💡Okay, so I have this problem about football game attendances on different days, and I need to figure out how many more people attended than expected. Let me try to break this down step by step.First, the problem says there were 80 people at the game on Saturday. That seems straightforward. Then, on Monday, there were 20 fewer people than on Saturday. So, if Saturday had 80 people, Monday must have had 80 minus 20, which is 60 people. Okay, that makes sense.Next, on Wednesday, there were 50 more people than on Monday. Since Monday had 60 people, adding 50 to that gives us 110 people on Wednesday. Got it. So far, we have 80 on Saturday, 60 on Monday, and 110 on Wednesday.Now, on Friday, the number of people was the same as the combined total of Saturday and Monday. So, that would be 80 plus 60, which is 140 people on Friday. Alright, so Friday had 140 attendees.Let me summarize the attendances:- Saturday: 80- Monday: 60- Wednesday: 110- Friday: 140Now, the problem mentions that the expected total audience for the week was 350 people. I need to find out how many more people actually attended than expected.First, I should calculate the actual total attendance for the week. That means adding up the numbers for each day:- Saturday: 80- Monday: 60- Wednesday: 110- Friday: 140Adding these together: 80 + 60 = 140, 140 + 110 = 250, 250 + 140 = 390. So, the total attendance for the week was 390 people.The expected attendance was 350 people. To find out how many more people attended than expected, I subtract the expected number from the actual number:390 (actual) - 350 (expected) = 40.So, 40 more people attended the games than expected.Wait, let me double-check my calculations to make sure I didn't make any mistakes. Starting with Saturday: 80, Monday: 60, Wednesday: 110, Friday: 140. Adding them up: 80 + 60 is 140, plus 110 is 250, plus 140 is 390. Yes, that seems correct.Expected was 350, so 390 minus 350 is indeed 40. Okay, I think that's right. I don't see any errors in my reasoning or calculations.

💡Okay, so I have this equation here: |x - 1| = 1 - x. Hmm, absolute value equations can sometimes be tricky, but I think I remember that the absolute value of something is always non-negative. So, |x - 1| is going to be greater than or equal to zero. That means 1 - x must also be greater than or equal to zero because they're equal. So, 1 - x ≥ 0. If I solve that, I get x ≤ 1. Wait, let me make sure I'm doing this right. The absolute value |x - 1| equals 1 - x. So, another way to think about absolute value is that it's the distance from x to 1 on the number line. So, if the distance from x to 1 is equal to 1 - x, that means that 1 - x must be non-negative because distance can't be negative. So, 1 - x ≥ 0, which again gives x ≤ 1. But let me double-check. If x is less than or equal to 1, then x - 1 is less than or equal to zero. So, |x - 1| would be -(x - 1), which is 1 - x. So, that makes sense because |x - 1| = 1 - x. Let me test a value to see if this works. Let's say x = 0. Then |0 - 1| = | -1 | = 1, and 1 - 0 = 1. So, 1 = 1, which is correct. What if x = 1? Then |1 - 1| = 0, and 1 - 1 = 0. So, 0 = 0, which is also correct. What if x is greater than 1? Let's say x = 2. Then |2 - 1| = 1, and 1 - 2 = -1. So, 1 ≠ -1, which doesn't work. So, that confirms that x has to be less than or equal to 1. So, the correct answer should be option C: x ≤ 1.

