💡First, Janet starts with 10 action figures.She sells 6 of them, leaving her with 10 - 6 = 4 action figures.Then, she acquires 4 more action figures in better condition, increasing her collection to 4 + 4 = 8 action figures.Her brother gives her a collection that is twice the size of what she currently has. Therefore, her brother gives her 2 * 8 = 16 action figures.Finally, Janet's total number of action figures is 8 (her current collection) + 16 (her brother's collection) = 24 action figures.

💡Okay, so I have this problem where I need to find the ratio of the sum of the odd divisors of N to the sum of the even divisors of N. N is given as 34 multiplied by 34, multiplied by 63, multiplied by 270. Hmm, that sounds a bit complicated, but I think I can break it down step by step.First, I remember that to find the sum of divisors, whether odd or even, it's helpful to have the prime factorization of the number. So, I should start by prime factorizing each of the components of N.Let me write down each number and factorize it:- 34: I know that 34 is 2 times 17. Both 2 and 17 are prime numbers, so that's straightforward.- 63: 63 is 7 times 9, and 9 is 3 squared. So, 63 is 3 squared times 7.- 270: Let's see, 270 divided by 2 is 135. 135 is 5 times 27, and 27 is 3 cubed. So, 270 is 2 times 3 cubed times 5.So, putting it all together, N is 34 times 34 times 63 times 270. Substituting the factorizations:- 34 is 2 times 17, so 34 times 34 is (2 times 17) squared, which is 2 squared times 17 squared.- 63 is 3 squared times 7.- 270 is 2 times 3 cubed times 5.Multiplying all these together, let's collect like terms:- For the prime number 2: we have 2 squared from the first 34 times 34, and another 2 from 270. So, total exponent for 2 is 2 + 1 = 3.- For the prime number 3: we have 3 squared from 63 and 3 cubed from 270. So, total exponent for 3 is 2 + 3 = 5.- For the prime number 5: we have 5 from 270, so exponent is 1.- For the prime number 7: we have 7 from 63, so exponent is 1.- For the prime number 17: we have 17 squared from the two 34s, so exponent is 2.So, the prime factorization of N is 2 cubed times 3 to the fifth power times 5 times 7 times 17 squared.Now, to find the sum of the odd divisors, I remember that odd divisors cannot have any factor of 2. So, I can ignore the 2 cubed part and just consider the rest of the prime factors.The formula for the sum of divisors is the product of (1 + p + p squared + ... + p^k) for each prime p raised to the kth power. So, for the odd part, which is 3^5 times 5^1 times 7^1 times 17^2, the sum of the odd divisors would be:(1 + 3 + 3^2 + 3^3 + 3^4 + 3^5) times (1 + 5) times (1 + 7) times (1 + 17 + 17^2).Let me compute each part step by step.First, for 3^5:1 + 3 + 9 + 27 + 81 + 243. Let me add these up:1 + 3 = 44 + 9 = 1313 + 27 = 4040 + 81 = 121121 + 243 = 364So, the sum for 3^5 is 364.Next, for 5^1:1 + 5 = 6For 7^1:1 + 7 = 8For 17^2:1 + 17 + 289. Let's compute that:1 + 17 = 1818 + 289 = 307So, the sum for 17^2 is 307.Now, multiplying all these together:364 * 6 * 8 * 307.Hmm, that seems like a big number, but maybe I don't need to compute it fully because I might just need the ratio.Wait, the problem asks for the ratio of the sum of the odd divisors to the sum of the even divisors. Maybe I can find a smarter way without computing the actual sums.I remember that the sum of all divisors is equal to the sum of the odd divisors multiplied by the sum of the divisors contributed by the power of 2. Since N has 2^3, the sum contributed by 2 is (1 + 2 + 4 + 8) = 15.So, the sum of all divisors is 15 times the sum of the odd divisors.Therefore, sum of all divisors = 15 * sum of odd divisors.But the sum of even divisors is just the total sum minus the sum of odd divisors. So:Sum of even divisors = 15 * sum of odd divisors - sum of odd divisors = 14 * sum of odd divisors.Therefore, the ratio of sum of odd divisors to sum of even divisors is sum of odd divisors : 14 * sum of odd divisors, which simplifies to 1:14.Wait, that seems too straightforward. Let me verify.Alternatively, I could compute the sum of odd divisors as I started earlier, which is 364 * 6 * 8 * 307. Let me compute that:First, 364 * 6 = 21842184 * 8 = 1747217472 * 307. Hmm, that's a big number, but maybe I don't need the exact value.But since the ratio is 1:14, as I found earlier, maybe that's correct.Alternatively, if I think about the sum of all divisors being 15 times the sum of odd divisors, then the sum of even divisors is 14 times the sum of odd divisors, so the ratio is 1:14.Yeah, that makes sense. So, the answer should be 1:14, which is option C.

