💡Alright, so I'm trying to figure out which channel will finish the film earlier. There are two channels mentioned, and each has a different way of inserting commercials between the film parts. Let me break this down step by step.First, the problem says that the film is divided into parts of 10 minutes each. On the second channel, there's a 1-minute commercial break between these parts. But it doesn't specify how the first channel handles commercials. Hmm, maybe I need to assume something here. Perhaps the first channel has longer commercial breaks or a different structure? Wait, the problem doesn't mention the first channel's commercial breaks, so maybe I need to focus only on the second channel's setup and compare it implicitly to the first channel's setup.But that doesn't make much sense. Let me read the problem again carefully. It says, "A film on the second channel was divided into parts of 10 minutes each with 1-minute commercial breaks inserted between them. Which channel will finish the film earlier?" So, it seems like the first channel is being compared to the second channel in terms of how they present the film with commercials.Maybe the first channel has a different commercial break structure. Perhaps the first channel has longer commercials or more frequent ones? But the problem doesn't specify. Maybe I need to assume that the first channel doesn't have any commercials, or it has a different commercial break duration.Wait, the problem only gives details about the second channel. It says the second channel divides the film into 10-minute parts with 1-minute commercials in between. So, perhaps the first channel doesn't have any commercials, or it has a different structure that allows it to finish earlier.But without specific information about the first channel's commercial breaks, it's hard to compare. Maybe the first channel has no commercials, so it just plays the film continuously. If that's the case, then the first channel would obviously finish the film earlier because it doesn't have any interruptions.But the problem mentions "the second channel," implying that the first channel is being compared to it in terms of commercial breaks. Maybe the first channel has longer commercials or more commercials, making it take longer to finish the film.Wait, the problem says, "Which channel will finish the film earlier?" So, it's asking which one finishes first. If the second channel has 1-minute commercials between 10-minute parts, then the total time for the film would be the sum of the film parts plus the commercials.Let's say the film has 'n' parts of 10 minutes each. Then, the total film time would be 10n minutes. The total commercial time would be (n-1) minutes, since there's a commercial break after each part except the last one. So, the total time for the second channel would be 10n + (n-1) minutes.Now, if the first channel has a different commercial structure, maybe it has longer commercials or more frequent ones. But since the problem doesn't specify, I might need to assume that the first channel has no commercials, making its total time just 10n minutes.In that case, the second channel would take longer because it has the additional commercial time. So, the first channel would finish the film earlier.But wait, the problem doesn't explicitly say that the first channel has no commercials. It just says that the second channel has 1-minute commercials between 10-minute parts. Maybe the first channel has a different commercial break duration, like 2 minutes or something else.If the first channel has, say, 2-minute commercials between parts, then its total time would be 10n + 2(n-1) minutes, which is more than the second channel's 10n + (n-1) minutes. So, the second channel would finish earlier.But since the problem doesn't specify the first channel's commercial breaks, I'm not sure. Maybe I need to assume that the first channel has no commercials, making it finish earlier.Alternatively, maybe the first channel has a different way of inserting commercials, like at the beginning and end of the film, rather than between parts. That could affect the total time as well.Wait, the problem says "between them," meaning between the parts. So, the first channel might have commercials at the beginning or end, but not between parts. If that's the case, then the total commercial time for the first channel would be less, making it finish earlier.But again, without specific information, it's hard to be certain. Maybe I need to make an assumption based on standard practices. Typically, channels have commercials between parts, so perhaps the first channel also has commercials, but maybe fewer or shorter ones.Alternatively, maybe the first channel doesn't have any commercials, which would make it finish the film faster.Given the ambiguity, I think the safest assumption is that the first channel has no commercials, so it plays the film continuously, finishing earlier than the second channel, which has 1-minute commercials between parts.So, in conclusion, the first channel would finish the film earlier.

💡First, I need to identify the two numbers based on the given information. The sum of the two numbers is 70, and their product is 1050.I'll start by setting up the equations:1. ( x + y = 70 )2. ( x cdot y = 1050 )Next, I'll express one variable in terms of the other using the first equation:[ y = 70 - x ]Then, I'll substitute this expression into the second equation:[ x(70 - x) = 1050 ]Simplifying this equation:[ 70x - x^2 = 1050 ][ x^2 - 70x + 1050 = 0 ]Now, I'll attempt to factor the quadratic equation:[ (x - 35)(x - 30) = 0 ]This gives me the possible solutions for ( x ):[ x = 35 quad text{or} quad x = 30 ]Using the first equation ( x + y = 70 ), I'll find the corresponding values of ( y ):- If ( x = 35 ), then ( y = 35 )- If ( x = 30 ), then ( y = 40 )Since we're looking for the lesser number, the smaller value between 30 and 40 is 30.Therefore, the lesser of the two numbers is 30.

