💡First, determine the number of students wearing each color of shirt:- **Blue shirts:** 40% of 1,000 students = 0.40 * 1,000 = 400 students- **Red shirts:** 25% of 1,000 students = 0.25 * 1,000 = 250 students- **Green shirts:** 20% of 1,000 students = 0.20 * 1,000 = 200 studentsNext, calculate the number of students with each pattern for red and green shirts:- **Red shirts:** - **Polka dots:** 40% of 250 students = 0.40 * 250 = 100 students - **No pattern:** Remaining = 250 - (50 + 100) = 100 students- **Green shirts:** - **No pattern:** Remaining = 200 - (50 + 50) = 100 studentsNow, find out how many students wear additional accessories:- **Red shirts with polka dots and a hat:** 15% of 100 students = 0.15 * 100 = 15 students- **Green shirts with no pattern and a scarf:** 10% of 100 students = 0.10 * 100 = 10 studentsFinally, add the two groups together:15 (red with polka dots and hat) + 10 (green with no pattern and scarf) = 25 students

💡Okay, so I have this problem where I need to find the area of a triangle inscribed in a circle with a radius of 2 cm. The triangle has two angles given: π/3 and π/4. Hmm, let me think about how to approach this.First, I remember that in a circle, the triangle is inscribed, which means all its vertices lie on the circumference. The circle has a radius of 2 cm, so the diameter is 4 cm. I also recall that for a triangle inscribed in a circle, the side lengths can be related to the angles using the Law of Sines. The Law of Sines states that in any triangle, the ratio of the length of a side to the sine of the opposite angle is constant and equal to twice the radius of the circumscribed circle. So, the formula is:a / sin(A) = b / sin(B) = c / sin(C) = 2RWhere a, b, c are the sides opposite angles A, B, C respectively, and R is the radius of the circumscribed circle.Given that, since the radius R is 2 cm, 2R would be 4 cm. So, each side can be found by multiplying 4 cm by the sine of the opposite angle.But wait, I only have two angles given: π/3 and π/4. I need the third angle to use the Law of Sines. Since the sum of angles in a triangle is π radians, the third angle should be π - π/3 - π/4. Let me calculate that.π is approximately 3.1416, but I can work in exact terms. So, π - π/3 - π/4. Let me find a common denominator for the fractions. The denominators are 3 and 4, so the least common denominator is 12.π/3 is equal to 4π/12, and π/4 is equal to 3π/12. So, adding those together: 4π/12 + 3π/12 = 7π/12. Therefore, the third angle is π - 7π/12, which is 5π/12. So, the three angles are π/3, π/4, and 5π/12.Now, I can label the sides opposite these angles. Let me denote side a opposite angle A = π/3, side b opposite angle B = π/4, and side c opposite angle C = 5π/12.Using the Law of Sines:a = 2R * sin(A) = 4 * sin(π/3)b = 4 * sin(π/4)c = 4 * sin(5π/12)I need to compute these sine values.First, sin(π/3) is a known value. π/3 is 60 degrees, and sin(60°) is √3/2. So, a = 4 * (√3/2) = 2√3 cm.Next, sin(π/4) is also a known value. π/4 is 45 degrees, and sin(45°) is √2/2. So, b = 4 * (√2/2) = 2√2 cm.Now, sin(5π/12) is a bit trickier. 5π/12 is 75 degrees. I can use the sine addition formula because 75° is 45° + 30°. The sine addition formula is:sin(A + B) = sin A cos B + cos A sin BSo, sin(75°) = sin(45° + 30°) = sin(45°)cos(30°) + cos(45°)sin(30°)I know that sin(45°) = √2/2, cos(30°) = √3/2, cos(45°) = √2/2, and sin(30°) = 1/2.Plugging these in:sin(75°) = (√2/2)(√3/2) + (√2/2)(1/2) = (√6)/4 + (√2)/4 = (√6 + √2)/4Therefore, sin(5π/12) = (√6 + √2)/4So, side c = 4 * [(√6 + √2)/4] = √6 + √2 cm.Alright, so now I have all three sides:a = 2√3 cmb = 2√2 cmc = √6 + √2 cmNow, I need to find the area of the