💡First, I need to determine the total distance that needs to be covered for the two trains to completely pass each other. This is the sum of their lengths.Next, I'll convert the speeds of both trains from kilometers per hour to meters per second to ensure the units are consistent.Since the trains are moving towards each other, their relative speed is the sum of their individual speeds.Finally, I'll calculate the time it takes for the trains to clear each other by dividing the total distance by the relative speed.

💡Okay, so I have this problem where I need to find the value of ( y ) given two equations involving ( x ) and ( y ). The equations are:( x = 2 + frac{1}{y} ) ( y = 3 + frac{1}{x} )Hmm, both equations are kind of linked together, meaning ( x ) is defined in terms of ( y ) and vice versa. I think I need to find a way to substitute one into the other to solve for either ( x ) or ( y ).Let me start by looking at the first equation: ( x = 2 + frac{1}{y} ). Maybe I can express ( x ) in terms of ( y ) and then plug that into the second equation. Alternatively, I could solve for ( y ) in terms of ( x ) from the first equation and substitute into the second. Let me try both approaches and see which one works better.First, solving the first equation for ( y ):( x = 2 + frac{1}{y} ) Subtract 2 from both sides: ( x - 2 = frac{1}{y} ) Take the reciprocal of both sides: ( y = frac{1}{x - 2} )Okay, so ( y ) is expressed in terms of ( x ). Now, let's plug this into the second equation:( y = 3 + frac{1}{x} ) Substitute ( y ) from above: ( frac{1}{x - 2} = 3 + frac{1}{x} )Hmm, now I have an equation with just ( x ). Let me solve this for ( x ). First, I'll write it out:( frac{1}{x - 2} = 3 + frac{1}{x} )To eliminate the fractions, I can multiply both sides by ( x(x - 2) ), which is the common denominator. Let's do that:( x(x - 2) cdot frac{1}{x - 2} = x(x - 2) cdot left(3 + frac{1}{x}right) )Simplify the left side:( x = x(x - 2) cdot 3 + x(x - 2) cdot frac{1}{x} )Simplify each term on the right side:First term: ( x(x - 2) cdot 3 = 3x(x - 2) ) Second term: ( x(x - 2) cdot frac{1}{x} = (x - 2) )So now the equation becomes:( x = 3x(x - 2) + (x - 2) )Let me expand the right side:First, expand ( 3x(x - 2) ): ( 3x^2 - 6x )So the equation is now:( x = 3x^2 - 6x + x - 2 )Combine like terms on the right side:( x = 3x^2 - 5x - 2 )Now, bring all terms to one side to set the equation to zero:( 0 = 3x^2 - 5x - 2 - x ) Wait, that's not quite right. Let me subtract ( x ) from both sides:( 0 = 3x^2 - 5x - 2 - x ) Combine like terms:( 0 = 3x^2 - 6x - 2 )So, the quadratic equation is:( 3x^2 - 6x - 2 = 0 )Now, I can solve this quadratic equation using the quadratic formula. The quadratic formula is:( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} )Where ( a = 3 ), ( b = -6 ), and ( c = -2 ). Plugging these values in:( x = frac{-(-6) pm sqrt{(-6)^2 - 4 cdot 3 cdot (-2)}}{2 cdot 3} ) Simplify:( x = frac{6 pm sqrt{36 + 24}}{6} ) ( x = frac{6 pm sqrt{60}}{6} )Simplify ( sqrt{60} ). Since ( 60 = 4 times 15 ), ( sqrt{60} = 2sqrt{15} ). So:( x = frac{6 pm 2sqrt{15}}{6} )We can factor out a 2 in the numerator:( x = frac{2(3 pm sqrt{15})}{6} ) Simplify by dividing numerator and denominator by 2:( x = frac{3 pm sqrt{15}}{3} )So, ( x ) has two possible solutions:( x = frac{3 + sqrt{15}}{3} ) or ( x = frac{3 - sqrt{15}}{3} )Simplify these:( x = 1 + frac{sqrt{15}}{3} ) or ( x = 1 - frac{sqrt{15}}{3} )Now, since ( x ) and ( y ) are non-zero, we need to make sure that these solutions don't result in division by zero in the original equations. Let's check both solutions.First, let's take ( x = 1 + frac{sqrt{15}}{3} ). Plugging this into the expression for ( y ) we had earlier:( y = frac{1}{x - 2} )So,( y = frac{1}{left(1 + frac{sqrt{15}}{3}right) - 2} ) Simplify the denominator:( 1 - 2 + frac{sqrt{15}}{3} = -1 + frac{sqrt{15}}{3} )So,( y = frac{1}{-1 + frac{sqrt{15}}{3}} )To simplify this, let's rationalize the denominator. Multiply numerator and denominator by 3:( y = frac{3}{-3 + sqrt{15}} )Multiply numerator and denominator by the conjugate of the denominator, which is ( -3 - sqrt{15} ):( y = frac{3(-3 - sqrt{15})}{(-3 + sqrt{15})(-3 - sqrt{15})} )Calculate the denominator:( (-3)^2 - (sqrt{15})^2 = 9 - 15 = -6 )So,( y = frac{-9 - 3sqrt{15}}{-6} )Divide numerator and denominator by -3:( y = frac{3 + sqrt{15}}{2} )So, that's one possible value for ( y ): ( frac{3 + sqrt{15}}{2} )Now, let's check the other solution for ( x ): ( x = 1 - frac{sqrt{15}}{3} )Again, plug this into ( y = frac{1}{x - 2} ):( y = frac{1}{left(1 - frac{sqrt{15}}{3}right) - 2} ) Simplify the denominator:( 1 - 2 - frac{sqrt{15}}{3} = -1 - frac{sqrt{15}}{3} )So,( y = frac{1}{-1 - frac{sqrt{15}}{3}} )Again, rationalize the denominator by multiplying numerator and denominator by 3:( y = frac{3}{-3 - sqrt{15}} )Multiply numerator and denominator by the conjugate of the denominator, which is ( -3 + sqrt{15} ):( y = frac{3(-3 + sqrt{15})}{(-3 - sqrt{15})(-3 + sqrt{15})} )Calculate the denominator:( (-3)^2 - (sqrt{15})^2 = 9 - 15 = -6 )So,( y = frac{-9 + 3sqrt{15}}{-6} )Divide numerator and denominator by -3:( y = frac{3 - sqrt{15}}{2} )So, the other possible value for ( y ) is ( frac{3 - sqrt{15}}{2} )Now, let's check if both solutions are valid. Remember, ( x ) and ( y ) are non-zero, so we need to ensure that neither ( x ) nor ( y ) is zero in either case.First, for ( y = frac{3 + sqrt{15}}{2} ):Since ( sqrt{15} ) is approximately 3.872, so:( y approx frac{3 + 3.872}{2} = frac{6.872}{2} approx 3.436 )Which is non-zero, so that's valid.For ( y = frac{3 - sqrt{15}}{2} ):( y approx frac{3 - 3.872}{2} = frac{-0.872}{2} approx -0.436 )Which is also non-zero, so that's valid as well.Now, let's check if these ( y ) values satisfy the original equations.First, take ( y = frac{3 + sqrt{15}}{2} approx 3.436 )From the first equation: ( x = 2 + frac{1}{y} approx 2 + frac{1}{3.436} approx 2 + 0.291 approx 2.291 )But earlier, we found ( x = 1 + frac{sqrt{15}}{3} approx 1 + 1.291 approx 2.291 ), which matches. So, this solution is consistent.Now, check the second equation: ( y = 3 + frac{1}{x} approx 3 + frac{1}{2.291} approx 3 + 0.436 approx 3.436 ), which matches our ( y ) value. So, this solution is valid.Now, check the other solution: ( y = frac{3 - sqrt{15}}{2} approx -0.436 )From the first equation: ( x = 2 + frac{1}{y} approx 2 + frac{1}{-0.436} approx 2 - 2.291 approx -0.291 )But earlier, we found ( x = 1 - frac{sqrt{15}}{3} approx 1 - 1.291 approx -0.291 ), which matches. So, this solution is