💡Okay, so I have this problem where I need to prove that for functions p(x) and q(x), given that p(0) = q(0) > 0 and p'(x) * sqrt(q'(x)) = sqrt(2) for any x in [0, 1], then p(x) + 2q(x) > 3x for all x in [0, 1]. Hmm, okay, let me try to unpack this.First, I know that p(0) equals q(0), and both are greater than zero. So at x=0, both p and q have the same positive value. That's a starting point. Now, the condition p'(x) * sqrt(q'(x)) = sqrt(2) is interesting. It relates the derivatives of p and q. Maybe I can express one derivative in terms of the other?Let me write that down: p'(x) * sqrt(q'(x)) = sqrt(2). If I solve for q'(x), I can square both sides to get rid of the square root. So, squaring both sides gives (p'(x))^2 * q'(x) = 2. Therefore, q'(x) = 2 / (p'(x))^2. Okay, so q'(x) is expressed in terms of p'(x). That might be useful later.Now, the goal is to show that p(x) + 2q(x) > 3x for all x in [0, 1]. Maybe I can define a new function, say f(x) = p(x) + 2q(x) - 3x, and then show that f(x) is always positive on [0, 1]. To do that, I can analyze the behavior of f(x). If I can show that f(x) is non-decreasing and f(0) > 0, then f(x) will stay positive.Let me compute the derivative of f(x). f'(x) = p'(x) + 2q'(x) - 3. If I can show that f'(x) >= 0 for all x in [0, 1], then f(x) is non-decreasing. Since f(0) = p(0) + 2q(0) - 0 = 3p(0) > 0, because p(0) = q(0) > 0, then f(x) will remain positive.So, f'(x) = p'(x) + 2q'(x) - 3. I already have q'(x) in terms of p'(x): q'(x) = 2 / (p'(x))^2. Let me substitute that into f'(x):f'(x) = p'(x) + 2*(2 / (p'(x))^2) - 3 = p'(x) + 4 / (p'(x))^2 - 3.Now, I need to show that this expression is non-negative. Let me denote y = p'(x). Then, f'(x) becomes y + 4/y^2 - 3. So, I need to show that y + 4/y^2 - 3 >= 0 for all y > 0 (since p'(x) must be positive because q'(x) is positive as well, given that sqrt(q'(x)) is in the original condition).Hmm, how can I analyze this expression? Maybe I can consider it as a function of y and find its minimum. Let me define g(y) = y + 4/y^2 - 3. I need to find the minimum of g(y) for y > 0.To find the critical points, I can take the derivative of g(y) with respect to y:g'(y) = 1 - 8/y^3.Set g'(y) = 0:1 - 8/y^3 = 0 => 8/y^3 = 1 => y^3 = 8 => y = 2.So, the critical point is at y = 2. Let me check the second derivative to confirm if it's a minimum:g''(y) = 24/y^4, which is always positive for y > 0. So, y = 2 is a point of local minimum.Now, evaluate g(y) at y = 2:g(2) = 2 + 4/(2)^2 - 3 = 2 + 4/4 - 3 = 2 + 1 - 3 = 0.So, the minimum value of g(y) is 0, achieved at y = 2. Therefore, for all y > 0, g(y) >= 0. Hence, f'(x) >= 0 for all x in [0, 1].Since f'(x) >= 0, f(x) is non-decreasing on [0, 1]. We already know that f(0) = 3p(0) > 0. Therefore, f(x) >= f(0) > 0 for all x in [0, 1]. Thus, p(x) + 2q(x) - 3x > 0, which implies p(x) + 2q(x) > 3x for all x in [0, 1].Wait, let me double-check my steps. I defined f(x) correctly, computed its derivative, substituted q'(x) in terms of p'(x), transformed it into a function of y, found its minimum, and concluded that f'(x) is non-negative. Then, since f(0) is positive, f(x) remains positive. Yeah, that seems solid.Alternatively, maybe I can use the AM-GM inequality on the terms p'(x) and 2q'(x). Let me see:We have p'(x) * sqrt(q'(x)) = sqrt(2). Let me denote a = p'(x), b = sqrt(q'(x)). Then, ab = sqrt(2). So, b = sqrt(2)/a.Then, p'(x) + 2q'(x) = a + 2b^2 = a + 2*(2/a^2) = a + 4/a^2. Wait, that's the same expression as before. So, it's consistent.Alternatively, maybe I can use the AM-GM inequality on the terms a, a, and 2b^2. Let me try:By AM-GM, (a + a + 2b^2)/3 >= (a * a * 2b^2)^(1/3). So, (2a + 2b^2)/3 >= (2a^2b^2)^(1/3). But I'm not sure if that helps directly.Wait, let me think again. Since ab = sqrt(2), then b = sqrt(2)/a. So, p'(x) + 2q'(x) = a + 4/a^2. To minimize this, as I did before, we can set derivative to zero and find the minimum at a=2, giving the minimum value of 0. So, p'(x) + 2q'(x) >= 3, which is exactly what we need for f'(x) >= 0.So, both methods lead to the same conclusion. Therefore, I'm confident that the proof is correct.**Final Answer**boxed{p(x) + 2 q(x) > 3 x}