💡Alright, so I've got this problem here: three different positive integers a, b, and c add up to 133. Also, the sum of any two of these numbers is a perfect square. I need to find the values of a, b, and c. Hmm, okay, let me try to figure this out step by step.First, let's write down what we know. We have three numbers a, b, and c, all positive integers, and they're all different. Their sum is 133, so:a + b + c = 133.Also, the sum of any two of them is a perfect square. That means:a + b = perfect square,a + c = perfect square,b + c = perfect square.So, each pair adds up to a square number. Let me denote these perfect squares as follows:Let’s say:a + b = x²,a + c = y²,b + c = z²,where x, y, z are integers greater than or equal to 1.Now, if I add all three equations together, I get:(a + b) + (a + c) + (b + c) = x² + y² + z².Simplifying the left side:2a + 2b + 2c = x² + y² + z².But from the original equation, a + b + c = 133, so 2a + 2b + 2c = 266. Therefore:x² + y² + z² = 266.So, now I need to find three perfect squares that add up to 266. Let me think about what perfect squares are possible here.First, let's list the perfect squares less than 266:1² = 1,2² = 4,3² = 9,4² = 16,5² = 25,6² = 36,7² = 49,8² = 64,9² = 81,10² = 100,11² = 121,12² = 144,13² = 169,14² = 196,15² = 225,16² = 256.Okay, so the squares go up to 256, but since we need three squares adding to 266, let's see what combinations might work.I need to find x², y², z² such that x² + y² + z² = 266.Let me try starting with the largest possible square less than 266, which is 225 (15²). If I take 225, then the remaining two squares need to add up to 266 - 225 = 41.Looking for two squares that add up to 41. Let's see:36 + 5 = 41, but 5 isn't a perfect square. 25 + 16 = 41, which are 5² and 4². So, 225 + 25 + 16 = 266.So, one possible set is 15², 5², and 4².Alternatively, let's see if there are other combinations. Maybe 196 (14²). Then, 266 - 196 = 70. Now, looking for two squares that add up to 70.64 + 6 = 70, but 6 isn't a square. 49 + 21 = 70, 21 isn't a square. 36 + 34 = 70, 34 isn't a square. 25 + 45 = 70, 45 isn't a square. 16 + 54 = 70, 54 isn't a square. 9 + 61 = 70, 61 isn't a square. 4 + 66 = 70, 66 isn't a square. 1 + 69 = 70, 69 isn't a square. So, no luck with 196.How about 169 (13²). Then, 266 - 169 = 97. Looking for two squares that add to 97.81 + 16 = 97, which are 9² and 4². So, 169 + 81 + 16 = 266. That's another set: 13², 9², 4².Another option: 144 (12²). Then, 266 - 144 = 122. Looking for two squares adding to 122.121 + 1 = 122, which are 11² and 1². So, 144 + 121 + 1 = 266. That's another set: 12², 11², 1².Wait, but 1² is just 1, which is a perfect square, but when we get back to the original numbers a, b, c, they need to be positive integers. Let me check if that would cause any issues. If one of the squares is 1, then one of the sums a + c or b + c would be 1, which would mean one of the numbers is 1 - the other, but since all numbers are positive, that would require one of them to be 0, which isn't allowed. So, that set might not be valid because it would result in a zero or negative number. So, maybe that combination is invalid.Similarly, let's check the other combinations.So, we have two possible sets of squares:1. 225, 25, 16 (15², 5², 4²)2. 