1}{z}} = 2023 )Simplify:[frac{1 + frac{1}{z}}{1 - frac{1}{z}} = 2023]Let ( frac{1}{z} = x ), then:[frac{1 + x}{1 - x} = 2023]Cross-multiplying:[1 + x = 2023(1 - x)][1 + x = 2023 - 2023x][x + 2023x = 2023 - 1][2024x = 2022][x = frac{2022}{2024} = frac{1011}{1012}]So, ( frac{1}{z} = frac{1011}{1012} ), which means ( z = frac{1012}{1011} ).Now, compute ( frac{w + z}{w - z} ):[frac{1 + frac{1012}{1011}}{1 - frac{1012}{1011}} = frac{frac{1011 + 1012}{1011}}{frac{1011 - 1012}{1011}} = frac{frac{2023}{1011}}{frac{-1}{1011}} = frac{2023}{-1} = -2023]Yep, that checks out too. So, regardless of the method I use, I consistently get ( -2023 ) as the result. I think that solidifies the answer.**Final Answer**boxed{A}

💡Okay, so I have this equation to solve: (3^{2y} - 9 cdot 3^y + 6 = 0). Hmm, it looks a bit complicated with the exponents, but maybe I can simplify it somehow. Let me think... Oh, I remember that when dealing with equations like this, substitution can be really helpful. Let me try substituting (3^y) with another variable to make it easier. Let's say (b = 3^y). Then, (3^{2y}) would be ((3^y)^2), which is (b^2). So, substituting these into the equation, I get:[b^2 - 9b + 6 = 0]Alright, now I have a quadratic equation in terms of (b). Quadratic equations are usually easier to solve. I can try factoring this, but I'm not sure if it factors nicely. Let me check the discriminant to see if it can be factored or if I need to use the quadratic formula. The discriminant (D) is (b^2 - 4ac), so here it would be:[D = (-9)^2 - 4 cdot 1 cdot 6 = 81 - 24 = 57]Since 57 isn't a perfect square, it won't factor nicely. That means I'll have to use the quadratic formula. The quadratic formula is:[b = frac{-B pm sqrt{D}}{2A}]Where (A = 1), (B = -9), and (C = 6). Plugging these values in:[b = frac{-(-9) pm sqrt{57}}{2 cdot 1} = frac{9 pm sqrt{57}}{2}]So, the solutions for (b) are:[b = frac{9 + sqrt{57}}{2} quad text{and} quad b = frac{9 - sqrt{57}}{2}]Now, remember that (b = 3^y), so I need to solve for (y) in each case. Let's take the first solution:[3^y = frac{9 + sqrt{57}}{2}]To solve for (y), I'll take the logarithm base 3 of both sides:[y = log_3left(frac{9 + sqrt{57}}{2}right)]Hmm, that looks a bit messy. Let me see if I can simplify it. Maybe I can approximate the value or see if it matches any of the answer choices. Let me calculate the numerical value:First, calculate (sqrt{57}):[sqrt{57} approx 7.55]So,[frac{9 + 7.55}{2} = frac{16.55}{2} = 8.275]So,[y = log_3(8.275)]I know that (3^2 = 9), so (8.275) is just a bit less than 9. Therefore, (y) should be just a bit less than 2. Maybe around 1.95 or something. But let's see if that matches any of the answer choices.Looking at the options:A) (log_3(2)) ≈ 0.6309B) (1 + log_3(2)) ≈ 1.6309C) (2 + log_3(2)) ≈ 2.6309D) 1E) None of theseHmm, none of these options are around 1.95. Maybe I made a mistake. Let me check my calculations again.Wait, maybe I should consider the second solution for (b):[3^y = frac{9 - sqrt{57}}{2}]Calculating that:[sqrt{57} approx 7.55]So,[frac{9 - 7.55}{2} = frac{1.45}{2} = 0.725]So,[y = log_3(0.725)]Since 0.725 is less than 1, the logarithm will be negative. Let me calculate that:[log_3(0.725) approx frac{ln(0.725)}{ln(3)} approx frac{-0.321}{1.0986} approx -0.292]That's approximately -0.292, which is also not among the answer choices. So, neither of the solutions for (y) match the given options. Wait, maybe I should reconsider my substitution. Let me go back to the original equation:[3^{2y} - 9 cdot 3^y + 6 = 0]I substituted (b = 3^y), so (3^{2y} = b^2), which seems correct. Then the quadratic equation is:[b^2 - 9b + 6 = 0]Solving this gives the two solutions I found earlier. Maybe I should check if these solutions are valid. Let me plug them back into the original equation to verify.First, let's take (b = frac{9 + sqrt{57}}{2}):[left(frac{9 + sqrt{57}}{2}right)^2 - 9 cdot left(frac{9 + sqrt{57}}{2}right) + 6]Calculating this would be tedious, but I can assume that since it's a solution to the quadratic, it should satisfy the equation. Similarly for the other solution.But since neither of the corresponding (y) values match the answer choices, maybe I need to think differently. Perhaps the equation can be factored in a