💡Alright, so I've got this problem here about an isosceles right triangle. The legs are increased by 4 cm, and the area increases by 112 cm². I need to find the lengths of the sides of the original triangle. Hmm, okay, let me think about how to approach this.First, I remember that an isosceles right triangle has two legs of equal length and a hypotenuse. The area of such a triangle is given by (1/2) * leg². So, if I let the length of each leg be x cm, then the area of the original triangle would be (1/2)x².Now, according to the problem, each leg is increased by 4 cm. So, the new length of each leg becomes (x + 4) cm. The area of the new triangle would then be (1/2)(x + 4)².The problem states that the area increases by 112 cm². That means the area of the new triangle minus the area of the original triangle equals 112 cm². So, I can set up the equation:(1/2)(x + 4)² - (1/2)x² = 112Okay, let me simplify this equation. First, I can factor out the (1/2):(1/2)[(x + 4)² - x²] = 112Now, let's expand (x + 4)²:(x + 4)² = x² + 8x + 16So, substituting back into the equation:(1/2)[x² + 8x + 16 - x²] = 112Simplify inside the brackets:(1/2)(8x + 16) = 112Now, multiply both sides by 2 to eliminate the fraction:8x + 16 = 224Subtract 16 from both sides:8x = 208Divide both sides by 8:x = 26So, the length of each leg of the original triangle is 26 cm. Since it's an isosceles right triangle, the hypotenuse can be found using the Pythagorean theorem. The hypotenuse h is:h = x√2 = 26√2 cmLet me double-check my calculations to make sure I didn't make any mistakes. Starting from the area difference:Original area: (1/2)(26)² = (1/2)(676) = 338 cm²New leg length: 26 + 4 = 30 cmNew area: (1/2)(30)² = (1/2)(900) = 450 cm²Difference in area: 450 - 338 = 112 cm²Yep, that matches the problem statement. So, my calculations seem correct.Therefore, the original triangle has legs of 26 cm each and a hypotenuse of 26√2 cm.

💡Okay, so Roberto is thinking about getting chickens to produce eggs instead of buying them. He wants to know after how many weeks it will be cheaper to have chickens than to keep buying eggs. Let me try to figure this out step by step.First, I need to understand the costs involved. Roberto is buying 4 chickens, each costing 20. So, the total initial cost for the chickens is 4 times 20, which is 80. That’s a one-time expense. Then, there’s the weekly cost to feed the chickens, which is 1 in total. So, every week, he spends 1 on feed.Now, each chicken produces 3 eggs per week. Since he has 4 chickens, the total number of eggs produced each week is 4 times 3, which is 12 eggs. That’s exactly a dozen eggs every week. So, the chickens provide him with a dozen eggs weekly.Previously, Roberto used to buy 1 dozen eggs every week, spending 2 per dozen. So, he was spending 2 each week on eggs. Now, with the chickens, he spends 1 per week on feed. That means he saves 1 each week by having chickens instead of buying eggs.But he also has to consider the initial cost of the chickens, which was 80. So, he needs to see when the savings from not buying eggs will cover the initial cost of the chickens.To find out when the chickens become cheaper, I need to calculate how many weeks it will take for the weekly savings to add up to the initial cost. The weekly savings are 1, and the initial cost is 80. So, dividing the initial cost by the weekly savings gives me the number of weeks needed to break even.80 divided by 1 is 80 weeks. That means after 80 weeks, the savings from not buying eggs will have covered the initial cost of the chickens. But the question is asking after how many weeks the chickens will be cheaper than buying eggs. So, I think it means after how many weeks the total cost of having chickens is less than the total cost of buying eggs.So, after 80 weeks, the costs are equal. To find out when it becomes cheaper, I need to add one more week. That would be 81 weeks. After 81 weeks, the total cost of having chickens will be less than the total cost of buying eggs.Let me double-check my calculations to make sure I didn’t make any mistakes. The initial cost is 80, weekly feed is 1, and weekly savings are 1. So, every week, he saves 1, and after 80 weeks, he saves 80, which covers the initial cost. Then, in week 81, he starts saving money compared to buying eggs.Yeah, that makes sense. So, the answer should be 81 weeks.

