answer：First, identify the total amount spent and the number of candy bars purchased.Next, calculate the total cost of the candy bars by multiplying the number of candy bars by the cost per candy bar.Subtract the total cost of the candy bars from the total amount spent to find the total cost of the soft drinks.Finally, divide the total cost of the soft drinks by the number of soft drinks to determine the cost per soft drink.

answer：First, I note that Carolyn starts with 47 marbles and ends with 5 marbles after sharing some with Diana.To find out how many marbles Carolyn shared with Diana, I subtract the number of marbles she has at the end from the number she started with.So, 47 marbles minus 5 marbles equals 42 marbles.Therefore, Carolyn shared 42 marbles with Diana.

answer：Alright, so I have this problem where I need to find a time between 2 and 3 o'clock that corresponds to a time between 6 and 7 o'clock such that the positions of the clock hands are the same, but with the hands swapped. Hmm, that sounds a bit tricky, but let me try to break it down step by step.First, I know that clock problems often involve figuring out the angles of the hour and minute hands. The hour hand moves at a certain rate, and the minute hand moves faster. Maybe I can represent their positions mathematically.Let's denote the time between 2 and 3 o'clock as 2 hours plus some minutes, say x minutes. Similarly, the time between 6 and 7 o'clock would be 6 hours plus y minutes. But since the hands are swapped, the positions of the hour and minute hands at these two times should be the same, just swapped.So, at 2 o'clock plus x minutes, the hour hand has moved past 2, and the minute hand is at x minutes. At 6 o'clock plus y minutes, the hour hand has moved past 6, and the minute hand is at y minutes. But since the hands are swapped, the hour hand at 2:x should be where the minute hand is at 6:y, and vice versa.Let me think about how to express the positions of the hands. The hour hand moves 30 degrees per hour (since 360 degrees / 12 hours = 30 degrees per hour) and 0.5 degrees per minute (since 30 degrees per hour / 60 minutes = 0.5 degrees per minute). The minute hand moves 6 degrees per minute (since 360 degrees / 60 minutes = 6 degrees per minute).So, at 2:x, the hour hand is at 2*30 + 0.5*x = 60 + 0.5x degrees, and the minute hand is at 6x degrees.At 6:y, the hour hand is at 6*30 + 0.5*y = 180 + 0.5y degrees, and the minute hand is at 6y degrees.Now, since the hands are swapped, the hour hand at 2:x should equal the minute hand at 6:y, and the minute hand at 2:x should equal the hour hand at 6:y.So, setting up the equations:60 + 0.5x = 6y ...(1)6x = 180 + 0.5y ...(2)Now, I have two equations with two variables, x and y. I can solve this system of equations to find x and y.From equation (1):60 + 0.5x = 6yLet's solve for y:6y = 60 + 0.5xy = (60 + 0.5x)/6y = 10 + (0.5x)/6y = 10 + x/12 ...(3)Now, substitute y from equation (3) into equation (2):6x = 180 + 0.5y6x = 180 + 0.5*(10 + x/12)6x = 180 + 5 + x/246x = 185 + x/24Now, let's get rid of the fraction by multiplying both sides by 24:24*6x = 24*185 + 24*(x/24)144x = 4440 + xNow, subtract x from both sides:144x - x = 4440143x = 4440x = 4440 / 143x ≈ 31.0559 minutesSo, x is approximately 31.0559 minutes. That means the time between 2 and 3 o'clock is approximately 2:31.0559.Now, let's find y using equation (3):y = 10 + x/12y = 10 + (4440/143)/12y = 10 + (4440)/(143*12)y = 10 + 4440/1716Simplify 4440/1716:Divide numerator and denominator by 12:4440 ÷ 12 = 3701716 ÷ 12 = 143So, y = 10 + 370/143y ≈ 10 + 2.5874y ≈ 12.5874 minutesSo, y is approximately 12.5874 minutes. That means the time between 6 and 7 o'clock is approximately 6:12.5874.Let me check if these times make sense. At 2:31.0559, the hour hand is a bit past 2, and the minute hand is at 31.0559 minutes. At 6:12.5874, the hour hand is a bit past 6, and the minute hand is at 12.5874 minutes. If I swap the hands, the hour hand at 2:31 should be where the minute hand is at 6:12.5874, and vice versa. It seems consistent.I think I've got it. The times are approximately 2:31.0559 and 6:12.5874. To express them more precisely, I can convert the decimal parts of the minutes into seconds.For x ≈ 31.0559 minutes:0.0559 minutes * 60 seconds/minute ≈ 3.354 secondsSo, x ≈ 31 minutes and 3.354 seconds.For y ≈ 12.5874 minutes:0.5874 minutes * 60 seconds/minute ≈ 35.244 secondsSo, y ≈ 12 minutes and 35.244 seconds.Therefore, the exact times are:Between 2 and 3 o'clock: 2 hours, 31 minutes, and approximately 3.35 seconds.Between 6 and 7 o'clock: 6 hours, 12 minutes, and approximately 35.24 seconds.I think that's the solution. It took a bit of setting up the equations and solving the system, but it makes sense now.

