💡Okay, so I need to figure out how much Emily's mom spent on school supplies for each child: Emily, Sophie, and Sam. She bought glue sticks, pencils, and erasers, and then split them among the three kids. I know the total number of each item she bought and the price per item. I also know how many of each item Emily and Sophie received, but not Sam's share. So, I guess I need to figure out how many supplies Sam got and then calculate the cost for each child.First, let's list out what I know:- Total glue sticks bought: 27 packs- Price per glue stick: 1.00- Total pencils bought: 40 pencils- Price per pencil: 0.50- Total erasers bought: 15 erasers- Price per eraser: 0.75Emily received:- 9 packs of glue sticks- 18 pencils- 5 erasersSophie received:- 12 packs of glue sticks- 14 pencils- 4 erasersSam's supplies aren't mentioned, so I need to calculate how many of each item he got by subtracting what Emily and Sophie received from the total.Let me start with glue sticks. Total glue sticks are 27. Emily got 9, and Sophie got 12. So, 9 + 12 = 21. That means Sam got 27 - 21 = 6 packs of glue sticks.Next, pencils. Total pencils are 40. Emily got 18, and Sophie got 14. So, 18 + 14 = 32. Therefore, Sam got 40 - 32 = 8 pencils.Lastly, erasers. Total erasers are 15. Emily got 5, and Sophie got 4. So, 5 + 4 = 9. That means Sam got 15 - 9 = 6 erasers.Now that I know how many supplies each child got, I can calculate the cost for each.Starting with Emily:- Glue sticks: 9 packs * 1.00 = 9.00- Pencils: 18 pencils * 0.50 = 9.00- Erasers: 5 erasers * 0.75 = 3.75Adding these up: 9.00 + 9.00 + 3.75 = 21.75So, Emily's total cost is 21.75.Now for Sophie:- Glue sticks: 12 packs * 1.00 = 12.00- Pencils: 14 pencils * 0.50 = 7.00- Erasers: 4 erasers * 0.75 = 3.00Adding these up: 12.00 + 7.00 + 3.00 = 22.00Sophie's total cost is 22.00.Finally, for Sam:- Glue sticks: 6 packs * 1.00 = 6.00- Pencils: 8 pencils * 0.50 = 4.00- Erasers: 6 erasers * 0.75 = 4.50Adding these up: 6.00 + 4.00 + 4.50 = 14.50Sam's total cost is 14.50.So, to summarize:- Emily: 21.75- Sophie: 22.00- Sam: 14.50I think that's it! I just need to make sure I didn't make any calculation errors. Let me double-check the numbers quickly.For glue sticks:- Emily: 9 * 1 = 9- Sophie: 12 * 1 = 12- Sam: 6 * 1 = 6Total glue sticks cost: 9 + 12 + 6 = 27, which matches the total bought.For pencils:- Emily: 18 * 0.50 = 9- Sophie: 14 * 0.50 = 7- Sam: 8 * 0.50 = 4Total pencils cost: 9 + 7 + 4 = 20, which matches the total bought.For erasers:- Emily: 5 * 0.75 = 3.75- Sophie: 4 * 0.75 = 3.00- Sam: 6 * 0.75 = 4.50Total erasers cost: 3.75 + 3.00 + 4.50 = 11.25, which matches the total bought.Everything adds up correctly. So, the costs per child are accurate.