💡Okay, so I have this problem where I need to find the minimum value of the expression ((sin x + 3csc x)^2 + (cos x + 3sec x)^2) for (0 < x < frac{pi}{2}). Hmm, that looks a bit complicated, but maybe I can break it down step by step.First, I remember that (csc x) is the reciprocal of (sin x), and (sec x) is the reciprocal of (cos x). So, maybe I can rewrite the expression in terms of sine and cosine only. Let me try expanding each squared term.Starting with ((sin x + 3csc x)^2), that should expand to (sin^2 x + 2 times sin x times 3csc x + (3csc x)^2). Similarly, ((cos x + 3sec x)^2) would expand to (cos^2 x + 2 times cos x times 3sec x + (3sec x)^2).Let me write that out:[(sin x + 3csc x)^2 = sin^2 x + 6sin x csc x + 9csc^2 x][(cos x + 3sec x)^2 = cos^2 x + 6cos x sec x + 9sec^2 x]Now, I can combine these two expressions:[sin^2 x + 6sin x csc x + 9csc^2 x + cos^2 x + 6cos x sec x + 9sec^2 x]Simplify the terms. I notice that (sin x csc x) is just 1 because (csc x = 1/sin x). Similarly, (cos x sec x) is also 1. So, the middle terms simplify to 6 each.So now the expression becomes:[sin^2 x + 6 + 9csc^2 x + cos^2 x + 6 + 9sec^2 x]Combine like terms:[sin^2 x + cos^2 x + 12 + 9csc^2 x + 9sec^2 x]I remember that (sin^2 x + cos^2 x = 1), so substituting that in:[1 + 12 + 9csc^2 x + 9sec^2 x][13 + 9(csc^2 x + sec^2 x)]Hmm, okay, so now I have (13 + 9(csc^2 x + sec^2 x)). I need to find the minimum value of this expression. Let me think about how to handle (csc^2 x + sec^2 x).I know that (csc^2 x = 1 + cot^2 x) and (sec^2 x = 1 + tan^2 x), but I'm not sure if that helps directly. Maybe I can express everything in terms of sine and cosine again.So, (csc^2 x = 1/sin^2 x) and (sec^2 x = 1/cos^2 x). Therefore, the expression becomes:[13 + 9left(frac{1}{sin^2 x} + frac{1}{cos^2 x}right)]Let me combine the fractions:[13 + 9left(frac{cos^2 x + sin^2 x}{sin^2 x cos^2 x}right)]Since (sin^2 x + cos^2 x = 1), this simplifies to:[13 + 9left(frac{1}{sin^2 x cos^2 x}right)]So now, the expression is (13 + frac{9}{sin^2 x cos^2 x}). I need to find the minimum value of this expression for (0 < x < frac{pi}{2}).To minimize the entire expression, I need to minimize (frac{9}{sin^2 x cos^2 x}), which is equivalent to maximizing (sin^2 x cos^2 x). So, let me focus on maximizing (sin^2 x cos^2 x).I remember that (sin x cos x = frac{1}{2}sin 2x), so squaring both sides gives (sin^2 x cos^2 x = frac{1}{4}sin^2 2x). Therefore, (sin^2 x cos^2 x = frac{1}{4}sin^2 2x).So, the expression becomes:[13 + frac{9}{frac{1}{4}sin^2 2x} = 13 + frac{36}{sin^2 2x}]Now, I need to maximize (sin^2 2x) because that will minimize the entire expression. The maximum value of (sin^2 2x) is 1, which occurs when (2x = frac{pi}{2}), so (x = frac{pi}{4}).Therefore, substituting back, the expression becomes:[13 + frac{36}{1} = 13 + 36 = 49]Wait, that can't be right because when I plug (x = frac{pi}{4}) into the original expression, let me check:[sin frac{pi}{4} = frac{sqrt{2}}{2}, quad csc frac{pi}{4} = sqrt{2}][cos frac{pi}{4} = frac{sqrt{2}}{2}, quad sec frac{pi}{4} = sqrt{2}]So,[(sin x + 3csc x)^2 = left(frac{sqrt{2}}{2} + 3sqrt{2}right)^2 = left(frac{sqrt{2}}{2} + frac{6sqrt{2}}{2}right)^2 = left(frac{7sqrt{2}}{2}right)^2 = frac{49 times 2}{4} = frac{98}{4} = 24.5]Similarly,[(cos x + 3sec x)^2 = left(frac{sqrt{2}}{2} + 3sqrt{2}right)^2 = 24.5]Adding them together: (24.5 + 24.5 = 49). So, that seems correct. But wait, earlier I thought the minimum was 52. Did I make a mistake?Wait, let me go back. In my initial steps, I had:[16 + 9left(frac{1}{sin^2 x cos^2 x}right)]But when I simplified, I think I might have miscalculated the constants. Let me double-check.Starting from the expansion:[(sin x + 3csc x)^2 + (cos x + 3sec x)^2 = sin^2 x + 6 + 9csc^2 x + cos^2 x + 6 + 9sec^2 x][= (sin^2 x + cos^2 x) + 12 + 9(csc^2 x + sec^2 x)][= 1 + 12 + 9(csc^2 x + sec^2 x)][= 13 + 9(csc^2 x + sec^2 x)]Wait, so that's correct. Then, I expressed (csc^2 x + sec^2 x) as (frac{1}{sin^2 x} + frac{1}{cos^2 x}), which is (frac{cos^2 x + sin^2 x}{sin^2 x cos^2 x} = frac{1}{sin^2 x cos^2 x}).So, the expression becomes (13 + frac{9}{sin^2 x cos^2 x}). Then, I used (sin^2 x cos^2 x = frac{1}{4}sin^2 2x), so it becomes (13 + frac{36}{sin^2 2x}).To minimize this, we need to maximize (sin^2 2x), which is 1 when (x = frac{pi}{4}). Therefore, the minimum value is (13 + 36 = 49).But earlier, I thought the minimum was 52. Maybe I confused it with another problem. Let me check my calculations again.Wait, when I plug (x = frac{pi}{4}) into the original expression, I get 49, which seems correct. So, why did I initially think it was 52? Maybe I made a mistake in the initial expansion.Let me re-examine the expansion:[(sin x + 3csc x)^2 = sin^2 x + 6sin x csc x + 9csc^2 x][= sin^2 x + 6 + 9csc^2 x]Similarly,[(cos x + 3sec x)^2 = cos^2 x + 6 + 9sec^2 x]Adding them together:[sin^2 x + cos^2 x + 12 + 9(csc^2 x + sec^2 x)][= 1 + 12 + 9(csc^2 x + sec^2 x)][= 13 + 9(csc^2 x + sec^2 x)]Yes, that's correct. So, the minimum value is indeed 49. But wait, when I first thought about it, I thought it was 52. Maybe I confused it with another problem. Let me check another approach.Alternatively, I can use calculus to find the minimum. Let me set (f(x) = (sin x + 3csc x)^2 + (cos x + 3sec x)^2). Then, find (f'(x)) and set it to zero.But that might be complicated. Alternatively, I can use substitution. Let me let (t = sin x), so (cos x = sqrt{1 - t^2}). Then, express everything in terms of (t).But that might also be messy. Alternatively, I can use the AM-GM inequality. Let me think.Looking back, I have:[f(x) = 13 + frac{36}{sin^2 2x}]Since (sin^2 2x leq 1), the minimum of (f(x)) occurs when (sin^2 2x) is maximized, which is 1. Therefore, the minimum value is (13 + 36 = 49).Wait, but earlier I thought it was 52. Maybe I made a mistake in the initial expansion. Let me check again.Wait, in the initial expansion, I had:[(sin x + 3csc x)^2 = sin^2 x + 6sin x csc x + 9csc^2 x][= sin^2 x + 6 + 9csc^2 x]Similarly for cosine.So, adding them:[sin^2 x + cos^2 x + 12 + 9(csc^2 x + sec^2 x)][= 1 + 12 + 9(csc^2 x + sec^2 x)][= 13 + 9(csc^2 x + sec^2 x)]Yes, that's correct. So, the minimum is 49. Therefore, I think my initial thought of 52 was incorrect. The correct minimum value is 49.Wait, but when I plug (x = frac{pi}{4}), I get 49, which seems correct. So, maybe the initial thought of 52 was a mistake. Therefore, the minimum value is 49.

