💡First, I identify the total number of adults and the cost per meal. There are 2 adults, and each meal costs 3.Next, I calculate the total cost for the adults' meals by multiplying the number of adults by the cost per meal: 2 adults * 3 = 6.Then, I subtract the cost of the adults' meals from the total bill to find out how much was spent on the children's meals: 21 (total bill) - 6 (adults' meals) = 15.Finally, I determine the number of children by dividing the remaining amount by the cost per child's meal: 15 / 3 = 5 children.

💡Okay, so I have this problem where I need to compare the sizes of three expressions: a, b, and c. The expressions are defined as follows: a is the natural logarithm of sin x, b is sin x itself, and c is e raised to the power of sin x. The variable x is between 0 and pi/2, which is like between 0 and 90 degrees in radians. First, I should recall what the sine function does in this interval. Since x is between 0 and pi/2, sin x is going to be between 0 and 1. It starts at 0 when x is 0, increases to 1 when x is pi/2, and is positive throughout this interval. So, sin x is a positive number less than 1. That means b is between 0 and 1.Now, let's think about a, which is the natural logarithm of sin x. I remember that the natural logarithm function, ln(y), is only defined for positive y, which is fine here because sin x is positive. Also, I recall that ln(y) is negative when y is between 0 and 1. For example, ln(1) is 0, and as y approaches 0, ln(y) goes to negative infinity. So, since sin x is between 0 and 1, ln(sin x) will be negative. Therefore, a is less than 0.Next, let's consider c, which is e raised to the power of sin x. The exponential function e^y is always positive, and it's increasing. Since sin x is between 0 and 1, e^{sin x} will be between e^0 and e^1, which is between 1 and e. So, c is greater than 1 but less than e.Putting this all together: a is negative, b is between 0 and 1, and c is greater than 1. So, a is the smallest, followed by b, and then c is the largest. Therefore, the order from smallest to largest is a < b < c.I should double-check if there are any special cases or if my reasoning might be flawed. For example, when x approaches 0, sin x approaches 0, so ln(sin x) approaches negative infinity, which makes a very small. When x approaches pi/2, sin x approaches 1, so ln(sin x) approaches 0, which is still less than b, which is approaching 1. Similarly, c approaches e^1, which is about 2.718, so it's definitely larger than 1. Another way to think about it is to consider specific values. Let's take x = pi/6, which is 30 degrees. Then sin(pi/6) is 0.5. So, a = ln(0.5) which is approximately -0.693, b = 0.5, and c = e^{0.5} which is approximately 1.648. So, indeed, a < b < c.What if x is something else, like pi/4? Sin(pi/4) is sqrt(2)/2, approximately 0.707. Then a = ln(0.707) which is about -0.347, b = 0.707, and c = e^{0.707} which is approximately 2.028. Again, a < b < c.So, it seems consistent across different values of x in the given interval. Therefore, I'm confident that the order is a < b < c.**Final Answer**The sizes of (a), (b), and (c) are in the order boxed{a < b < c}.