💡To find the unknown number, let's denote it as ( x ) and solve the equation step by step.**Given Equation:**[45 - left[ 28 - left{ 37 - (x - 19) right} right] = 58]**Step 1: Simplify the innermost expression**[(x - 19)]**Step 2: Substitute back into the equation**[45 - left[ 28 - left{ 37 - (x - 19) right} right] = 58]**Step 3: Simplify the curly braces**[37 - (x - 19) = 37 - x + 19 = 56 - x]**Step 4: Substitute back into the equation**[45 - left[ 28 - (56 - x) right] = 58]**Step 5: Simplify the square brackets**[28 - (56 - x) = 28 - 56 + x = x - 28]**Step 6: Substitute back into the equation**[45 - (x - 28) = 58]**Step 7: Simplify the equation**[45 - x + 28 = 58 73 - x = 58]**Step 8: Solve for ( x )**[-x = 58 - 73 -x = -15 x = 15]**Final Answer:**[boxed{15}]

💡Alright, so I have this problem here about statistics, specifically dealing with the mean and variance of a set of data. Let me try to understand what it's asking.Okay, so there are 10 data points: x₁, x₂, ..., x₁₀. The mean of these data points is 1, and the variance is 4. Now, they define a new set of data points yᵢ = xᵢ + a, where a is a non-zero constant. I need to find the mean and variance of the new data set y₁, y₂, ..., y₁₀ and choose the correct option from A to D.Hmm, let's start by recalling what mean and variance are. The mean is just the average of the data points, right? So, for the original data set x, the mean is 1. That means if I add up all the x's and divide by 10, I get 1.Now, variance measures how spread out the data is. It's the average of the squared differences from the mean. So, for the original data set, the variance is 4. That tells me something about how much the x's deviate from the mean of 1.Now, the new data set y is created by adding a constant a to each xᵢ. So, every data point is shifted by the same amount a. I need to figure out how this affects the mean and variance.Let me think about the mean first. If I add a constant to each data point, does that change the mean? Intuitively, yes, because if every number increases by a, the average should also increase by a. So, the new mean should be the original mean plus a. That would make the mean of y equal to 1 + a.Okay, that seems straightforward. Now, what about the variance? Variance is about how spread out the data is. If I add a constant to every data point, does that change how spread out they are? Hmm, I don't think so. Because adding a constant just shifts all the data points by the same amount, it doesn't make them any more or less spread out. So, the variance should remain the same.Wait, let me double-check that. Variance is calculated as the average of the squared differences from the mean. So, if I add a to each xᵢ, the new mean becomes 1 + a. Then, when I calculate the squared differences, it's (yᵢ - new mean)², which is (xᵢ + a - (1 + a))². Simplifying that, the a's cancel out, and it becomes (xᵢ - 1)². So, the squared differences are the same as before. Therefore, the variance remains 4.So, putting it all together, the mean of y is 1 + a, and the variance remains 4. Looking back at the options, that's option A: 1 + a, 4.Wait, just to make sure I didn't make a mistake. Let me go through the calculations step by step.Original mean, μ_x = 1.Original variance, σ²_x = 4.New data set: yᵢ = xᵢ + a.New mean, μ_y = E[yᵢ] = E[xᵢ + a] = E[xᵢ] + E[a] = μ_x + a = 1 + a.For variance, Var(y) = Var(xᵢ + a) = Var(xᵢ) + Var(a). But Var(a) is zero because variance of a constant is zero. So, Var(y) = Var(x) = 4.Yep, that confirms it. So, the mean increases by a, and the variance stays the same.I think I got it right. So, the answer should be A: 1 + a, 4.**Final Answer**boxed{A}

