💡Okay, so I've got this game with 7 coins, all showing heads up. My goal is to flip some coins so that each pair of adjacent coins has different faces up. The rule is that in each move, I must flip two adjacent coins. I need to figure out the minimum number of moves required to win the game.First, let me understand the problem better. All coins start as heads. I need to end up with a configuration where no two adjacent coins are the same. That means the final arrangement should alternate between heads and tails. Since there are 7 coins, which is an odd number, the sequence will start and end with the same face. For example, it could be H, T, H, T, H, T, H or T, H, T, H, T, H, T.Now, each move allows me to flip two adjacent coins. Flipping a coin changes its state from heads to tails or vice versa. So, each move affects two coins. I need to find the minimum number of such moves to reach the desired alternating pattern.Let me consider the initial state: all heads. So, the sequence is H, H, H, H, H, H, H. I need to turn this into an alternating sequence. Let's pick one of the possible alternating sequences, say H, T, H, T, H, T, H. To get from H, H, H, H, H, H, H to H, T, H, T, H, T, H, I need to flip certain coins. But I can't flip individual coins; I have to flip two adjacent coins at a time. So, I need to figure out a sequence of adjacent flips that will result in the desired pattern.Let me think about how flipping two adjacent coins affects the sequence. If I flip two heads, they become two tails. If I flip a head and a tail, they become a tail and a head. If I flip two tails, they become two heads.Since all coins start as heads, any flip will turn two adjacent heads into two tails. So, my first move will definitely create two tails somewhere in the sequence. Let's say I flip the first two coins. Now the sequence becomes T, T, H, H, H, H, H.But now, the first two coins are both tails, which is not allowed in the final configuration. So, I need to fix that. Maybe I can flip the second and third coins. Flipping the second and third coins would turn the second tail into a head and the third head into a tail. So, the sequence becomes T, H, T, H, H, H, H.Now, the first three coins are T, H, T, which is good. But the fourth coin is still a head, which is adjacent to a tail. So, I need to flip the third and fourth coins. Flipping them would turn the third tail into a head and the fourth head into a tail. The sequence becomes T, H, H, T, H, H, H.Hmm, now the third and fourth coins are H and T, which is good, but the second and third coins are both H, which is bad. So, I need to flip the second and third coins again. Flipping them would turn the second H into a T and the third H into a T. The sequence becomes T, T, T, T, H, H, H.Wait, now the first four coins are all tails, which is worse. Maybe I'm approaching this the wrong way. Let me try a different sequence of moves.Starting again: H, H, H, H, H, H, H.First, flip coins 2 and 3: H, T, T, H, H, H, H.Now, coins 2 and 3 are both tails. That's not good. Maybe I should flip coins 1 and 2 first: T, T, H, H, H, H, H.Then flip coins 3 and 4: T, T, T, T, H, H, H.Still not good. Maybe I need to flip coins 1 and 2, then coins 2 and 3, then coins 3 and 4, and so on, creating a wave of flips.Let me try that:1. Flip 1 and 2: T, T, H, H, H, H, H.2. Flip 2 and 3: T, H, T, H, H, H, H.3. Flip 3 and 4: T, H, H, T, H, H, H.4. Flip 4 and 5: T, H, H, H, T, H, H.5. Flip 5 and 6: