💡Okay, so I have this problem where two people, A and B, are running around a circular track. They start from diametrically opposite points, which means they're half a lap apart to begin with. They run at constant speeds but in opposite directions. The problem gives me two specific meeting points: the first time they meet after B has run 150 yards, and the second time they meet when A is 90 yards short of completing a lap. I need to find the circumference of the track.Alright, let's break this down. First, since they start diametrically opposite, the distance between them is half the circumference. Let me denote the circumference as C. So, half the circumference is C/2.When they start running towards each other (since they're moving in opposite directions), their first meeting will happen after they've covered the distance between them, which is C/2. The time it takes for them to meet the first time can be calculated based on their speeds.Let me denote the speed of A as v_A and the speed of B as v_B. Since they meet after covering C/2 together, the time taken, t1, can be expressed as:t1 = (C/2) / (v_A + v_B)But we also know that when they meet the first time, B has traveled 150 yards. So, the distance B covers is 150 yards, which is equal to v_B * t1. Similarly, the distance A covers is (C/2 - 150) yards, which is equal to v_A * t1.So, from this, I can write:v_B * t1 = 150v_A * t1 = C/2 - 150Therefore, the ratio of their speeds is:v_A / v_B = (C/2 - 150) / 150That's one equation.Now, moving on to the second meeting. They meet again after some more time. By the time they meet the second time, A is 90 yards short of completing the lap. So, A has run (C - 90) yards in total. Similarly, B would have run some distance as well.But wait, how much distance do they cover between the first and second meeting? Since they're moving in opposite directions, the distance between meetings is again C/2, right? Because after the first meeting, they have to cover the entire circumference relative to each other to meet again. Wait, no, actually, in opposite directions, the time between meetings is the same as the time it takes for them to cover the circumference relative to each other.Wait, maybe I need to think differently. Let me consider the total distance each has run by the second meeting.From the start to the first meeting, A has run (C/2 - 150) yards, and B has run 150 yards.From the first meeting to the second meeting, they again cover the circumference relative to each other, so the total distance covered between the first and second meeting is C. Therefore, the time taken for the second meeting after the first is t2, where:t2 = C / (v_A + v_B)But in this time, A runs v_A * t2 and B runs v_B * t2.So, the total distance A has run by the second meeting is:(C/2 - 150) + v_A * t2 = C - 90Similarly, the total distance B has run by the second meeting is:150 + v_B * t2But since they meet again, the total distance covered by both between the first and second meeting is C. So, v_A * t2 + v_B * t2 = CWhich is consistent with t2 = C / (v_A + v_B)So, let's write the equation for A:(C/2 - 150) + v_A * (C / (v_A + v_B)) = C - 90Similarly, for B:150 + v_B * (C / (v_A + v_B)) = ?But maybe it's better to express everything in terms of the ratio of their speeds.From the first meeting, we have:v_A / v_B = (C/2 - 150) / 150Let me denote this ratio as k:k = (C/2 - 150) / 150So, v_A = k * v_BNow, let's consider the total distance A has run by the second meeting:A has run (C - 90) yards.This is equal to the distance A ran before the first meeting plus the distance A ran after the first meeting:(C/2 - 150) + v_A * t2 = C - 90Similarly, the distance B has run by the second meeting is:150 + v_B * t2But since they meet again, the total distance covered by both between the first and second meeting is C:v_A * t2 + v_B * t2 = CSo, t2 = C / (v_A + v_B)But we can express t2 in terms of k:t2 = C / (k * v_B + v_B) = C / (v_B (k + 1))So, t2 = C / (v_B (k + 1))Now, let's plug this into A's total distance:(C/2 - 150) + v_A * (C / (v_B (k + 1))) = C - 90But v_A = k * v_B, so:(C/2 - 150) + k * v_B * (C / (v_B (k + 1))) = C - 90Simplify:(C/2 - 150) + k * C / (k + 1) = C - 90Now, let's substitute k with (C/2 - 150)/150:Let me denote k = (C/2 - 150)/150So, k = (C/2 - 150)/150Therefore, k + 1 = (C/2 - 150)/150 + 1 = (C/2 - 150 + 150)/150 = C/2 / 150 = C / 300So, k + 1 = C / 300Similarly, k = (C/2 - 150)/150So, let's plug back into the equation:(C/2 - 150) + [(C/2 - 150)/150] * C / (C / 300) = C - 90Simplify the second term:[(C/2 - 150)/150] * C / (C / 300) = [(C/2 - 150)/150] * (C * 300 / C) = [(C/2 - 150)/150] * 300Simplify:[(C/2 - 150)/150] * 300 = 2*(C/2 - 150) = C - 300So, the equation becomes:(C/2 - 150) + (C - 300) = C - 90Combine like terms:C/2 - 150 + C - 300 = C - 90Combine C terms:(3C/2) - 450 = C - 90Subtract C from both sides:C/2 - 450 = -90Add 450 to both sides:C/2 = 360Multiply both sides by 2:C = 720So, the circumference of the track is 720 yards.

