💡Alright, so I've got this problem about high-speed trains between Zhengzhou and Beijing. It says that China's high-speed rail is the best in the world, which I guess is true because I've heard it's super fast and efficient. The distance between Zhengzhou and Beijing is about 700 km. Okay, that's a good distance, not too short, not too long.The problem mentions that taking a high-speed train from Zhengzhou to Beijing takes 3.6 hours less than taking an express train. Hmm, so the high-speed train is faster, which makes sense. It also says that the average speed of the high-speed train is 2.8 times that of the express train. Interesting, so if the express train's speed is x km/h, then the high-speed train's speed is 2.8x km/h.Alright, so we need to set up an equation based on this information. Let's break it down step by step. First, let's define the variables:- Let x be the average speed of the express train in km/h.- Then, the average speed of the high-speed train is 2.8x km/h.The distance between the two cities is 700 km. So, the time it takes for each train to cover this distance can be calculated using the formula:[text{Time} = frac{text{Distance}}{text{Speed}}]So, for the express train, the time taken would be:[text{Time}_{text{express}} = frac{700}{x}]And for the high-speed train, the time taken would be:[text{Time}_{text{high-speed}} = frac{700}{2.8x}]Now, according to the problem, the high-speed train takes 3.6 hours less than the express train. So, the difference between the time taken by the express train and the high-speed train is 3.6 hours. That gives us the equation:[text{Time}_{text{express}} - text{Time}_{text{high-speed}} = 3.6]Substituting the expressions we have for the times:[frac{700}{x} - frac{700}{2.8x} = 3.6]Let me double-check that. If I subtract the time of the high-speed train from the express train, I should get the time difference, which is 3.6 hours. Yeah, that makes sense.Looking at the options provided:A: (frac{700}{x} - frac{700}{2.8x} = 3.6)B: (frac{700}{2.8x} - frac{700}{x} = 3.6)C: (frac{700 times 2.8}{x} - frac{700}{x} = 3.6)D: (frac{700}{2.8x} = 3.6 - frac{700}{x})So, comparing my equation with the options, option A matches exactly. Option B has the subtraction reversed, which would give a negative time difference, which doesn't make sense in this context. Option C seems to be multiplying 700 by 2.8, which isn't something we did in our calculation. Option D rearranges the equation but doesn't match the original setup.Therefore, the correct equation should be option A.

💡First, I need to determine how many magazine subscriptions Maggie sold to each person.She sold 4 subscriptions to her parents, 1 subscription to her grandfather, and 2 subscriptions to her next-door neighbor. Additionally, she sold twice that amount to another neighbor, which means she sold 4 subscriptions to the other neighbor.Next, I will add up all the subscriptions she sold:4 (parents) + 1 (grandfather) + 2 (next-door neighbor) + 4 (another neighbor) = 11 subscriptions in total.Since Maggie earns 5.00 for each subscription she sells, I will multiply the total number of subscriptions by the earnings per subscription:11 subscriptions × 5.00 per subscription = 55.00.Therefore, Maggie earned a total of 55.00.

💡Alright, so I've got this problem here where I need to prove that for positive numbers (a, b, c, d) with (a leq b leq c leq d) and (a + b + c + d geq 1), the expression (a^2 + 3b^2 + 5c^2 + 7d^2) is at least 1. Hmm, okay, let's break this down step by step.First, I know that all the variables are positive and ordered such that each is at least as big as the previous one. Also, their sum is at least 1. The expression we need to evaluate is a weighted sum of their squares, with weights increasing as 1, 3, 5, 7. That seems like the weights are increasing linearly, which might be significant.I wonder if there's a way to relate the sum of the variables to the sum of their weighted squares. Maybe using some inequality like Cauchy-Schwarz or AM-QM? Let me think.The Cauchy-Schwarz inequality says that for sequences of real numbers, the square of the sum is less than or equal to the product of the sums of squares. But here, we have a weighted sum, so maybe