T, H, H, H, H, T, H.6. Flip 6 and 7: T, H, H, H, H, H, T.Now, the sequence is T, H, H, H, H, H, T. That's not alternating. I need to fix the middle coins.Maybe I need to flip some coins again. Let's see:7. Flip 3 and 4: T, H, T, T, H, H, T.8. Flip 4 and 5: T, H, T, H, T, H, T.Now, the sequence is T, H, T, H, T, H, T. That's alternating! So, it took me 8 moves.But wait, is there a way to do it in fewer moves? Maybe I'm being inefficient.Let me think about the problem differently. Each flip affects two coins. To achieve the alternating pattern, each coin (except the first and last) needs to be flipped an odd number of times if it's in an even position, or an even number of times if it's in an odd position, or something like that.Alternatively, maybe I can model this as a system of equations. Each coin needs to be in a certain state, and each move affects two coins. So, I can set up equations where each move is a variable, and the sum of moves affecting each coin equals the desired change.But that might be complicated. Maybe there's a simpler way.I remember that in similar problems, the minimum number of moves is often related to the number of coins divided by 2 or something like that. Since there are 7 coins, maybe it's 4 moves?Let me try to find a sequence of 4 moves that achieves the alternating pattern.Starting with H, H, H, H, H, H, H.1. Flip 1 and 2: T, T, H, H, H, H, H.2. Flip 3 and 4: T, T, T, T, H, H, H.3. Flip 5 and 6: T, T, T, T, T, T, H.4. Flip 6 and 7: T, T, T, T, T, H, T.Now, the sequence is T, T, T, T, T, H, T. That's not alternating. Hmm.Maybe a different sequence:1. Flip 2 and 3: H, T, T, H, H, H, H.2. Flip 4 and 5: H, T, T, T, T, H, H.3. Flip 6 and 7: H, T, T, T, T, T, T.4. Flip 3 and 4: H, T, H, H, T, T, T.5. Flip 5 and 6: H, T, H, H, H, T, T.6. Flip 4 and 5: H, T, H, T, H, T, T.7. Flip 6 and 7: H, T, H, T, H, H, H.8. Flip 5 and 6: H, T, H, T, T, T, H.9. Flip 7 and... Wait, there's no 8th coin. Maybe this approach isn't working.Perhaps I need to think about the parity of the number of flips each coin undergoes. Each coin needs to be flipped a certain number of times to change from H to T or vice versa.In the initial state, all coins are H. The target is an alternating sequence, which means coins 1, 3, 5, 7 are H, and coins 2, 4, 6 are T. So, coins 2, 4, 6 need to be flipped once, and coins 1, 3, 5, 7 need to remain H, meaning they shouldn't be flipped or flipped an even number of times.But since each move flips two adjacent coins, flipping a coin affects its neighbor. So, it's not straightforward.Maybe I can represent this as a system where each move affects two coins, and I need to find a combination of moves that results in the desired flips.Let me denote the coins as C1, C2, C3, C4, C5, C6, C7.Each move can be represented as flipping two adjacent coins:Move 1: C1 and C2Move 2: C2 and C3Move 3: C3 and C4Move 4: C4 and C5Move 5: C5 and C6Move 6: C6 and C7I need to find a set of moves such that:- C2, C4, C6 are flipped once (to change from H to T)- C1, C3, C5, C7 are flipped zero times or an even number of times (to remain H)Let me set up equations for each coin:C1: flipped by Move 1C2: flipped by Move 1 and Move 2C3: flipped by Move 2 and Move 3C4: flipped by Move 3 and Move 4C5: flipped by Move 4 and Move 5C6: flipped by Move 5 and Move 6C7: flipped by Move 6We need:C1: 0 flips (mod 2)C2: 1 flip (mod 2)C3: 0 flips (mod 2)C4: 1 flip (mod 2)C5: 0 flips (mod 2)C6: 1 flip (mod 2)C7: 0 flips (mod 2)Let