answer:To solve this problem, we first introduce a new function g(x) defined as g(x) = frac{f(x)}{x}. The derivative of g(x), denoted as g'(x), can be calculated using the quotient rule or by recognizing it as a product of f(x) and x^{-1}. This gives us: [g'(x) = frac{xf'(x) - f(x)}{x^2}.] Given that xf'(x) - f(x) < 0 for x > 0, we can infer that g'(x) < 0 for x > 0. This means that g(x) is a decreasing function on the interval (0, +infty). Next, we evaluate the given expressions for a, b, and c in terms of g(x): - a = g(2^{0.2}), - b = g(0.2^2), - c = g(log_{2}5). To compare a, b, and c, we need to compare their corresponding x values: 2^{0.2}, 0.2^2, and log_{2}5. First, note that log_{2}5 > log_{2}4 = 2 because 5 > 4. This gives us log_{2}5 > 2. Second, we observe that 1 < 2^{0.2} < 2. This is because 2^{0.2} is the fifth root of 2, which is greater than 1 but less than 2. Third, we have 0.2^2 = 0.04, which is clearly less than 1, and thus less than both 2^{0.2} and log_{2}5. Putting these observations together, we have log_{2}5 > 2^{0.2} > 0.2^2, which implies that g(log_{2}5) < g(2^{0.2}) < g(0.2^2) due to the decreasing nature of g(x) on (0, +infty). Therefore, we conclude that c < a < b. This corresponds to choice C. Thus, the correct answer is boxed{C}.

answer:1. First, we note that each term of the given product can be simplified using the factorization: [ left(1 - frac{1}{k^2}right) = left(1 - frac{1}{k}right)left(1 + frac{1}{k}right) = frac{k-1}{k} cdot frac{k+1}{k} ] 2. Applying this to each term in the product: [ left(1 - frac{1}{4}right)left(1 - frac{1}{9}right)left(1 - frac{1}{16}right) cdots left(1 - frac{1}{2021^2}right)left(1 - frac{1}{2022^2}right) ] we have: [ left(1 - frac{1}{4}right) = frac{2 cdot 3}{2 cdot 2}, quad left(1 - frac{1}{9}right) = frac{2 cdot 4}{3 cdot 3}, quad left(1 - frac{1}{16}right) = frac{3 cdot 5}{4 cdot 4}, ldots, left(1 - frac{1}{2022^2}right) = frac{2021 cdot 2023}{2022 cdot 2022} ] 3. Therefore, the entire product equals: [ frac{1}{2} cdot frac{3}{2} cdot frac{2}{3} cdot frac{4}{3} cdot frac{3}{4} cdot frac{5}{4} cdots frac{2020}{2021} cdot frac{2022}{2021} cdot frac{2021}{2022} cdot frac{2023}{2022} ] 4. Observe that most of the fractions will cancel out, since each term ( frac{k+1}{k} ) will be followed by ( frac{k}{k+1} ). Explicitly, the sequence will look like this: [ frac{1}{2} cdot frac{3}{2} cdot frac{2}{3} cdots frac{k+1}{k} cdot frac{k}{k+1} cdots frac{2022}{2023} ] 5. After canceling out all intermediate terms, only the first term ( frac{1}{2} ) and the last term ( frac{2023}{2022} ) will remain: [ frac{1}{2} cdot frac{2023}{2022} ] 6. Simplify the remaining product: [ frac{1}{2} cdot frac{2023}{2022} = frac{2023}{4044} ] # Conclusion: The result of the given product is: [ boxed{frac{2023}{4044}} ]

answer:To find the length of the other side of the right triangle, we can use the Pythagorean theorem, which states that in a right triangle, the square of the length of the hypotenuse (c) is equal to the sum of the squares of the lengths of the other two sides (a and b). The formula is: c^2 = a^2 + b^2 In this case, we know the length of the hypotenuse (c) is 17 and the length of one of the other sides (a) is 15. We need to find the length of the other side (b). We can rearrange the formula to solve for b: b^2 = c^2 - a^2 Now we can plug in the values we know: b^2 = 17^2 - 15^2 b^2 = 289 - 225 b^2 = 64 To find the length of side b, we take the square root of 64: b = √64 b = 8 So, the length of the other side of the right triangle is boxed{8} units.

answer:To find the maximum number of small boxes that can fit into the larger box, we need to calculate the volume of both the larger box and the smaller box, and then divide the volume of the larger box by the volume of the smaller box. The volume ( V ) of a rectangular box can be calculated using the formula: [ V = text{length} times text{width} times text{height} ] Let's denote the unknown length of the larger box as ( L ). The volume of the larger box is: [ V_{text{large}} = L times 14 times 16 ] The volume of the smaller box is: [ V_{text{small}} = 3 times 7 times 2 ] Now we calculate ( V_{text{small}} ): [ V_{text{small}} = 3 times 7 times 2 = 42 ] To find the maximum number of small boxes that can fit into the larger box, we divide the volume of the larger box by the volume of the smaller box: [ text{Number of small boxes} = frac{V_{text{large}}}{V_{text{small}}} ] We are given that the maximum number of small boxes is 64, so: [ 64 = frac{L times 14 times 16}{42} ] Now we solve for ( L ): [ 64 times 42 = L times 14 times 16 ] [ 2688 = L times 224 ] [ L = frac{2688}{224} ] [ L = 12 ] Therefore, the length of the first dimension of the larger box is boxed{12} inches.