💡Okay, so I have this problem here where I need to find the intersection of two sets, A and B. Let me try to understand what each set is first.Set A is defined as {x | x > l}. Hmm, wait, is that "l" as in the letter L or is it a typo? Maybe it's supposed to be a number. Let me check the problem again. Oh, it says "x > l". Maybe "l" is a variable or a constant. Since it's not defined, perhaps it's a typo and should be "1"? Because in the options, I see numbers like 1 and 3. Maybe it's supposed to be x > 1. That would make sense. So I'll assume set A is {x | x > 1}. Set B is given as {x | x² - 2x - 3 < 0}. Okay, so I need to solve this quadratic inequality. Let me recall how to solve quadratic inequalities. First, I should find the roots of the quadratic equation x² - 2x - 3 = 0. To find the roots, I can use the quadratic formula: x = [2 ± √(4 + 12)] / 2 = [2 ± √16] / 2 = [2 ± 4]/2. So the roots are (2 + 4)/2 = 3 and (2 - 4)/2 = -1. So the quadratic can be factored as (x - 3)(x + 1) = 0. Now, to solve the inequality (x - 3)(x + 1) < 0, I need to determine where the quadratic expression is negative. The critical points are x = -1 and x = 3, which divide the number line into three intervals: (-∞, -1), (-1, 3), and (3, ∞). I can test each interval to see where the product is negative. Let's pick a test point in each interval:1. For x < -1, say x = -2: (-2 - 3)(-2 + 1) = (-5)(-1) = 5, which is positive.2. For -1 < x < 3, say x = 0: (0 - 3)(0 + 1) = (-3)(1) = -3, which is negative.3. For x > 3, say x = 4: (4 - 3)(4 + 1) = (1)(5) = 5, which is positive.So the inequality x² - 2x - 3 < 0 holds true for x in (-1, 3). Therefore, set B is (-1, 3).Now, set A is {x | x > 1}, which is the interval (1, ∞). To find A ∩ B, I need the overlap between (1, ∞) and (-1, 3). Visualizing the two intervals on the number line, (1, ∞) starts at 1 and goes to infinity, while (-1, 3) starts at -1 and ends at 3. The overlap between these two intervals is from 1 to 3. So the intersection A ∩ B is (1, 3).Looking at the options provided:A: ∅ (empty set)B: (1, +∞)C: (-1, 3)D: (1, 3)The correct answer is D: (1, 3).I think that makes sense. I solved the quadratic inequality correctly, found the intervals, and then found their intersection. I didn't make any calculation errors, and the logic seems sound. So I'm confident that D is the right choice.**Final Answer**boxed{D}

💡First, I need to determine how much Y is paid per week. I know that the total payment to both X and Y is Rs. 560.I'm told that X is paid 120% of what Y is paid. To make this easier, I'll let the amount paid to Y be represented by Y.Since X is paid 120% of Y, I can express X's payment as 1.2Y.Now, I can set up an equation that represents the total payment:Y (for Y) + 1.2Y (for X) = 560Combining like terms, this simplifies to:2.2Y = 560To find Y, I'll divide both sides of the equation by 2.2:Y = 560 / 2.2Calculating this gives:Y = 254.545454545...Since we're dealing with currency, I'll round this to the nearest whole number:Y ≈ Rs. 255Therefore, Y is paid approximately Rs. 255 per week.

💡First, I need to establish the relationships between the length, width, and height of the hall based on the given proportions. The width is half of the length, and the height is one-third of the width.Next, I'll express the width and height in terms of the length using these relationships. This will allow me to write the volume formula in terms of a single variable, the length.Once I have the volume expressed in terms of the length, I can set up the equation using the given volume of 600 cubic meters and solve for the length. After finding the length, I'll use the established relationships to determine the width and height.Finally, I'll calculate the difference between the length, width, and height by subtracting the width and height from the length. This will give me the desired difference.