💡First, I need to determine the normal cost of a movie ticket. Let's denote the cost of the movie ticket as ( T ).According to the problem:- A bucket of popcorn costs ( T - 3 ) dollars.- A drink costs ( (T - 3) + 1 = T - 2 ) dollars.- A box of candy costs half as much as a drink, so it costs ( frac{T - 2}{2} ) dollars.The deal costs 20 and saves a customer 2, so the normal cost of buying all these items separately would be ( 20 + 2 = 22 ) dollars.Now, I'll add up the normal costs of all the items:- Movie ticket: ( T )- Popcorn: ( T - 3 )- Drink: ( T - 2 )- Candy: ( frac{T - 2}{2} )The sum of these costs should equal the normal total cost of 22:[T + (T - 3) + (T - 2) + frac{T - 2}{2} = 22]To eliminate the fraction, I'll multiply the entire equation by 2:[2T + 2(T - 3) + 2(T - 2) + (T - 2) = 44]Expanding and combining like terms:[2T + 2T - 6 + 2T - 4 + T - 2 = 44][7T - 12 = 44]Adding 12 to both sides:[7T = 56]Finally, dividing both sides by 7:[T = 8]Therefore, the normal cost of a movie ticket is 8.

💡Okay, so I have this problem with two propositions, p and q, and I need to figure out which of the given options (A, B, C, D) is true. Let me try to understand each proposition step by step.First, proposition p says: For all x less than 1, log base 3 of x is greater than 0. Hmm, logarithms can be tricky. I remember that log base a of b is greater than 0 if b is greater than 1 when a is greater than 1. Since 3 is greater than 1, log base 3 of x will be greater than 0 only if x is greater than 1. But here, x is less than 1. So, for x less than 1, log base 3 of x should actually be negative, right? Because if x is between 0 and 1, the logarithm of x with base greater than 1 is negative. And if x is less than or equal to 0, the logarithm isn't even defined. So, p seems to be false because it's claiming that log base 3 of x is greater than 0 for all x less than 1, which isn't true.Now, proposition q says: There exists some real number x0 such that x0 squared is greater than or equal to 2 raised to the power of x0. Okay, so I need to find at least one real number where this inequality holds. Let me think of some numbers. If x0 is 2, then 2 squared is 4, and 2 raised to 2 is also 4. So, 4 is equal to 4, which satisfies the inequality. Therefore, q is true because there exists at least one x0 (like 2) that makes the inequality hold.Now, looking at the options:A: p OR q. Since p is false and q is true, p OR q is true because OR only needs one of them to be true.B: NOT p AND NOT q. NOT p would be true because p is false, and NOT q would be false because q is true. So, true AND false is false.C: p OR NOT q. p is false, NOT q is false, so false OR false is false.D: p AND q. p is false and q is true, so false AND true is false.So, the only true proposition is A: p OR q.Wait, let me double-check. For p, I concluded it's false because log base 3 of x is negative when x is less than 1. That seems right. For q, I found x0=2 works, so q is true. Then, A is p OR q, which is false OR true, which is true. The others don't hold. Yeah, I think A is correct.

