answer：First, I need to determine how much Melody will read tomorrow for each of her classes. I'll start with her English class, where she plans to read one-third of the 60 pages assigned. Calculating one-third of 60 gives me 20 pages.Next, for her Math class, Melody needs to read 35.5 pages and intends to cover 46.5% of them. To find 46.5% of 35.5, I'll multiply 35.5 by 0.465, which equals approximately 16.5075 pages. Since she can't read a fraction of a page, I'll round this up to 17 pages.Moving on to her History class, Melody has 48 pages to read and plans to go through three-eighths of them. Calculating three-eighths of 48 results in 18 pages.For her Chinese class, she needs to cover 25 chapters and aims to read 40% of them. Finding 40% of 25 gives me 10 chapters.In her Art class, Melody has 22 illustrations to go through and plans to review 27.3% of them. Calculating 27.3% of 22 results in approximately 6.006 illustrations. Rounding this to the nearest whole number, she will read 6 illustrations.Lastly, for her Audio book class, Melody needs to listen for 150 minutes and intends to cover 75% of that time. Calculating 75% of 150 minutes gives me 112.5 minutes. Rounding this to the nearest whole number, she will spend 113 minutes on her audio book.Summarizing all these calculations, Melody will read 20 pages for English, 17 pages for Math, 18 pages for History, 10 chapters for Chinese, 6 illustrations for Art, and 113 minutes for her Audio book tomorrow.

answer：Okay, so I have this problem about constructing an open-topped rectangular pool. The volume needs to be 8 cubic meters, and the depth is 2 meters. The cost for the bottom is 120 yuan per square meter, and the walls are 80 yuan per square meter. I need to find the minimum cost to build this pool. Hmm, let's break this down step by step.First, let me visualize the pool. It's open-topped, so there's no cover on top. It's a rectangular prism, meaning it has a length, width, and height (which is the depth here). The depth is given as 2 meters, so that's fixed. The volume is length × width × depth, which is 8 cubic meters. Since depth is 2, that means length × width × 2 = 8. So, length × width = 4. That's useful because it relates the length and width.Now, I need to figure out the cost. The cost depends on the area of the bottom and the area of the walls. The bottom is just the length times the width, which we already know is 4 square meters. But wait, no, actually, since the depth is 2 meters, the bottom area is length × width, and the walls are the sides. There are four walls: two with area length × depth and two with area width × depth.So, the total area for the bottom is length × width, and the total area for the walls is 2 × (length × depth) + 2 × (width × depth). Since depth is 2, that simplifies to 2 × (length × 2) + 2 × (width × 2) = 4 × length + 4 × width.Now, the cost for the bottom is 120 yuan per square meter, so the cost for the bottom is 120 × (length × width). The cost for the walls is 80 yuan per square meter, so the cost for the walls is 80 × (4 × length + 4 × width). Therefore, the total cost is 120 × (length × width) + 80 × (4 × length + 4 × width).But we already know that length × width = 4, so the cost for the bottom is 120 × 4 = 480 yuan. So, the total cost is 480 + 80 × (4 × length + 4 × width). Let's simplify that expression. 