169, 81, 16 (13², 9², 4²)Let me check if these can lead to valid a, b, c.Starting with the first set: 15²=225, 5²=25, 4²=16.So, we have:a + b = 225,a + c = 25,b + c = 16.Wait, hold on, if a + c = 25 and b + c = 16, then subtracting these two equations:(a + c) - (b + c) = 25 - 16 => a - b = 9.But we also have a + b = 225. So, we have:a + b = 225,a - b = 9.Adding these two equations:2a = 234 => a = 117.Then, b = 225 - a = 225 - 117 = 108.Then, from a + c =25, c =25 - a =25 - 117 = -92.Wait, that can't be. c is negative, which is not allowed. So, this combination leads to a negative number, which is invalid. So, this set of squares is invalid.Hmm, so maybe the first set doesn't work. Let's try the second set: 13²=169, 9²=81, 4²=16.So, we have:a + b = 169,a + c = 81,b + c = 16.Again, subtracting the second and third equations:(a + c) - (b + c) = 81 - 16 => a - b = 65.We also have a + b = 169.So, adding these two equations:(a + b) + (a - b) = 169 + 65 => 2a = 234 => a = 117.Then, b = 169 - a = 169 - 117 = 52.Then, from a + c =81, c =81 - a =81 - 117 = -36.Again, negative number. Hmm, that's not valid either. So, this set also leads to a negative c.Wait, so both of these sets lead to negative numbers. Maybe I made a mistake in assigning which square corresponds to which pair.Perhaps, instead of assigning a + b =225, a + c=25, b + c=16, maybe I should assign differently.Let me try another approach. Let's consider that the largest square is a + b, the next is a + c, and the smallest is b + c. But maybe I need to assign the squares differently.Wait, in the first case, when I assigned a + b=225, a + c=25, b + c=16, that led to c being negative. Maybe if I assign the largest square to a different pair.Let me try assigning a + c=225, a + b=25, b + c=16. Let's see.So, a + c=225,a + b=25,b + c=16.Subtracting the second equation from the first:(a + c) - (a + b) =225 -25 => c - b=200.From the third equation, b + c=16.So, we have:c - b=200,b + c=16.Adding these two equations:2c=216 => c=108.Then, from b + c=16, b=16 -108= -92.Again, negative. Not valid.Alternatively, assign a + b=25, a + c=16, b + c=225.So, a + b=25,a + c=16,b + c=225.Subtracting second from first:(a + b) - (a + c)=25 -16 => b - c=9.From third equation, b + c=225.So, adding:(b - c) + (b + c)=9 +225 => 2b=234 => b=117.Then, c=225 -117=108.From a + b=25, a=25 -117= -92. Again, negative. So, no good.So, seems like this set of squares doesn't work because it leads to negative numbers. Maybe I need to try another set of squares.Wait, earlier I had another set: 144, 121, 1. But that led to one of the numbers being zero or negative. Let me check.So, 144=12², 121=11², 1=1².Assigning:a + b=144,a + c=121,b + c=1.But b + c=1, which would mean b and c are both 0 or 1, but since they are positive integers, they can't be zero. So, one of them would have to be 1 and the other 0, which isn't allowed. So, invalid.Alternatively, assign:a + b=121,a + c=144,b + c=1.But again, b + c=1 is invalid.Alternatively, assign:a + b=1,a + c=121,b + c=144.But a + b=1 would mean a and b are 1 and 0, which is invalid.So, that set is invalid.Hmm, so maybe the only other set is 169, 81, 16, but that also led to negative numbers when I assigned them as a + b=169, a + c=81, b + c=16.Wait, maybe I should try a different assignment. Let's see.Suppose a + b=169,a + c=16,b + c=81.So, subtracting the second equation from the first:(a + b) - (a + c)=169 -16 => b - c=153.From the third equation, b + c=81.So, adding:(b - c) + (b + c)=153 +81 => 2b=234 => b=117.Then, c=81 -117= -36. Again, negative.Alternatively, assign a + b=16,a + c=169,b + c=81.Subtracting first from second:(a + c) - (a + b)=169 -16 => c - b=153.From third