💡Okay, so I have this function f(x) = 2 sin(2x + π/6), and I need to answer three questions about it. Let me take them one by one.First, I need to find the amplitude and the smallest positive period of f(x). Hmm, I remember that for a sine function of the form A sin(Bx + C), the amplitude is |A| and the period is 2π / |B|. So in this case, A is 2, so the amplitude should be 2. That seems straightforward. Now, for the period. The function inside the sine is 2x + π/6, so B is 2. Therefore, the period should be 2π divided by 2, which is π. So the smallest positive period is π. Okay, that makes sense because the coefficient inside the sine function affects how quickly the function repeats. A higher coefficient would make the function repeat more often, hence a smaller period.Moving on to the second part. I need to find the range of f(x) when x is in the interval [0, π/2]. So, I know that the sine function normally has a range of [-1, 1], but since it's multiplied by 2, the range should be [-2, 2]. However, since we're only looking at a specific interval for x, the actual range might be different.Let me think about how the argument of the sine function behaves as x goes from 0 to π/2. The argument is 2x + π/6. When x is 0, the argument is π/6. When x is π/2, the argument is 2*(π/2) + π/6 = π + π/6 = 7π/6. So, the argument of the sine function goes from π/6 to 7π/6 as x goes from 0 to π/2.Now, I need to figure out the maximum and minimum values of sin(θ) where θ ranges from π/6 to 7π/6. Let me recall the unit circle. At θ = π/6, sin(θ) is 1/2. As θ increases to π/2, sin(θ) increases to 1. Then, as θ goes from π/2 to 3π/2, sin(θ) decreases from 1 to -1. But in our case, θ only goes up to 7π/6, which is π/6 past π. So, at 7π/6, sin(θ) is -1/2.So, the sine function reaches its maximum at θ = π/2, which is 1, and its minimum at θ = 7π/6, which is -1/2. Therefore, the range of sin(2x + π/6) over x in [0, π/2] is [-1/2, 1]. Since f(x) is 2 times that, the range of f(x) should be [-1, 2]. Wait, let me check that: 2*(-1/2) is -1, and 2*1 is 2. Yep, that seems right.Okay, so the range is from -1 to 2.Now, the third part: determine the intervals of monotonic decrease when x is in [-π, π]. Monotonic decrease means the function is going down as x increases. For the sine function, it's decreasing in certain intervals. Let me recall that the sine function increases from -π/2 to π/2 and decreases from π/2 to 3π/2, and this pattern repeats every 2π.But since our function is 2 sin(2x + π/6), the period is π, so the behavior repeats every π. So, I need to find where the function is decreasing within one period and then see how that translates over the interval [-π, π].First, let's find the general intervals where f(x) is decreasing. The derivative of f(x) is f'(x) = 2*cos(2x + π/6)*2 = 4 cos(2x + π/6). So, f'(x) is positive when cos(2x + π/6) is positive, and negative when cos(2x + π/6) is negative. Therefore, f(x) is decreasing when cos(2x + π/6) < 0.So, let's solve cos(2x + π/6) < 0. The cosine function is negative in the intervals (π/2 + 2πk, 3π/2 + 2πk) for any integer k. Therefore, 2x + π/6 must lie in (π/2 + 2πk, 3π/2 + 2πk).Let me solve for x:π/2 + 2πk < 2x + π/6 < 3π/2 + 2πkSubtract π/6 from all parts:π/2 - π/6 + 2πk < 2x < 3π/2 - π/6 + 2πkSimplify the fractions:π/2 is 3π/6, so 3π/6 - π/6 = 2π/6 = π/3.Similarly, 3π/2 is 9π/6, so 9π/6 - π/6 = 8π/6 = 4π/3.So, we have:π/3 + 2πk < 2x < 4π/3 + 2πkDivide all parts by 2:π/6 + πk < x < 2π/3 + πkSo, the function f(x) is decreasing on intervals (π/6 + πk, 2π/3 + πk) for any integer k.Now, we need to find all such intervals within x ∈ [-π, π].Let's find the values of k such that the intervals (π/6 + πk, 2π/3 + πk) overlap with [-π, π].Let me list possible k values:For k = -1:Interval is (π/6 - π, 2π/3 - π) = (-5π/6, -π/3)For k = 0:Interval is (π/6, 2π/3)For k = 1:Interval is (π/6 + π, 2π/3 + π) = (7π/6, 5π/3)But 7π/6 is approximately 3.665, which is greater than π (≈3.1416), so this interval doesn't overlap with [-π, π].Similarly, for k = -2:Interval is (π/6 - 2π, 2π/3 - 2π) = (-11π/6, -4π/3)But -11π/6 is approximately -5.7596, which is less than -π, so this interval doesn't overlap with [-π, π].Therefore, the only intervals within [-π, π] where f(x) is decreasing are (-5π/6, -π/3) and (π/6, 2π/3).Wait, let me double-check:For k = -1: (-5π/6 ≈ -2.618, -π/3 ≈ -1.047). Both are within [-π ≈ -3.1416, π ≈ 3.1416].For k = 0: (π/6 ≈ 0.523, 2π/3 ≈ 2.094). Also within [-π, π].For