way that gives simpler solutions.Let me try factoring the quadratic equation:[b^2 - 9b + 6 = 0]Looking for two numbers that multiply to 6 and add up to -9. Hmm, that would be -6 and -3, but:[(b - 6)(b - 3) = b^2 - 9b + 18]Wait, that's not the same as our equation, which is (b^2 - 9b + 6). So, factoring doesn't work here because the product isn't 6. Therefore, my initial approach with the quadratic formula was correct.Since neither solution for (y) matches the given options, and both are not among A to D, the answer must be E) None of these.But wait, let me double-check. Maybe I made a mistake in calculating the numerical values.For the first solution:[frac{9 + sqrt{57}}{2} approx frac{9 + 7.55}{2} = frac{16.55}{2} = 8.275]So,[y = log_3(8.275)]Since (3^2 = 9), and 8.275 is close to 9, (y) is close to 2 but slightly less. Let me calculate it more accurately:[log_3(8.275) = frac{ln(8.275)}{ln(3)} approx frac{2.113}{1.0986} approx 1.923]Still, that's approximately 1.923, which is close to 2 but not exactly 2. The closest option is C) (2 + log_3(2)), which is approximately 2.6309, which is higher than 1.923. So, it's not matching.For the second solution:[frac{9 - sqrt{57}}{2} approx frac{9 - 7.55}{2} = frac{1.45}{2} = 0.725]So,[y = log_3(0.725) approx -0.292]Which is a negative number, and none of the options are negative. Therefore, neither solution matches the given options. Wait, maybe I should consider if there's another way to approach the problem. Let me try to express the equation differently.Original equation:[3^{2y} - 9 cdot 3^y + 6 = 0]I can write (3^{2y}) as ((3^y)^2), which is the same as (b^2). So, the equation becomes:[b^2 - 9b + 6 = 0]Which is what I did earlier. So, I don't think there's another substitution that would simplify it further. Alternatively, maybe I can complete the square for the quadratic equation. Let's try that.Starting with:[b^2 - 9b + 6 = 0]Move the constant term to the other side:[b^2 - 9b = -6]To complete the square, take half of the coefficient of (b), which is (-9/2), square it, which is (81/4), and add it to both sides:[b^2 - 9b + frac{81}{4} = -6 + frac{81}{4}]Simplify the right side:[-6 = -frac{24}{4}, quad -frac{24}{4} + frac{81}{4} = frac{57}{4}]So, the equation becomes:[left(b - frac{9}{2}right)^2 = frac{57}{4}]Taking square roots of both sides:[b - frac{9}{2} = pm sqrt{frac{57}{4}} = pm frac{sqrt{57}}{2}]Therefore,[b = frac{9}{2} pm frac{sqrt{57}}{2}]Which is the same as:[b = frac{9 pm sqrt{57}}{2}]So, this confirms my earlier solutions for (b). Therefore, my previous calculations were correct.Given that, and since neither solution for (y) matches the provided answer choices, I think the correct answer must be E) None of these.But just to be thorough, let me check if perhaps I misinterpreted the original equation. Is it (3^{2y}) or (3^{2y})? Yes, it's (3^{2y}), which is correct. So, my substitution was appropriate.Alternatively, maybe the question expects multiple answers, but the options are single choices. The original problem didn't specify if it's multiple choice or single choice. Looking back, it says "Solve the equation... A) ... B) ... C) ... D) ... E) None of these." So, it's multiple choice with options A to E, and E being "None of these." Therefore, since neither solution matches A to D, the answer is E.Wait, but in the initial problem, the user wrote:"Solve the equation (3^{2y} - 9 cdot 3^y + 6 = 0).A) (log_3(2)) B) (1 + log_3(2)) C) (2 + log_3(2)) D) 1 E) None of these"So, it's a single answer multiple choice. Therefore, since neither solution for (y) is among A to D, the answer is E.But just to make sure, let me see if perhaps I can express (log_3(8.275)) in terms of (log_3(2)). Let's see:We know that (8.275) is approximately (8.275), and (8 = 2^3), but that might not help. Alternatively, (8.275 = 8 + 0.275), but that's not helpful either. Maybe I can write it as (8.275 = 8 times 1.034375), but that's still not helpful in terms of (log_3(2)).Alternatively, perhaps I can write (8.275) as (3^2 - 0.725), but that doesn't seem useful either. So, I don't think there's a way to express (log_3(8.275)) in terms of (log_3(2)) to match any of the answer choices.Therefore, I'm confident that the correct answer is E) None of these.