💡First, I notice that the expression (100^2 - 2 times 100 times 99 + 99^2) resembles the form of a squared binomial. Specifically, it matches the pattern (a^2 - 2ab + b^2), which is the expansion of ((a - b)^2).By identifying (a) as 100 and (b) as 99, I can rewrite the expression as ((100 - 99)^2).Calculating the difference inside the parentheses, (100 - 99) equals 1.Finally, squaring this result gives (1^2 = 1).Therefore, the value of the expression is 1.

💡First, identify the events:- Event ( A ): The number is odd.- Event ( B ): The number does not exceed 3.Determine the probabilities:- ( P(A) = frac{3}{6} = frac{1}{2} )- ( P(B) = frac{3}{6} = frac{1}{2} )- ( P(A cap B) = frac{2}{6} = frac{1}{3} )Apply the formula for the union of two events:[P(A cup B) = P(A) + P(B) - P(A cap B) = frac{1}{2} + frac{1}{2} - frac{1}{3} = frac{2}{3}]Thus, the probability ( P(A cup B) ) is ( frac{2}{3} ).

💡Okay, so I have this limit problem to solve: the limit as x approaches π/6 of (sin x) raised to the power of (6x/π). Hmm, that seems a bit tricky at first glance, but let me break it down step by step.First, I remember that when dealing with limits of the form [f(x)]^{g(x)}, one common approach is to use logarithms to simplify the expression. Maybe I can take the natural logarithm of the function, find the limit, and then exponentiate the result to get the final answer. That might make things easier.So, let me denote the function as y = (sin x)^{6x/π}. If I take the natural logarithm of both sides, I get ln y = (6x/π) * ln(sin x). Now, I can find the limit of ln y as x approaches π/6, and then exponentiate to find the limit of y.Alright, let's compute the limit of ln y first. That is, lim_{x→π/6} (6x/π) * ln(sin x). I can split this into two parts: lim_{x→π/6} (6x/π) and lim_{x→π/6} ln(sin x). Starting with lim_{x→π/6} (6x/π). If I plug in x = π/6, I get (6*(π/6))/π = (π)/π = 1. So, that part is straightforward.Now, for lim_{x→π/6} ln(sin x). I know that sin(π/6) is 1/2, so ln(1/2) is ln(1) - ln(2) = 0 - ln(2) = -ln(2). So, the limit of ln(sin x) as x approaches π/6 is -ln(2).Putting it all together, the limit of ln y is 1 * (-ln(2)) = -ln(2). Therefore, the limit of y is e^{-ln(2)}. Wait, e^{-ln(2)} can be simplified. Remember that e^{ln(a)} = a, so e^{-ln(2)} = 1/e^{ln(2)} = 1/2. So, the limit of (sin x)^{6x/π} as x approaches π/6 is 1/2.But just to make sure I didn't make any mistakes, let me verify each step again. First, taking the natural logarithm was a good move because it transforms the exponent into a multiplier, making it easier to handle. Then, splitting the limit into two parts is valid because both limits exist. Calculating lim_{x→π/6} (6x/π) gives 1, which seems correct. For lim_{x→π/6} ln(sin x), since sin(π/6) is indeed 1/2, and the natural log of 1/2 is -ln(2), that part checks out too. Multiplying 1 by -ln(2) gives -ln(2), and exponentiating that gives e^{-ln(2)} = 1/2. That all makes sense. I think I've got it right. The limit is 1/2.