80 × (4 × length + 4 × width) = 80 × 4 × (length + width) = 320 × (length + width). So, total cost is 480 + 320 × (length + width).Now, we need to minimize this cost. Since length × width = 4, we can express width in terms of length. Let's let length = l, so width = 4 / l. Then, the total cost becomes 480 + 320 × (l + 4 / l). So, we need to minimize the expression 320 × (l + 4 / l).To minimize this, I can use calculus or maybe the AM-GM inequality. Let me try both methods to see which one is easier.First, using calculus. Let me define the function C(l) = 320 × (l + 4 / l). To find the minimum, I can take the derivative of C with respect to l, set it equal to zero, and solve for l.The derivative C'(l) = 320 × (1 - 4 / l²). Setting this equal to zero: 320 × (1 - 4 / l²) = 0. So, 1 - 4 / l² = 0. That means 1 = 4 / l², so l² = 4, which gives l = 2 or l = -2. Since length can't be negative, l = 2 meters.So, when l = 2, width = 4 / 2 = 2 meters. Therefore, the pool is a square in terms of length and width, both 2 meters. Now, plugging back into the cost equation: C = 480 + 320 × (2 + 4 / 2) = 480 + 320 × (2 + 2) = 480 + 320 × 4 = 480 + 1280 = 1760 yuan.Alternatively, using the AM-GM inequality. For positive numbers a and b, the arithmetic mean is greater than or equal to the geometric mean. So, (a + b)/2 ≥ √(ab). Let me apply this to l and 4 / l.So, (l + 4 / l)/2 ≥ √(l × 4 / l) = √4 = 2. Therefore, l + 4 / l ≥ 4. The minimum occurs when l = 4 / l, which means l² = 4, so l = 2. Again, we get l = 2, width = 2.Therefore, the minimum cost is 1760 yuan.Wait, let me double-check my calculations. The bottom area is 2 × 2 = 4, so 4 × 120 = 480. The walls: each pair of walls is 2 × 2 and 2 × 2, so each pair is 4, and there are two pairs, so total walls area is 8. 8 × 80 = 640. So total cost is 480 + 640 = 1120. Wait, that's different from 1760. Did I make a mistake?Oh, no, wait. The walls are actually four walls: two with area length × depth and two with area width × depth. So, if length and width are both 2, then each wall is 2 × 2 = 4. So, four walls: 4 × 4 = 16. So, walls area is 16. 16 × 80 = 1280. Then, bottom is 4 × 120 = 480. So total cost is 480 + 1280 = 1760. Okay, that matches my earlier calculation. So, my initial double-check was incorrect because I miscalculated the walls area.So, yes, the minimum cost is 1760 yuan.

answer：Alright, so I have this problem where I'm given a cubic equation: (x^3 + ax^2 + bx + 48 = 0). They tell me that (-1 - 5sqrt{3}) is a root of this equation, and that both (a) and (b) are rational numbers. I need to find the value of (a).Hmm, okay. Since the coefficients (a) and (b) are rational, I remember something about conjugate roots. If a polynomial has rational coefficients and a root that's irrational, then its conjugate must also be a root. So, in this case, since (-1 - 5sqrt{3}) is a root, the other irrational root should be its conjugate, which is (-1 + 5sqrt{3}). That makes sense because multiplying these two roots should give me a rational number, which will help in keeping the coefficients rational.So, now I know two roots: (-1 - 5sqrt{3}) and (-1 + 5sqrt{3}). Since it's a cubic equation, there must be a third root, let's call it (r), which is also rational because the coefficients are rational. Using Vieta's formulas, which relate the coefficients of a polynomial to sums and products of its roots, I can find the sum and product of the roots. For a cubic equation (x^3 + ax^2 + bx + c = 0), the sum of the roots is (-a), the sum of the products of the roots two at a time is (b), and the product of all three roots is (-c).Given that, let's write down what we know:1. The product of all three roots is (-48) because (c = 48).2. The two irrational roots are (-1 - 5sqrt{3}) and (-1 + 5sqrt{3}).First, let me find the product of the two irrational roots. That should help me find the third root.Calculating the product:[(-1 - 5sqrt{3})(-1 + 5sqrt{3}) = (-1)^2 - (5sqrt{3})^2 = 1 - 75 = -74]So, the product of these two roots is (-74). Now, since the product of all three roots is (-48), I can find the third root by dividing (-48) by the product of the two irrational roots:[r = frac{-48}{-74} = frac{24}{37}]So, the third root is (frac{24}{37}).Now, I can find the sum of the roots, which is (-a). Let's compute the sum:[(-1 - 5sqrt{3}) + (-1 + 5sqrt{3}) + frac{24}{37}]Simplifying this:- The (-5sqrt{3}) and (+5sqrt{3}) cancel each other out.- The constants: (-1 -1 = -2)- So, the sum becomes: (-2 + frac{24}{37})To add these together, I need a common denominator. Let's convert (-2) to a fraction with denominator 37:[-2 = -frac{74}{37}]So, adding (-frac{74}{37} + frac{24}{37}):[-frac{74}{37} + frac{24}{37} = -frac{50}{37}]Therefore, the sum of the roots is (-frac{50}{37}).But according to Vieta's formula, the sum of the roots is (-a). So:[-a = -frac{50}{37}]Multiplying both sides by (-1):[a = frac{50}{37}]So, the value of (a) is (frac{50}{37}).Wait, let me double-check my calculations to make sure I didn't make any mistakes. First, the product of the two irrational roots:[(-1 - 5sqrt{3})(-1 + 5sqrt{3}) = (-1)^2 - (5sqrt{3})^2 = 1 - 75 = -74]That seems correct.Then, the third root:[r = frac{-48}{-74} = frac{24}{37}]Yes, that's right.Sum of the roots:[(-1 - 5sqrt{3}) + (-1 + 5sqrt{3}) + frac{24}{37} = -2 + frac{24}{37}]Which is:[-frac{74}{37} + frac{24}{37} = -frac{50}{37}]So, (-a = -frac{50}{37}) implies (a = frac{50}{37}). That all checks out.I think I did everything correctly. I used the fact that irrational roots come in conjugate pairs when coefficients are rational, calculated the product of those roots, found the third root, then used Vieta's formula to find the sum of the roots, which gave me (a). Just to be thorough, maybe I can reconstruct the polynomial and see if it fits.If the roots are (-1 - 5sqrt{3}), (-1 + 5sqrt{3}), and (frac{24}{37}), then the polynomial can be written as:[(x - (-1 - 5sqrt{3}))(x - (-1 + 5sqrt{3}))(x - frac{24}{37}) = 0]Simplify the first two factors:[(x + 1 + 5sqrt{3})(x + 1 - 5sqrt{3}) = (x + 1)^2 - (5sqrt{3})^2 = x^2 + 2x + 1 - 75 = x^2 + 2x - 74]So, the polynomial is:[(x^2 + 2x - 74)(x - frac{24}{37}) = 0]Let me multiply this out:First, multiply (x^2) by each term in the second factor:[x^2 cdot x = x^3][x^2 cdot (-frac{24}{37}) = -frac{24}{37}x^2]Then, multiply (2x) by each term:[2x cdot x = 2x^2][2x cdot (-frac{24}{37}) = -frac{48}{37}x]Finally, multiply (-74) by each term:[-74 cdot x = -74x][-74 cdot (-frac{24}{37}) = frac{1776}{37} = 48]Now, combine all these terms:[x^3 - frac{24}{37}x^2 + 2x^2 - frac{48}{37}x - 74x + 48]Combine like terms:- For (x^3): (x^3)- For (x^2): (-frac{24}{37}x^2 + 2x^2 = (-frac{24}{37} + frac{74}{37})x^2 = frac{50}{37}x^2)- For (x): (-frac{48}{37}x - 74x = (-frac{48}{37} - frac{2738}{37})x = -frac{2786}{37}x)- Constants: (48)So, the polynomial becomes:[x^3 + frac{50}{37}x^2 - frac{2786}{37}x + 48 = 0]Comparing this with the original equation (x^3 + ax^2 + bx + 48 = 0), we can see that:- (a = frac{50}{37})- (b = -frac{2786}{37})Since the problem only asks for (a), and we've confirmed that (a = frac{50}{37}), I can be confident that this is the correct answer.I think I've covered all the steps and double-checked my work, so I'm pretty sure that (a = frac{50}{37}) is correct.**Final Answer**The value of (a) is boxed{dfrac{50}{37}}.