💡First, I need to determine the side length of the original square. Since the perimeter is 200 units and a square has four equal sides, I can divide the perimeter by 4 to find the side length.Next, I know that each smaller square has a side length that is half of the original square's side length. So, I'll calculate the side length of the smaller squares by dividing the original side length by 2.The rectangle formed in the division has a width equal to the side length of the smaller squares and a length equal to the side length of the original square. To find the perimeter of the rectangle, I'll use the formula for the perimeter of a rectangle, which is twice the sum of its length and width.Finally, I'll plug in the values I've calculated into the formula to find the perimeter of the rectangle.

💡To determine the weight of the new person, let's analyze the situation step by step.1. **Initial Setup:** - There are 10 persons with an average weight. - When a new person replaces one of them, the average weight increases by 5 kg.2. **Understanding the Change:** - The increase in average weight implies that the total weight of the group has increased. - The increase in total weight is due to the difference between the new person's weight and the weight of the person they replaced.3. **Calculating the Increase in Total Weight:** - The average weight increases by 5 kg for 10 persons. - Therefore, the total increase in weight is ( 5 , text{kg} times 10 = 50 , text{kg} ).4. **Determining the New Person's Weight:** - The person being replaced weighed 60 kg. - The increase in total weight (50 kg) is the difference between the new person's weight and 60 kg. - Thus, the new person's weight is ( 60 , text{kg} + 50 , text{kg} = 110 , text{kg} ).

💡Okay, so I need to figure out the smaller angle between the clock hands at 3:15. Hmm, let's see. I remember that clocks are circles, and circles have 360 degrees. So, each hour mark must represent a certain number of degrees. Since there are 12 hours on a clock, each hour would be 360 divided by 12, which is 30 degrees. So, every hour, the hour hand moves 30 degrees. Now, the minute hand moves around the clock too. There are 60 minutes in an hour, so each minute must be 360 divided by 60, which is 6 degrees per minute. That means every minute, the minute hand moves 6 degrees. At 3:15, the minute hand is at the 15-minute mark. So, 15 minutes times 6 degrees per minute would be 90 degrees. That makes sense because 15 minutes is a quarter of an hour, and a quarter of 360 degrees is 90 degrees. So, the minute hand is at 90 degrees.Now, what about the hour hand? At 3:00, the hour hand is exactly at the 3, which is 90 degrees from the top (since 3 times 30 degrees is 90 degrees). But it's not exactly 3:00; it's 3:15. So, the hour hand has moved a bit past the 3. How much has it moved?Since the hour hand moves 30 degrees per hour, it must move 30 degrees divided by 60 minutes, which is 0.5 degrees per minute. So, in 15 minutes, the hour hand moves 15 times 0.5 degrees, which is 7.5 degrees. Therefore, at 3:15, the hour hand is at 90 degrees plus 7.5 degrees, which is 97.5 degrees. Now, to find the angle between the two hands, I subtract the smaller angle from the larger angle. The minute hand is at 90 degrees, and the hour hand is at 97.5 degrees. So, the difference is 97.5 degrees minus 90 degrees, which is 7.5 degrees. Wait, but I need to make sure that this is the smaller angle. Since a clock is a circle, the total around is 360 degrees. So, the other angle would be 360 degrees minus 7.5 degrees, which is 352.5 degrees. Clearly, 7.5 degrees is the smaller angle. So, the smaller angle formed by the hands of the clock at 3:15 is 7.5 degrees. I think that makes sense, but let me double-check. If it's 3:15, the minute hand is at 90 degrees, and the hour hand is just a little past 3, which is also 90 degrees. So, the hour hand has moved a bit, making the angle between them just a small fraction. 7.5 degrees seems right because it's a quarter of an hour, and the hour hand moves half a degree per minute, so 15 minutes would be 7.5 degrees. Yeah, I'm pretty confident now that 7.5 degrees is the correct answer.