equation, b + c=81.So, adding:(c - b) + (b + c)=153 +81 => 2c=234 => c=117.Then, b=81 -117= -36. Again, negative.Alternatively, assign a + b=81,a + c=169,b + c=16.Subtracting first from second:(a + c) - (a + b)=169 -81 => c - b=88.From third equation, b + c=16.Adding:(c - b) + (b + c)=88 +16 => 2c=104 => c=52.Then, b=16 -52= -36. Negative again.So, no luck with that set either.Wait, maybe I need to consider that the squares don't have to be in any particular order. Maybe I need to try different combinations.Let me think. Maybe the squares are 121, 81, 64.121 +81 +64=266. Yes, that adds up.So, 11²=121, 9²=81, 8²=64.Let me try assigning these.So, a + b=121,a + c=81,b + c=64.Let me subtract the second equation from the first:(a + b) - (a + c)=121 -81 => b - c=40.From the third equation, b + c=64.So, adding:(b - c) + (b + c)=40 +64 => 2b=104 => b=52.Then, c=64 -52=12.From a + b=121, a=121 -52=69.So, a=69, b=52, c=12.Let me check if these satisfy all conditions.First, a + b + c=69 +52 +12=133. Good.Now, check the sums:a + b=69 +52=121=11², perfect square.a + c=69 +12=81=9², perfect square.b + c=52 +12=64=8², perfect square.Perfect! All conditions are satisfied, and all numbers are positive integers, and they are all different.So, the solution is a=69, b=52, c=12.But wait, let me check if there are other possible sets of squares that could lead to different solutions.Earlier, I had 169, 81, 16, but that led to negative numbers. Maybe another combination.Wait, let's see, are there other triplets of squares that add up to 266?Let me think. 144 + 100 + 22=266, but 22 isn't a square. 144 + 81 + 41=266, 41 isn't a square. 121 + 100 + 45=266, 45 isn't a square. 121 + 81 + 64=266, which we already did.Another combination: 100 + 100 + 66=266, but 66 isn't a square, and also, we need distinct squares? Wait, the problem doesn't specify that the squares have to be distinct, just that a, b, c are different. So, maybe 100 + 100 + 66, but 66 isn't a square. So, no.Alternatively, 100 + 81 + 85=266, 85 isn't a square. 100 + 64 + 102=266, 102 isn't a square.Wait, 64 + 64 + 138=266, 138 isn't a square. 49 + 49 + 168=266, 168 isn't a square.Hmm, seems like the only triplet of squares that add up to 266 without leading to negative numbers is 121, 81, 64.Therefore, the only solution is a=69, b=52, c=12.I think that's the answer.**Final Answer**The values of (a), (b), and (c) are (boxed{69}), (boxed{52}), and (boxed{12}).

💡Alright, so I have this complex number problem here, and I need to find the value of θ. Let me try to understand what's given and what I need to do step by step.First, the complex number z is given as z = cosθ + i sinθ, where 0 < θ < π. Okay, that looks familiar—it's like the polar form of a complex number on the unit circle since the magnitude is 1. So, z lies somewhere on the unit circle in the complex plane, specifically in the upper half because θ is between 0 and π.Next, ω is defined as ω = (1 - (ẑ)^4)/(1 + z^4). Hmm, ẑ is the complex conjugate of z, right? Since z = cosθ + i sinθ, then ẑ should be cosθ - i sinθ. So, ẑ^4 would be (cosθ - i sinθ)^4. I remember that when you raise a complex number in polar form to a power, you can use De Moivre's theorem, which says (cosθ + i sinθ)^n = cos(nθ) + i sin(nθ). So, applying that, ẑ^4 should be cos(4θ) - i sin(4θ), right?Similarly, z^4 would be cos(4θ) + i sin(4θ). So, substituting these into ω, we have:ω = [1 - (cos4θ - i sin4θ)] / [1 + (cos4θ + i sin4θ)]Let me simplify the numerator