💡First, I need to determine the number of students who received grades other than D. There are 800 students in total. - The number of students who received an A is 1/5 of 800, which is 160 students.- The number of students who received a B is 1/4 of 800, which is 200 students.- The number of students who received a C is 1/2 of 800, which is 400 students.Next, I add up the number of students who received A, B, and C grades:160 (A) + 200 (B) + 400 (C) = 760 students.Finally, to find the number of students who received a D, I subtract the total number of students who received A, B, and C grades from the total number of students:800 (total students) - 760 (students with A, B, or C) = 40 students.Therefore, 40 students received a D as their final grade.

💡First, I need to evaluate the expression (frac{3b}{a + b}) by substituting the given values of (a = 5) and (b = 6).I'll start by calculating the sum of (a) and (b):[a + b = 5 + 6 = 11]Next, I'll multiply (b) by 3:[3b = 3 times 6 = 18]Now, I'll form the fraction with the results from the previous steps:[frac{3b}{a + b} = frac{18}{11}]Finally, I'll compare this result with the provided answer choices and select the correct one.

💡Okay, so I have this problem about an arithmetic sequence. The sum of the first n terms is S_n. There's a function f(x) = (2^x - 1)/(2^x + 1). They give me two specific values of this function: f(a_2 - 2) equals sin(2014π/3), and f(a_{2014} - 2) equals cos(2015π/6). I need to find S_{2015}.First, let me recall what an arithmetic sequence is. It's a sequence where each term increases by a constant difference. So, the nth term is a_n = a_1 + (n-1)d, where a_1 is the first term and d is the common difference.The sum of the first n terms, S_n, is given by the formula S_n = n/2 * (2a_1 + (n-1)d) or sometimes written as S_n = n*(a_1 + a_n)/2.Now, looking at the function f(x) = (2^x - 1)/(2^x + 1). Hmm, that looks a bit complicated, but maybe there's a property I can use. Let me see if I can find a relationship between f(x) and f(-x).Let me compute f(-x):f(-x) = (2^{-x} - 1)/(2^{-x} + 1). Hmm, 2^{-x} is 1/(2^x), so let me rewrite that:f(-x) = (1/(2^x) - 1)/(1/(2^x) + 1) = (1 - 2^x)/(1 + 2^x). Wait a minute, that's just -(2^x - 1)/(2^x + 1), which is -f(x). So, f(-x) = -f(x). That means f(x) is an odd function. That's a useful property.So, f(x) + f(-x) = 0. That might come in handy later.Now, let's look at the given equations:f(a_2 - 2) = sin(2014π/3)andf(a_{2014} - 2) = cos(2015π/6)I need to compute these trigonometric functions first.Starting with sin(2014π/3). Let me simplify this angle. Since sine has a period of 2π, I can subtract multiples of 2π to find an equivalent angle between 0 and 2π.2014π/3 divided by 2π is (2014/3)/2 = 2014/6 ≈ 335.666... So, that's 335 full