💡Okay, so I have this problem where I need to figure out how many different ways five students can line up in a row under three different sets of requirements. Let me take each requirement one by one and try to work through them step by step.**Problem (1): A and B must be adjacent.**Alright, so I need to count the number of permutations where A and B are next to each other. Hmm, how do I approach this? I remember something about treating A and B as a single unit or "block." If I consider A and B together as one unit, then effectively, I'm arranging four items instead of five: the AB block and the other three students.So, the number of ways to arrange four items is 4 factorial, which is 4! = 24. But wait, within the AB block, A and B can switch places. That means for each of those 24 arrangements, there are 2 possibilities: AB or BA. So, I need to multiply 24 by 2 to account for both arrangements. Let me write that down: 4! * 2 = 24 * 2 = 48. So, there are 48 ways where A and B are adjacent.**Problem (2): A and B are not adjacent.**Okay, this is the opposite of the first problem. Instead of A and B being next to each other, they should not be. How do I calculate this? I think the total number of permutations without any restrictions is 5! = 120. If I subtract the number of permutations where A and B are adjacent, that should give me the number of permutations where they are not adjacent.From the first problem, I know that there are 48 ways where A and B are adjacent. So, subtracting that from the total: 120 - 48 = 72. Therefore, there are 72 ways where A and B are not adjacent.Wait, let me think if there's another way to approach this. Maybe using the gap method. If I first arrange the other three students, that's 3! = 6 ways. These three students create four gaps where A and B can be placed: one before the first student, one between the first and second, one between the second and third, and one after the third student.So, there are 4 gaps and I need to place A and B in two of these gaps without them being adjacent. The number of ways to choose two gaps out of four is C(4,2) = 6. But since A and B are distinct, the order matters, so it's actually permutations: P(4,2) = 4*3 = 12.Then, multiply this by the number of ways to arrange the other three students: 6 * 12 = 72. Yep, same answer. So, that confirms it. There are 72 ways.**Problem (3): A is not in the middle, and B is not at either end.**This one seems a bit more complex. I need to count the number of permutations where A isn't in the middle position and B isn't at either end. Let me break it down.First, the total number of permutations without any restrictions is 5! = 120.Now, I need to subtract the cases where A is in the middle or B is at either end. But wait, I have to be careful about double-counting the cases where both A is in the middle and B is at either end.So, using the inclusion-exclusion principle:Number of valid permutations = Total permutations - (permutations with A in the middle + permutations with B at either end) + permutations where both A is in the middle and B is at either end.Let me compute each part.1. **Total permutations:** 5! = 120.2. **Permutations with A in the middle:** If A is fixed in the middle position, then the remaining four students can be arranged in 4! = 24 ways.3. **Permutations with B at either end:** There are two end positions. If B is fixed at one end, the remaining four students can be arranged in 4! = 24 ways. Since B can be at either end, it's 2 * 24 = 48.4. **Permutations where both A is in the middle and B is at either end:** If A is fixed in the middle and B is fixed at one end, then the remaining three students can be arranged in 3! = 6 ways. Since B can be at either end, it's 2 * 6 = 12.Putting it all together:Number of valid permutations = 120 - (24 + 48) + 12 = 120 - 72 + 12 = 60.So, there are 60 valid permutations where A is not in the middle and B is not at either end.Wait, let me double-check this. Another approach could be to count directly.First, figure out the number of positions available for A and B.- There are 5 positions: 1, 2, 3, 4, 5.- A cannot be in position 3 (the middle).- B cannot be in positions 1 or 5 (the ends).So, A can be in positions 1, 2, 4, or 5.B can be in positions 2, 3, or 4.But we also need to ensure that A and B are placed such that they don't violate each other's restrictions.Let me count the number of ways:1. Choose a position for A: 4 choices (positions 1, 2, 4, 5).2. For each choice of A, choose a position for B: - If A is in position 1: B can be in 2, 3, 4 (3 choices). - If A is in position 2: B can be in 3, 4 (since B can't be in position 1 or 5, and position 2 is taken by A). - If A is in position 4: B can be in 2, 3 (since position 4 is taken by A, and B can't be in 1 or 5). - If A is in position 5: B can be in 2, 3, 4 (3 choices).So, total choices for B:- A in 1: 3- A in 2: 2- A in 4: 2- A in 5: 3Total = 3 + 2 + 2 + 3 = 10.Then, for each of these 10 combinations of A and B positions, the remaining 3 students can be arranged in 3! = 6 ways.So, total permutations = 10 * 6 = 60.Yep, same answer. So, that's correct.**Final Answer**(1) boxed{48}(2) boxed{72}(3) boxed{60}

💡Okay, so I need to find the modulus of the complex number 3/4 minus 3i. Hmm, I remember that the modulus of a complex number is like its distance from the origin on the complex plane. It's calculated using the formula where you take the square root of the sum of the squares of the real part and the imaginary part. Let me write that down to make sure I get it right.So, if I have a complex number a + bi, the modulus is sqrt(a² + b²). In this case, the real part is 3/4 and the imaginary part is -3. Wait, is it -3 or just 3? I think it's -3 because the number is 3/4 minus 3i. So, the imaginary part is -3. But when I square it, the negative will go away, so it doesn't matter. Alright, let me plug the numbers into the formula. The modulus should be sqrt((3/4)² + (-3)²). Let me calculate each part step by step. First, (3/4) squared. 3 squared is 9, and 4 squared is 16, so that's 9/16. Next, (-3) squared is 9, right? Because negative times negative is positive. So, now I have sqrt(9/16 + 9). Wait, 9 is a whole number, and 9/16 is a fraction. I need to add them together. To add them, they should have the same denominator. So, I can write 9 as 144/16 because 9 times 16 is 144. So, now I have sqrt(9/16 + 144/16). Adding those together gives me 153/16. So, the modulus is sqrt(153/16). I can simplify that as sqrt(153) divided by sqrt(16). The square root of 16 is 4, so that's sqrt(153)/4.Now, can I simplify sqrt(153) any further? Let me see. 153 divided by 9 is 17, right? So, 153 is 9 times 17. That means sqrt(153) is sqrt(9*17), which is sqrt(9) times sqrt(17). The square root of 9 is 3, so that simplifies to 3*sqrt(17). Putting it all together, the modulus is 3*sqrt(17) divided by 4. So, the final answer should be 3√17 over 4. Let me just double-check my steps to make sure I didn't make any mistakes. First, I identified the real and imaginary parts correctly: 3/4 and -3. Then, I squared each part: (3/4)² is 9/16, and (-3)² is 9. I converted 9 to 144/16 to add it to 9/16, resulting in 153/16. Taking the square root gave me sqrt(153)/4, which I simplified by factoring out 9 from 153. That gave me 3*sqrt(17)/4. Everything seems to check out. I don't think I made any calculation errors. So, I'm confident that the modulus of 3/4 minus 3i is 3√17 over 4.

