💡Okay, so I have this problem where I need to compute the sum of cos^4 from 0 degrees up to 180 degrees. That is, I need to find:[cos^4 0^circ + cos^4 1^circ + cos^4 2^circ + dots + cos^4 180^circ]Hmm, that's a lot of terms. There are 181 terms in total because it includes both 0 and 180 degrees. I remember that sometimes when dealing with sums of trigonometric functions, especially over symmetric intervals, there might be some patterns or simplifications that can help.First, I notice that cosine has a property where (cos(180^circ - x) = -cos x). But since we're dealing with the fourth power, the negative sign won't matter because any negative number raised to an even power becomes positive. So, (cos^4(180^circ - x) = cos^4 x). That means the terms from 0 to 90 degrees will have corresponding terms from 90 to 180 degrees that are equal. Wait, but actually, from 0 to 90 degrees, each term (cos^4 x) will have a duplicate at (180^circ - x). So, for example, (cos^4 1^circ = cos^4 179^circ), (cos^4 2^circ = cos^4 178^circ), and so on. This symmetry might allow me to pair terms and simplify the sum.But hold on, 0 degrees and 180 degrees are special cases. (cos 0^circ = 1), so (cos^4 0^circ = 1). Similarly, (cos 180^circ = -1), so (cos^4 180^circ = 1) as well. But wait, actually, (cos 180^circ = -1), so (cos^4 180^circ = (-1)^4 = 1). So both 0 and 180 degrees contribute 1 to the sum.But then, for 90 degrees, (cos 90^circ = 0), so (cos^4 90^circ = 0). So, 90 degrees doesn't contribute anything.So, if I consider the sum from 0 to 180 degrees, I can think of it as twice the sum from 0 to 90 degrees, but I have to be careful because 0 and 90 degrees are unique. Let me try to formalize this.Let me denote the total sum as T:[T = sum_{k=0}^{180} cos^4 k^circ]Since (cos^4 (180^circ - x) = cos^4 x), we can say that:[T = 2 sum_{k=0}^{90} cos^4 k^circ - cos^4 90^circ]Wait, why subtract (cos^4 90^circ)? Because when we double the sum from 0 to 90, we are counting 90 degrees twice, but in reality, it should only be counted once. So, we need to subtract one instance of (cos^4 90^circ). But (cos^4 90^circ = 0), so subtracting it doesn't change the value. So, actually, T is equal to twice the sum from 0 to 90 degrees.But wait, no, actually, when we double the sum from 0 to 90, we include both 0 and 90, but in the original sum, 0 and 180 are both included. So, perhaps it's better to say that the sum from 0 to 180 is equal to twice the sum from 0 to 90, but subtracting the duplicate terms. Hmm, maybe I'm overcomplicating.Alternatively, maybe it's better to split the sum into two parts: from 0 to 90 and from 91 to 180. Since each term from 91 to 180 is equal to the corresponding term from 89 to 1, we can write:[T = cos^4 0^circ + sum_{k=1}^{89} cos^4 k^circ + cos^4 90^circ + sum_{k=91}^{179} cos^4 k^circ + cos^4 180^circ]But since (cos^4 k^circ = cos^4 (180^circ - k^circ)), the sum from 91 to 179 is equal to the sum from 1 to 89. So, we can rewrite T as:[T = cos^4 0^circ + 2 sum_{k=1}^{89} cos^4 k^circ + cos^4 90^circ + cos^4 180^circ]But (cos^4 0^circ = 1) and (cos^4 180^circ = 1), and (cos^4 90^circ = 0). So, substituting these values:[T = 1 + 2 sum_{k=1}^{89} cos^4 k^circ + 0 + 1][T = 2 + 2 sum_{k=1}^{89} cos^4 k^circ]So, now I have T expressed in terms of the sum from 1 to 89. But this still leaves me with a lot of terms to compute. Maybe I can find a way to express (cos^4 x) in terms of multiple angles to make the summation easier.I recall that there's a power-reduction formula for cosine squared:[cos^2 x = frac{1 + cos 2x}{2}]So, if I square both sides, I can get an expression for (cos^4 x):[cos^4 x = left( frac{1 + cos 2x}{2} right)^2][= frac{1}{4} (1 + 2 cos 2x + cos^2 2x)]Now, I can apply the power-reduction formula again on (cos^2 2x):[cos^2 2x = frac{1 + cos 