answer：Okay, so I have this problem here about vectors, and I need to find the magnitude of the sum of two vectors, a and b. Let me try to understand what the problem is asking and how to approach it step by step.First, let me write down the given information:- Vector a is (x, 1)- Vector b is (2, y)- Vector c is (-2, 2)We are told that vector a is perpendicular to vector c, and vector b is parallel to vector c. We need to find the magnitude of vector a + vector b.Alright, so I remember that two vectors are perpendicular if their dot product is zero. That should help me find the value of x. Also, if two vectors are parallel, one must be a scalar multiple of the other. That should help me find the value of y.Let me start with the first condition: vector a is perpendicular to vector c. So, their dot product should be zero.The dot product of a and c is calculated as:a · c = (x)(-2) + (1)(2) = -2x + 2Since they are perpendicular, this should equal zero:-2x + 2 = 0Let me solve for x:-2x + 2 = 0 -2x = -2 x = (-2)/(-2) x = 1Okay, so x is 1. That means vector a is (1, 1).Now, moving on to the second condition: vector b is parallel to vector c. If two vectors are parallel, one is a scalar multiple of the other. So, vector b must be some scalar multiple of vector c.Let me denote the scalar as k. So, vector b = k * vector c.In terms of components, that would be:(2, y) = k * (-2, 2) Which means: 2 = -2k y = 2kFrom the first equation, 2 = -2k, I can solve for k:2 = -2k k = 2 / (-2) k = -1Now that I have k, I can find y from the second equation:y = 2k y = 2*(-1) y = -2So, y is -2. Therefore, vector b is (2, -2).Now, I need to find the magnitude of vector a + vector b. First, let me compute vector a + vector b.Vector a is (1, 1) and vector b is (2, -2). Adding them component-wise:a + b = (1 + 2, 1 + (-2)) a + b = (3, -1)Now, I need to find the magnitude of this resulting vector. The magnitude of a vector (p, q) is given by the formula:|a + b| = sqrt(p^2 + q^2)Applying this to our vector (3, -1):|a + b| = sqrt(3^2 + (-1)^2) |a + b| = sqrt(9 + 1) |a + b| = sqrt(10)So, the magnitude of vector a + vector b is sqrt(10).Let me just double-check my steps to make sure I didn't make any mistakes.1. Calculated the dot product of a and c, set it to zero, solved for x. Got x = 1. That seems correct.2. For vector b being parallel to vector c, I set up the scalar multiple and solved for k, which gave me k = -1, and then y = -2. That also seems correct.3. Added vectors a and b: (1,1) + (2,-2) = (3,-1). That's straightforward.4. Calculated the magnitude: sqrt(3^2 + (-1)^2) = sqrt(9 + 1) = sqrt(10). That looks right.I think I've covered all the steps and didn't skip anything important. So, I'm confident that the magnitude of vector a + vector b is sqrt(10).**Final Answer**The magnitude of (overrightarrow{a} + overrightarrow{b}) is boxed{sqrt{10}}.

answer：Okay, so I have this problem where I need to show that for positive real numbers (a_1, a_2, ldots, a_n) with a product of 1, the product ((1 + a_1)(1 + a_2)ldots(1 + a_n)) is at least (2^n). I also need to find when equality holds. Hmm, let's see how I can approach this.First, I remember that inequalities often involve using well-known inequalities like AM-GM or Cauchy-Schwarz. Since the problem mentions the product of the numbers is 1, maybe AM-GM is a good starting point because it relates sums and products.Let me recall the AM-GM inequality: for any non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. In other words, for two numbers, (frac{x + y}{2} geq sqrt{xy}), with equality when (x = y). Maybe I can apply this to each term (1 + a_i).So, for each (a_i), I can consider (1) and (a_i). Applying AM-GM to these two numbers:[frac{1 + a_i}{2} geq sqrt{1 cdot a_i} = sqrt{a_i}]Multiplying both sides by 2 gives:[1 + a_i geq 2sqrt{a_i}]Okay, so each term (1 + a_i) is at least (2sqrt{a_i}). If I take the product of all these inequalities, I might get something useful.So, multiplying all the inequalities together:[(1 + a_1)(1 + a_2)ldots(1 + a_n) geq (2sqrt{a_1})(2sqrt{a_2})ldots(2sqrt{a_n})]Simplifying the right-hand side, I can factor out the 2's and combine the square roots:[(2sqrt{a_1})(2sqrt{a_2})ldots(2sqrt{a_n}) = 2^n cdot sqrt{a_1 a_2 ldots a_n}]But wait, the problem states that the product (a_1 a_2 ldots a_n = 1). So, the square root of that product is (sqrt{1} = 1). Therefore, the right-hand side simplifies to:[2^n cdot 1 = 2^n]So, putting it all together, we have:[(1 + a_1)(1 + a_2)ldots(1 + a_n) geq 2^n]That proves the inequality. Now, I need to find when equality holds. In the AM-GM inequality, equality holds if and only if all the numbers are equal. In this case, for each (i), (1 = a_i) because we applied AM-GM to (1) and (a_i). So, equality holds when each (a_i = 1).Let me double-check that. If all (a_i = 1), then the product (a_1 a_2 ldots a_n = 1^n = 1), which satisfies the given condition. Also, each term (1 + a_i = 2), so the product is (2^n), which matches the right-hand side. So yes, equality holds when all (a_i = 1).I think that's it. I used AM-GM on each pair (1) and (a_i), multiplied the inequalities, and used the given condition that the product of all (a_i) is 1. Equality occurs when each (a_i = 1). I don't see any mistakes in my reasoning, so I'm confident this is correct.**Final Answer**The inequality holds and equality occurs when all (a_i = 1). Thus, the result is (boxed{2^n}).

answer：First, I need to evaluate the floor function of 1.999. The floor function returns the greatest integer less than or equal to the given number. Since 1.999 is just below 2, the floor of 1.999 is 1.Next, I need to evaluate the ceiling function of 3.001. The ceiling function returns the smallest integer greater than or equal to the given number. Since 3.001 is just above 3, the ceiling of 3.001 is 4.Finally, I add the results of the two functions together: 1 + 4 equals 5.