consistent.Now, check the second equation: ( y = 3 + frac{1}{x} approx 3 + frac{1}{-0.291} approx 3 - 3.436 approx -0.436 ), which matches our ( y ) value. So, this solution is also valid.Therefore, both solutions are valid. However, looking back at the problem statement, it just asks for the value of ( y ). The options given are:A) ( frac{3}{2} - frac{sqrt{15}}{2} ) B) ( frac{3}{2} + frac{sqrt{15}}{2} ) C) ( frac{6 + sqrt{15}}{3} ) D) ( frac{6 - sqrt{15}}{3} )Wait, let me express our solutions in terms of halves:We found ( y = frac{3 + sqrt{15}}{2} ) and ( y = frac{3 - sqrt{15}}{2} )Looking at the options, option B is ( frac{3}{2} + frac{sqrt{15}}{2} ), which is the same as ( frac{3 + sqrt{15}}{2} ). Option A is ( frac{3}{2} - frac{sqrt{15}}{2} ), which is the same as ( frac{3 - sqrt{15}}{2} ).So, both options A and B are possible solutions. However, the problem doesn't specify any additional constraints, so both could be correct. But looking back at the problem statement, it just says "find the value of ( y )", without specifying which one. So, perhaps both are acceptable, but looking at the options, both A and B are present.Wait, but in the original equations, if ( y ) is negative, then ( x ) would also be negative, as we saw in the second solution. Let me check if that's acceptable.The problem states that ( x ) and ( y ) are non-zero numbers, so negative values are allowed. Therefore, both solutions are valid.But the options include both A and B, so perhaps the answer expects both? But in the original problem, it's a single answer. Hmm.Wait, looking back at the problem, it's presented as a multiple-choice question with options A to D. So, perhaps I need to see which of these options match our solutions.Our solutions are:( y = frac{3 + sqrt{15}}{2} ) which is option B, and ( y = frac{3 - sqrt{15}}{2} ) which is option A.So, both A and B are correct. But in the problem, it's presented as a single answer. Maybe I made a mistake somewhere.Wait, let me check my calculations again.Starting from:( x = 2 + frac{1}{y} ) ( y = 3 + frac{1}{x} )I solved for ( y ) in terms of ( x ) and substituted into the second equation, leading to a quadratic equation in ( x ). Solving that gave me two solutions for ( x ), which in turn gave me two solutions for ( y ).So, both solutions are mathematically valid. However, perhaps in the context of the problem, only one solution is acceptable. Let me think.If ( y ) is negative, then ( x = 2 + frac{1}{y} ) would be less than 2, and since ( y ) is negative, ( x ) would be less than 2. Then, ( y = 3 + frac{1}{x} ) would be 3 plus a negative number (since ( x ) is negative), so ( y ) would be less than 3. But in our second solution, ( y ) is negative, so that's consistent.Alternatively, if ( y ) is positive, then ( x ) is greater than 2, and ( y = 3 + frac{1}{x} ) would be greater than 3, which is consistent with our first solution.So, both solutions are valid in different contexts. However, the problem doesn't specify any constraints on the sign of ( y ), so both are possible.But looking back at the options, both A and B are present. So, perhaps the answer expects both? But in the original problem, it's a single answer. Hmm.Wait, perhaps I made a mistake in the substitution step. Let me double-check.Starting from:( x = 2 + frac{1}{y} ) ( y = 3 + frac{1}{x} )I solved for ( y ) in terms of ( x ): ( y = frac{1}{x - 2} )Then substituted into the second equation:( frac{1}{x - 2} = 3 + frac{1}{x} )Multiplying both sides by ( x(x - 2) ):( x = 3x(x - 2) + (x - 2) )Expanding:( x = 3x^2 - 6x + x - 2 )Simplifying:( x = 3x^2 - 5x - 2 )Bringing all terms to one side:( 3x^2 - 6x - 2 = 0 )That seems correct.Solving the quadratic:( x = frac{6 pm sqrt{36 + 24}}{6} = frac{6 pm sqrt{60}}{6} = frac{6 pm 2sqrt{15}}{6} = 1 pm frac{sqrt{15}}{3} )That's correct.Then, substituting back to find ( y ):For ( x = 1 + frac{sqrt{15}}{3} ), ( y = frac{3 + sqrt{15}}{2} ) For ( x = 1 - frac{sqrt{15}}{3} ), ( y = frac{3 - sqrt{15}}{2} )So, both solutions are correct.But the problem asks for "the value of ( y )", implying a single answer. However, since both are valid, perhaps the answer expects both, but in the options, both A and B are present. So, maybe the answer is both A and B, but since it's a multiple-choice question, perhaps only one is correct.Wait, looking back at the problem, it's presented as a single answer with options A to D. So, perhaps I need to check if both solutions are present in the options or if one is extraneous.Wait, in the options, both A and B are present, so perhaps both are correct. But in the original problem, it's a single answer. Hmm.Alternatively, perhaps I made a mistake in the substitution step. Let me try a different approach.Instead of solving for ( y ) in terms of ( x ) from the first equation, let me solve for ( x ) in terms of ( y ) and substitute into the second equation.From the first equation: ( x = 2 + frac{1}{y} )Substitute this into the second equation:( y = 3 + frac{1}{2 + frac{1}{y}} )Simplify the denominator:( 2 + frac{1}{y} = frac{2y + 1}{y} )So,( y = 3 + frac{1}{frac{2y + 1}{y}} = 3 + frac{y}{2y + 1} )Now, the equation is:( y = 3 + frac{y}{2y + 1} )Let me solve this for ( y ). First, subtract 3 from both sides:( y - 3 = frac{y}{2y + 1} )Multiply both sides by ( 2y + 1 ):( (y - 3)(2y + 1) = y )Expand the left side:( y cdot 2y + y cdot 1 - 3 cdot 2y - 3 cdot 1 = y ) ( 2y^2 + y - 6y - 3 = y ) Combine like terms:( 2y^2 - 5y - 3 = y )Bring all terms to one side:( 2y^2 - 5y - 3 - y = 0 ) ( 2y^2 - 6y - 3 = 0 )Now, solve this quadratic equation using the quadratic formula:( y = frac{-b pm sqrt{b^2 - 4ac}}{2a} ) Where ( a = 2 ), ( b = -6 ), ( c = -3 )So,( y = frac{6 pm sqrt{36 + 24}}{4} = frac{6 pm sqrt{60}}{4} = frac{6 pm 2sqrt{15}}{4} )Simplify:( y = frac{3 pm sqrt{15}}{2} )So, the solutions are ( y = frac{3 + sqrt{15}}{2} ) and ( y = frac{3 - sqrt{15}}{2} ), which matches our earlier solutions.Therefore, both solutions are correct. However, the problem asks for "the value of ( y )", and both A and B are present in the options. So, perhaps the answer is both A and B, but since it's a multiple-choice question, it's likely that both are acceptable, but the options are separate.Wait, looking back at the options:A) ( frac{3}{2} - frac{sqrt{15}}{2} ) B) ( frac{3}{2} + frac{sqrt{15}}{2} ) C) ( frac{6 + sqrt{15}}{3} ) D) ( frac{6 - sqrt{15}}{3} )Wait, option C is ( frac{6 + sqrt{15}}{3} ), which simplifies to ( 2 + frac{sqrt{15}}{3} ), which is not the same as our solutions. Similarly, option D is ( frac{6 - sqrt{15}}{3} ), which simplifies to ( 2 - frac{sqrt{15}}{3} ), which is also not the same as our solutions.Wait a minute, so our solutions are ( frac{3 pm sqrt{15}}{2} ), which are options A and B. So, perhaps the answer is both A and B, but since it's a single answer question, maybe I need to check if both are correct.But in the problem statement, it's a single answer, so perhaps only one is correct. Wait, but both are correct. Maybe the problem expects both, but in the options, both A and B are present. So, perhaps the answer is both A and B, but since it's a multiple-choice question, it's likely that only one is correct.Wait, perhaps I made a mistake in the substitution step. Let me try another approach.Let me assume that ( x ) and ( y ) are positive. Then, ( y = frac{3 - sqrt{15}}{2} ) would be negative, which contradicts the assumption. Therefore, if we assume ( x ) and ( y ) are positive, then only ( y = frac{3 + sqrt{15}}{2} ) is valid, which is option B.But the problem doesn't specify that ( x ) and ( y ) are positive, only that they are non-zero. So, both solutions are valid.However, looking at the options, both A and B are present, so perhaps the answer is both A and B. But since it's a single answer question, maybe the answer is both, but the options are separate.Wait, perhaps the problem expects both solutions, but in the options, both A and B are present. So, perhaps the answer is both A and B, but since it's a multiple-choice question, it's likely that only one is correct.Wait, perhaps I need to check which solution satisfies both original equations.Let me take ( y = frac{3 + sqrt{15}}{2} ) and plug it back into the first equation:( x = 2 + frac{1}{y} = 2 + frac{2}{3 + sqrt{15}} )Rationalize the denominator:( frac{2}{3 + sqrt{15}} times frac{3 - sqrt{15}}{3 - sqrt{15}} = frac{2(3 - sqrt{15})}{9 - 15} = frac{6 - 2sqrt{15}}{-6} = -1 + frac{sqrt{15}}{3} )So, ( x = 2 - 1 + frac{sqrt{15}}{3} = 1 + frac{sqrt{15}}{3} ), which matches our earlier solution.Now, check the second equation with ( y = frac{3 + sqrt{15}}{2} ):( y = 3 + frac{1}{x} = 3 + frac{1}{1 + frac{sqrt{15}}{3}} )Simplify the denominator:( 1 + frac{sqrt{15}}{3} = frac{3 + sqrt{15}}{3} )So,( frac{1}{x} = frac{3}{3 + sqrt{15}} )Rationalize:( frac{3}{3 + sqrt{15}} times frac{3 - sqrt{15}}{3 - sqrt{15}} = frac{9 - 3sqrt{15}}{9 - 15} = frac{9 - 3sqrt{15}}{-6} = -frac{3}{2} + frac{sqrt{15}}{2} )So,( y = 3 + left(-frac{3}{2} + frac{sqrt{15}}{2}right) = frac{3}{2} + frac{sqrt{15}}{2} ), which matches our solution.Now, check the other solution ( y = frac{3 - sqrt{15}}{2} ):From the first equation:( x = 2 + frac{1}{y} = 2 + frac{2}{3 - sqrt{15}} )Rationalize:( frac{2}{3 - sqrt{15}} times frac{3 + sqrt{15}}{3 + sqrt{15}} = frac{6 + 2sqrt{15}}{9 - 15} = frac{6 + 2sqrt{15}}{-6} = -1 - frac{sqrt{15}}{3} )So, ( x = 2 - 1 - frac{sqrt{15}}{3} = 1 - frac{sqrt{15}}{3} ), which matches our earlier solution.Now, check the second equation with ( y = frac{3 - sqrt{15}}{2} ):( y = 3 + frac{1}{x} = 3 + frac{1}{1 - frac{sqrt{15}}{3}} )Simplify the denominator:( 1 - frac{sqrt{15}}{3} = frac{3 - sqrt{15}}{3} )So,( frac{1}{x} = frac{3}{3 - sqrt{15}} )Rationalize:( frac{3}{3 - sqrt{15}} times frac{3 + sqrt{15}}{3 + sqrt{15}} = frac{9 + 3sqrt{15}}{9 - 15} = frac{9 + 3sqrt{15}}{-6} = -frac{3}{2} - frac{sqrt{15}}{2} )So,( y = 3 + left(-frac{3}{2} - frac{sqrt{15}}{2}right) = frac{3}{2} - frac{sqrt{15}}{2} ), which matches our solution.Therefore, both solutions are valid. However, the problem asks for "the value of ( y )", and both A and B are present in the options. So, perhaps the answer is both A and B, but since it's a multiple-choice question, it's likely that only one is correct.Wait, but in the original problem, it's presented as a single answer. So, perhaps the answer is both A and B, but since it's a multiple-choice question, it's likely that only one is correct.Alternatively, perhaps the problem expects only the positive solution, which would be option B.Given that, I think the answer is option B: ( frac{3}{2} + frac{sqrt{15}}{2} )

💡Alright, so I have this probability problem here about rolling three dice. Let me try to figure it out step by step. First, the problem says: Three dice are rolled once each. Event A is "the three numbers are all different," and event B is "at most one odd number appears." I need to find the probability P(A ∩ B), which is the probability that both events A and B happen at the same time.Okay, let's break this down. I think I need to calculate the number of favorable outcomes where both A and B occur, and then divide that by the total number of possible outcomes when rolling three dice.The total number of possible outcomes when rolling three dice is 6 × 6 × 6, which is 216. That makes sense because each die has 6 faces, and the rolls are independent.Now, let's think about event A: "the three numbers are all different." So, for event A, we need all three dice to show different numbers. How do we calculate the number of ways this can happen?Well, for the first die, there are 6 possible outcomes. For the second die, since it needs to be different from the first, there are 5 possible outcomes. And for the third die, it needs to be different from both the first and the second, so there are 4 possible outcomes. So, the number of favorable outcomes for event A is 6 × 5 × 4 = 120.Wait, but hold on, the problem isn't just asking for event A; it's asking for the intersection of A and B. So, I need to consider both events happening together.Event B is "at most one odd number appears." That means either zero odd numbers or one odd number. So, in the context of three dice, we can have either all three numbers even, or exactly one number odd and the other two even.But since we're also considering event A, which requires all three numbers to be different, we need to find the number of outcomes where all three numbers are different and there are at most one odd number.So, let's break this down further. We need two cases:1. All three numbers are even and all different.2. Exactly one number is odd, and the other two are even, and all three numbers are different.Let's calculate each case separately.First, case 1: All three numbers are even and all different. The even numbers on a die are 2, 4, and 6. So, there are three even numbers. Since we need all three dice to show different even numbers, we can think of this as permutations of the three even numbers.The number of ways to arrange three distinct even numbers is 3! = 6.Okay, so case 1 gives us 6 favorable outcomes.Now, case 