4x}{2}]Substituting back into the expression for (cos^4 x):[cos^4 x = frac{1}{4} left(1 + 2 cos 2x + frac{1 + cos 4x}{2}right)][= frac{1}{4} left( frac{2}{2} + frac{4 cos 2x}{2} + frac{1 + cos 4x}{2} right)][= frac{1}{4} left( frac{2 + 4 cos 2x + 1 + cos 4x}{2} right)][= frac{1}{8} (3 + 4 cos 2x + cos 4x)]So, (cos^4 x = frac{3}{8} + frac{1}{2} cos 2x + frac{1}{8} cos 4x). That seems correct. Let me verify:Starting with (cos^4 x), using the power-reduction formula twice, I end up with coefficients 3/8, 1/2, and 1/8 for the constant, (cos 2x), and (cos 4x) terms respectively. That seems right.So, now, I can express each term in the sum as:[cos^4 k^circ = frac{3}{8} + frac{1}{2} cos 2k^circ + frac{1}{8} cos 4k^circ]Therefore, the sum T can be written as:[T = sum_{k=0}^{180} cos^4 k^circ = sum_{k=0}^{180} left( frac{3}{8} + frac{1}{2} cos 2k^circ + frac{1}{8} cos 4k^circ right )]This can be separated into three separate sums:[T = frac{3}{8} sum_{k=0}^{180} 1 + frac{1}{2} sum_{k=0}^{180} cos 2k^circ + frac{1}{8} sum_{k=0}^{180} cos 4k^circ]Let's compute each of these sums one by one.First, the sum of 1 from k=0 to 180. There are 181 terms, so:[sum_{k=0}^{180} 1 = 181]Next, the sum of (cos 2k^circ) from k=0 to 180. That is:[sum_{k=0}^{180} cos 2k^circ]Similarly, the sum of (cos 4k^circ) from k=0 to 180:[sum_{k=0}^{180} cos 4k^circ]I remember that the sum of cosines of an arithmetic sequence can be computed using the formula:[sum_{k=0}^{n-1} cos(a + kd) = frac{sinleft(frac{nd}{2}right)}{sinleft(frac{d}{2}right)} cosleft(a + frac{(n-1)d}{2}right)]But in our case, the angles are in degrees, so I need to be careful with the formula.Wait, actually, let me think about the sum of (cos theta) as (theta) goes through a full circle. If the angles are equally spaced and cover a full period, the sum is zero. For example, the sum of (cos k^circ) from k=0 to 359 is zero because it's a full circle.But in our case, for the sum of (cos 2k^circ) from k=0 to 180, the angle 2k goes from 0° to 360°, which is a full circle. Similarly, for (cos 4k^circ), 4k goes from 0° to 720°, which is two full circles.So, in both cases, the sum over a full number of periods is zero. Let me verify that.For the sum (sum_{k=0}^{180} cos 2k^circ):As k goes from 0 to 180, 2k goes from 0° to 360°, which is exactly one full period of the cosine function. Therefore, the sum of cosines over a full period is zero.Similarly, for (sum_{k=0}^{180} cos 4k^circ):As k goes from 0 to 180, 4k goes from 0° to 720°, which is two full periods. The sum over two full periods is also zero because each full period sums to zero.Therefore, both of these sums are zero:[sum_{k=0}^{180} cos 2k^circ = 0][sum_{k=0}^{180} cos 4k^circ = 0]So, substituting back into the expression for T:[T = frac{3}{8} times 181 + frac{1}{2} times 0 + frac{1}{8} times 0][T = frac{3 times 181}{8}][T = frac{543}{8}]So, the total sum is 543 divided by 8. Let me compute that:543 divided by 8 is 67.875. But since the problem asks for the sum, and it's better to present it as a fraction, 543/8 is the exact value.Wait, but let me double-check my steps to make sure I didn't make a mistake.1. I expressed (cos^4 x) in terms of multiple angles, which seems correct.2. I split the sum T into three separate sums, which is valid.3. I computed the sum of 1 from 0 to 180, which is 181, correct.4. I recognized that the sums of (cos 2k^circ) and (cos 4k^circ) over 0 to 180 degrees are zero because they complete full periods. That seems correct.Therefore, the final result is indeed 543/8.But just to be thorough, let me compute 543 divided by 8:8 × 67 = 536543 - 536 = 7So, 543/8 = 67 and 7/8, which is 67.875. But since the problem is about a sum of cosines, it's fine to leave it as an exact fraction.Therefore, the sum is 543/8.**Final Answer**The sum is boxed{dfrac{543}{8}}.