I need to adjust for that.Alternatively, maybe I can use the fact that (a leq b leq c leq d) to find some relationship between the variables. Since each subsequent variable is at least as big as the previous, perhaps I can bound each variable in terms of the next one.Let me try to express (a^2 + 3b^2 + 5c^2 + 7d^2) in a way that relates to ((a + b + c + d)^2), which we know is at least 1. Expanding ((a + b + c + d)^2) gives (a^2 + b^2 + c^2 + d^2 + 2(ab + ac + ad + bc + bd + cd)). Hmm, that's a lot of cross terms.But in our expression, we have coefficients 1, 3, 5, 7 for the squares. Maybe I can write (a^2 + 3b^2 + 5c^2 + 7d^2) as (a^2 + b^2 + c^2 + d^2 + 2b^2 + 4c^2 + 6d^2). That might help in relating it to the square of the sum.So, (a^2 + 3b^2 + 5c^2 + 7d^2 = (a^2 + b^2 + c^2 + d^2) + 2(b^2 + 2c^2 + 3d^2)). Now, the first part is just the sum of squares, and the second part is twice the sum of (b^2, 2c^2, 3d^2).I need to find a way to relate (b^2 + 2c^2 + 3d^2) to the cross terms in the square of the sum. Maybe I can find an inequality that bounds (b^2 + 2c^2 + 3d^2) from below by some combination of the cross terms.Given that (a leq b leq c leq d), perhaps I can use the ordering to establish some inequalities. For example, since (a leq b), then (ab leq b^2). Similarly, (ac leq c^2), (ad leq d^2), and so on. Maybe I can bound each cross term by a multiple of the square terms.Let me try that. Since (a leq b), then (ab leq b^2). Similarly, (ac leq c^2), (ad leq d^2), (bc leq c^2), (bd leq d^2), and (cd leq d^2). So, each cross term is less than or equal to the square of the larger variable.But wait, in our expression, we have (b^2 + 2c^2 + 3d^2). Maybe I can find a way to express the cross terms in terms of these.Looking back at the expansion of ((a + b + c + d)^2), we have (2(ab + ac + ad + bc + bd + cd)). If I can show that (b^2 + 2c^2 + 3d^2) is greater than or equal to the sum of these cross terms, then adding that to the sum of squares would give us something greater than or equal to ((a + b + c + d)^2).But is that true? Let's see. Let's compare (b^2 + 2c^2 + 3d^2) with (ab + ac + ad + bc + bd + cd).Given that (a leq b leq c leq d), we can say that:- (ab leq b^2)- (ac leq c^2)- (ad leq d^2)- (bc leq c^2)- (bd leq d^2)- (cd leq d^2)So, adding these up:(ab + ac + ad + bc + bd + cd leq b^2 + c^2 + d^2 + c^2 + d^2 + d^2 = b^2 + 2c^2 + 3d^2)Yes! That works. So, (ab + ac + ad + bc + bd + cd leq b^2 + 2c^2 + 3d^2). Therefore, (2(ab + ac + ad + bc + bd + cd) leq 2(b^2 + 2c^2 + 3d^2)).Therefore, going back to our expression:(a^2 + 3b^2 + 5c^2 + 7d^2 = (a^2 + b^2 + c^2 + d^2) + 2(b^2 + 2c^2 + 3d^2) geq (a^2 + b^2 + c^2 + d^2) + 2(ab + ac + ad + bc + bd + cd) = (a + b + c + d)^2)So, we've established that (a^2 + 3b^2 + 5c^2 + 7d^2 geq (a + b + c + d)^2). Since (a + b + c + d geq 1), squaring both sides gives ((a + b + c + d)^2 geq 1). Therefore, combining these two results, we have (a^2 + 3b^2 + 5c^2 + 7d^2 geq 1), which is exactly what we needed to prove.Wait, let me double-check that step where I compared (ab + ac + ad + bc + bd + cd) to (b^2 + 2c^2 + 3d^2). I used the fact that each cross term is less than or equal to the square of the larger variable. Since (a leq b leq c leq d), each cross term (ab, ac, ad, bc, bd, cd) is indeed less than or equal to (b^2, c^2, d^2) respectively. So, when I sum them up, I get that the total cross terms are less than or equal to (b^2 + 2c^2 + 3d^2). That seems correct.Also, the step where I expanded (a^2 + 3b^2 + 5c^2 + 7d^2) into (a^2 + b^2 + c^2 + d^2 + 2(b^2 + 2c^2 + 3d^2)) is straightforward algebra, so that's solid.Therefore, the proof seems to hold. I think I've covered all the necessary steps and verified each part logically. I don't see any gaps or errors in the reasoning.**Final Answer**boxed{1}