2: Exactly one number is odd, and the other two are even, with all three numbers being different. Let's figure this out.First, we need to choose which one of the three dice shows the odd number. There are 3 choices for this (since there are three dice).Next, for the chosen die that shows an odd number, there are three possible odd numbers: 1, 3, and 5. So, 3 choices here.Then, for the remaining two dice, they need to show even numbers, and they also need to be different from each other. The even numbers available are 2, 4, and 6. Since we need two different even numbers, the number of ways to choose and arrange them is the number of permutations of 3 even numbers taken 2 at a time, which is 3 × 2 = 6.So, putting this together, for case 2, the number of favorable outcomes is 3 (choices for which die is odd) × 3 (choices for the odd number) × 6 (arrangements for the even numbers) = 3 × 3 × 6 = 54.Wait, hold on, that seems a bit high. Let me double-check.We have 3 choices for which die is odd. For each of those, 3 choices for the odd number. Then, for the two even numbers, since they have to be different, it's 3 choices for the first even number and 2 choices for the second, so 3 × 2 = 6. So, yes, 3 × 3 × 6 = 54. That seems correct.So, case 2 gives us 54 favorable outcomes.Now, adding both cases together: case 1 has 6 outcomes, case 2 has 54 outcomes, so in total, the number of favorable outcomes for A ∩ B is 6 + 54 = 60.Therefore, the probability P(A ∩ B) is the number of favorable outcomes divided by the total number of possible outcomes, which is 60 / 216.Simplifying that fraction: both numerator and denominator are divisible by 12. 60 ÷ 12 = 5, and 216 ÷ 12 = 18. So, 60 / 216 simplifies to 5 / 18.Hmm, looking at the answer choices, option C is 5/18, so that must be the correct answer.Wait, just to make sure I didn't make a mistake in my calculations. Let me go through it again.Total outcomes: 6^3 = 216. That's correct.For event A ∩ B, we have two cases:1. All even and all different: 3! = 6. That seems right because there are three even numbers, and we need to arrange them in three dice.2. Exactly one odd, two evens, all different: - Choose which die is odd: 3 choices.- Choose the odd number: 3 choices (1, 3, 5).- Choose two different even numbers and arrange them on the remaining two dice: 3 choices for the first even, 2 for the second, so 3 × 2 = 6.Multiply them all together: 3 × 3 × 6 = 54. That seems correct.Total favorable: 6 + 54 = 60.Probability: 60 / 216 = 5 / 18. Yep, that's correct.So, I think I did this right. It's always good to double-check, especially when dealing with probability problems because it's easy to miss a case or overcount.Another way to think about it is to consider the possible combinations for event B. Since B is "at most one odd number," it includes two scenarios: zero odd numbers and one odd number.For zero odd numbers (all even), as we calculated, there are 3! = 6 ways.For one odd number, we have to choose which die is odd, which odd number it is, and then arrange the two different even numbers on the other dice. That gives us 3 × 3 × 6 = 54, as before.Adding them together gives 60, which is 5/18 of the total 216 outcomes.I think that's solid. I don't see any mistakes in my reasoning, so I feel confident that the answer is 5/18, which is option C.**Final Answer**boxed{C}

💡Alright, so I have this geometry problem here about an equilateral triangle ABC. On each side of this triangle, they've constructed these other triangles outwardly—AB'C, CA'B, and BC'A. This forms a hexagon AB'CA'BC'. The problem states that each of the angles A'BC', C'AB', and B'CA' is greater than 120 degrees. Also, there are these side equalities: AB' equals AC', BC' equals BA', and CA' equals CB'. I need to prove that the segments AB', BC', and CA' can form a triangle.Hmm, okay. So, first, let me try to visualize this. We have an equilateral triangle ABC, which means all sides are equal, and all angles are 60 degrees. Then, on each side, we're constructing another triangle outwardly. So, for example, on side AB, we construct triangle AB'C, which is outside of ABC. Similarly, on BC, we construct CA'B, and on CA, we construct BC'A. This forms a hexagon AB'CA'BC'.Now, each of these angles A'BC', C'AB', and B'CA' is greater than 120 degrees. That seems significant because in an equilateral triangle, all angles are 60 degrees, so these constructed angles are more than double that. Also, the side equalities: AB' = AC', BC' = BA', and CA' = CB'. So, each of these segments is equal to another segment in a different part of the hexagon.I need to prove that AB', BC', and CA' can form a triangle. To form a triangle, the sum of any two sides must be greater than the third side. So, I need to show that AB' + BC' > CA', AB' + CA' > BC', and BC' + CA' > AB'.But since the problem mentions that each of the angles A'BC', C'AB', and B'CA' is greater than 120 degrees, maybe I can use the Law of Cosines or something related to triangle inequalities involving angles.Wait, let me think about the rotation idea. Since ABC is equilateral, maybe rotating one of the triangles could help. For example, if I rotate triangle AB'C by 60 degrees around point A, point C would move to point B because ABC is equilateral. Then, point B' would move to some new point, say B''. This might create a new triangle where I can apply the triangle inequality.So, if I rotate AB'C by 60 degrees counterclockwise around A, point C goes to B, and point B' goes to B''. Then, triangle AC'B'' is formed. In this triangle, AC' is equal to AC', which is given. AB'' is equal to AB' because it's just a rotation. The angle between AC' and AB'' would be more than 60 degrees because the original angle C'AB' was more than 120 degrees.Wait, why is the angle more than 60 degrees? Because the original angle at A'BC' is more than 120 degrees, so when we rotate, the angle at A would be related. Maybe I need to think about how the rotation affects the angles.Alternatively, maybe I can use vectors or coordinate geometry. Let me assign coordinates to the points. Let me place point A at (0, 0), point B at (1, 0), and point C at (0.5, sqrt(3)/2) since it's an equilateral triangle. Then, I can try to find coordinates for points A', B', and C' based on the given conditions.But that might get complicated. Maybe there's a simpler way. Since AB' = AC', BC' = BA', and CA' = CB', perhaps these segments form a triangle with equal sides? Wait, no, because the segments are of different lengths, but each is