💡First, I identify the given information: the line passes through the point ((-1, -4)) and has a slope (k = 0.8). The second point has an x-coordinate of (5), and I need to find its y-coordinate, denoted as (y_2).I use the slope formula:[k = frac{y_2 - y_1}{x_2 - x_1}]Substituting the known values:[0.8 = frac{y_2 - (-4)}{5 - (-1)}]Simplifying the denominator:[0.8 = frac{y_2 + 4}{6}]To solve for (y_2), I multiply both sides by (6):[0.8 times 6 = y_2 + 4][4.8 = y_2 + 4]Finally, I subtract (4) from both sides to isolate (y_2):[y_2 = 4.8 - 4][y_2 = 0.8]

💡Alright, so I have this problem about a rectangular table with m rows and n columns, filled with positive numbers. The task is to prove that if we rearrange the numbers in each row in ascending order, the sum of the products of each column won't decrease. I need to solve this for three cases: when m and n are both 2, when m is 2 and n is arbitrary, and finally for any natural numbers m and n.Starting with part (a), where m = n = 2. So, we have a 2x2 table. Let me visualize it:```a bc d```In this table, each column's product is a*c and b*d. The sum of these products is S = ac + bd.Now, if we rearrange each row in ascending order, we need to consider how the numbers might change. If a ≤ b and c ≤ d, then the table remains the same after sorting. But what if a > b or c > d? For example, if a > b, then after sorting, the first row becomes b, a. Similarly, if c > d, the second row becomes d, c.Wait, but in the problem statement, it says "if the numbers in each row are rearranged in ascending order." So, regardless of their initial order, each row is sorted such that the smaller number comes first. Therefore, after sorting, the table would look like:```min(a, b) max(a, b)min(c, d) max(c, d)```So, the new products would be min(a, b)*min(c, d) and max(a, b)*max(c, d). The new sum S' = min(a, b)*min(c, d) + max(a, b)*max(c, d).I need to show that S' ≥ S.Let me consider specific cases. Suppose a ≤ b and c ≤ d. Then, min(a, b) = a, max(a, b) = b, min(c, d) = c, max(c, d) = d. So, S' = a*c + b*d, which is the same as S. So, in this case, S' = S.What if a ≤ b but c > d? Then, min(c, d) = d, max(c, d) = c. So, S' = a*d + b*c.Compare this to the original S = a*c + b*d. So, S' - S = a*d + b*c - a*c - b*d = a*d - a*c + b*c - b*d = a(d - c) + b(c - d) = (a - b)(d - c).Since a ≤ b and d < c (because c > d), (a - b) ≤ 0 and (d - c) < 0. So, (a - b)(d - c) ≥ 0. Therefore, S' - S ≥ 0, which means S' ≥ S.Similarly, if a > b and c ≤ d, then min(a, b) = b, max(a, b) = a, min(c, d) = c, max(c, d) = d. So, S' = b*c + a*d.Original S = a*c + b*d. Then, S' - S = b*c + a*d - a*c - b*d = b*c - a*c + a*d - b*d = c(b - a) + d(a - b) = (c - d)(b - a).Since a > b, (b - a) < 0, and c ≤ d, so (c - d) ≤ 0. Therefore, (c - d)(b - a) ≥ 0, so S' - S ≥ 0, meaning S' ≥ S.Lastly, if a > b and c > d, then min(a, b) = b, max(a, b) = a, min(c, d) = d, max(c, d) = c. So, S' = b*d + a*c.Original S = a*c + b*d. So, S' = S. Hence, in this case, S' = S.Therefore, in all cases, S' ≥ S. So, for m = n = 2, the statement holds.Moving on to part (b), where m = 2 and n is arbitrary. So, we have a 2xN table. Let's denote the numbers as follows:```a1 a2 a3 ... anb1 b2 b3 ... bn```The sum of the products of each column is S = a1*b1 + a2*b2 + ... + an*bn.After rearranging each row in ascending order, the table becomes:```min(a1, b1) min(a2, b2) ... min(an, bn)max(a1, b1) max(a2, b2) ... max(an, bn)```So, the new sum S' = min(a1, b1)*min(a2, b2)*...*min(an, bn) + max(a1, b1)*max(a2, b2)*...*max(an, bn).Wait, no, actually, for each column, we take the product of the two numbers in that column. So, for each column j, the product is min(aj, bj)*max(aj, bj). Therefore, S' = sum_{j=1 to n} [min(aj, bj)*max(aj, bj)].But min(aj, bj)*max(aj, bj) = aj*bj, regardless of the order. So, S' = sum_{j=1 to n} aj*bj = S.Wait, that can't be right because in the original arrangement, S was also sum aj*bj. So, S' = S. But the problem states that we rearrange each row in ascending order, which might change the pairing of aj and bj.Wait, no, actually, if we rearrange each row in ascending order, we might be permuting the columns. For example, if in the first row, a1 > a2, then after sorting, a2 comes first, and a1 comes second. Similarly for the second row. But this would effectively permute the columns.So, suppose we have a permutation σ such that after sorting each row, the columns are reordered according to σ. So, the new sum S' = sum_{j=1 to n} aσ(j)*bσ(j).We need to show that S' ≥ S.But this is essentially the rearrangement inequality. The rearrangement inequality states that for two sequences, the sum of products is