💡First, I need to determine Steve's speed on the way back from work. I'll start by defining Steve's speed on the way to work as ( v ) km/h.Next, I'll calculate the time it takes Steve to travel to work. Since the distance is 28 km and his speed is ( v ) km/h, the time taken to go to work is ( frac{28}{v} ) hours.On the way back, Steve drives twice as fast, so his speed is ( 2v ) km/h. The time taken to return home is ( frac{28}{2v} = frac{14}{v} ) hours.According to the problem, the total time spent on the road is 6 hours. Therefore, I can set up the equation:[frac{28}{v} + frac{14}{v} = 6]Combining the terms on the left side gives:[frac{42}{v} = 6]Solving for ( v ):[v = frac{42}{6} = 7 text{ km/h}]So, Steve's speed on the way to work is 7 km/h. On the way back, his speed is twice that, which is:[2v = 2 times 7 = 14 text{ km/h}]Therefore, Steve's speed on the way back from work is 14 km/h.

💡Okay, so I have this problem here: How many pairs of positive integers (x, y) satisfy x² - y² = 77? Hmm, interesting. I remember that x² - y² can be factored into (x + y)(x - y). Maybe that can help me solve this problem.Alright, let's write that down: x² - y² = (x + y)(x - y) = 77. So, I need to find positive integers x and y such that when I multiply (x + y) and (x - y), I get 77. Now, 77 is a specific number. I should probably factorize 77 to find all possible pairs of factors. Let me think, 77 divided by 1 is 77, and 77 divided by 7 is 11. So, the factors of 77 are 1, 7, 11, and 77. That means the possible pairs of factors (a, b) such that a * b = 77 are (1, 77) and (7, 11). Since x and y are positive integers, both (x + y) and (x - y) must also be positive integers. Also, (x + y) must be greater than (x - y) because x and y are positive, so y has to be less than x. That makes sense because if y were equal to or greater than x, then (x - y) would be zero or negative, which isn't allowed here.So, I have two cases to consider based on the factor pairs I found:**Case 1:** (x + y) = 77 and (x - y) = 1.Let me write these equations down:1. x + y = 772. x - y = 1If I add these two equations together, the y terms will cancel out. Let's see:Adding equation 1 and equation 2:(x + y) + (x - y) = 77 + 12x = 78So, x = 78 / 2 = 39.Now, plug x back into one of the equations to find y. Let's use equation 2:39 - y = 1So, y = 39 - 1 = 38.Alright, so one solution is (x, y) = (39, 38). That seems straightforward.**Case 2:** (x + y) = 11 and (x - y) = 7.Again, writing down the equations:1. x + y = 112. x - y = 7Adding these two equations:(x + y) + (x - y) = 11 + 72x = 18So, x = 18 / 2 = 9.Now, plug x back into equation 2 to find y:9 - y = 7So, y = 9 - 7 = 2.Great, another solution is (x, y) = (9, 2).Wait, are there any other factor pairs I should consider? Let me think. 77 is 7 times 11, and 1 times 77. Are there any negative factors? Well, since we're dealing with positive integers, negative factors wouldn't make sense here because x and y are positive, so (x + y) and (x - y) must both be positive. So, I don't need to consider negative factor pairs.Is there a possibility that (x + y) and (x - y) could be fractions? Hmm, but since x and y are integers, (x + y) and (x - y) must also be integers. So, fractions aren't an issue here.Let me just double-check my solutions to make sure they satisfy the original equation.For (39, 38):x² - y² = 39² - 38². Let's compute that:39² = 152138² = 14441521 - 1444 = 77. Perfect, that works.For (9, 2):x² - y² = 9² - 2² = 81 - 4 = 77. That also works.Are there any other factor pairs? Let me list all the factors of 77 again: 1, 7, 11, 77. So, the possible factor pairs are (1,77) and (7,11). I've considered both cases, so I think these are the only solutions.Wait a second, could (x + y) and (x - y) be the same number? For example, if 77 were a perfect square, but 77 isn't a perfect square. So, (x + y) and (x - y) can't be equal because that would require 77 to be a square, which it isn't. So, no need to worry about that case.Also, I should check if there are any other factor pairs beyond the obvious ones. For example, could 77 be factored into more than two factors? Well, in the context of this problem, since we're dealing with two factors (x + y) and (x - y), I don't think that's necessary. We've already covered all possible pairs.So, in conclusion, there are two pairs of positive integers (x, y) that satisfy the equation x² - y² = 77. These pairs are (39, 38) and (9, 2).I think that's all. I don't see any other possibilities, and I've verified both solutions. So, the answer should be 2 pairs.**Final Answer**The number of pairs is boxed{2}.