💡Okay, so I have this function ( f(x) = sin^2x - sqrt{3}cos xcos left(x+ frac{pi}{2}right) ), and I need to find the interval where it's monotonically increasing on the interval ([0, frac{pi}{2}]). Hmm, okay, let's break this down step by step.First, I remember that to find where a function is increasing or decreasing, I need to look at its derivative. If the derivative is positive, the function is increasing; if it's negative, the function is decreasing. So, my first step is to find ( f'(x) ).But before I jump into taking the derivative, maybe I can simplify the function ( f(x) ) a bit. Let's see. The function has two terms: ( sin^2x ) and ( -sqrt{3}cos xcos left(x+ frac{pi}{2}right) ). Maybe I can simplify the second term using some trigonometric identities.I recall that ( cosleft(x + frac{pi}{2}right) ) can be simplified. There's an identity that says ( cosleft(x + frac{pi}{2}right) = -sin x ). Let me verify that. Yes, because ( cosleft(theta + frac{pi}{2}right) = -sintheta ). So, substituting that in, the second term becomes ( -sqrt{3}cos x cdot (-sin x) ), which simplifies to ( sqrt{3}cos x sin x ).So now, the function ( f(x) ) simplifies to ( sin^2x + sqrt{3}cos x sin x ). That's better. Now, maybe I can combine these terms or use another identity to make it even simpler.Looking at ( sin^2x ), I remember the identity ( sin^2x = frac{1 - cos 2x}{2} ). Let me apply that. So, substituting that in, the function becomes ( frac{1 - cos 2x}{2} + sqrt{3}cos x sin x ).Now, looking at the second term ( sqrt{3}cos x sin x ), I recall that ( sin 2x = 2sin x cos x ), so ( sin x cos x = frac{sin 2x}{2} ). Therefore, ( sqrt{3}cos x sin x = frac{sqrt{3}}{2}sin 2x ).So, substituting that back in, the function becomes ( frac{1}{2} - frac{cos 2x}{2} + frac{sqrt{3}}{2}sin 2x ). Let me write that as ( frac{1}{2} + frac{sqrt{3}}{2}sin 2x - frac{1}{2}cos 2x ).Hmm, this looks like it can be expressed as a single sine function with a phase shift. I remember that ( asintheta + bcostheta = Rsin(theta + phi) ), where ( R = sqrt{a^2 + b^2} ) and ( phi = arctanleft(frac{b}{a}right) ) or something like that. Let me try that.In this case, the coefficients are ( frac{sqrt{3}}{2} ) for sine and ( -frac{1}{2} ) for cosine. So, ( a = frac{sqrt{3}}{2} ) and ( b = -frac{1}{2} ). Therefore, ( R = sqrt{left(frac{sqrt{3}}{2}right)^2 + left(-frac{1}{2}right)^2} = sqrt{frac{3}{4} + frac{1}{4}} = sqrt{1} = 1 ).Okay, so ( R = 1 ). Now, to find ( phi ), I think it's ( arctanleft(frac{b}{a}right) ). So, ( phi = arctanleft(frac{-1/2}{sqrt{3}/2}right) = arctanleft(-frac{1}{sqrt{3}}right) ). I know that ( arctanleft(-frac{1}{sqrt{3}}right) = -frac{pi}{6} ) because ( tanleft(-frac{pi}{6}right) = -frac{1}{sqrt{3}} ).Therefore, the expression ( frac{sqrt{3}}{2}sin 2x - frac{1}{2}cos 2x ) can be written as ( sinleft(2x - frac{pi}{6}right) ). So, the entire function ( f(x) ) becomes ( frac{1}{2} + sinleft(2x - frac{pi}{6}right) ).Alright, so now ( f(x) = sinleft(2x - frac{pi}{6}right) + frac{1}{2} ). That seems simpler. Now, to find where this function is increasing, I need to find where its derivative is positive.Let's compute the derivative ( f'(x) ). The derivative of ( sin(u) ) is ( cos(u) cdot u' ), so ( f'(x) = cosleft(2x - frac{pi}{6}right) cdot 2 + 0 ) (since the derivative of a constant is zero). So, ( f'(x) = 2cosleft(2x - frac{pi}{6}right) ).We need to find where ( f'(x) > 0 ). So, ( 2cosleft(2x - frac{pi}{6}right) > 0 ). Dividing both sides