answer：Okay, so I have this problem here about sequences and products, and I need to prove that the product from i=1 to 2019 of (1 + 1/b_i) is less than 7/4. Hmm, let me try to unpack this step by step.First, the problem defines two sequences: {a_n} and {b_n}. The sequence {a_n} is given by the sum from i=1 to n of 1/i!, which I recognize as the partial sums of the series expansion for e. So, a_n is approaching e as n increases, right? But here, we're dealing with finite sums, so a_n is just the sum up to 1/n!.Then, the sequence {b_n} is defined as (n+1)! times a_n. So, b_n = (n+1)! * a_n. That seems like it's scaling the partial sum a_n by (n+1)!.The goal is to show that the product from i=1 to 2019 of (1 + 1/b_i) is less than 7/4. That seems a bit abstract at first, but maybe I can find a pattern or a telescoping product here.Let me write down what I know:1. a_n = sum_{i=1}^n 1/i!2. b_n = (n+1)! * a_nSo, I can express b_n in terms of a_n. Maybe I can also express b_n in terms of b_{n-1} or some other recursive relation. Let me try that.Starting with b_n = (n+1)! * a_n. Since a_n = a_{n-1} + 1/n!, substituting that in:b_n = (n+1)! * (a_{n-1} + 1/n!) = (n+1)! * a_{n-1} + (n+1)! * 1/n!Simplify the second term: (n+1)! / n! = n+1. So, the second term is (n+1).So, b_n = (n+1)! * a_{n-1} + (n+1). But wait, (n+1)! * a_{n-1} is actually b_{n-1} * (n+1), because b_{n-1} = n! * a_{n-1}. So, (n+1)! * a_{n-1} = (n+1) * n! * a_{n-1} = (n+1) * b_{n-1}.Therefore, b_n = (n+1) * b_{n-1} + (n+1). Factor out (n+1):b_n = (n+1)(b_{n-1} + 1)That's a nice recursive formula. So, b_n = (n+1)(b_{n-1} + 1). Maybe this can help me express 1 + 1/b_i in terms of b_i and b_{i-1}.Let me see. If I have 1 + 1/b_i, maybe I can relate this to b_i and b_{i-1}.From the recursive formula:b_i = (i+1)(b_{i-1} + 1)So, solving for (b_{i-1} + 1):b_{i-1} + 1 = b_i / (i+1)Therefore, 1 + 1/b_i = (b_i + 1)/b_i = (b_i + 1)/b_i.Wait, but from above, b_{i-1} + 1 = b_i / (i+1). So, b_i + 1 = b_{i+1}/(i+2)? Hmm, maybe not directly.Wait, let me think again. If b_i = (i+1)(b_{i-1} + 1), then (b_{i-1} + 1) = b_i / (i+1). So, 1 + 1/b_i = (b_i + 1)/b_i. But I don't see a direct relation yet.Alternatively, maybe I can express 1 + 1/b_i in terms of b_i and b_{i-1}.From the recursive formula:b_i = (i+1)(b_{i-1} + 1)So, (b_{i-1} + 1) = b_i / (i+1)Therefore, 1 + 1/b_i = (b_i + 1)/b_i = (b_i + 1)/b_iBut I don't see how this helps yet. Maybe I need to write the product as a telescoping product.Let me write out the product:Product_{i=1}^{2019} (1 + 1/b_i) = Product_{i=1}^{2019} (b_i + 1)/b_iSo, that's equal to (b_1 + 1)/b_1 * (b_2 + 1)/b_2 * ... * (b_{2019} + 1)/b_{2019}Hmm, if I can express each (b_i + 1)/b_i in terms of b_{i+1} or something, maybe it will telescope.From the recursive formula, we have:b_i = (i+1)(b_{i-1} + 1)So, (b_{i-1} + 1) = b_i / (i+1)Therefore, (b_i + 1) = ?Wait, let's think about (b_i + 1). From the recursive formula:b_{i+1} = (i+2)(b_i + 1)So, (b_i + 1) = b_{i+1}/(i+2)Therefore, (b_i + 1)/b_i = (b_{i+1}/(i+2)) / b_i = b_{i+1}/( (i+2) b_i )So, each term in the product is b_{i+1}/( (i+2) b_i )Therefore, the entire product becomes:Product_{i=1}^{2019} [ b_{i+1} / ( (i+2) b_i ) ]This is a telescoping product because the b_i terms will cancel out.Let's write it out:= [ b_2 / (3 b_1) ] * [ b_3 / (4 b_2) ] * [ b_4 / (5 b_3) ] * ... * [ b_{2020} / (2021 b_{2019}) ]Notice that each b_i in the numerator cancels with the b_i in the denominator of the next term. So, all the b terms cancel except for the first denominator and the last numerator.Thus, the product simplifies to:= b_{2020} / ( (3 * 4 * 5 * ... * 2021) b_1 )Now, let's compute this.First, b_1 = (1+1)! a_1 = 2! * (1/1!) = 2 * 1 = 2.So, b_1 = 2.Next, the denominator has the product from 3 to 2021, which is 2021! / (2!) = 2021! / 2.Wait, because 3*4*...