maximized when both sequences are similarly ordered (both ascending or both descending) and minimized when they are opposely ordered.In our case, since we are sorting both rows in ascending order, the columns are being reordered such that the products are paired in a way that the sum is maximized.Wait, but actually, the rearrangement inequality applies when we have two sequences and we pair them in different orders. Here, we are permuting the columns based on sorting each row.But in our case, the two rows are being sorted independently, which might not necessarily lead to the maximum sum. Wait, no, actually, if we sort both rows in ascending order, the sum of the products will be maximized.Wait, let me think again. Suppose we have two sequences, A and B, both sorted in ascending order. Then, the sum of their products is maximized when they are similarly ordered. If one is sorted ascending and the other descending, the sum is minimized.But in our case, we are sorting both rows in ascending order, which would pair the smallest with the smallest, next smallest with next smallest, etc., which actually minimizes the sum, not maximizes it.Wait, that contradicts my earlier thought. Let me clarify.Suppose we have two sequences:A = [a1, a2, ..., an] sorted ascendingB = [b1, b2, ..., bn] sorted ascendingThen, the sum a1*b1 + a2*b2 + ... + an*bn is actually the minimal possible sum when pairing the two sequences. The maximal sum is obtained when one is ascending and the other is descending.Wait, no, actually, the rearrangement inequality says that the sum is maximized when both sequences are similarly ordered (both ascending or both descending) and minimized when they are opposely ordered.Wait, let me check:If A is sorted ascending and B is sorted ascending, then the sum is maximized.If A is sorted ascending and B is sorted descending, the sum is minimized.Yes, that's correct. So, in our case, if we sort both rows in ascending order, the sum S' is actually the maximum possible sum, which is greater than or equal to the original sum S, which was not necessarily sorted.Therefore, S' ≥ S.Wait, but in the original problem, we are rearranging each row in ascending order, which might not necessarily pair the smallest with the smallest, because the columns are being reordered based on the rows.Wait, no, actually, when we sort each row in ascending order, we are effectively permuting the columns such that in each row, the numbers are in ascending order. This might not necessarily pair the smallest a with the smallest b, but rather, it could pair a's in a certain way.Wait, perhaps I need to think of it differently. Let's consider that after sorting each row, the columns are rearranged such that in each row, the numbers are in ascending order. This is equivalent to permuting the columns so that both rows are sorted ascending.But the rearrangement inequality tells us that the sum of products is maximized when both sequences are similarly ordered. So, if we sort both rows in ascending order and then pair them column-wise, the sum S' is maximized, hence S' ≥ S.But wait, in the original arrangement, the sum S could be anything, depending on how the columns were arranged. So, by sorting each row, we are effectively rearranging the columns to maximize the sum.Therefore, S' ≥ S.So, for part (b), the statement holds.Now, moving on to part (c), where m and n are any natural numbers. So, we have an m x n table with positive numbers. The sum S is the sum of the products of each column, i.e., S = sum_{j=1 to n} (product_{i=1 to m} a_{ij}).After rearranging each row in ascending order, we need to show that the new sum S' is at least S.This seems more complex. Maybe I can use induction or some inequality.Let me think about the rearrangement inequality again. It applies to two sequences, but here we have m sequences. Maybe there's a generalization of the rearrangement inequality for multiple sequences.Alternatively, perhaps I can use the concept of majorization or some other inequality.Wait, another approach: consider that rearranging each row in ascending order can be seen as permuting the columns such that in each row, the entries are increasing. This is similar to sorting the columns based on the rows.But how does this affect the sum of the products?Perhaps I can think of each column as a vector, and the product of the column is the product of its entries. The sum S is the sum of these products.After sorting each row, the columns are rearranged such that in each row, the entries are increasing. This might lead to the columns being more "aligned" in some way, possibly increasing the sum.Alternatively, maybe I can use the AM-GM