💡Alright, so I've got this probability problem about an island called Nevezennya. The island has an equal number of hares and rabbits. Hares make mistakes in one out of every four cases, and rabbits make mistakes in one out of every three cases. A creature comes to the center of the island and says, "I am not a hare!" Then, after thinking, it says, "I am not a rabbit." I need to find the probability that this creature is actually a hare.Hmm, okay, let's break this down. First, since the population is equally divided, the prior probability that the creature is a hare is 1/2, and the same for a rabbit. That seems straightforward.Now, the creature makes two statements: "I am not a hare" and "I am not a rabbit." I need to figure out the probability that it's a hare given these two statements.Let me think about hares and rabbits. Hares make mistakes 25% of the time, so if a hare says something, there's a 25% chance it's wrong. Similarly, rabbits make mistakes 33.33% of the time, so if a rabbit says something, there's a 33.33% chance it's wrong.So, if the creature is a hare, the first statement "I am not a hare" would be a lie because it's actually a hare. Since hares make mistakes 25% of the time, the probability that the hare lies here is 1/4.For the second statement, "I am not a rabbit," if the creature is a hare, this statement is actually true because hares are not rabbits. So, the probability that the hare correctly states this is 3/4, since it only makes mistakes 1/4 of the time.On the other hand, if the creature is a rabbit, the first statement "I am not a hare" is true because rabbits are not hares. Since rabbits make mistakes 1/3 of the time, the probability that the rabbit correctly states this is 2/3.For the second statement, "I am not a rabbit," if the creature is a rabbit, this is a lie. So, the probability that the rabbit lies here is 1/3.Now, I need to calculate the joint probability of both statements given that the creature is a hare and given that it's a rabbit.For the hare:- Probability of lying on the first statement: 1/4- Probability of telling the truth on the second statement: 3/4- So, joint probability: (1/4) * (3/4) = 3/16For the rabbit:- Probability of telling the truth on the first statement: 2/3- Probability of lying on the second statement: 1/3- So, joint probability: (2/3) * (1/3) = 2/9Now, using Bayes' theorem, I need to find the probability that the creature is a hare given both statements. That would be the joint probability for the hare divided by the total joint probability (hare plus rabbit).First, let's calculate the total joint probability:- For hare: 3/16 * 1/2 = 3/32- For rabbit: 2/9 * 1/2 = 1/9Wait, why are we multiplying by 1/2? Oh, right, because the prior probability of being a hare or rabbit is 1/2 each.So, total joint probability is 3/32 + 1/9. To add these, I need a common denominator. The least common multiple of 32 and 9 is 288.Converting:- 3/32 = 27/288- 1/9 = 32/288So, total joint probability is 27/288 + 32/288 = 59/288.Now, the probability that the creature is a hare given both statements is (3/32) divided by (59/288). Let's compute that.First, 3/32 divided by 59/288 is the same as 3/32 * 288/59.Simplify:- 288 divided by 32 is 9.- So, 3 * 9 = 27- Therefore, 27/59.So, the probability that the creature is a hare is 27/59.Wait, let me double-check my calculations to make sure I didn't make a mistake.Starting from the joint probabilities:- Hare: (1/4)*(3/4) = 3/16- Rabbit: (2/3)*(1/3) = 2/9Then, multiplying by prior probabilities:- Hare: 3/16 * 1/2 = 3/32- Rabbit: 2/9 * 1/2 = 1/9Convert to common denominator:- 3/32 = 27/288- 1/9 = 32/288- Total: 59/288Then, the probability is (3/32) / (59/288) = (3/32)*(288/59) = (3*9)/59 = 27/59.Yes, that seems correct. So, the probability that the creature is a hare is 27/59.