triangle. There are a few ways to find the area of a triangle when you know the sides and angles. One common method is using the formula:Area = (1/2)ab sin CWhere a and b are two sides, and C is the included angle between them.Alternatively, since I know all three sides, I could use Heron's formula, which is:Area = √[s(s - a)(s - b)(s - c)]Where s is the semi-perimeter, s = (a + b + c)/2But Heron's formula might be more complicated here because the sides involve square roots, which could make the calculations messy. Alternatively, since I know two sides and the included angle, maybe I can use the formula with two sides and the sine of the included angle.Wait, but do I know the included angle? Let me see. The sides a, b, c correspond to angles A, B, C. So, if I take sides a and b, the included angle would be angle C, which is 5π/12. Alternatively, if I take sides a and c, the included angle would be angle B, which is π/4, or sides b and c, the included angle would be angle A, which is π/3.So, perhaps it's easier to use the formula with two sides and the included angle. Let me choose sides a and b, with included angle C.So, Area = (1/2) * a * b * sin CPlugging in the values:Area = (1/2) * (2√3) * (2√2) * sin(5π/12)Simplify:First, multiply the constants:(1/2) * 2√3 * 2√2 = (1/2) * 4 * √6 = 2√6Now, multiply by sin(5π/12), which we already found is (√6 + √2)/4So, Area = 2√6 * (√6 + √2)/4Simplify:2√6 divided by 4 is (√6)/2So, Area = (√6)/2 * (√6 + √2)Multiply out:(√6 * √6)/2 + (√6 * √2)/2 = (6)/2 + (√12)/2 = 3 + (2√3)/2 = 3 + √3So, the area is 3 + √3 cm².Wait, let me double-check my steps to make sure I didn't make a mistake.1. Found the third angle correctly: π - π/3 - π/4 = 5π/12. That seems right.2. Applied the Law of Sines correctly:a = 4 sin(π/3) = 2√3b = 4 sin(π/4) = 2√2c = 4 sin(5π/12) = √6 + √2Yes, that seems correct.3. Then, for the area, I used the formula (1/2)ab sin C, where a and b are sides, and C is the included angle.Plugged in a = 2√3, b = 2√2, and sin C = (√6 + √2)/4Calculations:(1/2) * 2√3 * 2√2 = (1/2) * 4√6 = 2√6Then, 2√6 * (√6 + √2)/4 = (2√6)(√6 + √2)/4Simplify numerator:2√6 * √6 = 2 * 6 = 122√6 * √2 = 2√12 = 2 * 2√3 = 4√3So, numerator is 12 + 4√3Divide by 4: (12 + 4√3)/4 = 3 + √3Yes, that matches. So, the area is 3 + √3 cm².Alternatively, I can check using another method, maybe Heron's formula, to see if I get the same result.First, compute the semi-perimeter s:s = (a + b + c)/2 = (2√3 + 2√2 + √6 + √2)/2Combine like terms:2√3 + (2√2 + √2) + √6 = 2√3 + 3√2 + √6So, s = (2√3 + 3√2 + √6)/2Now, Heron's formula is √[s(s - a)(s - b)(s - c)]Compute each term:s - a = (2√3 + 3√2 + √6)/2 - 2√3 = (2√3 + 3√2 + √6 - 4√3)/2 = (-2√3 + 3√2 + √6)/2s - b = (2√3 + 3√2 + √6)/2 - 2√2 = (2√3 + 3√2 + √6 - 4√2)/2 = (2√3 - √2 + √6)/2s - c = (2√3 + 3√2 + √6)/2 - (√6 + √2) = (2√3 + 3√2 + √6 - 2√6 - 2√2)/2 = (2√3 + √2 - √6)/2So, now, the product s(s - a)(s - b)(s - c) is:[(2√3 + 3√2 + √6)/2] * [(-2√3 + 3√2 + √6)/2] * [(2√3 - √2 + √6)/2] * [(2√3 + √2 - √6)/2]This looks really complicated. Maybe it's better to stick with the previous method since Heron's formula is getting too messy.Alternatively, another way to find the area is using the formula:Area = (abc)/(4R)Where a, b, c are the sides, and R is the radius of the circumscribed circle.Let me try this.We have a = 2√3, b = 2√2, c = √6 + √2, and R = 2.So, Area = (2√3 * 2√2 * (√6 + √2))/(4*2) = (4√6 * (√6 + √2))/8Simplify numerator:4√6 * √6 = 4 * 6 = 244√6 * √2 = 4√12 = 4 * 2√3 = 8√3So, numerator is 24 + 8√3Divide by 8: (24 + 8√3)/8 = 3 + √3Same result as before. So, that confirms the area is indeed 3 + √3 cm².Therefore, I'm confident that the area of the triangle is 3 + √3 cm².