by 2, we get ( cosleft(2x - frac{pi}{6}right) > 0 ).Now, when is cosine positive? Cosine is positive in the intervals ( (-frac{pi}{2} + 2kpi, frac{pi}{2} + 2kpi) ) for any integer ( k ). So, we need ( 2x - frac{pi}{6} ) to lie within one of these intervals.Let me write that as:( -frac{pi}{2} + 2kpi < 2x - frac{pi}{6} < frac{pi}{2} + 2kpi ).Let's solve for ( x ). First, add ( frac{pi}{6} ) to all parts:( -frac{pi}{2} + frac{pi}{6} + 2kpi < 2x < frac{pi}{2} + frac{pi}{6} + 2kpi ).Simplify the left side:( -frac{pi}{2} + frac{pi}{6} = -frac{3pi}{6} + frac{pi}{6} = -frac{2pi}{6} = -frac{pi}{3} ).Simplify the right side:( frac{pi}{2} + frac{pi}{6} = frac{3pi}{6} + frac{pi}{6} = frac{4pi}{6} = frac{2pi}{3} ).So, the inequality becomes:( -frac{pi}{3} + 2kpi < 2x < frac{2pi}{3} + 2kpi ).Divide all parts by 2:( -frac{pi}{6} + kpi < x < frac{pi}{3} + kpi ).Now, we need to find the values of ( x ) in the interval ( [0, frac{pi}{2}] ) that satisfy this inequality. Let's consider ( k = 0 ) first:( -frac{pi}{6} < x < frac{pi}{3} ).But since ( x ) must be in ( [0, frac{pi}{2}] ), the lower bound becomes 0 instead of ( -frac{pi}{6} ). So, for ( k = 0 ), the interval where ( f'(x) > 0 ) is ( [0, frac{pi}{3}] ).What about ( k = 1 )? Let's check:( -frac{pi}{6} + pi < x < frac{pi}{3} + pi ),which simplifies to:( frac{5pi}{6} < x < frac{4pi}{3} ).But ( frac{5pi}{6} ) is approximately 2.618, which is greater than ( frac{pi}{2} ) (approximately 1.571). So, this interval doesn't overlap with our domain ( [0, frac{pi}{2}] ).Similarly, for ( k = -1 ):( -frac{pi}{6} - pi < x < frac{pi}{3} - pi ),which simplifies to:( -frac{7pi}{6} < x < -frac{2pi}{3} ).This is entirely negative and doesn't overlap with our domain either.Therefore, the only interval within ( [0, frac{pi}{2}] ) where ( f'(x) > 0 ) is ( [0, frac{pi}{3}] ).To double-check, let's consider the behavior of ( f'(x) ) at the endpoints and in between. At ( x = 0 ), ( f'(0) = 2cosleft(-frac{pi}{6}right) = 2cosleft(frac{pi}{6}right) = 2 times frac{sqrt{3}}{2} = sqrt{3} > 0 ). So, the function is increasing at ( x = 0 ).At ( x = frac{pi}{3} ), ( f'left(frac{pi}{3}right) = 2cosleft(frac{2pi}{3} - frac{pi}{6}right) = 2cosleft(frac{pi}{2}right) = 2 times 0 = 0 ). So, the derivative is zero here, indicating a possible maximum or minimum.For ( x ) just above ( frac{pi}{3} ), say ( x = frac{pi}{3} + epsilon ), where ( epsilon ) is a small positive number, let's compute ( f'(x) ):( f'left(frac{pi}{3} + epsilonright) = 2cosleft(2left(frac{pi}{3} + epsilonright) - frac{pi}{6}right) = 2cosleft(frac{2pi}{3} + 2epsilon - frac{pi}{6}right) = 2cosleft(frac{pi}{2} + 2epsilonright) ).Since ( cosleft(frac{pi}{2} + 2epsilonright) = -sin(2epsilon) approx -2epsilon ) for small ( epsilon ), which is negative. So, the derivative becomes negative just after ( frac{pi}{3} ), meaning the function starts decreasing.Therefore, the function ( f(x) ) is increasing from ( x = 0 ) up to ( x = frac{pi}{3} ), and then it starts decreasing. So, the interval of monotonic increase is indeed ( [0, frac{pi}{3}] ).I think that's thorough enough. I considered simplifying the function, taking the derivative, solving the inequality, checking the intervals, and even verifying the behavior around the critical point. I don't see any mistakes in my reasoning, so I'm confident in the answer.**Final Answer**The interval of monotonic increase of ( f(x) ) on ( left[0, frac{pi}{2}right] ) is boxed{left[0, frac{pi}{3}right]}.