me denote the moves as variables M1, M2, M3, M4, M5, M6, where each Mi is 1 if the move is performed, 0 otherwise.Then, the equations become:C1: M1 ≡ 0 mod 2C2: M1 + M2 ≡ 1 mod 2C3: M2 + M3 ≡ 0 mod 2C4: M3 + M4 ≡ 1 mod 2C5: M4 + M5 ≡ 0 mod 2C6: M5 + M6 ≡ 1 mod 2C7: M6 ≡ 0 mod 2Now, let's solve this system step by step.From C1: M1 ≡ 0 ⇒ M1 = 0From C7: M6 ≡ 0 ⇒ M6 = 0From C2: M1 + M2 ≡ 1 ⇒ 0 + M2 ≡ 1 ⇒ M2 = 1From C3: M2 + M3 ≡ 0 ⇒ 1 + M3 ≡ 0 ⇒ M3 = 1From C4: M3 + M4 ≡ 1 ⇒ 1 + M4 ≡ 1 ⇒ M4 = 0From C5: M4 + M5 ≡ 0 ⇒ 0 + M5 ≡ 0 ⇒ M5 = 0From C6: M5 + M6 ≡ 1 ⇒ 0 + 0 ≡ 1 ⇒ 0 ≡ 1, which is a contradiction.Hmm, that's a problem. It seems there's no solution with M1=0 and M6=0. Maybe I need to allow M1 or M6 to be 1.Wait, but C1 and C7 need to remain H, so they should not be flipped. Therefore, M1 and M6 must be 0. But that leads to a contradiction in C6. So, perhaps it's impossible to achieve the desired configuration with these constraints.But that can't be right because I know it's possible to achieve an alternating sequence with a certain number of moves. Maybe I made a mistake in setting up the equations.Let me double-check. The target is to have C2, C4, C6 flipped once, and C1, C3, C5, C7 flipped zero times. But since each move affects two coins, it's impossible to flip only C2, C4, C6 without affecting others. Therefore, maybe the target is not exactly that, but rather to have an alternating sequence, which could start with T instead of H.Wait, the problem says "each pair of adjacent coins has different faces up." It doesn't specify starting with H or T. So, the target could be either H, T, H, T, H, T, H or T, H, T, H, T, H, T.If I choose the latter, starting with T, then C1 needs to be T, which means it needs to be flipped once. Similarly, C3, C5, C7 need to be flipped once, and C2, C4, C6 need to be flipped once as well.Wait, no. If I start with T, the sequence is T, H, T, H, T, H, T. So, C1, C3, C5, C7 are T, and C2, C4, C6 are H. But initially, all are H, so C1, C3, C5, C7 need to be flipped once, and C2, C4, C6 need to remain H, meaning they shouldn't be flipped or flipped an even number of times.So, let's set up the equations again with this target:C1: flipped onceC2: not flippedC3: flipped onceC4: not flippedC5: flipped onceC6: not flippedC7: flipped onceSo, the equations are:C1: M1 ≡ 1C2: M1 + M2 ≡ 0C3: M2 + M3 ≡ 1C4: M3 + M4 ≡ 0C5: M4 + M5 ≡ 1C6: M5 + M6 ≡ 0C7: M6 ≡ 1Now, let's solve this system.From C1: M1 = 1From C2: M1 + M2 ≡ 0 ⇒ 1 + M2 ≡ 0 ⇒ M2 = 1From C3: M2 + M3 ≡ 1 ⇒ 1 + M3 ≡ 1 ⇒ M3 = 0From C4: M3 + M4 ≡ 0 ⇒ 0 + M4 ≡ 0 ⇒ M4 = 0From C5: M4 + M5 ≡ 1 ⇒ 0 + M5 ≡ 1 ⇒ M5 = 1From C6: M5 + M6 ≡ 0 ⇒ 1 + M6 ≡ 0 ⇒ M6 = 1From C7: M6 = 1So, the solution is M1=1, M2=1, M3=0, M4=0, M5=1, M6=1.Therefore, the moves are:1. Flip 1 and 22. Flip 2 and 33. Flip 5 and 64. Flip 6 and 7So, four moves in total.Let me verify this:Starting sequence: H, H, H, H, H, H, HAfter Move 1 (flip 1 and 2): T, T, H, H, H, H, HAfter Move 2 (flip 2 and 3): T, H, T, H, H, H, HAfter Move 3 (flip 5 and 6): T, H, T, H, T, T, HAfter Move 4 (flip 6 and 7): T, H, T, H, T, H, TYes, that's the desired alternating sequence. So, it took four moves.I think this is the minimum number of moves because any fewer moves wouldn't allow us to flip all the necessary coins without overlapping flips that might undo previous changes. Each move affects two coins, and by strategically choosing which pairs to flip, we can achieve the desired pattern in four moves.