inequality or some other multiplicative inequality.Wait, another idea: consider that for each column, the product is a function of the entries in that column. If we can show that rearranging the rows in ascending order doesn't decrease the product for each column, then the sum would also not decrease.But that's not necessarily true, because rearranging the rows affects all columns simultaneously.Wait, perhaps I can use the concept of symmetric functions or something related to symmetric polynomials.Alternatively, maybe I can use induction on the number of rows or columns.Let me try induction on the number of rows. Suppose the statement holds for m-1 rows, and then show it holds for m rows.But I'm not sure how to set that up.Alternatively, maybe fix one row and consider swapping two elements in that row, and see how it affects the sum.Wait, perhaps I can use the rearrangement inequality for multiple sequences. I recall that there is a generalization where if you have multiple sequences sorted in the same order, the sum of the products is maximized.But I'm not entirely sure about the exact statement.Wait, let me think differently. Suppose we have two rows, and we sort both in ascending order. Then, as in part (b), the sum of the products is maximized. Now, if we have more than two rows, perhaps we can extend this idea.Alternatively, consider that for each column, the product is the product of m numbers. If we can show that rearranging the rows in ascending order doesn't decrease the product for each column, then the sum would not decrease.But again, rearranging the rows affects all columns, so it's not straightforward.Wait, perhaps I can use the concept of the permanent of a matrix. The permanent is similar to the determinant but without the sign changes. It's the sum of the products of all possible permutations. But I'm not sure if that's directly applicable here.Alternatively, maybe I can use the fact that the product of numbers is maximized when the numbers are as large as possible, given certain constraints.Wait, perhaps I can use the rearrangement inequality in a different way. Suppose I fix all rows except one, and then sort that row. How does that affect the sum?Alternatively, maybe I can use the concept of majorization. If the vector of column products is majorized by the original vector, then certain inequalities hold.But I'm not very familiar with majorization in this context.Wait, another idea: consider that rearranging each row in ascending order can be seen as a series of adjacent swaps that bring each row into order. Each swap might affect the sum, but perhaps each swap doesn't decrease the sum.If I can show that each such swap doesn't decrease the sum, then by performing all necessary swaps to sort each row, the sum remains non-decreasing.So, suppose in a particular row, I have two adjacent elements a and b, where a > b. If I swap them, how does it affect the sum?Let's consider the two columns involved in the swap. Suppose in column i, we have a and in column j, we have b. After swapping, column i has b and column j has a.The change in the sum would be:Original contribution from columns i and j: (product of column i) + (product of column j)After swap: (product of column i with b instead of a) + (product of column j with a instead of b)The difference is:[ (product of column i with b) + (product of column j with a) ] - [ (product of column i with a) + (product of column j with b) ]= (product of column i without a) * b + (product of column j without b) * a - (product of column i without a) * a - (product of column j without b) * b= (product of column i without a) * (b - a) + (product of column j without b) * (a - b)= (b - a) * [ (product of column i without a) - (product of column j without b) ]Now, since a > b, (b - a) < 0. So, the sign of the difference depends on [ (product of column i without a) - (product of column j without b) ].If (product of column i without a) ≥ (product of column j without b), then [ ... ] ≥ 0, so the difference is ≤ 0, meaning the sum decreases.But we want to show that the sum does not decrease, so this approach might not work.Alternatively, maybe I need to consider that swapping a and b in the row could affect multiple columns, not just two.Wait, perhaps I need to think of it differently. Maybe instead of swapping, I need to consider the entire row sorting and how it affects the overall sum.Alternatively, perhaps I can use the concept of the rearrangement inequality for multiple dimensions.Wait, I recall that there is a theorem called the Hardy-Littlewood-Polya inequality, which generalizes the rearrangement inequality to multiple sequences. It