💡Okay, so I have this trigonometry problem here: (4cos 50^circ - tan 40^circ). And I need to figure out which of the options A to D it equals. The options are (sqrt{2}), (sqrt{3}), (frac{sqrt{2} + sqrt{3}}{2}), and (2sqrt{2} - 1). Hmm, let's see how to approach this.First, I know that (tan theta = frac{sin theta}{cos theta}), so maybe I can rewrite the tangent part in terms of sine and cosine. That might make it easier to combine with the cosine term. So, let me try that.So, (4cos 50^circ - tan 40^circ) becomes (4cos 50^circ - frac{sin 40^circ}{cos 40^circ}). Now, I have two terms: one is (4cos 50^circ) and the other is (-frac{sin 40^circ}{cos 40^circ}). To combine these, I might need a common denominator. The denominator for the second term is (cos 40^circ), so maybe I can write the first term with the same denominator.Let me try that. So, (4cos 50^circ) can be written as (frac{4cos 50^circ cos 40^circ}{cos 40^circ}). That way, both terms have the same denominator. So now, the expression becomes:[frac{4cos 50^circ cos 40^circ - sin 40^circ}{cos 40^circ}]Okay, so now I have a single fraction. Let me see if I can simplify the numerator. I remember that there are product-to-sum identities in trigonometry that might help here. The product (cos A cos B) can be expressed as (frac{1}{2}[cos(A+B) + cos(A-B)]). Let me apply that to (4cos 50^circ cos 40^circ).So, (4cos 50^circ cos 40^circ = 4 times frac{1}{2}[cos(50^circ + 40^circ) + cos(50^circ - 40^circ)] = 2[cos 90^circ + cos 10^circ]).I know that (cos 90^circ = 0), so this simplifies to (2 times cos 10^circ = 2cos 10^circ). So, the numerator becomes (2cos 10^circ - sin 40^circ).So now, the expression is:[frac{2cos 10^circ - sin 40^circ}{cos 40^circ}]Hmm, okay. Let me see if I can simplify (sin 40^circ). I remember that (sin(90^circ - theta) = cos theta), so (sin 40^circ = cos 50^circ). But I'm not sure if that helps here. Alternatively, maybe I can express (sin 40^circ) in terms of (cos 10^circ) or something else.Wait, I also remember that (sin 40^circ) can be written using the sine addition formula. Since (40^circ = 30^circ + 10^circ), maybe I can expand that.So, (sin(30^circ + 10^circ) = sin 30^circ cos 10^circ + cos 30^circ sin 10^circ). I know the exact values for (sin 30^circ) and (cos 30^circ): (sin 30^circ = frac{1}{2}) and (cos 30^circ = frac{sqrt{3}}{2}). So, substituting those in:[sin 40^circ = frac{1}{2}cos 10^circ + frac{sqrt{3}}{2}sin 10^circ]So, substituting back into the numerator:[2cos 10^circ - left(frac{1}{2}cos 10^circ + frac{sqrt{3}}{2}sin 10^circright) = 2cos 10^circ - frac{1}{2}cos 10^circ - frac{sqrt{3}}{2}sin 10^circ]Let me combine like terms:[2cos 10^circ - frac{1}{2}cos 10^circ = left(2 - frac{1}{2}right)cos 10^circ = frac{3}{2}cos 10^circ]So, the numerator becomes:[frac{3}{2}cos 10^circ - frac{sqrt{3}}{2}sin 10^circ]Hmm, this looks like it could be expressed as a single trigonometric function using the formula for (acos theta + bsin theta). I remember that (acos theta + bsin theta = Rcos(theta - phi)), where (R = sqrt{a^2 + b^2}) and (tan phi = frac{b}{a}). Let me try that.Here, (a = frac{3}{2}) and (b = -frac{sqrt{3}}{2}). So, (R = sqrt{left(frac{3}{2}right)^2 + left(-frac{sqrt{3}}{2}right)^2} = sqrt{frac{9}{4} + frac{3}{4}} = sqrt{frac{12}{4}} = sqrt{3}).Then, (tan phi = frac{b}{a} = frac{-frac{sqrt{3}}{2}}{frac{3}{2}} = -frac{sqrt{3}}{3}). So, (phi = -30^circ) because (tan(-30^circ) = -frac{sqrt{3}}{3}).Therefore, the numerator can be written as:[sqrt{3}cosleft(10^circ - (-30^circ)right) = sqrt{3}cos(40^circ)]So, now the entire expression becomes:[frac{sqrt{3}cos 40^circ}{cos 40^circ} = sqrt{3}]Wow, that worked out! So, the original expression simplifies to (sqrt{3}), which is option B.But let me double-check my steps to make sure I didn't make any mistakes. Starting from the beginning, I converted the tangent to sine over cosine, found a common denominator, used the product-to-sum identity, expanded the sine of a sum, combined like terms, and then expressed the result as a single cosine function. Each step seems logical and I don't see any calculation errors. So, I think the answer is indeed (sqrt{3}).**Final Answer**boxed{B}