equal to another segment.Wait, actually, AB' = AC', BC' = BA', and CA' = CB'. So, AB' = AC', BC' = BA', which is equal to AB', so BC' = AB'. Similarly, CA' = CB', which is equal to BC', so CA' = BC'. Therefore, all three segments AB', BC', and CA' are equal in length.Wait, hold on. If AB' = AC', and BC' = BA', which is AB', so BC' = AB'. Similarly, CA' = CB', which is BC', so CA' = BC'. Therefore, AB' = BC' = CA'. So, all three segments are equal. Therefore, they can form an equilateral triangle.But wait, the problem says "can form a triangle," not necessarily an equilateral one. But if all three sides are equal, then yes, they can form an equilateral triangle. So, is that the case?Wait, but the problem doesn't state that AB' = BC' = CA', but rather AB' = AC', BC' = BA', and CA' = CB'. So, AB' = AC', BC' = BA' = AB', and CA' = CB' = BC'. Therefore, AB' = BC' = CA'. So, all three are equal.Therefore, they can form an equilateral triangle. So, the conclusion is that they can form a triangle, specifically an equilateral one.But wait, let me double-check. The problem says "can form a triangle," so even if they are equal, it's still a triangle. So, maybe the problem is more straightforward than I thought.But then, why does the problem mention that each of the angles A'BC', C'AB', and B'CA' is greater than 120 degrees? That seems like it's important, but if all the segments are equal, maybe it's just a given condition to ensure that the triangles constructed are such that the segments are equal.Alternatively, maybe my initial assumption is wrong, and AB', BC', and CA' are not necessarily equal. Let me think again.Given AB' = AC', BC' = BA', and CA' = CB'. So, AB' = AC', BC' = BA', which is AB', so BC' = AB'. Similarly, CA' = CB', which is BC', so CA' = BC'. Therefore, AB' = BC' = CA'. So, they are all equal. Therefore, they can form an equilateral triangle.But then, why the mention of angles greater than 120 degrees? Maybe it's a red herring, or maybe it's to ensure that the triangles constructed are such that the segments are equal.Alternatively, maybe I'm misapplying the equalities. Let me parse it again: AB' = AC', BC' = BA', and CA' = CB'. So, AB' = AC', BC' = BA', and CA' = CB'. So, AB' = AC', BC' = BA', which is AB', so BC' = AB'. Similarly, CA' = CB', which is BC', so CA' = BC'. Therefore, AB' = BC' = CA'.Therefore, all three segments are equal, so they can form an equilateral triangle. Therefore, the conclusion is straightforward.But maybe I'm missing something. Let me think about the hexagon AB'CA'BC'. Each of the angles A'BC', C'AB', and B'CA' is greater than 120 degrees. So, in the hexagon, these angles are all greater than 120 degrees. That might affect the lengths of the sides.Wait, but if AB' = AC', BC' = BA', and CA' = CB', then regardless of the angles, the sides are equal. So, even if the angles are greater than 120 degrees, the sides AB', BC', and CA' are equal, so they can form a triangle.Alternatively, maybe the problem is more complex, and the equalities AB' = AC', etc., are not directly implying that AB' = BC' = CA', but perhaps through some geometric transformations or triangle congruence.Wait, let me think about triangle AB'C. It's constructed outwardly on side AB. Similarly, triangle CA'B is constructed on BC, and triangle BC'A is constructed on CA.Given that AB' = AC', BC' = BA', and CA' = CB', perhaps these triangles are congruent or similar in some way.Wait, if AB' = AC', then triangle AB'C is isosceles with AB' = AC'. Similarly, triangle CA'B has BC' = BA', so it's also isosceles. And triangle BC'A has CA' = CB', so it's isosceles as well.Therefore, each of these constructed triangles is isosceles with two equal sides.Given that, maybe the base angles of these isosceles triangles are equal. So, for triangle AB'C, angles at B' and C are equal. Similarly for the others.But since ABC is equilateral, all sides are equal, so the base of each isosceles triangle is equal. Therefore, the equal sides AB' = AC', BC' = BA', and CA' = CB' are equal to each other.Wait, so AB' = AC', BC' = BA', and CA' = CB'. So, AB' = AC', and BC' = BA', which is AB', so BC' = AB'. Similarly, CA' = CB', which is BC', so CA' = BC'. Therefore, AB' = BC' = CA'.Therefore, all three segments are equal, so they can form an equilateral triangle.But then, why the mention of angles greater than 120 degrees? Maybe it's to ensure that the triangles are constructed outwardly in such a way that the segments are equal. If the angles were less than or equal to 120 degrees, maybe the segments wouldn't be equal.Alternatively, maybe the problem is more about the triangle inequality, and the angles being greater than 120 degrees ensures that the sum of two sides is greater than the third.Wait, but if all sides are equal, then the triangle inequality is automatically satisfied. So, maybe the problem is just to recognize that the segments are equal, hence forming an equilateral triangle.But perhaps I'm oversimplifying. Let me try to think differently.Suppose I don't assume that AB' = BC' = CA', but instead, use the given equalities and the angles to prove the triangle inequality.Given that AB' = AC', BC' = BA', and CA' = CB', let's denote AB' = AC' = x, BC' = BA' = y, and CA' = CB' = z.Wait, but from the given equalities, AB' = AC', BC' = BA', and CA' = CB'. So, AB' = AC' = x, BC' = BA' = y, and CA' = CB' = z.But then, since BA' = BC' = y, and CA' = CB' = z, and AB' = AC' = x, we can see that x = y = z.Wait, because BA' = BC' = y, and CA' = CB' = z, but BA' is equal to BC', which is y, and CA' is equal to CB', which is z. But also, AB' = AC' = x. So, unless x = y = z, which would make all segments equal.Therefore, x = y = z, so AB' = BC' = CA', hence they can form an equilateral triangle.But again, this seems too straightforward, and the problem mentions the angles being greater than 120 degrees, which might be a crucial piece of information that I'm not utilizing.Alternatively, maybe the problem is not as straightforward, and the segments AB', BC', and CA' are not necessarily equal, but the given equalities and angles allow us to prove the triangle inequality.Wait, let me think about the triangle inequality. To form a triangle, the sum of any two sides must be greater than the third side. So, I need to show that AB' + BC' > CA', AB' + CA' > BC', and BC' + CA' > AB'.Given that AB' = AC', BC' = BA', and CA' = CB', maybe I can express these inequalities in terms of the sides of the original triangle or the constructed triangles.Alternatively, maybe using vectors