💡Okay, so I have to figure out the truth values for these compound statements: "p or q," "p and q," and "not p" for two different sets of propositions. Let me take it step by step.Starting with the first set:1. **p: The equation (x^2 + 1 = 0) has real roots.** Hmm, I remember that for a quadratic equation (ax^2 + bx + c = 0), the discriminant is (b^2 - 4ac). If the discriminant is positive, there are two real roots; if it's zero, there's one real root; and if it's negative, the roots are complex or imaginary. In this case, the equation is (x^2 + 1 = 0), so (a = 1), (b = 0), and (c = 1). Plugging into the discriminant: (0^2 - 4(1)(1) = -4). Since the discriminant is negative, there are no real roots. So, p is false.2. **q: The two roots of the equation (x^2 - 1 = 0) are equal.** Let's solve (x^2 - 1 = 0). Factoring, we get ((x - 1)(x + 1) = 0), so the roots are (x = 1) and (x = -1). These roots are not equal; they are distinct. Therefore, q is false.Wait, that doesn't seem right. The problem says "the two roots are equal." But in this case, they are not equal. So, q is false. Hmm, but earlier I thought q was true because the roots are real. Maybe I confused something. Let me double-check.No, the statement specifically says "the two roots are equal." Since 1 and -1 are not equal, q is indeed false.Okay, so for the first set:- p is false- q is falseNow, let's evaluate the compound statements:- **p or q**: This is true if either p is true or q is true. Since both are false, "p or q" is false.- **p and q**: This is true only if both p and q are true. Since both are false, "p and q" is false.- **not p**: This is the negation of p. Since p is false, "not p" is true.Wait, but earlier I thought q was true. Did I make a mistake? Let me check again.The equation (x^2 - 1 = 0) has roots 1 and -1, which are not equal. So, q is false. Okay, so my initial thought was wrong. I thought q was true because the roots are real, but the statement is about equality, not reality. So, q is false.Therefore, for the first set:- p or q: False- p and q: False- not p: TrueNow, moving on to the second set:1. **p: The base angles of an isosceles triangle are equal.** I recall that in an isosceles triangle, the base angles are equal. That's one of the fundamental properties. So, p is true.2. **q: An isosceles triangle is an acute triangle.** Hmm, an acute triangle has all angles less than 90 degrees. An isosceles triangle can be acute, right, or obtuse depending on the angles. For example, an isosceles triangle with two 80-degree angles and one 20-degree angle is acute. But if the two equal angles are 100 degrees each, then it's obtuse. So, q is not necessarily true. Therefore, q is false.So, for the second set:- p is true- q is falseNow, evaluating the compound statements:- **p or q**: This is true if either p is true or q is true. Since p is true, "p or q" is true.- **p and q**: This is true only if both p and q are true. Since q is false, "p and q" is false.- **not p**: This is the negation of p. Since p is true, "not p" is false.Wait, but I thought q was false because an isosceles triangle isn't necessarily acute. Is that correct? Let me think again.Yes, an isosceles triangle can be acute, right, or obtuse. So, it's not always acute. Therefore, q is false.Therefore, for the second set:- p or q: True- p and q: False- not p: FalseI think that's it. Let me summarize:For the first set:- p or q: False- p and q: False- not p: TrueFor the second set:- p or q: True- p and q: False- not p: FalseI hope I didn't mix up anything. It's important to carefully check each statement, especially the meaning of "equal roots" and the nature of isosceles triangles.

