💡First, I need to determine the total cost of the large bottles by multiplying the number of large bottles by the price per large bottle.Next, I'll calculate the total number of bottles by adding the number of large bottles to the number of small bottles.Then, I'll calculate the total cost for all the bottles by multiplying the total number of bottles by the average price per bottle.After that, I'll set up an equation where the total cost of the large bottles plus the total cost of the small bottles equals the total cost calculated using the average price.I'll solve this equation for the price per small bottle by isolating the variable and performing the necessary calculations.Finally, I'll present the approximate price per small bottle as the solution.

💡Okay, so I have this problem here about sets and conditions. Let me try to understand it step by step. We have two sets, A and B. Set A is defined as {1, m²}, and set B is {2, 4}. The question is asking about the condition "m=2" in relation to the statement "A ∩ B = {4}". The options are whether "m=2" is a necessary and sufficient condition, just necessary, just sufficient, or neither.First, I need to recall what necessary and sufficient conditions mean. A necessary condition is something that must be true for another statement to be true, but it alone doesn't guarantee it. A sufficient condition means that if it's true, then the other statement must also be true, but there might be other ways for that statement to be true as well. If something is both necessary and sufficient, then it's the exact condition needed—no more, no less.So, let's break down the problem. We need to find out if "m=2" is necessary, sufficient, both, or neither for the intersection of A and B to be {4}.Let me first figure out what A ∩ B = {4} means. The intersection of sets A and B is the set of elements that are in both A and B. So, for A ∩ B to be {4}, the number 4 must be in both A and B, and there should be no other common elements.Given that B is {2, 4}, it already contains 4, so for 4 to be in A, A must also contain 4. Since A is {1, m²}, that means m² must be 4. So, m² = 4. Solving this, m can be either 2 or -2 because both 2² and (-2)² equal 4.Now, the condition given is "m=2". If m is 2, then A becomes {1, 4}, and B is {2, 4}. The intersection A ∩ B is indeed {4}, which satisfies the condition. So, "m=2" is a sufficient condition because it guarantees that A ∩ B = {4}.But is "m=2" a necessary condition? That is, is it the only way for A ∩ B to be {4}? Well, as I found earlier, m could also be -2, because (-2)² is also 4. So, if m is -2, A would still be {1, 4}, and the intersection with B would still be {4}. Therefore, "m=2" is not the only way to satisfy A ∩ B = {4}; "m=-2" also works.So, "m=2" is sufficient because it ensures the intersection is {4}, but it's not necessary because there's another value of m that also works. Therefore, "m=2" is a sufficient but not necessary condition.Let me double-check to make sure I didn't miss anything. If m is 2, A is {1,4}, intersection with B is {4}. If m is -2, A is still {1,4}, intersection with B is {4}. If m is something else, say 3, then A would be {1,9}, and the intersection with B would be empty, which doesn't satisfy the condition. Similarly, if m is 0, A is {1,0}, intersection with B is empty. So, only when m² is 4 does the intersection become {4}, which happens when m is 2 or -2.Therefore, "m=2" is one of the possible solutions, but not the only one. So, it's sufficient because it works, but not necessary because it's not the only way.I think that makes sense. So, the answer should be that "m=2" is a sufficient but not necessary condition, which corresponds to option C.

💡First, I need to determine the distance Joann traveled by multiplying her average speed by the time she rode her bike.Joann's distance = 14 miles per hour × 4 hours = 56 miles.Next, I need to find the average speed Fran must ride to cover the same 56 miles in 2 hours. I'll set up the equation:Fran's speed × 2 hours = 56 miles.Solving for Fran's speed:Fran's speed = 56 miles ÷ 2 hours = 28 miles per hour.Therefore, Fran needs to ride at an average speed of 28 miles per hour.