💡Okay, so I have this problem where Vasya writes all natural numbers from 1 to 2014 on a board, and then Petya erases 1006 of them. I need to prove that among the remaining numbers, there will be two such that one is a divisor of the other. Hmm, interesting. Let me try to figure this out step by step.First, let me understand the problem. We have numbers from 1 to 2014, which is a total of 2014 numbers. Petya erases 1006 of them, so that leaves us with 2014 - 1006 = 1008 numbers. So, we're left with 1008 numbers, and I need to show that among these, there are two numbers where one divides the other.I remember something about the Pigeonhole Principle, which might be useful here. The principle states that if you have more pigeons than pigeonholes, at least two pigeons must share a pigeonhole. Maybe I can apply this idea somehow.Let me think about the structure of the numbers from 1 to 2014. Each number can be expressed as 2^k * m, where m is an odd number. For example, 12 can be written as 2^2 * 3, where 3 is odd. Similarly, 15 is 2^0 * 15, since 15 is already odd. So, every number has a unique odd part and a power of two part.Now, if I consider the odd parts of these numbers, how many unique odd numbers are there between 1 and 2014? Well, the odd numbers are 1, 3, 5, ..., up to 2013. Let me calculate how many that is. Since 2014 divided by 2 is 1007, there are 1007 odd numbers in this range.So, there are 1007 unique odd numbers. Now, if I have 1008 numbers left on the board after Petya erases 1006, each of these numbers has an odd part. By the Pigeonhole Principle, since there are more numbers (1008) than unique odd parts (1007), at least two of these numbers must share the same odd part.Let me denote these two numbers as a and b. So, both a and b have the same odd part, say m. Then, we can write a = 2^k * m and b = 2^l * m, where k and l are non-negative integers. Without loss of generality, let's assume that k < l. Then, a = 2^k * m and b = 2^l * m. Since k < l, 2^k divides 2^l, which means that a divides b.Therefore, among the remaining numbers, there must be two numbers where one is a divisor of the other. That seems to make sense.Wait, let me double-check my reasoning. I considered the odd parts of the numbers and used the Pigeonhole Principle to show that two numbers must share the same odd part. Then, since they differ by powers of two, one must divide the other. That seems correct.Is there another way to think about this? Maybe considering the binary representations or something else? Hmm, not sure. The approach I took seems straightforward and relies on the fundamental theorem of arithmetic, where each number can be uniquely expressed as a product of primes, specifically separating the power of two and the odd part.Another thought: if I didn't use the Pigeonhole Principle, could I approach this problem differently? Perhaps by induction or some other combinatorial method? But induction might be complicated here because the range is quite large. The Pigeonhole Principle seems the most efficient way.Also, considering that 2014 is an even number, and we're dealing with divisors, it's natural to separate the numbers into their odd and even components. By focusing on the odd parts, we can apply the Pigeonhole Principle effectively.So, to summarize my thought process:1. Recognize that each number can be written as 2^k * m, where m is odd.2. Note that there are 1007 unique odd numbers between 1 and 2014.3. After erasing 1006 numbers, 1008 remain.4. By the Pigeonhole Principle, at least two of these must share the same odd part.5. These two numbers will then be of the form 2^k * m and 2^l * m, so one divides the other.This seems solid. I don't see any flaws in this reasoning. It's a classic application of the Pigeonhole Principle in number