k = 1: (7π/6 ≈ 3.665, 5π/3 ≈ 5.236). Both are outside of [-π, π].Similarly, for k = -2: (-11π/6 ≈ -5.759, -4π/3 ≈ -4.188). Both are outside of [-π, π].So yes, only k = -1 and k = 0 give intervals within [-π, π].Therefore, the intervals of monotonic decrease are (-5π/6, -π/3) and (π/6, 2π/3).Wait, but the question says "intervals of monotonic decrease when x ∈ [-π, π]". So, I should present them as closed intervals or open? Since the function is decreasing on open intervals where the derivative is negative, but at the endpoints, the derivative is zero, so the function is neither increasing nor decreasing there. So, it's safer to present them as open intervals.But sometimes, in calculus, when we talk about intervals of increase or decrease, we include the endpoints if the function is defined there. So, perhaps I should write them as closed intervals. Hmm.Wait, let me think. The function is defined on [-π, π], and at the endpoints of these intervals, the derivative is zero, meaning those points are either maxima or minima. So, at x = -5π/6, x = -π/3, x = π/6, and x = 2π/3, the derivative is zero. So, the function is decreasing just after -5π/6 and just before -π/3, and similarly for the other interval.Therefore, the intervals where the function is decreasing are (-5π/6, -π/3) and (π/6, 2π/3).But let me check if these intervals are correct by plugging in some test points.For example, take x = -π/2, which is approximately -1.5708. Is this in (-5π/6 ≈ -2.618, -π/3 ≈ -1.047)? Yes, it is. Let me compute f'(x) at x = -π/2.f'(x) = 4 cos(2x + π/6). So, 2*(-π/2) + π/6 = -π + π/6 = -5π/6. So, cos(-5π/6) = cos(5π/6) = -√3/2. Therefore, f'(-π/2) = 4*(-√3/2) = -2√3 < 0. So, the function is decreasing there, which is consistent.Similarly, take x = 0, which is in (π/6 ≈ 0.523, 2π/3 ≈ 2.094)? No, x=0 is less than π/6. So, let's take x = π/4 ≈ 0.785, which is in (π/6, 2π/3). Compute f'(π/4):2*(π/4) + π/6 = π/2 + π/6 = 2π/3. So, cos(2π/3) = -1/2. Therefore, f'(π/4) = 4*(-1/2) = -2 < 0. So, decreasing there as well.Good, that seems correct.So, summarizing:1. Amplitude is 2, period is π.2. Range is [-1, 2].3. Intervals of decrease are (-5π/6, -π/3) and (π/6, 2π/3).I think that's it. I should probably write the final answers in the required format.

💡Alright, so I'm trying to figure out which mathematical concept Hua Luogeng used in his optimal selection method. The options are the golden ratio, mean, mode, and median. Hmm, okay, let's break this down.First, who is Hua Luogeng? I remember he's a famous Chinese mathematician, but I'm not too familiar with his specific contributions. Maybe I should look that up. Oh, right, he worked on things like number theory and popularizing mathematical methods. Okay, so he's known for making complex math more accessible.Now, the question is about the optimal selection method. I think this might be related to optimization techniques, like finding the best option among several choices. That sounds a bit like the golden section search, which I've heard is a method for finding the extremum (minimum or maximum) of a unimodal function. Wait, is that related to the golden ratio?Let me recall: the golden ratio is approximately 1.618 and has interesting properties in geometry and art. It's often associated with aesthetically pleasing proportions. But how does that tie into optimization?Oh, I think the golden section search uses the golden ratio to determine the points at which to evaluate the function. It's a technique that reduces the interval of uncertainty by a constant factor at each step, which is based on the golden ratio. That makes sense because the golden ratio has a self-similar property, which could be useful in iterative methods.So, if Hua Luogeng popularized the optimal selection method, and one of the methods uses the golden ratio, then the answer should be A: Golden ratio.But wait, let me make sure. The other options are mean, mode, and median. The mean is the average, mode is the most frequent value, and median is the middle value. These are all measures of central tendency, but they don't seem directly related to optimization methods like the golden section search.I think I'm on the right track with the golden ratio. It's specifically used in optimization techniques, which aligns with the idea of optimal selection. The other options are more about describing data rather than selecting the optimal point.Yeah, I'm pretty confident now that the answer is A: Golden ratio.