or complex numbers could help. Let me try to assign coordinates.Let me place point A at (0, 0), point B at (1, 0), and point C at (0.5, sqrt(3)/2). Then, I can try to find coordinates for points A', B', and C' based on the given conditions.But this might be time-consuming. Alternatively, maybe I can use the Law of Cosines in the triangles involved.Given that angle A'BC' is greater than 120 degrees, and AB' = AC', maybe I can relate the sides using the Law of Cosines.Wait, in triangle A'BC', angle at B is greater than 120 degrees, and sides BA' and BC' are equal (since BA' = BC'). Therefore, triangle A'BC' is isosceles with BA' = BC', and angle at B is greater than 120 degrees.In such a triangle, the base A'C' can be found using the Law of Cosines:A'C'^2 = BA'^2 + BC'^2 - 2 * BA' * BC' * cos(angle A'BC')Since BA' = BC' = y, and angle A'BC' > 120 degrees, cos(angle A'BC') < cos(120 degrees) = -0.5.Therefore, A'C'^2 > y^2 + y^2 - 2 * y * y * (-0.5) = 2y^2 + y^2 = 3y^2.So, A'C' > y * sqrt(3).Similarly, in triangle C'AB', angle at A is greater than 120 degrees, and sides AC' and AB' are equal (AC' = AB' = x). So, triangle C'AB' is isosceles with AC' = AB', and angle at A > 120 degrees.Using the Law of Cosines:C'B'^2 = AC'^2 + AB'^2 - 2 * AC' * AB' * cos(angle C'AB')= x^2 + x^2 - 2x^2 cos(angle C'AB')Since angle C'AB' > 120 degrees, cos(angle C'AB') < -0.5.Therefore, C'B'^2 > 2x^2 - 2x^2*(-0.5) = 2x^2 + x^2 = 3x^2.So, C'B' > x * sqrt(3).Similarly, in triangle B'CA', angle at C is greater than 120 degrees, and sides CB' and CA' are equal (CB' = CA' = z). So, triangle B'CA' is isosceles with CB' = CA', and angle at C > 120 degrees.Using the Law of Cosines:B'A'^2 = CB'^2 + CA'^2 - 2 * CB' * CA' * cos(angle B'CA')= z^2 + z^2 - 2z^2 cos(angle B'CA')Since angle B'CA' > 120 degrees, cos(angle B'CA') < -0.5.Therefore, B'A'^2 > 2z^2 - 2z^2*(-0.5) = 2z^2 + z^2 = 3z^2.So, B'A' > z * sqrt(3).Now, I have that A'C' > y * sqrt(3), C'B' > x * sqrt(3), and B'A' > z * sqrt(3).But from the given equalities, AB' = AC' = x, BC' = BA' = y, and CA' = CB' = z.So, A'C' > y * sqrt(3), but A'C' is a side in the hexagon, which is part of triangle A'BC'. Similarly, C'B' > x * sqrt(3), and B'A' > z * sqrt(3).But I'm not sure how this helps me directly with the triangle inequality for AB', BC', and CA'.Wait, maybe I can relate these lengths to the sides of the original triangle ABC.Since ABC is equilateral, all its sides are equal, say of length a.Then, in triangle AB'C, which is constructed outwardly on AB, AB' = AC' = x, and BC is the base of the original triangle, which is length a.Similarly, in triangle CA'B, BC' = BA' = y, and CA is the base, length a.In triangle BC'A, CA' = CB' = z, and AB is the base, length a.So, in each case, the base of the constructed triangle is a side of the original equilateral triangle, length a, and the two equal sides are x, y, z respectively.Therefore, in triangle AB'C, sides AB' = AC' = x, and base BC = a.Similarly, in triangle CA'B, sides BC' = BA' = y, and base CA = a.In triangle BC'A, sides CA' = CB' = z, and base AB = a.Therefore, in each constructed triangle, we have an isosceles triangle with two equal sides (x, y, z) and base a.Given that, maybe I can express x, y, z in terms of a and the angles.Wait, in triangle AB'C, which is isosceles with AB' = AC' = x, and base BC = a, the angle at B' is angle AB'C.But the problem mentions angle A'BC' is greater than 120 degrees. Wait, angle A'BC' is in the hexagon, which is at point B, between points A' and C'.Wait, maybe I need to consider triangle A'BC', which is part of the hexagon.In triangle A'BC', sides BA' = BC' = y, and angle at B is greater than 120 degrees. So, as I calculated earlier, A'C' > y * sqrt(3).Similarly, in triangle C'AB', sides AC' = AB' = x, angle at A is greater than 120 degrees, so C'B' > x * sqrt(3).In triangle B'CA', sides CB' = CA' = z, angle at C is greater than 120 degrees, so B'A' > z * sqrt(3).But I'm not sure how this helps me with the triangle inequality for AB', BC', and CA'.Wait, maybe I can consider the triangle formed by AB', BC', and CA'. If I can show that the sum of any two is greater than the third, then they can form a triangle.Given that AB' = AC' = x, BC' = BA' = y, and CA' = CB' = z.So, I need to show that x + y > z, x + z > y, and y + z > x.But from the given equalities, x = AC', y = BA', z = CB'.Wait, but BA' = BC' = y, and CB' = CA' = z. So, BA' = BC' = y, and CB' = CA' = z.Therefore, BA' = BC' = y, and CB' = CA' = z.So, in triangle BA'C, sides BA' = BC' = y, and CA' = z.Wait, but triangle BA'C is part of the hexagon. Hmm.Alternatively, maybe I can use the fact that in triangle A'BC', A'C' > y * sqrt(3), and since A'C' is a side in the hexagon, which is also part of triangle AB'C, which has sides AB' = AC' = x.Wait, this is getting convoluted. Maybe I need a different approach.Let me think about the rotation idea again. If I rotate triangle AB'C by 60 degrees around point A, point C goes to point B, and point B' goes to some point B''. Then, triangle AC'B'' is formed, with AC' = AC', AB'' = AB', and angle at A being more than 60 degrees.In triangle AC'B'', since angle at A is more than 60 degrees, the side opposite to this angle, which is C'B'', is longer than the other sides.Therefore, C'B'' > AC' and C'B'' > AB''.But AC' = x, AB'' = AB' = x, so C'B'' > x.But C'B'' is equal to BC' because of the rotation. Wait, no, because when we rotate, the length BC' is preserved.Wait, maybe not. Let me think carefully.When we rotate triangle AB'C by 60 degrees around A, point B' moves to B'', and point C moves to B. So, the segment CB' is rotated to BB''.But CB' = CA' = z, so BB'' = z.But BB'' is the segment from B to B'', which is the rotated position of B'.But I'm not sure if that helps.Alternatively, maybe I can consider triangle AB''C'. Since AB'' = AB' = x, AC' = x, and angle at A is more than 60 degrees, then by the Law of Cosines, B''C' > x.But B''C' is equal to BC' because of the rotation? Wait, not necessarily.Alternatively, maybe I can use the triangle inequality in triangle B''C'B.Wait, I'm getting confused. Maybe I need to draw a diagram.But since I can't draw, I'll try to imagine it. After rotating AB'C by 60 degrees around A, point B' goes to B'', and point C goes to B. So, triangle AB''C is formed, which is congruent to triangle AB'C.But I'm not sure.Alternatively, maybe I can use complex numbers. Let me assign complex coordinates to the points.Let me place point A at 0, point B at 1, and point C at e^(iπ/3) = 0.5 + i*(sqrt(3)/2).Then, point