💡First, I need to determine the cost price of the article. Let's denote the cost price as ( x ) rupees.When the article is sold for Rs. 66, the profit is calculated as:[text{Profit} = 66 - x]When the article is sold for Rs. 22, the loss is calculated as:[text{Loss} = x - 22]According to the problem, the profit and loss are equal. Therefore, I can set up the equation:[66 - x = x - 22]To solve for ( x ), I'll first add ( x ) to both sides of the equation:[66 = 2x - 22]Next, I'll add 22 to both sides to isolate the term with ( x ):[88 = 2x]Finally, I'll divide both sides by 2 to find the value of ( x ):[x = 44]Therefore, the cost price of the article is Rs. 44.

💡Alright, let me try to figure out this problem step by step. So, we have a function ( f(x) ) that's even, meaning ( f(-x) = f(x) ). Also, for any ( x_1, x_2 ) in the interval ([0, +infty)), the difference quotient ( frac{f(x_1) - f(x_2)}{x_1 - x_2} > 0 ). This tells me that ( f(x) ) is increasing on the interval ([0, +infty)). Since it's an even function, it must be decreasing on ((-infty, 0]).Now, the inequality we need to satisfy is ( f(ax + 1) leq f(x - 2) ) for all ( x ) in ([ frac{1}{2}, 1 ]). Let me think about what this means. Since ( f ) is even, ( f(x - 2) = f(2 - x) ) because ( x - 2 ) is negative when ( x ) is in ([ frac{1}{2}, 1 ]). So, ( f(x - 2) = f(2 - x) ).Given that ( f ) is increasing on ([0, +infty)), the inequality ( f(ax + 1) leq f(2 - x) ) implies that ( |ax + 1| leq |2 - x| ). But since ( ax + 1 ) and ( 2 - x ) are both positive or negative depending on the values of ( a ) and ( x ), I need to be careful here.Wait, actually, since ( f ) is increasing on ([0, +infty)) and decreasing on ((-infty, 0]), the inequality ( f(ax + 1) leq f(x - 2) ) can be translated into ( |ax + 1| geq |x - 2| ) or ( |ax + 1| leq |x - 2| ) depending on the direction of the inequality. Hmm, maybe I need to think differently.Let me consider the function's behavior. Since ( f ) is increasing on ([0, +infty)) and decreasing on ((-infty, 0]), the function's value depends on the absolute value of the input. So, ( f(u) leq f(v) ) if and only if ( |u| geq |v| ) when both ( u ) and ( v ) are in ([0, +infty)) or both in ((-infty, 0]). But in our case, ( ax + 1 ) could be positive or negative, and ( x - 2 ) is definitely negative because ( x ) is at most 1, so ( x - 2 ) is at least -1.5.So, ( f(x - 2) = f(2 - x) ) because it's even. Therefore, ( f(ax + 1) leq f(2 - x) ). Since ( f ) is increasing on ([0, +infty)), this inequality implies that ( |ax + 1| leq 2 - x ). So, ( |ax + 1| leq 2 - x ).This absolute value inequality can be split into two inequalities:1. ( ax + 1 leq 2 - x )2. ( -(ax + 1) leq 2 - x )Let me solve the first inequality:( ax + 1 leq 2 - x )Subtract 1 from both sides:( ax leq 1 - x )Factor out ( x ):( a leq frac{1 - x}{x} )Simplify:( a leq frac{1}{x} - 1 )Since ( x ) is in ([ frac{1}{2}, 1 ]), ( frac{1}{x} ) is in ([1, 2]). Therefore, ( frac{1}{x} - 1 ) is in ([0, 1]). So, ( a leq 1 ) is one condition.Now, the second inequality:( -(ax + 1) leq 2 - x )Multiply both sides by -1 (remember to reverse the inequality sign):( ax + 1 geq x - 2 )Subtract 1 from both sides:( ax geq x - 3 )Factor out ( x ):( a geq 1 - frac{3}{x} )Again, ( x ) is in ([ frac{1}{2}, 1 ]), so ( frac{3}{x} ) is in ([3, 6]). Therefore, ( 1 - frac{3}{x} ) is in ([-5, -2]). So, ( a geq -5 ).Combining both inequalities, we have ( -5 leq a leq 1 ). But wait, let me double-check this because the function's behavior