💡Okay, so I have this question about compass and straightedge constructions, specifically about the basis for constructing an angle equal to a known angle. The options are ASA, SAS, SSS, and AAS. Hmm, I remember these are all congruence theorems for triangles, but I need to figure out which one applies to this particular construction.First, let me recall what each of these abbreviations stands for:- **ASA (Angle-Side-Angle)**: If two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent. - **SAS (Side-Angle-Side)**: If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent. - **SSS (Side-Side-Side)**: If all three sides of one triangle are congruent to all three sides of another triangle, then the triangles are congruent. - **AAS (Angle-Angle-Side)**: If two angles and a non-included side of one triangle are congruent to two angles and the corresponding non-included side of another triangle, then the triangles are congruent.Now, the question is about constructing an angle equal to a known angle using just a compass and straightedge. I think this process involves creating a triangle that is congruent to the original triangle, which would mean the angles would naturally be equal as well.When constructing an angle equal to a known angle, I believe the steps involve:1. Drawing a ray with one endpoint.2. Using the compass to measure the distance between two points on the original angle.3. Transferring that distance to the new ray to create a corresponding point.4. Using the compass again to replicate another distance, ensuring the triangles formed are congruent.5. Finally, connecting the points to form the new angle.This seems to rely on ensuring that all corresponding sides are equal, which would fall under the SSS congruence theorem. Because if all three sides are equal, then the angles must be equal too.Wait, but could it be SAS? Because in the construction, you might be transferring two sides and the included angle. Hmm, but actually, in constructing an angle, you're not directly transferring the angle itself but ensuring that the sides are equal, which would imply the angle is equal because of the SSS theorem.Let me think of an example. Suppose I have angle ABC, and I want to construct an angle DEF equal to ABC. I would:1. Draw a ray DE.2. With the compass, measure the length of AB and transfer it to DE, marking point F such that DF = AB.3. Then, measure the length of BC with the compass and transfer it to a new ray from F, ensuring that the distance matches.4. The intersection of these arcs would give me the point to complete the angle.In this process, I'm ensuring that all corresponding sides are equal, which aligns with SSS.Alternatively, if I were using SAS, I would need to know two sides and the included angle, but in this case, I'm constructing the angle, so I don't have the angle to begin with; I'm using the sides to define the angle.Therefore, it seems more consistent with SSS because the construction is based on matching all three sides to ensure congruence, which in turn ensures the angles are equal.I think I'm confident now that the correct answer is SSS.

💡First, I need to determine the volume of the cube and the rectangular box. The cube has an edge length of 15 cm, so its volume is (15^3 = 3375) cm³. The rectangular box has dimensions of 10 cm x 5 cm x 8 cm, so its volume is (10 times 5 times 8 = 400) cm³.Next, I'll add the volumes of the cube and the box to find the total volume of water displaced: (3375 + 400 = 3775) cm³.The base of the vessel has dimensions of 20 cm x 15 cm, giving it an area of (20 times 15 = 300) cm².Finally, to find the rise in water level, I'll divide the total displaced volume by the base area: (frac{3775}{300} approx 12.58) cm.