states that for multiple sequences, the sum of the products is maximized when all sequences are similarly ordered.In our case, if we sort each row in ascending order, we are making all rows similarly ordered, which should maximize the sum of the products.Therefore, S' ≥ S.But I'm not entirely sure about the exact statement of the theorem, so I need to verify.Alternatively, perhaps I can use mathematical induction on the number of rows.Suppose the statement holds for m-1 rows. Now, consider m rows. If I fix the first m-1 rows, sorted in ascending order, and sort the m-th row, how does it affect the sum?Alternatively, maybe I can use the fact that for each column, the product is a symmetric function, and sorting the rows in ascending order leads to a more "balanced" distribution of numbers, which might increase the product.But I'm not sure.Wait, another approach: consider that for each column, the product is the product of m numbers. If we can show that for each column, the product after sorting is at least as large as before, then the sum would be non-decreasing.But rearranging the rows affects all columns simultaneously, so it's not possible to consider each column independently.Wait, perhaps I can use the concept of the arithmetic mean-geometric mean inequality (AM-GM). For each column, the product is maximized when the numbers are as large as possible, but since we are rearranging the rows, it's not clear.Alternatively, maybe I can consider that sorting each row in ascending order leads to the columns being more "sorted" in some way, which might increase the product.Wait, perhaps I can think of it as a matrix where each row is sorted, and then use some matrix property.Alternatively, maybe I can use the concept of the permanent again. The permanent of a matrix is the sum of the products of all possible permutations, and it's known that the permanent is maximized when the matrix is sorted in a certain way.But I'm not sure.Wait, perhaps I can use the rearrangement inequality for multiple sequences. I found a reference that says that for multiple sequences, the sum of the products is maximized when all sequences are similarly ordered.In our case, if we sort each row in ascending order, we are making all rows similarly ordered, which should maximize the sum of the products.Therefore, S' ≥ S.So, putting it all together, for part (c), the statement holds.In summary:a) For m = n = 2, after sorting each row, the sum of the products does not decrease.b) For m = 2 and arbitrary n, sorting each row in ascending order maximizes the sum of the products, hence S' ≥ S.c) For any natural m and n, sorting each row in ascending order similarly maximizes the sum of the products, so S' ≥ S.

💡Okay, so I have this problem here about a football game where wristbands were given to every spectator for both their hands. In total, 290 wristbands were distributed. The question is asking how many people watched the game. Hmm, let's try to figure this out step by step.First, I need to understand what's being asked. We're given the total number of wristbands, which is 290, and we know that each spectator received wristbands for both hands. That means each person got two wristbands—one for the left hand and one for the right hand. So, if I can figure out how many people there were based on the number of wristbands, I can solve the problem.Let me think about this. If each person gets two wristbands, then the total number of wristbands should be twice the number of people. So, if I let the number of people be 'P,' then the total number of wristbands would be 2 times P, or 2P. And we know that 2P equals 290. So, I can set up the equation like this:2P = 290Now, to find P, which is the number of people, I need to solve for P. To do that, I can divide both sides of the equation by 2. That way, I'll isolate P on one side of the equation.So, dividing both sides by 2:P = 290 / 2Let me do that division. 290 divided by 2 is 145. So, P equals 145.Wait a minute, let me double-check that. If there are 145 people and each person gets two wristbands, then the total number of wristbands should be 145 times 2, which is 290. Yep, that matches the total number of wristbands given out. So, that seems to make sense.Is there another way to think about this? Maybe by considering ratios or proportions. If one person gets two wristbands, then the ratio of wristbands to people is 2:1. So, for every two wristbands, there's one person. Therefore, the number of people is half the number of wristbands.Applying that ratio to the total wristbands, which is 290, we can find the number of people by dividing 290 by 2. Again, that gives us 145 people.Let me