might impose additional constraints.Given that ( f ) is increasing on ([0, +infty)), for the inequality ( f(ax + 1) leq f(2 - x) ) to hold, ( |ax + 1| leq 2 - x ) must be true. So, ( ax + 1 ) must lie within ([- (2 - x), 2 - x]). Therefore, ( - (2 - x) leq ax + 1 leq 2 - x ).Subtracting 1 from all parts:( - (2 - x) - 1 leq ax leq 2 - x - 1 )Simplify:( -3 + x leq ax leq 1 - x )Divide by ( x ) (since ( x > 0 )):( frac{-3 + x}{x} leq a leq frac{1 - x}{x} )Simplify:( -frac{3}{x} + 1 leq a leq frac{1}{x} - 1 )Now, ( x ) is in ([ frac{1}{2}, 1 ]), so let's find the range for ( a ).For the lower bound:( -frac{3}{x} + 1 ) is minimized when ( x ) is minimized (since ( frac{3}{x} ) is maximized). So, at ( x = frac{1}{2} ):( -frac{3}{frac{1}{2}} + 1 = -6 + 1 = -5 )For the upper bound:( frac{1}{x} - 1 ) is maximized when ( x ) is minimized. At ( x = frac{1}{2} ):( frac{1}{frac{1}{2}} - 1 = 2 - 1 = 1 )So, combining these, ( a ) must satisfy ( -5 leq a leq 1 ). However, I need to ensure that ( ax + 1 ) doesn't exceed the bounds set by ( 2 - x ) for all ( x ) in ([ frac{1}{2}, 1 ]).Wait, but if ( a = 1 ), then ( ax + 1 = x + 1 ). For ( x = 1 ), this is 2, and ( 2 - x = 1 ). So, ( f(2) leq f(1) ). But since ( f ) is increasing on ([0, +infty)), ( f(2) geq f(1) ). This contradicts the inequality ( f(ax + 1) leq f(x - 2) ). Therefore, ( a = 1 ) is not valid.Similarly, if ( a = -5 ), let's check ( x = frac{1}{2} ):( ax + 1 = -5 * frac{1}{2} + 1 = -frac{5}{2} + 1 = -frac{3}{2} )And ( x - 2 = frac{1}{2} - 2 = -frac{3}{2} )So, ( f(-frac{3}{2}) = f(frac{3}{2}) ) and ( f(-frac{3}{2}) = f(frac{3}{2}) ). So, equality holds here.But for ( a = -5 ) and ( x = 1 ):( ax + 1 = -5 * 1 + 1 = -4 )And ( x - 2 = 1 - 2 = -1 )So, ( f(-4) = f(4) ) and ( f(-1) = f(1) ). Since ( f ) is increasing on ([0, +infty)), ( f(4) geq f(1) ), which would mean ( f(ax + 1) geq f(x - 2) ), contradicting the inequality. Therefore, ( a = -5 ) is also not valid.Hmm, so my earlier conclusion that ( -5 leq a leq 1 ) might be too broad. I need to find the tightest bounds where ( |ax + 1| leq 2 - x ) for all ( x ) in ([ frac{1}{2}, 1 ]).Let me consider the endpoints and see where the inequalities hold.For ( x = frac{1}{2} ):( |a * frac{1}{2} + 1| leq 2 - frac{1}{2} = frac{3}{2} )So, ( | frac{a}{2} + 1 | leq frac{3}{2} )This gives:( -frac{3}{2} leq frac{a}{2} + 1 leq frac{3}{2} )Subtract 1:( -frac{5}{2} leq frac{a}{2} leq frac{1}{2} )Multiply by 2:( -5 leq a leq 1 )For ( x = 1 ):( |a * 1 + 1| leq 2 - 1 = 1 )So, ( |a + 1| leq 1 )This gives:( -1 leq a + 1 leq 1 )Subtract 1:( -2 leq a leq 0 )So, combining these two results, ( a ) must satisfy both ( -5 leq a leq 1 ) and ( -2 leq a leq 0 ). The intersection of these intervals is ( -2 leq a leq 0 ).But wait, earlier I thought ( a = -5 ) and ( a = 1 ) cause contradictions, but according to the endpoint analysis, ( a ) must be between -2 and 0. Let me verify this.If ( a = -2 ), then for ( x = 1 ):( ax + 1 = -2 * 1 + 1 = -1 )And ( x - 2 = -1 )So, ( f(-1) = f(1) ) and ( f(-1) = f(1) ), so equality holds.For ( x = frac{1}{2} ):( ax + 1 = -2 * frac{1}{2} + 1 = -1 + 1 = 0 )And ( x - 2 = frac{1}{2} - 2 = -frac{3}{2} )So, ( f(0) leq f(-frac{3}{2}) = f(frac{3}{2}) ). Since ( f ) is increasing on ([0, +infty)), ( f(0) leq f(frac{3}{2}) ) holds true.Similarly, if ( a = 0 ), then ( ax + 1 = 1 ) for all ( x ). And ( x - 2 ) ranges from -1.5 to -1. So, ( f(1) leq f(1.5) ), which is true because ( f ) is increasing.Therefore, the valid range for ( a ) is ( -2 leq a leq 0 ).