💡Okay, so I have this function f(x) given by f(x) = (1 - cos(2ωx))/2 + √3 sin(ωx) cos(ωx). The problem says that the smallest positive period of this function is π, and ω is greater than 0. I need to find the analytic expression of the function and then determine its range when x is in the interval [-π/12, π/2].First, I think I should simplify the function f(x). It looks like there are some trigonometric identities I can use here. Let me recall that cos(2θ) = 1 - 2 sin²θ, so maybe I can rewrite the first term. The first term is (1 - cos(2ωx))/2. If I use the identity cos(2θ) = 1 - 2 sin²θ, then 1 - cos(2θ) = 2 sin²θ. So, substituting θ = ωx, we get 1 - cos(2ωx) = 2 sin²(ωx). Therefore, (1 - cos(2ωx))/2 = sin²(ωx). So, the first term simplifies to sin²(ωx).Now, the second term is √3 sin(ωx) cos(ωx). I remember that sin(2θ) = 2 sinθ cosθ, so sinθ cosθ = (1/2) sin(2θ). Applying this identity, we get √3 sin(ωx) cos(ωx) = (√3/2) sin(2ωx).So, putting it all together, f(x) becomes sin²(ωx) + (√3/2) sin(2ωx). Hmm, that's a bit simpler, but maybe I can combine these terms into a single sine or cosine function. I think using the identity for combining sine terms might help here.Let me write f(x) as:f(x) = sin²(ωx) + (√3/2) sin(2ωx)But wait, sin²(ωx) can also be expressed using a double-angle identity. Since sin²θ = (1 - cos(2θ))/2, so sin²(ωx) = (1 - cos(2ωx))/2. So, substituting back, f(x) becomes:f(x) = (1 - cos(2ωx))/2 + (√3/2) sin(2ωx)Hmm, that's actually the original expression. Maybe I should try a different approach. Perhaps express the entire function as a single sine or cosine function with a phase shift.Let me consider the expression:A sin(2ωx) + B cos(2ωx) + CLooking at f(x), I have:f(x) = (1 - cos(2ωx))/2 + (√3/2) sin(2ωx)Which can be rewritten as:f(x) = 1/2 - (1/2) cos(2ωx) + (√3/2) sin(2ωx)So, that's in the form:f(x) = C + A sin(2ωx) + B cos(2ωx)Where C = 1/2, A = √3/2, and B = -1/2.I know that A sinθ + B cosθ can be written as R sin(θ + φ), where R = √(A² + B²) and φ is the phase shift. Let me compute R:R = √[(√3/2)² + (-1/2)²] = √[(3/4) + (1/4)] = √[1] = 1So, R = 1. Now, to find φ:tanφ = B/A = (-1/2)/(√3/2) = -1/√3So, φ = arctan(-1/√3). Since tan(π/6) = 1/√3, and the angle is negative, φ = -π/6.Therefore, A sinθ + B cosθ = sin(θ - π/6). So, substituting back:f(x) = 1/2 + sin(2ωx - π/6)So, f(x) simplifies to sin(2ωx - π/6) + 1/2.Now, the problem states that the smallest positive period of f(x) is π. The period of sin(2ωx - π/6) is 2π/(2ω) = π/ω. Since the function f(x) is just a sine function shifted vertically, its period is the same as that of the sine term. Therefore, the period of f(x) is π/ω. We are told this period is π, so:π/ω = π ⇒ ω = 1So, ω is 1. Therefore, f(x) becomes:f(x) = sin(2x - π/6) + 1/2Alright, that answers part (1). Now, moving on to part (2), where I need to find the range of f(x) when x is in the interval [-π/12, π/2].First, let's write f(x) again:f(x) = sin(2x - π/6) + 1/2To find the range, I need to determine the maximum and minimum values of f(x) over the given interval. Since sine functions oscillate between -1 and 1, the term sin(2x - π/6) will vary between -1 and 1, so f(x) will vary between -1 + 1/2 = -1/2 and 1 + 1/2 = 3/2. However, this is the general range, but since x is restricted to a specific interval, the actual maximum and minimum could be different.So, I need to find the maximum and minimum of sin(2x - π/6) when x is in [-π/12, π/2]. Let me find the corresponding interval for the argument of the sine function, which is 2x - π/6.When x = -π/12:2x - π/6 = 2*(-π/12) - π/6 = -π/6 - π/6 = -π/3When x = π/2:2x - π/6 = 2*(π/2) - π/6 = π - π/6 = 5π/6So, the argument 2x - π/6 ranges from -π/3 to 5π/6 as x goes from -π/12 to π/2.Now, let's analyze the sine function over the interval [-π/3, 5π/6]. The sine function reaches its maximum at π/2 and its minimum at -π/2. But in our interval, the maximum occurs at π/2, which is within [-π/3, 5π/6], and the minimum occurs at -π/3, since -π/3 is approximately -1.047, and the sine function is increasing from -π/3 to π/2.Wait, let me verify that. The sine function increases from -π/2 to π/2, so in the interval [-π/3, 5π/6], the maximum is at π/2, and the minimum is at -π/3 because -π/3 is less than π/2, and the sine function is increasing in that interval.Wait, actually, the sine function is increasing from -π/2 to π/2, which is from -1.5708 to 1.5708. Our interval is from -π/3 (-1.047) to 5π/6 (2.618). So, the sine function increases from -π/3 to π/2, reaching maximum at π/2, and then decreases from π/2 to 5π/6.Therefore, the maximum value of sin(θ) in this interval is at θ = π/2, which is 1. The minimum value is at θ = -π/3, which is