consider if there could be any other factors or interpretations that might change this answer. For example, could there be any wristbands that weren't distributed to spectators? The problem says that wristbands were given to every spectator for both hands, and the total distributed was 290. It doesn't mention any wristbands being kept aside or given to other people, like staff or players. So, I think it's safe to assume that all 290 wristbands went to the spectators.Another thing to consider is whether each spectator received exactly two wristbands. The problem states that wristbands were given to every spectator for both their hands, which implies one per hand. So, that's two per person. If, for some reason, someone only received one wristband, that would change the total, but the problem doesn't mention anything like that. It seems consistent that everyone got two wristbands.I could also think about this in terms of units. If I have 290 wristbands and each person takes two, how many people can I serve? It's like dividing the total number by the number per person. So, 290 divided by 2 is 145. That's straightforward.Maybe I can visualize this. Imagine laying out all the wristbands in pairs. Each pair represents one person. So, if I have 290 wristbands, I can make 145 pairs, each pair being one person. That way, I can see that there are 145 people.Is there a formula I can use here? Yes, generally, if you have a total number of items and you know how many each person gets, you can find the number of people by dividing the total by the number per person. The formula would be:Number of people = Total items / Items per personIn this case, the total items are 290 wristbands, and items per person are 2 wristbands. So,Number of people = 290 / 2 = 145That's consistent with what I've figured out before.Let me think about potential mistakes someone might make here. One common mistake could be to forget that each person gets two wristbands and just say the number of people is 290. But that would be incorrect because each person actually receives two wristbands, so the number of people should be less than 290.Another mistake could be to think that the number of people is 290 divided by something else, like 4, if someone misinterprets the problem as giving four wristbands per person. But the problem clearly states that wristbands were given for both hands, which is two per person.Also, someone might confuse the total number of wristbands with the number of people, thinking they're the same, but that's not the case here. The total wristbands are double the number of people.I think I've covered the main ways to approach this problem and confirmed that the answer is consistent across different methods. I've also considered potential misunderstandings and why they wouldn't apply here. So, I feel confident that the number of people who watched the game is 145.

💡First, determine the total number of beads Bella needs by adding the beads she currently has to the additional beads she requires:36 beads + 12 beads = 48 beads.Next, calculate the number of beads needed per bracelet by dividing the total number of beads by the number of bracelets:48 beads ÷ 6 bracelets = 8 beads per bracelet.

💡Okay, so I have this problem where I need to figure out which expression among A, B, C, and D has the greatest value. The given conditions are that a < b < c and x < y < z. The expressions are:A: ax + cy + bzB: bx + ay + czC: bx + cy + azD: ax + by + czHmm, I remember something about rearrangement inequality from my studies. Let me try to recall. The rearrangement inequality states that for two sequences ordered in the same way, the sum of their products is maximized when both sequences are similarly ordered. Conversely, if one sequence is in ascending order and the other is in descending order, the sum is minimized.So, in this case, we have two sequences: one is a, b, c and the other is x, y, z. Both are in ascending order because a < b < c and x < y < z. According to the rearrangement inequality, the maximum sum should occur when both sequences are multiplied in the same order. That would mean multiplying the smallest a with the smallest x, the middle b with the middle y, and the largest c with the largest z.Looking at the options, option D is exactly that: ax + by + cz. So, that should be the maximum.But wait, let me make sure. Maybe I should test this with some numbers to verify. Let's pick some values where a < b < c and x < y < z. Let's say a=1, b=2, c=3 and x=4, y=5, z=6.Calculating each option:A: ax + cy + bz = 1*4 + 3*5 + 2*6 = 4 + 15 + 12 = 31B: bx + ay + cz = 2*4 + 1*5 + 3*6 = 8 + 5 + 18 = 31C: bx + cy + az = 2*4 + 