💡Okay, so I have this economics problem here, and I need to figure it out step by step. Let me try to understand what it's asking.First, the problem defines something called the marginal function. It says that for a function f(x), the marginal function Mf(x) is equal to f(x + 1) minus f(x). Hmm, that makes sense. I think in economics, marginal usually refers to the change when you increase the quantity by one unit. So, marginal profit would be the additional profit from producing one more unit, right?Alright, moving on. The company has a revenue function R(x) which is 3000x minus 20x squared, and a cost function C(x) which is 500x plus 4000. Both of these are in yuan, and x is a natural number greater than zero, meaning they can't produce a fraction of a unit. Also, they don't produce more than 100 units per month. So, x is between 1 and 100.The first part of the problem asks me to find the profit function P(x) and its marginal profit function MP(x). Okay, profit is generally revenue minus cost, so P(x) should be R(x) minus C(x). Let me write that down.So, P(x) = R(x) - C(x) = (3000x - 20x²) - (500x + 4000). Let me simplify that.First, distribute the negative sign to both terms in C(x): 3000x - 20x² - 500x - 4000. Now, combine like terms. 3000x minus 500x is 2500x. So, P(x) = -20x² + 2500x - 4000. That looks right.Now, for the marginal profit function MP(x). According to the definition, MP(x) is P(x + 1) minus P(x). So, I need to compute P(x + 1) first.Let me substitute x + 1 into P(x):P(x + 1) = -20(x + 1)² + 2500(x + 1) - 4000.Let me expand that step by step. First, (x + 1) squared is x² + 2x + 1. So, multiplying by -20: -20x² - 40x - 20.Then, 2500 times (x + 1) is 2500x + 2500.So, putting it all together:P(x + 1) = (-20x² - 40x - 20) + (2500x + 2500) - 4000.Now, combine like terms:-20x² remains.For the x terms: -40x + 2500x = 2460x.For the constants: -20 + 2500 - 4000 = (-20 - 4000) + 2500 = (-4020) + 2500 = -1520.So, P(x + 1) = -20x² + 2460x - 1520.Now, to find MP(x), subtract P(x) from P(x + 1):MP(x) = P(x + 1) - P(x) = (-20x² + 2460x - 1520) - (-20x² + 2500x - 4000).Let me distribute the negative sign:= -20x² + 2460x - 1520 + 20x² - 2500x + 4000.Now, combine like terms:-20x² + 20x² cancels out.2460x - 2500x = -40x.-1520 + 4000 = 2480.So, MP(x) = -40x + 2480.Wait, that seems straightforward. So, the marginal profit function is linear, decreasing as x increases, which makes sense because the profit function is quadratic with a negative coefficient on x², so it's a downward opening parabola, meaning the marginal profit decreases as x increases.Okay, so that's part (1). I think I did that correctly. Let me double-check my calculations.Starting with P(x) = R(x) - C(x) = (3000x - 20x²) - (500x + 4000) = 3000x - 20x² - 500x - 4000 = 2500x - 20x² - 4000. Yep, that's correct.Then, P(x + 1) = -20(x + 1)² + 2500(x + 1) - 4000.Calculating (x + 1)²: x² + 2x + 1. Multiply by -20: -20x² - 40x - 20.2500(x + 1) = 2500x + 2500.So, P(x + 1) = (-20x² - 40x - 20) + (2500x + 2500) - 4000.Combine terms: -20x² + (2500x - 40x) + (-20 + 2500 - 4000).Which is -20x² + 2460x - 1520. Correct.Subtracting P(x): (-20x² + 2460x - 1520) - (-20x² + 2500x - 4000) = (-20x² + 2460x - 1520) + 20x² - 2500x + 4000.Simplify: (-20x² + 20x²) + (2460x - 2500x) + (-1520 + 4000) = 0 - 40x + 2480. So, MP(x) = 2480 - 40x. Yep, that's correct.Alright, so part (1) is done. Now, moving on to part (2): Find the difference between the maximum value of the profit function and the maximum value of the marginal profit function.Hmm, okay. So, I need to find the maximum of P(x) and the maximum of MP(x), then subtract them.First, let's find the maximum of the profit function P(x). P(x) is a quadratic function: -20x² + 2500x - 4000. Since the coefficient of x² is negative, it opens downward, so its maximum is at the vertex.The vertex of a quadratic ax² + bx + c is at x = -b/(2a). So, here, a = -20, b = 2500.So, x = -2500/(2*(-20)) = -2500/(-40) = 62.5.But