sin(-π/3) = -√3/2.Therefore, sin(2x - π/6) ranges from -√3/2 to 1. Therefore, f(x) = sin(2x - π/6) + 1/2 ranges from (-√3/2 + 1/2) to (1 + 1/2), which is (1 - √3)/2 to 3/2.Wait, let me compute that:Minimum value: sin(2x - π/6) = -√3/2, so f(x) = -√3/2 + 1/2 = (1 - √3)/2.Maximum value: sin(2x - π/6) = 1, so f(x) = 1 + 1/2 = 3/2.Therefore, the range of f(x) over x ∈ [-π/12, π/2] is [(1 - √3)/2, 3/2].But let me double-check if the sine function actually attains these values within the interval.At x = -π/12, 2x - π/6 = -π/3, and sin(-π/3) = -√3/2, so f(x) = -√3/2 + 1/2 = (1 - √3)/2.At x = π/3, 2x - π/6 = 2*(π/3) - π/6 = 2π/3 - π/6 = π/2, so sin(π/2) = 1, so f(x) = 1 + 1/2 = 3/2.Therefore, these are indeed the minimum and maximum values within the interval. So, the range is [(1 - √3)/2, 3/2].I think that's it. So, summarizing:(1) The analytic expression is f(x) = sin(2x - π/6) + 1/2.(2) The range of f(x) when x ∈ [-π/12, π/2] is [(1 - √3)/2, 3/2].**Final Answer**(1) The analytic expression of the function is boxed{sinleft(2x - frac{pi}{6}right) + frac{1}{2}}.(2) The range of the function is boxed{left[ frac{1 - sqrt{3}}{2}, frac{3}{2} right]}.

💡Okay, so I have this problem about Newton's Law of Cooling. Let me try to understand what it's asking. It says that the cooling rate of a body in the air is proportional to the difference between the temperature of the body, which is T, and the air temperature, which is T₀. They want me to determine the law of temperature change of the body over time. The experiment is conducted at T₀ = 20°C, and the body cools from 100°C to 60°C in 20 minutes.Alright, so first, I remember that Newton's Law of Cooling is a way to describe how the temperature of an object changes over time as it cools down to the ambient temperature. The formula, if I recall correctly, is something like the rate of change of temperature is proportional to the difference between the object's temperature and the surrounding temperature.Let me write that down:d(T)/dt = k(T - T₀)Where:- d(T)/dt is the rate of change of temperature,- k is the proportionality constant,- T is the temperature of the object at time t,- T₀ is the ambient temperature.Given that T₀ is 20°C, so I can plug that in:d(T)/dt = k(T - 20)Now, I need to solve this differential equation to find T as a function of time. This looks like a linear differential equation, and I think I can solve it by separation of variables.So, let's separate the variables:d(T)/(T - 20) = k dtNow, integrate both sides:∫ d(T)/(T - 20) = ∫ k dtThe left side integral is straightforward. The integral of 1/(T - 20) dT is ln|T - 20| + C₁, where C₁ is the constant of integration. The right side integral is k times t plus another constant, C₂.So, putting it together:ln|T - 20| = kt + CWhere C is the combined constant (C₂ - C₁). To make it simpler, I can write it as:T - 20 = e^(kt + C) = e^C * e^(kt)Let me denote e^C as another constant, say, C₀. So:T - 20 = C₀ e^(kt)Therefore, T(t) = 20 + C₀ e^(kt)Now, I need to find the constants C₀ and k. For that, I have initial conditions. The problem says that the body cools from 100°C to 60°C in 20 minutes. So, at time t = 0, T = 100°C, and at t = 20 minutes, T = 60°C.Let's use the first condition: t = 0, T = 100.Plugging into the equation:100 = 20 + C₀ e^(k*0)Since e^0 = 1, this simplifies to:100 = 20 + C₀So, C₀ = 80.Now, the equation becomes:T(t) = 20 + 80 e^(kt)Next, use the second condition: t = 20, T = 60.Plugging into the equation:60 = 20 + 80 e^(k*20)Subtract 20 from both sides:40 = 80 e^(20k)Divide both sides by 80:0.5 = e^(20k)Now, take the natural logarithm of both sides:ln(0.5) = 20kSo,k = ln(0.5) / 20I know that ln(0.5) is equal to -ln(2), so:k = -ln(2) / 20Now, substitute k back into the equation for T(t):T(t) = 20 + 80 e^(-ln(2)/20 * t)I can simplify this further. Remember that e^(ln(a)) = a, so e^(-ln(2)) = 1/2. Therefore,T(t) = 20 + 80 (1/2)^(t/20)So, the temperature of the body at any time t is given by:T(t) = 20 + 80 (1/2)^(t/20)This means that every 20 minutes, the temperature difference from the ambient temperature halves. So, starting from 80°C above ambient (100°C), after 20 minutes, it's 40°C above ambient (60°C), and so on.Let me double-check my steps to make