B' is constructed on AB, so AB' = AC'. Since AB is from 0 to 1, and AC is from 0 to e^(iπ/3). So, AC' is equal to AB', which is a segment on AB.Wait, no, AB' is constructed outwardly on AB, so AB' is a triangle outwardly on AB, meaning that B' is outside the original triangle ABC.Similarly, AC' is constructed outwardly on AC, so C' is outside ABC.Given that, maybe I can express B' and C' as complex numbers.But this might be complicated.Alternatively, maybe I can use vectors. Let me denote vectors AB, AC, etc.But I'm not sure.Wait, maybe I can consider the triangle inequality in the hexagon.In the hexagon AB'CA'BC', the sides are AB', B'C, CA', A'B, BC', and C'A.But I'm not sure.Alternatively, maybe I can consider the triangle formed by AB', BC', and CA' and use the given equalities and angles to show that the triangle inequality holds.Given that AB' = AC' = x, BC' = BA' = y, and CA' = CB' = z.So, I need to show that x + y > z, x + z > y, and y + z > x.But since x = AC', y = BA', z = CB', and from the hexagon, maybe there are relations between these.Wait, in the hexagon, we have sides AB', B'C, CA', A'B, BC', and C'A.But I'm not sure.Alternatively, maybe I can use the fact that in triangle A'BC', A'C' > y * sqrt(3), and since A'C' is a side in the hexagon, which is also part of triangle AB'C, which has sides AB' = AC' = x.Wait, but A'C' is a different side.Alternatively, maybe I can consider that in triangle AB'C, which is isosceles with AB' = AC' = x, and base BC = a.Using the Law of Cosines, BC^2 = AB'^2 + AC'^2 - 2 * AB' * AC' * cos(angle BAC').But angle BAC' is part of the hexagon, which is angle C'AB', which is greater than 120 degrees.Wait, angle C'AB' is at point A, between points C' and B'.So, in triangle AB'C, angle at A is angle BAC', which is equal to angle C'AB' in the hexagon, which is greater than 120 degrees.Therefore, in triangle AB'C, angle at A is greater than 120 degrees, sides AB' = AC' = x, and base BC = a.So, using the Law of Cosines:a^2 = x^2 + x^2 - 2x^2 cos(angle BAC')= 2x^2 (1 - cos(angle BAC'))Since angle BAC' > 120 degrees, cos(angle BAC') < cos(120 degrees) = -0.5.Therefore, 1 - cos(angle BAC') > 1 - (-0.5) = 1.5.So, a^2 > 2x^2 * 1.5 = 3x^2.Therefore, a > x * sqrt(3).Similarly, in triangle CA'B, which is isosceles with BC' = BA' = y, and base CA = a.Angle at C is angle A'CB', which is greater than 120 degrees.Using the Law of Cosines:a^2 = y^2 + y^2 - 2y^2 cos(angle A'CB')= 2y^2 (1 - cos(angle A'CB'))Since angle A'CB' > 120 degrees, cos(angle A'CB') < -0.5.Therefore, 1 - cos(angle A'CB') > 1.5.So, a^2 > 3y^2.Therefore, a > y * sqrt(3).Similarly, in triangle BC'A, which is isosceles with CA' = CB' = z, and base AB = a.Angle at B is angle B'AC', which is greater than 120 degrees.Using the Law of Cosines:a^2 = z^2 + z^2 - 2z^2 cos(angle B'AC')= 2z^2 (1 - cos(angle B'AC'))Since angle B'AC' > 120 degrees, cos(angle B'AC') < -0.5.Therefore, 1 - cos(angle B'AC') > 1.5.So, a^2 > 3z^2.Therefore, a > z * sqrt(3).So, from all three, we have:a > x * sqrt(3),a > y * sqrt(3),a > z * sqrt(3).Therefore, x < a / sqrt(3),y < a / sqrt(3),z < a / sqrt(3).But I need to relate x, y, z to each other.Wait, from the given equalities, AB' = AC' = x,BC' = BA' = y,CA' = CB' = z.But BA' = BC' = y,and CB' = CA' = z.So, BA' = BC' = y,and CB' = CA' = z.Therefore, BA' = BC' = y,and CB' = CA' = z.So, in triangle BA'C, sides BA' = BC' = y,and CA' = z.Similarly, in triangle CB'A, sides CB' = CA' = z,and BA' = y.But I'm not sure.Alternatively, maybe I can consider the triangle formed by AB', BC', and CA'.Let me denote the lengths as AB' = x, BC' = y, and CA' = z.I need to show that x + y > z, x + z > y, and y + z > x.But from the given equalities, x = AC', y = BA', z = CB'.But BA' = BC' = y,and CB' = CA' = z.So, BA' = BC' = y,and CB' = CA' = z.Therefore, in triangle BA'C, sides BA' = BC' = y,and CA' = z.So, in triangle BA'C, sides y, y, z.Similarly, in triangle CB'A, sides z, z, y.But I'm not sure.Alternatively, maybe I can use the triangle inequality in triangle BA'C.In triangle BA'C, sides BA' = y, BC' = y, and CA' = z.Therefore, by triangle inequality:y + y > z,so 2y > z,and y + z > y,so z > 0,and y + z > y,same as above.Similarly, in triangle CB'A, sides CB' = z, CA' = z, and BA' = y.So, triangle inequality gives:z + z > y,so 2z > y,and z + y > z,so y > 0,and z + y > z,same as above.So, from triangle BA'C, we have 2y > z,and from triangle CB'A, we have 2z > y.Similarly, in triangle AB'C, which is isosceles with AB' = AC' = x,and base BC = a.From earlier, we have a > x * sqrt(3).But I also know that in triangle AB'C, sides AB' = AC' = x,and base BC = a.So, by triangle inequality,x + x > a,so 2x > a,and x + a > x,so a > 0,and x + a > x,same as above.But from earlier, a > x * sqrt(3),and 2x > a,so combining these,x * sqrt(3) < a < 2x.Similarly, for y and z,a > y * sqrt(3),and 2y > a,so y * sqrt(3) < a < 2y,and similarly,z * sqrt(3) < a < 2z.Therefore,x * sqrt(3) < a < 2x,y * sqrt(3) < a < 2y,z * sqrt(3) < a < 2z.Therefore,x > a / 2,y > a / 2,z > a / 2,andx < a / sqrt(3),y < a / sqrt(3),z < a / sqrt(3).So, x, y, z are all between a/2 and a/sqrt(3).But since a/sqrt(3) is approximately 0.577a,and a/2 is 0.5a,so x, y, z are between 0.5a and ~0.577a.Therefore, x, y, z are all less than a/sqrt(3), which is less than a/2 * sqrt(3)/2, but not sure.Wait, but from the inequalities,x < a / sqrt(3),y < a / sqrt(3),z < a / sqrt(3),andx > a / 2,y > a / 2,z > a / 2.So, all three segments AB', BC', CA' are between a/2 and a/sqrt(3).But a/sqrt(3) is approximately 0.577a, which is greater than a/2 (0.5a).So, x, y, z are all in (0.5a, ~0.577a).Therefore, the sum of any two segments would be greater than a.But since each segment is greater than a/2, the sum of any two would be greater than a.But wait, we need to show that the sum of any two segments is greater than the third.So, for example, x + y > z.But since x > a/2, y > a/2,so x + y > a.But z < a / sqrt(3) ≈ 0.577a,so x + y > a > z.Therefore, x + y > z.Similarly, x + z > y,and y + z > x.Therefore, the triangle inequality holds.Therefore, segments AB', BC', and CA' can form a triangle.So, that's the conclusion.But let me recap.Given that AB' = AC' = x,BC' = BA' = y,and CA' = CB' = z.From the constructed triangles and the given angles greater than 120 degrees, we derived that each of x, y, z is between a/2 and a/sqrt(3).Therefore, the sum of any two of x, y, z is greater than a, which is greater than the third segment.Hence, the triangle inequality holds, and the segments can form a triangle.