💡Okay, so I have this problem here where I need to find the area enclosed by the curve ( y = cos x ) between ( x = -frac{pi}{2} ) and ( x = pi ), and the x-axis. The options given are A: 1, B: 2, C: 3, D: 4. Hmm, let me think about how to approach this.First, I remember that to find the area between a curve and the x-axis, I need to integrate the function over the given interval. But wait, I also recall that if the curve goes below the x-axis, the integral will give a negative area, so I need to take the absolute value of those parts to get the total area.So, let me sketch the graph of ( y = cos x ) to visualize it. Cosine starts at 1 when ( x = 0 ), goes down to 0 at ( x = frac{pi}{2} ), then to -1 at ( x = pi ). But wait, the interval is from ( -frac{pi}{2} ) to ( pi ). So, from ( -frac{pi}{2} ) to ( 0 ), cosine is positive, from 0 to ( frac{pi}{2} ) it's still positive, but from ( frac{pi}{2} ) to ( pi ), it becomes negative.Therefore, the area will have two parts: the area above the x-axis from ( -frac{pi}{2} ) to ( frac{pi}{2} ), and the area below the x-axis from ( frac{pi}{2} ) to ( pi ). Since area can't be negative, I'll take the absolute value of the integral for the negative part.Let me write down the integrals:1. From ( -frac{pi}{2} ) to ( frac{pi}{2} ), ( cos x ) is positive, so the integral is straightforward.2. From ( frac{pi}{2} ) to ( pi ), ( cos x ) is negative, so I'll take the negative of the integral to make it positive.So, the total area ( A ) is:[A = int_{-frac{pi}{2}}^{frac{pi}{2}} cos x , dx + left| int_{frac{pi}{2}}^{pi} cos x , dx right|]Let me compute each integral separately.First integral: ( int_{-frac{pi}{2}}^{frac{pi}{2}} cos x , dx )The antiderivative of ( cos x ) is ( sin x ). So,[int_{-frac{pi}{2}}^{frac{pi}{2}} cos x , dx = sinleft(frac{pi}{2}right) - sinleft(-frac{pi}{2}right)]I know that ( sinleft(frac{pi}{2}right) = 1 ) and ( sinleft(-frac{pi}{2}right) = -1 ). So,[1 - (-1) = 1 + 1 = 2]Okay, so the first part is 2.Second integral: ( int_{frac{pi}{2}}^{pi} cos x , dx )Again, the antiderivative is ( sin x ). So,[sin(pi) - sinleft(frac{pi}{2}right)]I know that ( sin(pi) = 0 ) and ( sinleft(frac{pi}{2}right) = 1 ). So,[0 - 1 = -1]But since this is the area below the x-axis, I take the absolute value, which is 1.Therefore, the total area is ( 2 + 1 = 3 ).Wait, but looking back at the original problem, the options are 1, 2, 3, 4. So, 3 is option C.But just to double-check, maybe I made a mistake in the limits or the signs.Let me verify the first integral:From ( -frac{pi}{2} ) to ( frac{pi}{2} ), cosine is positive, so integrating gives 2. That seems correct.Second integral:From ( frac{pi}{2} ) to ( pi ), cosine is negative, so the integral is negative, but taking absolute value gives 1. So, total area is 3.Alternatively, maybe I can think of it as the integral from ( -frac{pi}{2} ) to ( pi ) of the absolute value of ( cos x ). But integrating absolute values can be tricky because you have to split the integral at points where the function changes sign, which in this case is at ( frac{pi}{2} ).So, that's exactly what I did. So, I think my approach is correct.Alternatively, maybe the problem is simpler because it's symmetric or something? Let me think.From ( -frac{pi}{2} ) to ( frac{pi}{2} ), the integral is 2, as I found. Then from ( frac{pi}{2} ) to ( pi ), it's another 1, so total 3. So, yeah, 3 is the answer.I don't think I made any calculation errors. Let me just compute the integrals again.First integral:[int_{-frac{pi}{2}}^{frac{pi}{2}} cos x , dx = sinleft(frac{pi}{2}right) - sinleft(-frac{pi}{2}right) = 1 - (-1) = 2]Second integral:[int_{frac{pi}{2}}^{pi} cos x , dx = sin(pi) - sinleft(frac{pi}{2}right) = 0 - 1 = -1]Absolute value is 1. So, 2 + 1 = 3.Yes, that seems consistent. So, the answer should be 3, which is option C.Wait, but in the initial problem statement, the user wrote:"The area enclosed by the curve ( y = cos x ) ( left(-frac{pi}{2} < x < piright) ) and the x-axis is (　　) A: 1 B: 2 C: 3 D: 4"So, yeah, 3 is the correct answer.I think I'm confident with this. Maybe I can also recall that the integral of cosine over a full period is zero, but here we're not integrating over a full period, but from ( -frac{pi}{2} ) to ( pi ), which is ( frac{3pi}{2} ). But since we're taking absolute areas, it's different.Alternatively, maybe someone might think that the area is just 2, but that would be only from 0 to ( frac{pi}{2} ), but since we have the area from ( -frac{pi}{2} ) to ( frac{pi}{2} ) as 2, and then another 1 from ( frac{pi}{2} ) to ( pi ), so total 3.Yeah, I think that's solid. So, I'll go with option C: 3.**Final Answer**boxed{C}