💡Okay, so I have this inequality to prove for natural numbers ( n ):[frac{1 cdot 3 cdot 5 cdot ldots cdot (2n-1)}{2 cdot 4 cdot 6 cdot ldots cdot 2n} leq frac{1}{sqrt{2n+1}}]Hmm, let's see. I remember that the product of odd numbers divided by the product of even numbers can be expressed in terms of factorials. Maybe I can rewrite the left-hand side using factorials to simplify it.First, the numerator is the product of the first ( n ) odd numbers: ( 1 cdot 3 cdot 5 cdot ldots cdot (2n-1) ). I think this can be written as ( frac{(2n)!}{2^n n!} ). Let me check:The product of the first ( n ) odd numbers is equal to ( frac{(2n)!}{2^n n!} ). Yeah, that seems right because ( (2n)! ) is the product of all numbers from 1 to ( 2n ), and dividing by ( 2^n n! ) removes the even factors and accounts for the factorial in the denominator.Similarly, the denominator is the product of the first ( n ) even numbers: ( 2 cdot 4 cdot 6 cdot ldots cdot 2n ). This can be written as ( 2^n n! ) because each term is 2 times an integer from 1 to ( n ).So, putting it together, the left-hand side becomes:[frac{frac{(2n)!}{2^n n!}}{2^n n!} = frac{(2n)!}{(2^n n!)^2}]So, the inequality simplifies to:[frac{(2n)!}{(2^n n!)^2} leq frac{1}{sqrt{2n+1}}]Hmm, I think this expression ( frac{(2n)!}{(2^n n!)^2} ) is actually the central binomial coefficient divided by ( 2^{2n} ). Wait, no, the central binomial coefficient is ( binom{2n}{n} = frac{(2n)!}{(n!)^2} ), so our expression is ( frac{binom{2n}{n}}{2^{2n}} ).So, the inequality becomes:[frac{binom{2n}{n}}{2^{2n}} leq frac{1}{sqrt{2n+1}}]I've heard that the central binomial coefficient has some bounds related to ( frac{4^n}{sqrt{pi n}} ), but I'm not sure if that's directly applicable here.Alternatively, maybe I can use induction to prove this inequality. Let me try that approach.**Base Case:** For ( n = 1 ):Left-hand side: ( frac{1}{2} = 0.5 )Right-hand side: ( frac{1}{sqrt{3}} approx 0.577 )So, ( 0.5 leq 0.577 ), which is true.**Inductive Step:** Assume that for some ( k geq 1 ), the inequality holds:[frac{1 cdot 3 cdot 5 cdot ldots cdot (2k-1)}{2 cdot 4 cdot 6 cdot ldots cdot 2k} leq frac{1}{sqrt{2k+1}}]We need to show that:[frac{1 cdot 3 cdot 5 cdot ldots cdot (2k-1) cdot (2k+1)}{2 cdot 4 cdot 6 cdot ldots cdot 2k cdot 2(k+1)} leq frac{1}{sqrt{2(k+1)+1}} = frac{1}{sqrt{2k+3}}]Let me denote the left-hand side for ( n = k ) as ( L_k ), so:[L_k = frac{1 cdot 3 cdot 5 cdot ldots cdot (2k-1)}{2 cdot 4 cdot 6 cdot ldots cdot 2k}]Then, for ( n = k+1 ):[L_{k+1} = L_k cdot frac{2k+1}{2(k+1)}]According to the inductive hypothesis, ( L_k leq frac{1}{sqrt{2k+1}} ). Therefore,[L_{k+1} leq frac{1}{sqrt{2k+1}} cdot frac{2k+1}{2(k+1)} = frac{sqrt{2k+1}}{2(k+1)}]We need to show that:[frac{sqrt{2k+1}}{2(k+1)} leq frac{1}{sqrt{2k+3}}]Let's square both sides to eliminate the square roots:[left( frac{sqrt{2k+1}}{2(k+1)} right)^2 leq left( frac{1}{sqrt{2k+3}} right)^2]Simplifying:[frac{2k+1}{4(k+1)^2} leq frac{1}{2k+3}]Cross-multiplying (since all terms are positive):[(2k+1)(2k+3) leq 4(k+1)^2]Expanding both sides:Left-hand side:[(2k+1)(2k+3) = 4k^2 + 6k + 2k + 3 = 4k^2 + 8k + 3]Right-hand side:[4(k+1)^2 = 4(k^2 + 2k + 1) = 4k^2 + 8k + 4]So, we have:[4k^2 + 8k + 3 leq 4k^2 + 8k + 4]Subtracting ( 4k^2 + 8k ) from both sides:[3 leq 4]Which is true. Therefore, the inequality holds for ( n = k+1 ) if it holds for ( n = k ).By the principle of mathematical induction, the inequality is true for all natural numbers ( n ).Wait, but I just realized that in the inductive step, I assumed that squaring both sides preserves the inequality, which it does because both sides are positive. So, that should be fine.Alternatively, another approach could be using the Wallis formula or considering the product as a form of the beta function, but induction seems straightforward here.I think this proof works. I started with the base case, assumed it holds for ( n = k ), and then showed it holds for ( n = k+1 ) by manipulating the inequality and verifying that the resulting statement is true. Therefore, the original inequality is proven for all natural numbers ( n ).**Final Answer**boxed{frac{1 cdot 3 cdot 5 cdot ldots cdot(2n-1)}{2 cdot 4 cdot 6 cdot ldots cdot 2n} leq frac{1}{sqrt{2n+1}}}

💡First, I need to determine how many rooms the painter has left to paint. Since he will take 32 hours to complete the remaining rooms and each room takes 8 hours to paint, I can calculate the number of remaining rooms by dividing the total remaining hours by the hours per room.Next, I'll subtract the number of rooms he has left to paint from the total number of rooms to find out how many rooms he has already painted.