sure I didn't make any mistakes.1. I started with Newton's Law of Cooling: dT/dt = k(T - T₀). That seems right.2. I separated variables and integrated both sides. The integration looked correct, leading to ln|T - 20| = kt + C.3. Exponentiating both sides gave me T - 20 = C₀ e^(kt), which seems fine.4. Applied the initial condition T(0) = 100, leading to C₀ = 80. That makes sense because at t=0, the temperature is 100, so 100 - 20 = 80.5. Then, using T(20) = 60, I set up the equation 60 = 20 + 80 e^(20k), solved for k, and found k = -ln(2)/20. That seems correct because the temperature difference halves every 20 minutes, which aligns with the cooling process.6. Finally, substituting k back into the equation gave me T(t) = 20 + 80 (1/2)^(t/20). That looks good because it shows the exponential decay of the temperature difference.I think all the steps are correct. Maybe I should also consider units. The time is in minutes, so t is in minutes. The constant k has units of inverse minutes, which makes sense because it's a rate constant.Also, the final formula is in terms of (1/2)^(t/20), which is a common way to express half-life or half-decay processes. So, every 20 minutes, the excess temperature above ambient halves, which matches the given information that it cooled from 100°C to 60°C in 20 minutes (a decrease of 40°C, which is half of 80°C).So, I'm confident that the derived formula is correct.

💡Okay, so I have this problem here about negating a proposition. The proposition p is: For all x in real numbers, x is greater than sin(x). And I need to find the negation of p from the given options A to D. Hmm, let me think about how to approach this.First, I remember that when you negate a universal statement like "for all x, P(x)", the negation becomes "there exists an x such that not P(x)". So, in logical terms, the negation of ∀x P(x) is ∃x ¬P(x). That makes sense because if it's not true for all x, then there must be at least one x where it's false.So, applying that to the proposition p: For all x in ℝ, x > sin(x). The negation should be: There exists an x in ℝ such that x is not greater than sin(x). Now, what does "not greater than" mean? It means that x is less than or equal to sin(x). So, the negation would be: There exists an x₀ in ℝ such that x₀ ≤ sin(x₀).Looking at the options, option A says exactly that: There exists an x₀ in ℝ, x₀ ≤ sin(x₀). That seems correct. But let me double-check the other options to make sure I'm not missing something.Option B says: For all x in ℝ, x ≤ sin(x). Wait, that's a universal statement again, not a negation. The negation of a universal statement isn't another universal statement; it's an existential one. So B is not the correct negation.Option C says: There exists an x₀ in ℝ, x₀ < sin(x₀). Hmm, this is similar to A but uses a strict inequality. The original statement was x > sin(x), so the negation would include both x = sin(x) and x < sin(x). Therefore, using ≤ is more accurate because it covers both possibilities. So C is too restrictive because it only considers the case where x is strictly less than sin(x), not equal to.Option D is: For all x in ℝ, x < sin(x). Again, this is a universal statement, which isn't the negation. Plus, it's using a strict inequality, which isn't necessary. So D is also incorrect.Therefore, after analyzing all the options, A is the correct negation of p because it properly changes the universal quantifier to an existential one and uses the correct inequality that includes both less than and equal to.Just to make sure, let me think about specific values of x where x might be equal to or less than sin(x). For example, when x is 0, sin(0) is 0, so x = sin(x). That's a case where the inequality holds as equality. For negative values, say x = -π/2, sin(-π/2) is -1, and -π/2 is approximately -1.5708, which is less than -1. So, x < sin(x) in that case. Therefore, there are definitely points where x ≤ sin(x), which supports that the negation is correct.I think I've covered all the bases here. The key steps were understanding how to negate a universal statement, correctly interpreting the inequality, and verifying with specific examples. So, yeah, I'm confident that A is the right answer.