💡Alright, so I have this problem with two propositions, p and q, and I need to find the range of values for the real number a based on the truth values of p ∨ q and p ∧ q. Let me try to unpack this step by step.First, let's understand what each proposition means.Proposition p is: For all real numbers x, x² - 2x > a. So, this is saying that no matter what real number x I pick, when I compute x squared minus 2x, the result will always be greater than a. That means a has to be less than the minimum value of the quadratic function f(x) = x² - 2x.Proposition q is: There exists a real number x₀ such that x₀² + 2a x₀ + 2 - a = 0. This is saying that the quadratic equation x² + 2a x + (2 - a) = 0 has at least one real solution. For a quadratic equation to have real solutions, the discriminant must be non-negative. The discriminant D is given by D = (2a)² - 4 * 1 * (2 - a). So, D = 4a² - 8 + 4a. Simplifying that, D = 4a² + 4a - 8. For real solutions, D ≥ 0, so 4a² + 4a - 8 ≥ 0. Dividing both sides by 4, we get a² + a - 2 ≥ 0. Let me factor that quadratic: a² + a - 2 = (a + 2)(a - 1). So, the inequality becomes (a + 2)(a - 1) ≥ 0. This means that a is either less than or equal to -2 or greater than or equal to 1.Okay, so for proposition q to be true, a must be ≤ -2 or ≥ 1.Now, going back to proposition p: For all x, x² - 2x > a. Let's analyze this quadratic function f(x) = x² - 2x. Since the coefficient of x² is positive, it's a parabola opening upwards, so it has a minimum point. The vertex of this parabola is at x = -b/(2a) = 2/(2*1) = 1. Plugging x=1 into f(x), we get f(1) = 1 - 2 = -1. So, the minimum value of f(x) is -1. Therefore, for p to be true, a must be less than this minimum value, so a < -1.So, proposition p is true when a < -1, and false otherwise (i.e., when a ≥ -1).Now, the problem states that p ∨ q is true and p ∧ q is false. Let's recall what these mean.p ∨ q is true if at least one of p or q is true. p ∧ q is false if both p and q are not both true at the same time. So, combining these two, it means that exactly one of p or q is true, and the other is false. In other words, p and q must have opposite truth values.So, we have two cases:1. p is true and q is false.2. p is false and q is true.Let's analyze each case.Case 1: p is true and q is false.p is true when a < -1.q is false when a is not in the range where q is true. Since q is true when a ≤ -2 or a ≥ 1, q is false when -2 < a < 1.So, for p to be true and q to be false, a must satisfy both a < -1 and -2 < a < 1. The overlap of these two intervals is -2 < a < -1.Case 2: p is false and q is true.p is false when a ≥ -1.q is true when a ≤ -2 or a ≥ 1.So, for p to be false and q to be true, a must satisfy both a ≥ -1 and (a ≤ -2 or a ≥ 1). The overlap here is a ≥ 1 because a cannot be both ≥ -1 and ≤ -2 at the same time.So, combining both cases, the values of a are in the intervals (-2, -1) and [1, ∞).Wait, let me double-check this.In Case 1, p is true (a < -1) and q is false (-2 < a < 1). The intersection is indeed -2 < a < -1.In Case 2, p is false (a ≥ -1) and q is true (a ≤ -2 or a ≥ 1). The intersection here is a ≥ 1 because a cannot be both ≥ -1 and ≤ -2. So, yes, a ≥ 1.Therefore, the range of a is (-2, -1) union [1, ∞).I think that's correct. Let me just visualize it on a number line.For a < -1, p is true. But q is false only when a is between -2 and 1. So, overlapping these gives -2 < a < -1.For a ≥ -1, p is false. q is true when a is ≤ -2 or ≥ 1. Since a is already ≥ -1, the only overlapping part is a ≥ 1.So, combining both, a is between -2 and -1, or greater than or equal to 1.I think that's solid. I don't see any mistakes in the reasoning. So, the final answer should be a ∈ (-2, -1) ∪ [1, ∞).**Final Answer**The range of values for the real number ( a ) is boxed{(-2, -1) cup [1, +infty)}.