💡Okay, so I have this problem here where I need to figure out which pairs of functions are actually the same function. There are four options labeled A, B, C, and D. I need to go through each one step by step and see if they represent the same function or not. Let me start with Option A.**Option A: y₁ = (x² - 16)/(x - 4), y₂ = x + 4**Alright, so y₁ is a rational function, and y₂ is a linear function. At first glance, they don't look the same, but maybe I can simplify y₁. Let me try factoring the numerator:y₁ = (x² - 16)/(x - 4) = [(x - 4)(x + 4)]/(x - 4)Oh, I see, the (x - 4) terms cancel out, so y₁ simplifies to x + 4. But wait, there's a catch here. When I cancel out (x - 4), I have to remember that x can't be 4 because that would make the denominator zero in the original function. So, y₁ is equal to x + 4 but with the restriction that x ≠ 4. On the other hand, y₂ is just x + 4 without any restrictions. Hmm, so even though they simplify to the same expression, their domains are different. y₁ can't take x = 4, but y₂ can. That means they aren't exactly the same function because their domains differ. So, Option A is not the correct pair.**Option B: f(x) = x - 1, g(x) = √(x²) - 1**Alright, f(x) is straightforward—it's just a linear function. Now, g(x) has a square root, so let me simplify that:g(x) = √(x²) - 1 = |x| - 1Oh, because the square root of x squared is the absolute value of x. So, g(x) is |x| - 1. Now, f(x) is x - 1, which is different from |x| - 1. For example, if x is negative, f(x) would be negative minus 1, but g(x) would be positive minus 1. So, they don't give the same output for negative x values. Therefore, f(x) and g(x) are not the same function. So, Option B is also not correct.**Option C: f(x) = x² - 2x - 1, g(t) = t² - 2t + 1**Okay, f(x) is a quadratic function, and g(t) is also a quadratic function but with a different constant term. Let me write them out:f(x) = x² - 2x - 1g(t) = t² - 2t + 1Hmm, the coefficients of x² and x are the same, but the constants are different. So, f(x) has a constant term of -1, while g(t) has a constant term of +1. That means these two functions are not identical because their outputs will differ by 2 for any given input. For example, if x = 0, f(0) = -1 and g(0) = 1. So, they are not the same function. Therefore, Option C is not correct either.**Option D: f₁(x) = 1, f₂(x) = x⁰**Alright, f₁(x) is a constant function, always equal to 1. f₂(x) is x raised to the power of 0. I remember that any non-zero number raised to the power of 0 is 1. So, x⁰ = 1 for any x ≠ 0. But what about when x = 0? Is 0⁰ defined? Hmm, that's a tricky one. In some contexts, 0⁰ is considered undefined, but in others, it might be defined as 1 for convenience. Wait, in the context of functions, f₂(x) = x⁰ is typically defined as 1 for all x ≠ 0, and at x = 0, it's either undefined or sometimes taken as 1 depending on the convention. However, f₁(x) is always 1 for all x, including x = 0. So, if we consider the standard definition where x⁰ is 1 for x ≠ 0 and undefined at x = 0, then f₂(x) isn't exactly the same as f₁(x) because f₂(x) has a point where it's undefined. But, in many cases, especially in algebra, x⁰ is treated as 1 for all x, including x = 0, to maintain consistency. If that's the case, then f₂(x) is also always 1, just like f₁(x). So, depending on the convention, they might be considered the same function. I think in most mathematical contexts, especially when dealing with functions like this, x⁰ is defined as 1 for all x, including x = 0. Therefore, f₂(x) is equal to 1 for all x, just like f₁(x). So, they are the same function.**Conclusion:**After analyzing all four options, only Option D represents the same function because both f₁(x) and f₂(x) simplify to 1 for all x in their domain. The other options either have different domains or different expressions that lead to different outputs for some inputs.