*2021 = 2021! / (2!) = 2021! / 2.So, the denominator is (2021! / 2) * b_1 = (2021! / 2) * 2 = 2021!.Wait, let me check that again.Denominator is (3 * 4 * ... * 2021) * b_1.But 3 * 4 * ... * 2021 = 2021! / (2!) = 2021! / 2.So, denominator is (2021! / 2) * b_1.But b_1 = 2, so denominator becomes (2021! / 2) * 2 = 2021!.So, the entire product is b_{2020} / 2021!.Now, let's compute b_{2020}.b_{2020} = (2020 + 1)! * a_{2020} = 2021! * a_{2020}.Therefore, b_{2020} / 2021! = a_{2020}.So, the product simplifies to a_{2020}.Therefore, Product_{i=1}^{2019} (1 + 1/b_i) = a_{2020}.Wait, that's a big simplification. So, the product is equal to a_{2020}.But a_{2020} is the sum from i=1 to 2020 of 1/i!.We know that the sum from i=0 to infinity of 1/i! is e, so a_{2020} = e - 1 - 1/0! = e - 2? Wait, no.Wait, a_n is sum from i=1 to n of 1/i!, so a_n = e - sum from i=n+1 to infinity of 1/i! - 1/0!.Wait, actually, the full series is sum_{i=0}^infty 1/i! = e, so sum_{i=1}^infty 1/i! = e - 1.Therefore, a_n = sum_{i=1}^n 1/i! = e - 1 - sum_{i=n+1}^infty 1/i!.So, a_n = e - 1 - R_n, where R_n is the remainder term.Therefore, a_{2020} = e - 1 - R_{2020}.But since R_{2020} is positive, a_{2020} < e - 1.But e is approximately 2.718, so e - 1 is approximately 1.718, which is less than 7/4 = 1.75.Wait, but 1.718 is less than 1.75, so a_{2020} < e - 1 < 7/4.Therefore, the product is equal to a_{2020}, which is less than e - 1, which is less than 7/4.Hence, the product is less than 7/4.Wait, but let me check if a_{2020} is indeed less than 7/4.Compute a_{2020} = sum_{i=1}^{2020} 1/i!.We know that the series converges to e - 1, which is approximately 1.71828.7/4 is 1.75, so 1.71828 < 1.75, so yes, a_{2020} < 7/4.Therefore, the product is equal to a_{2020}, which is less than 7/4.Hence, the inequality holds.Wait, but let me double-check my steps because sometimes telescoping products can be tricky.Starting from the product:Product_{i=1}^{2019} (1 + 1/b_i) = Product_{i=1}^{2019} (b_i + 1)/b_iThen, using the recursive relation, each (b_i + 1) = b_{i+1}/(i+2)So, each term becomes b_{i+1}/( (i+2) b_i )Thus, the product becomes Product_{i=1}^{2019} b_{i+1}/( (i+2) b_i )Which telescopes to b_{2020}/( (3*4*...*2021) b_1 )As above, denominator is 2021! / 2 * 2 = 2021!Numerator is b_{2020} = 2021! a_{2020}Thus, the product is a_{2020}Since a_{2020} < e - 1 < 7/4, the product is less than 7/4.Yes, that seems correct.I think I've got it. The key was recognizing the telescoping nature of the product and expressing it in terms of a_{2020}, which is known to be less than e - 1, and since e - 1 is approximately 1.718, which is less than 1.75, the inequality holds.

answer：First, I recognize that the wire is divided into two parts, one of length ( a ) and the other of length ( b ). The piece of length ( a ) is bent into a square, and the piece of length ( b ) is bent into a regular hexagon.For the square, since all four sides are equal, the side length of the square is ( frac{a}{4} ). Therefore, the perimeter of the square is ( 4 times frac{a}{4} = a ).For the regular hexagon, all six sides are equal, so the side length of the hexagon is ( frac{b}{6} ). Consequently, the perimeter of the hexagon is ( 6 times frac{b}{6} = b ).According to the problem, the perimeters of the square and the hexagon are equal. This gives us the equation:[a = b]To find the ratio ( frac{a}{b} ), we substitute ( a ) with ( b ):[frac{a}{b} = frac{b}{b} = 1]Therefore, the ratio ( frac{a}{b} ) is 1.

answer：First, I need to determine the annual increase rate. The amount increases by 1/8th of itself each year, which is equivalent to 12.5%.Next, I'll use the compound interest formula to calculate the amount after two years. The formula is:Final Amount = Principal Amount × (1 + Rate of Increase)²Plugging in the values:Final Amount = 51200 × (1 + 0.125)²Final Amount = 51200 × (1.125)²Final Amount = 51200 × 1.265625Final Amount = 64800Therefore, after two years, the amount will be Rs. 64,800.