💡Alright, so I've got this problem here about General Mushtralkin trying to arrange his soldiers in rows of different sizes, and each time, the last row ends up incomplete with a specific number of soldiers left. The goal is to find the smallest number of soldiers that could cause this situation.Okay, let's break this down. The general tries rows of 2, 3, 4, 5, 6, 7, 8, 9, and 10 soldiers. Each time, the last row is incomplete, and the number of soldiers left over is 1, 2, 3, 4, 5, 6, 7, 8, and 9 respectively. So, when he tries to make rows of 2, there's 1 soldier left; when he tries rows of 3, there's 2 soldiers left, and so on up to rows of 10, where there's 9 soldiers left.Hmm, this seems like a problem related to division and remainders. Specifically, it sounds like we're dealing with congruences. For each row size, the number of soldiers modulo that row size gives the remainder, which is one less than the row size. So, for rows of 2, the number of soldiers modulo 2 is 1; for rows of 3, it's 2; and so on.Let me write this out mathematically. If N is the number of soldiers, then:N ≡ 1 mod 2 N ≡ 2 mod 3 N ≡ 3 mod 4 N ≡ 4 mod 5 N ≡ 5 mod 6 N ≡ 6 mod 7 N ≡ 7 mod 8 N ≡ 8 mod 9 N ≡ 9 mod 10Looking at these congruences, I notice a pattern. Each time, the remainder is one less than the modulus. That is, for modulus k, the remainder is k - 1. So, in general, we can say that N ≡ -1 mod k for each k from 2 to 10.This simplifies things because if N ≡ -1 mod k for all these k, then N + 1 must be divisible by each k. So, N + 1 is a common multiple of all the numbers from 2 to 10. To find the smallest such N, we need the least common multiple (LCM) of these numbers.Alright, so let's find the LCM of 2, 3, 4, 5, 6, 7, 8, 9, and 10. To do this, I'll break each number down into its prime factors:- 2 is prime: 2- 3 is prime: 3- 4 is 2²- 5 is prime: 5- 6 is 2 × 3- 7 is prime: 7- 8 is 2³- 9 is 3²- 10 is 2 × 5To find the LCM, I take the highest power of each prime that appears in the factorizations:- For 2: the highest power is 2³ (from 8)- For 3: the highest power is 3² (from 9)- For 5: the highest power is 5¹ (from 5 and 10)- For 7: the highest power is 7¹ (from 7)So, the LCM is 2³ × 3² × 5 × 7. Let's calculate that:2³ = 8 3² = 9 8 × 9 = 72 72 × 5 = 360 360 × 7 = 2520So, the LCM of numbers from 2 to 10 is 2520. This means that N + 1 = 2520, so N = 2520 - 1 = 2519.Let me double-check to make sure this works. If N = 2519, then:- 2519 ÷ 2 = 1259 with a remainder of 1 - 2519 ÷ 3 = 839 with a remainder of 2 - 2519 ÷ 4 = 629 with a remainder of 3 - 2519 ÷ 5 = 503 with a remainder of 4 - 2519 ÷ 6 = 419 with a remainder of 5 - 2519 ÷ 7 = 359 with a remainder of 6 - 2519 ÷ 8 = 314 with a remainder of 7 - 2519 ÷ 9 = 279 with a remainder of 8 - 2519 ÷ 10 = 251 with a remainder of 9Yes, that matches all the conditions given in the problem. So, the smallest number of soldiers is 2519.

💡First, I identify the amount of detergent used per pound of clothes, which is 2 ounces.Next, I determine the total amount of clothes to be washed, which is 9 pounds.To find the total ounces of detergent needed, I multiply the detergent per pound by the total pounds of clothes: 2 ounces/pound × 9 pounds.This calculation results in 18 ounces of detergent required to wash 9 pounds of clothes.