theory.Just to make sure, let me test this with a smaller example. Suppose we have numbers from 1 to 4. If Petya erases 2 numbers, leaving 2 numbers. Let's see:Numbers: 1, 2, 3, 4.If Petya erases, say, 2 and 4, leaving 1 and 3. Neither divides the other. Wait, that's a problem. Hmm, so in this case, it doesn't hold. But in our original problem, the number of erased numbers is exactly half, so maybe the smaller example isn't directly applicable.Wait, in the original problem, we have 2014 numbers, erase 1006, leaving 1008. In my smaller example, 4 numbers, erase 2, leaving 2. But in the original problem, the number of remaining numbers is just one more than the number of unique odd parts. So, in the smaller example, if I have numbers 1 to 4, unique odd parts are 1, 3, which is 2. If I have 3 numbers remaining, then by Pigeonhole, two must share the same odd part.Wait, in my previous smaller example, I only had 2 numbers remaining, which is equal to the number of unique odd parts. So, in that case, it's possible that they don't share the same odd part. But in the original problem, we have one more number than the number of unique odd parts, so it's guaranteed.So, in the smaller example, if I have numbers 1 to 4, and I erase 1 number, leaving 3. Then, among those 3, two must share the same odd part. Let's test that.Numbers: 1, 2, 3, 4.If I erase, say, 4, leaving 1, 2, 3. The odd parts are 1, 1, 3. So, 1 appears twice. So, 1 divides both 1 and 2. Similarly, if I erase 3, leaving 1, 2, 4. The odd parts are 1, 1, 1. So, 1 divides all.If I erase 2, leaving 1, 3, 4. The odd parts are 1, 3, 1. So, 1 appears twice, and 3 once. So, 1 divides both 1 and 4.So, in this case, it works. So, the key is that when the number of remaining numbers exceeds the number of unique odd parts, the Pigeonhole Principle ensures that at least two share the same odd part, leading to one dividing the other.Therefore, in the original problem, since we have 1008 numbers and only 1007 unique odd parts, it's guaranteed that two numbers share the same odd part, and thus one divides the other.I think this reasoning is correct. I don't see any issues with it. It's a neat application of the Pigeonhole Principle in number theory.

💡First, recall that in an arithmetic sequence, each term is obtained by adding a common difference ( d ) to the previous term. The general formula for the ( n )-th term of an arithmetic sequence is:[a_n = a_1 + (n - 1)d]Given:[a_3 = -5 quad text{and} quad a_7 = -1]Using the general formula for the third term:[a_3 = a_1 + 2d = -5]And for the seventh term:[a_7 = a_1 + 6d = -1]Set up a system of equations:[begin{cases}a_1 + 2d = -5 a_1 + 6d = -1end{cases}]Subtract the first equation from the second to eliminate ( a_1 ):[(a_1 + 6d) - (a_1 + 2d) = -1 - (-5) 4d = 4 d = 1]Now, substitute ( d = 1 ) back into the first equation to find ( a_1 ):[a_1 + 2(1) = -5 a_1 = -7]Finally, find ( a_5 ) using the general formula:[a_5 = a_1 + 4d = -7 + 4(1) = -3]

💡First, I need to understand the problem. We have a two-digit number, and when we interchange its digits, the difference between the original number and the new number is some value. Additionally, the difference between the two digits themselves is 9.To approach this, I'll denote the two digits of the number as ( x ) (the tens digit) and ( y ) (the units digit). The original number can be expressed as ( 10x + y ), and the number with interchanged digits is ( 10y + x ).The problem states that the difference between the two digits is 9, which gives me the equation:[ x - y = 9 ]Next, I need to find the difference between the original number and the number with interchanged digits:[ (10x + y) - (10y + x) ]Simplifying this expression:[ 10x + y - 10y - x = 9x - 9y ]Since I know from the equation ( x - y = 9 ), I can substitute 9 for ( x - y ) in the expression:[ 9x - 9y = 9(x - y) ][ 9x - 9y = 9(9) ][ 9x - 9y = 81 ]Therefore, the difference between the original number and the number with interchanged digits is 81.