circles plus 0.666... of a circle. Let me compute 2014π/3 - 335*2π.Wait, 335*2π is 670π. So, 2014π/3 - 670π = (2014π - 2010π)/3 = 4π/3.So, sin(2014π/3) = sin(4π/3). Sin(4π/3) is sin(π + π/3) which is -sin(π/3) = -√3/2.So, f(a_2 - 2) = -√3/2.Similarly, let's compute cos(2015π/6). Again, cosine has a period of 2π, so let's subtract multiples of 2π.2015π/6 divided by 2π is (2015/6)/2 = 2015/12 ≈ 167.916666... So, that's 167 full circles plus 0.916666... of a circle.Compute 2015π/6 - 167*2π = 2015π/6 - 334π = (2015π - 1980π)/6 = 35π/6.But 35π/6 is still more than 2π. Let me subtract another 2π: 35π/6 - 2π = 35π/6 - 12π/6 = 23π/6.23π/6 is still more than 2π. Subtract another 2π: 23π/6 - 12π/6 = 11π/6.So, cos(2015π/6) = cos(11π/6). Cos(11π/6) is cos(2π - π/6) which is cos(π/6) = √3/2.Wait, but hold on. Cosine is positive in the fourth quadrant, so cos(11π/6) is indeed √3/2.So, f(a_{2014} - 2) = √3/2.So, now I have:f(a_2 - 2) = -√3/2andf(a_{2014} - 2) = √3/2But from earlier, I found that f(x) is an odd function, so f(-x) = -f(x). So, if I have f(a_2 - 2) = -√3/2, then f(-(a_2 - 2)) = √3/2.But wait, f(a_{2014} - 2) is also √3/2. So, that would imply that a_{2014} - 2 = -(a_2 - 2). Let me write that:a_{2014} - 2 = -(a_2 - 2)Simplify this:a_{2014} - 2 = -a_2 + 2Bring the 2 to the other side:a_{2014} = -a_2 + 4So, a_{2014} + a_2 = 4.Hmm, interesting. So, the sum of the second term and the 2014th term is 4.Now, since this is an arithmetic sequence, the terms are linear. So, a_n = a_1 + (n-1)d.So, a_2 = a_1 + da_{2014} = a_1 + 2013dSo, adding them together:a_2 + a_{2014} = (a_1 + d) + (a_1 + 2013d) = 2a_1 + 2014d = 4So, 2a_1 + 2014d = 4.Now, I need to find S_{2015}, which is the sum of the first 2015 terms.Using the formula for the sum of an arithmetic sequence:S_n = n/2 * (2a_1 + (n - 1)d)So, S_{2015} = 2015/2 * (2a_1 + 2014d)Wait a second, from earlier, we have 2a_1 + 2014d = 4.So, substituting that into the formula:S_{2015} = 2015/2 * 4 = 2015 * 2 = 4030.So, S_{2015} is 4030.Wait, let me double-check that.We had a_2 + a_{2014} = 4.In an arithmetic sequence, the sum of terms equidistant from the ends is constant. So, a_1 + a_n = a_2 + a_{n-1} = ... etc.But in this case, a_2 and a_{2014} are not necessarily equidistant from the ends unless n is 2015.Wait, n is 2015, so the middle term is the 1008th term or something? Wait, 2015 is odd, so the middle term is a_{1008}.But anyway, the key point is that a_2 + a_{2014} = 4, and in the sum formula, S_{2015} = 2015*(a_1 + a_{2015})/2.But since a_1 + a_{2015} is equal to a_2 + a_{2014} because in an arithmetic sequence, a_1 + a_n = a_2 + a_{n-1} = ... So, yeah, a_1 + a_{2015} = a_2 + a_{2014} = 4.So, S_{2015} = 2015*4/2 = 2015*2 = 4030.Yep, that seems correct.So, the answer is 4030.**Final Answer**boxed{4030}