x has to be a natural number, so x can be 62 or 63. Since 62.5 is halfway, we need to check both x=62 and x=63 to see which gives a higher profit.Wait, but the company can produce up to 100 units, so 62 and 63 are both within the range.Let me compute P(62) and P(63).First, P(62):P(62) = -20*(62)^2 + 2500*62 - 4000.Compute 62 squared: 62*62. Let's see, 60^2 is 3600, 2*60*2=240, and 2^2=4, so (60+2)^2 = 3600 + 240 + 4 = 3844.So, -20*3844 = -76,880.2500*62: 2500*60=150,000; 2500*2=5,000; so total 155,000.Then, subtract 4000: 155,000 - 4,000 = 151,000.So, P(62) = -76,880 + 151,000 = 74,120 yuan.Now, P(63):63 squared is 3969.-20*3969 = -79,380.2500*63: 2500*60=150,000; 2500*3=7,500; total 157,500.Subtract 4000: 157,500 - 4,000 = 153,500.So, P(63) = -79,380 + 153,500 = 74,120 yuan.Wait, both P(62) and P(63) are equal? That's interesting. So, the maximum profit is 74,120 yuan, achieved at both x=62 and x=63.So, the maximum value of P(x) is 74,120.Now, moving on to the marginal profit function MP(x) = 2480 - 40x. This is a linear function with a negative slope, so it's decreasing. Therefore, its maximum value occurs at the smallest x, which is x=1.So, MP(1) = 2480 - 40*1 = 2480 - 40 = 2440 yuan.Wait, but hold on. Is that correct? Because marginal profit is the additional profit from producing one more unit. So, it's the change in profit when increasing x by 1. So, the maximum marginal profit would indeed be at the smallest x, since it's decreasing.But just to make sure, let's compute MP(1) and MP(2):MP(1) = 2480 - 40*1 = 2440.MP(2) = 2480 - 40*2 = 2400.Yes, it's decreasing by 40 each time. So, the maximum marginal profit is indeed 2440 yuan at x=1.So, now, the problem asks for the difference between the maximum value of the profit function and the maximum value of the marginal profit function.So, that would be 74,120 - 2,440 = ?Let me compute that: 74,120 minus 2,440.74,120 - 2,000 = 72,120.72,120 - 440 = 71,680.So, the difference is 71,680 yuan.Wait, but let me make sure I didn't make a mistake in the calculations.Wait, P(x) at x=62 and x=63 is 74,120. That seems correct.MP(x) at x=1 is 2440. Correct.Difference: 74,120 - 2,440 = 71,680.Yes, that seems right.But just to double-check, let me recalculate P(62):P(62) = -20*(62)^2 + 2500*62 - 4000.62 squared is 3844.-20*3844: 3844*20 is 76,880, so negative is -76,880.2500*62: 2500*60=150,000; 2500*2=5,000; total 155,000.155,000 - 4000 = 151,000.So, P(62) = -76,880 + 151,000 = 74,120. Correct.Similarly, P(63):63 squared is 3969.-20*3969 = -79,380.2500*63: 2500*60=150,000; 2500*3=7,500; total 157,500.157,500 - 4000 = 153,500.So, P(63) = -79,380 + 153,500 = 74,120. Correct.MP(1) = 2480 - 40*1 = 2440. Correct.Difference: 74,120 - 2,440 = 71,680. Correct.So, the difference is 71,680 yuan.Therefore, the answer to part (2) is 71,680.Wait, but hold on a second. The problem says "the difference between the maximum value of the profit function and the maximum value of the marginal profit function." So, is it 74,120 - 2,440 or 2,440 - 74,120? But since it's asking for the difference, it's the absolute value, so it's 71,680 regardless.But in the context, since the profit function's maximum is much larger, it's 71,680.So, I think that's correct.Just to recap:1. Profit function P(x) = -20x² + 2500x - 4000.2. Marginal profit function MP(x) = 2480 - 40x.3. Maximum P(x) is 74,120 at x=62 and x=63.4. Maximum MP(x) is 2,440 at x=1.5. Difference is 74,120 - 2,440 = 71,680.Yes, that all adds up.I think I've got it right.**Final Answer**The difference between the maximum value of the profit function and the maximum value of the marginal profit function is boxed{71680}.

💡First, identify the initial number of bales in the barn, which is 28.Next, note the total number of bales after Tim added more, which is 54.To find out how many bales Tim added, subtract the initial number of bales from the total number of bales.So, 54 minus 28 equals 26.Therefore, Tim stored 26 bales of hay in the barn.