💡Okay, so I have this expression to evaluate: tan 25° + tan 35° + √3 tan 25° tan 35°. Hmm, I remember there are some trigonometric identities that involve the tangent of sums or differences of angles. Maybe I can use one of those to simplify this expression.Let me recall the tangent addition formula: tan(A + B) = (tan A + tan B) / (1 - tan A tan B). That looks similar to the first two terms of my expression, tan 25° + tan 35°. If I set A = 25° and B = 35°, then tan(25° + 35°) = tan 60°, which I know is √3. So, tan 60° = √3 = (tan 25° + tan 35°) / (1 - tan 25° tan 35°).Wait, so that means (tan 25° + tan 35°) = √3 (1 - tan 25° tan 35°). Let me write that down:tan 25° + tan 35° = √3 (1 - tan 25° tan 35°)Hmm, interesting. Now, looking back at the original expression, I have tan 25° + tan 35° + √3 tan 25° tan 35°. If I substitute the expression I just found into this, I get:√3 (1 - tan 25° tan 35°) + √3 tan 25° tan 35°Let me distribute the √3 in the first term:√3 * 1 - √3 tan 25° tan 35° + √3 tan 25° tan 35°Oh, look at that! The -√3 tan 25° tan 35° and +√3 tan 25° tan 35° cancel each other out. So, I'm left with just √3.Wait, so does that mean the entire expression simplifies to √3? Let me double-check my steps to make sure I didn't make a mistake.First, I used the tangent addition formula correctly: tan(A + B) = (tan A + tan B) / (1 - tan A tan B). Then, I solved for tan A + tan B, which gave me tan A + tan B = tan(A + B)(1 - tan A tan B). Plugging in A = 25° and B = 35°, I got tan 25° + tan 35° = √3 (1 - tan 25° tan 35°). That seems right because tan 60° is indeed √3.Then, substituting back into the original expression, I had √3 (1 - tan 25° tan 35°) + √3 tan 25° tan 35°, which simplifies to √3. That makes sense because the terms involving tan 25° tan 35° cancel each other out.So, after carefully going through the steps, I think the value of the expression is √3. But wait, the problem didn't specify whether it wants the answer in terms of a numerical value or just in terms of √3. Since √3 is an exact value, I think that's acceptable.Just to be thorough, let me calculate the numerical values to confirm. Let's compute tan 25°, tan 35°, and then plug them into the original expression.First, tan 25° is approximately 0.4663, and tan 35° is approximately 0.7002. Then, tan 25° + tan 35° is approximately 0.4663 + 0.7002 = 1.1665. Next, √3 is approximately 1.732, so √3 tan 25° tan 35° is approximately 1.732 * 0.4663 * 0.7002.Calculating that: 0.4663 * 0.7002 ≈ 0.3266, and then 1.732 * 0.3266 ≈ 0.564. So, adding that to 1.1665 gives approximately 1.1665 + 0.564 ≈ 1.7305.And √3 is approximately 1.732, so 1.7305 is very close to √3, which confirms that my earlier conclusion was correct.Therefore, the value of the expression tan 25° + tan 35° + √3 tan 25° tan 35° is indeed √3.

💡Okay, so I have this problem here: The statement ( m notin (-3, -1) ) is a ( ) condition for the function ( f(x) = 3x + m ) not to have a zero within the interval ([0, 1]). The options are A: Sufficient but not necessary, B: Necessary but not sufficient, C: Both necessary and sufficient, D: Neither sufficient nor necessary.Alright, let me try to understand what this is asking. I need to figure out whether the condition ( m notin (-3, -1) ) is sufficient, necessary, both, or neither for ensuring that the function ( f(x) = 3x + m ) doesn't have a zero in the interval ([0, 1]).First, let's recall what it means for a function to have a zero in an interval. A zero of the function ( f(x) ) in the interval ([0, 1]) means there exists some ( x ) in ([0, 1]) such that ( f(x) = 0 ). So, we want to ensure that ( f(x) neq 0 ) for all ( x ) in ([0, 1]).Given ( f(x) = 3x + m ), this is a linear function with a slope of 3, which is positive, so it's increasing. The function will have a zero where ( 3x + m = 0 ), which is at ( x = -m/3 ).So, for the function not to have a zero in ([0, 1]), the value ( x = -m/3 ) should not lie within ([0, 1]). That means ( -m/3 ) should be less than 0 or greater than 1.Let me write that down:1. If ( -m/3 < 0 ), then ( m > 0 ).2. If ( -m/3 > 1 ), then ( m < -3 ).So, combining these two, the function ( f(x) = 3x + m ) does not have a zero in ([0, 1]) if ( m > 0 ) or ( m < -3 ).Wait, so ( m ) must be either greater than 0 or less than -3 for the function to not have a zero in ([0, 1]). That means the set of ( m ) values where the function doesn't have a zero is ( (-infty, -3) cup (0, infty) ).Now, the given condition is ( m notin (-3, -1) ). So, ( m ) is not in the open interval from -3 to -1. That means ( m ) can be less than or equal to -3, between -1 and 0, or greater than 0.But from earlier, we saw that the function doesn't have a zero in ([0, 1]) only when ( m > 0 ) or ( m < -3 ). So, if ( m ) is between -1 and 0, which is part of ( m notin (-3, -1) ), the function might still have a zero in ([0, 1]).Let me test this. Suppose ( m = -2 ), which is in ( (-3, -1) ). Then, ( f(x) = 3x - 2 ). Setting ( f(x) = 0 ), we get ( x = 2/3 ), which is approximately 0.666, and that's within ([0, 1]). So, ( m = -2 ) leads to a zero in the interval.What if ( m = -1 )? Then, ( f(x) = 3x - 1 ). Setting ( f(x) = 0 ), ( x = 1/3 ), which is also in ([0, 1]). So, ( m = -1 ) still gives a zero in the interval.If ( m = -4 ), which is less than -3, then ( f(x) = 3x - 4 ). Setting ( f(x) = 0 ), ( x = 4/3 ), which is approximately 1.333, outside of ([0, 1]). So, no zero in the interval.If ( m = 1 ), which is greater than 0, then ( f(x) = 3x + 1 ). Setting ( f(x) = 0 ), ( x = -1/3 ), which is negative, so again, no zero in ([0, 1]).So, from these examples, it seems that when ( m ) is in ( (-3, -1) ), the function does have a zero in ([0, 1]), and when ( m ) is outside of ( (-3, -1) ), specifically ( m < -3 ) or ( m > 0 ), the function does not have a zero in the interval.But wait, the condition given is ( m notin (-3, -1) ). So, ( m ) can be in ( (-infty, -3] cup [-1, infty) ). However, from our earlier analysis, the function does not have a zero in ([0, 1]) only when ( m in (-infty, -3) cup (0, infty) ).So, the condition ( m notin (-3, -1) ) includes some values where the function still has a zero in ([0, 1]). Specifically, when ( m ) is between -1 and 0, the function might still have a zero in the interval.For example, take ( m = -0.5 ). Then, ( f(x) = 3x - 0.5 ). Setting ( f(x) = 0 ), ( x = 0.5/3 approx 0.1667 ), which is within ([0, 1]). So, even though ( m notin (-3, -1) ), the function still has a zero in the interval.Therefore, the condition ( m notin (-3, -1) ) is not sufficient to guarantee that the function does not have a zero in ([0, 1]). However, if the function does not have a zero in ([0, 1]), then ( m ) must be outside of ( (-3, -1) ). So, the condition is necessary but not sufficient.Wait, let me make sure I'm not confusing necessary and sufficient here. A necessary condition must be satisfied for the statement to be true, but it alone doesn't guarantee the statement. A sufficient condition, if satisfied, guarantees the statement, but there might be other ways for the statement to be true.In this case, for the function not to have a zero in ([0, 1]), ( m ) must be outside of ( (-3, -1) ). So, ( m notin (-3, -1) ) is a necessary condition. However, just knowing that ( m notin (-3, -1) ) isn't enough to ensure the function doesn't have a zero in the interval because ( m ) could still be between -1 and 0, leading to a zero in the interval. Therefore, it's not sufficient.So, putting it all together, the condition ( m notin (-3, -1) ) is necessary but not sufficient for the function ( f(x) = 3x + m ) not to have a zero within the interval ([0, 1]). Therefore, the correct answer should be B: Necessary but not sufficient.