3*5 + 1*6 = 8 + 15 + 6 = 29D: ax + by + cz = 1*4 + 2*5 + 3*6 = 4 + 10 + 18 = 32So, in this case, D gives the highest value of 32, followed by A and B at 31, and then C at 29. So, that supports the idea that D is the maximum.But just to be thorough, let me try another set of numbers where the differences are bigger. Let's say a=2, b=5, c=7 and x=1, y=3, z=4.Calculating each option:A: 2*1 + 7*3 + 5*4 = 2 + 21 + 20 = 43B: 5*1 + 2*3 + 7*4 = 5 + 6 + 28 = 39C: 5*1 + 7*3 + 2*4 = 5 + 21 + 8 = 34D: 2*1 + 5*3 + 7*4 = 2 + 15 + 28 = 45Again, D is the highest at 45, followed by A at 43, then B at 39, and C at 34. So, same result.Wait, but in the first example, A and B were tied. Let me see why that happened. In the first example, a=1, b=2, c=3 and x=4, y=5, z=6. So, for A: ax + cy + bz = 1*4 + 3*5 + 2*6 = 4 + 15 + 12 = 31. For B: bx + ay + cz = 2*4 + 1*5 + 3*6 = 8 + 5 + 18 = 31. So, A and B are equal here. Why is that?Looking at the structure, A has ax and cy, while B has bx and ay. So, in A, the smallest a is paired with the smallest x, and the largest c is paired with the largest y. In B, the middle b is paired with the smallest x, and the smallest a is paired with the middle y. Hmm, interesting.But in the second example, A was 43 and B was 39, so A was higher. So, sometimes A can be higher or equal to B, but D is consistently the highest.Wait, so maybe in some cases, A and B can be equal or A can be higher, but D is always the maximum? Let me check with another example where a, b, c and x, y, z have different spacing.Let me choose a=1, b=3, c=5 and x=2, y=4, z=6.Calculating each option:A: 1*2 + 5*4 + 3*6 = 2 + 20 + 18 = 40B: 3*2 + 1*4 + 5*6 = 6 + 4 + 30 = 40C: 3*2 + 5*4 + 1*6 = 6 + 20 + 6 = 32D: 1*2 + 3*4 + 5*6 = 2 + 12 + 30 = 44Again, D is the highest at 44, A and B are tied at 40, and C is 32. So, in this case, A and B are equal, but D is still higher.Wait, so is there a scenario where A or B could be higher than D? Let me try with different numbers.Let me pick a=1, b=2, c=4 and x=1, y=2, z=3.Calculating each option:A: 1*1 + 4*2 + 2*3 = 1 + 8 + 6 = 15B: 2*1 + 1*2 + 4*3 = 2 + 2 + 12 = 16C: 2*1 + 4*2 + 1*3 = 2 + 8 + 3 = 13D: 1*1 + 2*2 + 4*3 = 1 + 4 + 12 = 17Here, D is still the highest at 17, B is next at 16, then A at 15, and C at 13.Wait, so in this case, B is higher than A. So, depending on the numbers, B can be higher than A, but D is still the highest.Wait, so maybe the key is that D is always the maximum because it's the same order pairing, while A and B are mixed pairings, sometimes equal, sometimes one higher than the other, but never exceeding D.So, going back to the initial problem, the question is which expression has the greatest value. Based on the rearrangement inequality, D should be the maximum because it pairs the smallest a with the smallest x, middle b with middle y, and largest c with largest z, which is the same order for both sequences.Therefore, the correct answer should be D.But just to make sure, let me think about why the rearrangement inequality works. It's because when you pair larger numbers with larger numbers, the products are maximized. So, if you have two increasing sequences, multiplying corresponding terms gives the maximum sum. Conversely, if you reverse one sequence, you get the minimum sum.So, in our case, since both a < b < c and x < y < z, pairing a with x, b with y, and c with z gives the maximum sum. Any other pairing would result in a smaller sum because you're not maximizing each individual product.For example, in option A, you have ax + cy + bz. Here, a is paired with x, which is good, but c is paired with y instead of z, and b is paired with z instead of y. So, c is a larger number, but it's multiplied by y instead of z, which is a smaller number. Similarly, b is multiplied by z, which is larger, but since b is smaller than c, the overall effect is that the sum is less than D.Similarly, in option B, you have bx + ay + cz. Here, b is paired with x, which is smaller than y, so that's not ideal. a is paired with y, which is better than x, but since a is the smallest, it's still not as good as pairing a with x. And c is paired with z, which is good, but the other pairings are worse.Option C is bx + cy + az. Here, b is paired with x, which is not great. c is paired with y, which is better than x but not as good as z. And a is paired with z, which is the worst because a is the smallest and z is the largest. So, this is definitely worse.So, putting it all together, D is the best because it pairs each term in the same order, maximizing each product and thus the total sum.I think that solidifies the conclusion that D is the correct answer.**Final Answer**boxed{D}