💡Okay, so I have this problem about two construction teams, A and B. They each have a certain number of workers. The problem gives me two scenarios:1. If team A transfers 90 workers to team B, then team B will have twice as many workers as team A.2. If team B transfers a certain number of workers to team A, then team A will have six times as many workers as team B.I need to find the minimum number of workers that team A originally had. Hmm, okay, let me try to break this down step by step.First, let's define some variables to represent the number of workers in each team. Let me call the number of workers in team A as "a" and in team B as "b". So, initially, team A has "a" workers and team B has "b" workers.Now, the first scenario says that if team A transfers 90 workers to team B, then team B will have twice as many workers as team A. Let me write that as an equation. After transferring 90 workers:- Team A will have (a - 90) workers.- Team B will have (b + 90) workers.According to the problem, at this point, team B has twice as many workers as team A. So, the equation becomes:b + 90 = 2 * (a - 90)Let me simplify this equation:b + 90 = 2a - 180Subtract 90 from both sides:b = 2a - 180 - 90b = 2a - 270Okay, so that's our first equation: b = 2a - 270.Now, moving on to the second scenario. If team B transfers a certain number of workers to team A, then team A will have six times as many workers as team B. Let me denote the number of workers transferred from team B to team A as "c". So, after transferring "c" workers:- Team A will have (a + c) workers.- Team B will have (b - c) workers.According to the problem, at this point, team A has six times as many workers as team B. So, the equation becomes:a + c = 6 * (b - c)Let me simplify this equation:a + c = 6b - 6cBring all terms to one side:a + c + 6c - 6b = 0a + 7c - 6b = 0So, that's our second equation: a + 7c = 6b.Now, I have two equations:1. b = 2a - 2702. a + 7c = 6bI can substitute the first equation into the second one to eliminate "b". Let's do that.From equation 1, b = 2a - 270. Substitute this into equation 2:a + 7c = 6*(2a - 270)Let me compute the right side:6*(2a - 270) = 12a - 1620So, the equation becomes:a + 7c = 12a - 1620Let me bring all terms to one side:a + 7c - 12a + 1620 = 0-11a + 7c + 1620 = 0Let me rearrange it:7c = 11a - 1620So, c = (11a - 1620)/7Hmm, okay. So, c is equal to (11a - 1620) divided by 7. Since the number of workers transferred, "c", must be a positive integer, this fraction must result in an integer. Therefore, (11a - 1620) must be divisible by 7.So, 11a - 1620 ≡ 0 mod 7Let me compute 11a mod 7 and 1620 mod 7.First, 11 mod 7 is 4, because 7*1=7, 11-7=4.So, 11a mod 7 is equivalent to 4a mod 7.Similarly, 1620 divided by 7: Let's compute 1620 ÷ 7.7*231 = 1617, so 1620 - 1617 = 3. So, 1620 mod 7 is 3.Therefore, the equation becomes:4a - 3 ≡ 0 mod 7Which simplifies to:4a ≡ 3 mod 7Now, I need to solve for "a" in this congruence equation.To solve 4a ≡ 3 mod 7, I can find the multiplicative inverse of 4 modulo 7.What number multiplied by 4 gives 1 mod 7? Let's see:4*2=8≡1 mod7So, the inverse of 4 mod7 is 2.Therefore, multiplying both sides by 2:a ≡ 3*2 mod7a ≡ 6 mod7So, a ≡6 mod7. That means a can be written as:a = 7k + 6, where k is an integer.But wait, let me check that. If a ≡6 mod7, then a =7k +6.But let me verify:If a =7k +6, then 4a =4*(7k +6)=28k +24.28k mod7 is 0, 24 mod7 is 3, because 7*3=21, 24-21=3.So, 4a mod7=3, which matches the equation 4a≡3 mod7.Okay, so that's correct.Therefore, a must be of the form 7k +6.But we need to find the minimum number of workers that team A originally had, so we need the smallest possible "a" such that all the conditions are satisfied.But we also have other constraints.First, let's note that the number of workers transferred, "c", must be a positive integer. So, c=(11a -1620)/7 must be positive.Therefore:11a -1620 >011a >1620a >1620/11Compute 1620 ÷11:11*147=1617, so 1620-1617=3, so 1620/11=147 +3/11≈147.27So, a must be greater than approximately 147.27, so a≥148.But since a=7k +6, and a must be at least 148, let's find the smallest k such that 7k +6≥148.Compute 7k +6≥1487k≥142k≥142/7≈20.2857Since k must be an integer, k≥21.Therefore, the smallest k is 21.So, a=7*21 +6=147 +6=153.So, a=153.Wait, let me check if this works.Compute b=2a -270=2*153 -270=306 -270=36.So, b=36.Now, check the first condition: If team A transfers 90 workers to team B, then team B will have twice as many as team A.After transfer:Team A:153 -90=63Team B:36 +90=126Is 126 equal to 2*63? Yes, 2*63=126. So, that works.Now, check the second condition: If team B transfers "c" workers to team A, then team A will have six times as many as team B.Compute c=(11a -1620)/7=(11*153 -1620)/7=(1683 -1620)/7=63/7=9.So, c=9.After transfer:Team A:153 +9=162Team B:36 -9=27Is 162 equal to 6*27? 6*27=162. Yes, that works.So, both conditions are satisfied when a=153, b=36, and c=9.Therefore, the minimum number of workers that team A originally had is 153.But wait, let me just make sure that there isn't a smaller "a" that also satisfies all conditions. Since we found that a must be of the form 7k +6 and k must be at least 21, the next possible value would be k=20, which gives a=7*20 +6=146. But 146 is less than 148, which we found earlier as the minimum a must be. So, a=146 would result in c=(11*146 -1620)/7=(1606 -1620)/7=(-14)/7=-2, which is negative, which doesn't make sense because you can't transfer a negative number of workers. So, a=146 is invalid.Therefore, the next possible value is a=153, which works. So, 153 is indeed the minimum number.I think that's it. So, the minimum number of workers team A originally had is 153.**Final Answer**The minimum number of workers that team A originally had is boxed{153}.