💡Okay, so I have this problem about three automated production lines. They all make the same products but have different productivity rates. The problem gives me some relationships between their combined productivities and how long it takes each line to complete a shift assignment. I need to figure out how many hours it takes the first line to complete its shift assignment.Let me try to break this down step by step.First, let's assign some variables to make things clearer. Let's say:- Let the time taken by the first line to complete its shift assignment be ( x ) hours.- Since the second line completes the same assignment 2 hours quicker than the first line, its time would be ( x - 2 ) hours.- Let the time taken by the third line be ( y ) hours.Now, productivity rates are usually the reciprocal of time taken, right? So, the productivity rates (in assignments per hour) would be:- First line: ( frac{1}{x} )- Second line: ( frac{1}{x - 2} )- Third line: ( frac{1}{y} )The problem states that the combined productivity of all three lines working together is 1.5 times greater than the combined productivity of the first and second lines working together. So, mathematically, that would be:[frac{1}{x} + frac{1}{x - 2} + frac{1}{y} = 1.5 left( frac{1}{x} + frac{1}{x - 2} right)]Let me simplify this equation. First, I can subtract ( 1.5 left( frac{1}{x} + frac{1}{x - 2} right) ) from both sides to get:[frac{1}{x} + frac{1}{x - 2} + frac{1}{y} - 1.5 left( frac{1}{x} + frac{1}{x - 2} right) = 0]Simplifying the left side:[left(1 - 1.5right) left( frac{1}{x} + frac{1}{x - 2} right) + frac{1}{y} = 0][-0.5 left( frac{1}{x} + frac{1}{x - 2} right) + frac{1}{y} = 0]Moving the term with ( -0.5 ) to the other side:[frac{1}{y} = 0.5 left( frac{1}{x} + frac{1}{x - 2} right)]So, that gives me an expression for ( frac{1}{y} ) in terms of ( x ). Let me write that as:[frac{1}{y} = frac{1}{2} left( frac{1}{x} + frac{1}{x - 2} right)]I can combine the terms inside the parentheses:[frac{1}{x} + frac{1}{x - 2} = frac{(x - 2) + x}{x(x - 2)} = frac{2x - 2}{x(x - 2)} = frac{2(x - 1)}{x(x - 2)}]So, substituting back into the equation for ( frac{1}{y} ):[frac{1}{y} = frac{1}{2} cdot frac{2(x - 1)}{x(x - 2)} = frac{x - 1}{x(x - 2)}]Therefore, ( y ) is the reciprocal of that:[y = frac{x(x - 2)}{x - 1}]Okay, so now I have ( y ) expressed in terms of ( x ). That's good progress.Next, the problem says that the first line's shift assignment can be completed 4 hours and 48 minutes faster by the second and third lines working together than by the first line alone. Let me convert 4 hours and 48 minutes into hours because our variables are in hours. 48 minutes is 48/60 = 0.8 hours, so total time difference is 4.8 hours.So, if the first line takes ( x ) hours alone, then the second and third lines together take ( x - 4.8 ) hours to complete the same assignment.Let me express this in terms of their combined productivity. The combined productivity of the second and third lines is ( frac{1}{x - 2} + frac{1}{y} ), so the time taken by them together is the reciprocal of that:[frac{1}{frac{1}{x - 2} + frac{1}{y}} = x - 4.8]So, substituting ( y ) from earlier:[frac{1}{frac{1}{x - 2} + frac{x - 1}{x(x - 2)}} = x - 4.8]Let me simplify the denominator:First, find a common denominator for the two terms in the denominator:[frac{1}{x - 2} + frac{x - 1}{x(x - 2)} = frac{x}{x(x - 2)} + frac{x - 1}{x(x - 2)} = frac{x + (x - 1)}{x(x - 2)} = frac{2x - 1}{x(x - 2)}]So, the equation becomes:[frac{1}{frac{2x - 1}{x(x - 2)}} = x - 4.8]Taking the reciprocal:[frac{x(x - 2)}{2x - 1} = x - 4.8]Now, let's solve for ( x ). Multiply both sides by ( 2x - 1 ):[x(x - 2) = (x - 4.8)(2x - 1)]Expanding both sides:Left side:[x^2 - 2x]Right side:[x cdot 2x + x cdot (-1) - 4.8 cdot 2x + (-4.8) cdot (-1)][2x^2 - x - 9.6x + 4.8][2x^2 - 10.6x + 4.8]So, putting it all together:[x^2 - 2x = 2x^2 - 10.6x + 4.8]Bring all terms to one side:[x^2 - 2x - 2x^2 + 10.6x - 4.8 = 0][- x^2 + 8.6x - 4.8 = 0]Multiply both sides by -1 to make the quadratic coefficient positive:[x^2 - 8.6x + 4.8 = 0]Hmm, this is a quadratic equation. Let me write it as:[x^2 - 8.6x + 4.8 = 0]To solve this quadratic equation, I can use the quadratic formula:[x = frac{-b pm sqrt{b^2 - 4ac}}{2a}]Where ( a = 1 ), ( b = -8.6 ), and ( c = 4.8 ).Calculating the discriminant:[b^2 - 4ac = (-8.6)^2 - 4 cdot 1 cdot 4.8 = 73.96 - 19.2 = 54.76]Square root of 54.76 is 7.4, because ( 7.4^2 = 54.76 ).So, plugging back into the quadratic formula:[x = frac{8.6 pm 7.4}{2}]Calculating both possibilities:First solution:[x = frac{8.6 + 7.4}{2} = frac{16}{2} = 8]Second solution:[x = frac{8.6 - 7.4}{2} = frac{1.2}{2} = 0.6]Now, let's consider the solutions. ( x = 8 ) hours and ( x = 0.6 ) hours. Since 0.6 hours is about 36 minutes, which seems too short for a shift assignment, especially considering that the second line takes 2 hours less than the first line. If the first line took 0.6 hours, the second line would take ( 0.6 - 2 = -1.4 ) hours, which doesn't make sense because time can't be negative.Therefore, the only feasible solution is ( x = 8 ) hours.Let me just verify this solution to make sure I didn't make any mistakes.If ( x = 8 ) hours, then:- The second line takes ( 8 - 2 = 6 ) hours.- The third line's time ( y ) is ( frac{8 times 6}{8 - 1} = frac{48}{7} approx 6.857 ) hours.Now, let's check the first condition: the combined productivity of all three lines is 1.5 times that of the first and second lines.Productivity of first and second lines together:[frac{1}{8} + frac{1}{6} = frac{3}{24} + frac{4}{24} = frac{7}{24} text{ assignments per hour}]1.5 times this is:[1.5 times frac{7}{24} = frac{10.5}{24} = frac{7}{16} text{ assignments per hour}]Productivity of all three lines together:[frac{1}{8} + frac{1}{6} + frac{1}{48/7} = frac{1}{8} + frac{1}{6} + frac{7}{48}]Convert all to 48 denominators:[frac{6}{48} + frac{8}{48} + frac{7}{48} = frac{21}{48} = frac{7}{16} text{ assignments per hour}]That matches, so the first condition is satisfied.Now, the second condition: the second and third lines together take 4.8 hours less than the first line alone.Time taken by second and third lines together:[frac{1}{frac{1}{6} + frac{7}{48}} = frac{1}{frac{8}{48} + frac{7}{48}} = frac{1}{frac{15}{48}} = frac{48}{15} = 3.2 text{ hours}]Time taken by first line alone is 8 hours. The difference is ( 8 - 3.2 = 4.8 ) hours, which is exactly what the problem states.So, both conditions are satisfied with ( x = 8 ) hours.Therefore, the time it takes the first line to complete its shift assignment is 8 hours.**Final Answer**The first line takes boxed{8} hours to complete its shift assignment.