and the denominator separately.Numerator: 1 - cos4θ + i sin4θDenominator: 1 + cos4θ + i sin4θSo, ω = (1 - cos4θ + i sin4θ) / (1 + cos4θ + i sin4θ)Now, I need to find the magnitude of ω, which is given as |ω| = √3 / 3. Also, the argument of ω is less than π/2. So, I need to compute |ω| and set it equal to √3 / 3, then solve for θ.To find |ω|, I can use the formula for the magnitude of a complex number: |a + ib| = √(a² + b²). But since ω is a quotient of two complex numbers, I can also use the property that |ω| = |numerator| / |denominator|.So, let's compute |numerator| and |denominator|.First, the numerator is (1 - cos4θ) + i sin4θ. So, its magnitude is √[(1 - cos4θ)² + (sin4θ)²].Let me expand that:(1 - cos4θ)² + (sin4θ)² = 1 - 2cos4θ + cos²4θ + sin²4θBut cos²x + sin²x = 1, so this simplifies to:1 - 2cos4θ + 1 = 2 - 2cos4θSo, |numerator| = √(2 - 2cos4θ) = √[2(1 - cos4θ)] = √[2 * 2sin²2θ] (since 1 - cos4θ = 2sin²2θ)Wait, let me check that identity. 1 - cos4θ = 2sin²2θ, yes, that's correct. So, substituting that in:√[2 * 2sin²2θ] = √[4sin²2θ] = 2|sin2θ|Since θ is between 0 and π, 2θ is between 0 and 2π, so sin2θ can be positive or negative. But since we're taking the magnitude, it's 2|sin2θ|.Similarly, the denominator is (1 + cos4θ) + i sin4θ. Its magnitude is √[(1 + cos4θ)² + (sin4θ)²].Expanding that:(1 + cos4θ)² + (sin4θ)² = 1 + 2cos4θ + cos²4θ + sin²4θAgain, cos²x + sin²x = 1, so this becomes:1 + 2cos4θ + 1 = 2 + 2cos4θSo, |denominator| = √(2 + 2cos4θ) = √[2(1 + cos4θ)] = √[2 * 2cos²2θ] (since 1 + cos4θ = 2cos²2θ)So, √[4cos²2θ] = 2|cos2θ|Again, since θ is between 0 and π, 2θ is between 0 and 2π, so cos2θ can be positive or negative. But since we're taking the magnitude, it's 2|cos2θ|.Therefore, |ω| = |numerator| / |denominator| = (2|sin2θ|) / (2|cos2θ|) = |tan2θ|Given that |ω| = √3 / 3, so |tan2θ| = √3 / 3So, tan2θ = ±√3 / 3Now, let's solve for 2θ.tan2θ = √3 / 3 or tan2θ = -√3 / 3We know that tan(π/6) = √3 / 3, so the solutions for 2θ in [0, 2π) are:2θ = π/6 + kπ, where k is an integer.So, 2θ = π/6, 7π/6, 13π/6, etc. But since θ is between 0 and π, 2θ is between 0 and 2π, so the possible solutions are 2θ = π/6, 7π/6.Thus, θ = π/12 or θ = 7π/12.But we also have the condition that arg ω < π/2. So, we need to check which of these θ values satisfy that condition.First, let's recall that ω = (1 - ẑ^4)/(1 + z^4). Let's express ω in terms of θ.We have z^4 = cos4θ + i sin4θ and ẑ^4 = cos4θ - i sin4θ.So, ω = [1 - (cos4θ - i sin4θ)] / [1 + (cos4θ + i sin4θ)] = [1 - cos4θ + i sin4θ] / [1 + cos4θ + i sin4θ]Let me write this as (a + ib)/(c + id), where a = 1 - cos4θ, b = sin4θ, c = 1 + cos4θ, d = sin4θ.To find the argument of ω, we can compute arg(ω) = arg(numerator) - arg(denominator).Alternatively, since ω is a quotient, arg(ω) = arg(numerator) - arg(denominator).But maybe it's easier to compute ω in terms of tan or something.Alternatively, let's rationalize ω by multiplying numerator and denominator by the conjugate of the denominator.So, ω = [ (1 - cos4θ + i sin4θ) * (1 + cos4θ - i sin4θ) ] / [ (1 + cos4θ + i sin4θ) * (1 + cos4θ - i sin4θ) ]The denominator becomes (1 + cos4θ)^2 + (sin4θ)^2 = 1 + 2cos4θ + cos²4θ + sin²4θ = 2 + 2cos4θ, as before.The numerator is (1 - cos4θ)(1 + cos4θ) + (1 - cos4θ)(-i sin4θ) + (i sin4θ)(1 + cos4θ) + (i sin4θ)(-i sin4θ)Wait, that might be messy. Alternatively, use the formula (a + ib)(c - id) = (ac + bd) + i(bc - ad).So, numerator:(1 - cos4θ)(1 + cos4θ) + (sin4θ)(sin4θ) + i [ (1 - cos4θ)(-sin4θ) + sin4θ(1 + cos4θ) ]Let's compute real and imaginary parts.Real part: (1 - cos4θ)(1 + cos4θ) + (sin4θ)^2 = 1 - cos²4θ + sin²4θBut 1 - cos²4θ = sin²4θ, so Real part = sin²4θ + sin²4θ = 2 sin²4θImaginary part: (1 - cos4θ)(-sin4θ) + sin4θ(1 + cos4θ) = -sin4θ + sin4θ cos4θ + sin4θ + sin4θ cos4θ = 2 sin4θ cos4θSo, numerator = 2 sin²4θ + i 2 sin4θ cos4θDenominator = 2 + 2 cos4θSo, ω = [2 sin²4θ + i 2 sin4θ cos4θ] / [2 + 2 cos4θ] = [sin²4θ + i sin4θ cos4θ] / [1 + cos4θ]We can factor sin4θ in the numerator:= sin4θ [sin4θ + i cos4θ] / [1 + cos4θ]Hmm, interesting. Let's see if we can simplify this further.Note that sin4θ + i cos4θ can be written as i (cos4θ - i sin4θ) = i ẑ^4, but I'm not sure if that helps.Alternatively, let's express sin4θ and cos4θ in terms of tan2θ.Wait, since we have tan2θ = ±√3 / 3, which we found earlier.Let me think about the argument of ω.arg(ω) = arctan(Imaginary part / Real part) = arctan[ (2 sin4θ cos4θ) / (2 sin²4θ) ] = arctan[ (cos4θ) / sin4θ ] = arctan(cot4θ) = arctan(tan(π/2 - 4θ)) = π/2 - 4θBut we have to be careful with the quadrant.Wait, let's see:Imaginary part is 2 sin4θ cos4θ = sin8θ (using double angle identity: sin2x = 2 sinx cosx, so sin8θ = 2 sin4θ cos4θ)Wait, no, wait: 2 sin4θ cos4θ = sin8θ, yes.And Real part is 2 sin²4θ.So, arg(ω) = arctan(sin8θ / (2 sin²4θ))But sin8θ = 2 sin4θ cos4θ, so:arg(ω) = arctan( (2 sin4θ cos4θ) / (2 sin²4θ) ) = arctan( cos4θ / sin4θ ) = arctan(cot4θ) = arctan(tan(π/2 - 4θ)) = π/2 - 4θBut this is only true if π/2 - 4θ is in the correct quadrant.Alternatively, since tan(arg(ω)) = cot4θ, which is tan(π/2 - 4θ), so arg(ω) = π/2 - 4θ + kπ.But we need to find the principal value, so we have to consider the signs of the real and imaginary parts.Given that |ω| = √3 / 3, which is positive, and arg ω < π/2, so ω is in the first quadrant.Therefore, both the real and imaginary parts of ω are positive.Looking back at the expression for ω:ω = [sin²4θ + i sin4θ cos4θ] / [1 + cos4θ]So, the real part is sin²4θ / (1 + cos4θ), and the imaginary part is sin4θ cos4θ / (1 + cos4θ)Since both real and imaginary parts are positive, we have:sin²4θ > 0 and sin4θ cos4θ > 0Which implies that sin4θ > 0 and cos4θ > 0, because sin²4θ is always non-negative, and for sin4θ cos4θ > 0, both sin4θ and cos4θ must be of the same sign.So, sin4θ > 0 and cos4θ > 0.Let's find the range of θ where this is true.Given θ is between 0 and π, so 4θ is between 0 and 4π.We need to find 4θ in the first quadrant, i.e., between 0 and π/2, or in the third quadrant, between π and 3π/2, but since sin4θ > 0 and cos4θ > 0, it's only in the first quadrant.Wait, no, in the first quadrant, both sin and cos are positive, and in the third quadrant, both are negative. But since we have sin4θ > 0 and cos4θ > 0, it's only in the first quadrant.So, 4θ must be in (0, π/2) or (2π, 5π/2), but since 4θ is between 0 and 4π, the possible intervals are:4θ ∈ (0, π/2) or 4θ ∈ (2π, 5π/2)But 5π/2 is 2.5π, which is less than 4π, so 4θ ∈ (0, π/2) ∪ (2π, 5π/2)But 4θ is between 0 and 4π, so 4θ ∈ (0, π/2) ∪ (2π, 5π/2)But 5π/2 is 2.5π, which is less than 4π, so 4θ ∈ (0, π/2) ∪ (2π, 5π/2)But 5π/2 is 2.5π, which is less than 4π, so 4θ ∈ (0, π/2) ∪ (2π, 5π/2)Wait, but 5π/2 is 2.5π, which is less than 4π, so 4θ can be in (0, π/2) or (2π, 5π/2). But 5π/2 is 2.5π, which is less than 4π, so 4θ can be in (0, π/2) or (2π, 5π/2).But 4θ is between 0 and 4π, so 4θ ∈ (0, π/2) ∪ (2π, 5π/2)But 5π/2 is 2.5π, which is less than 4π, so 4θ ∈ (0, π/2) ∪ (2π, 5π/2)Wait, but 5π/2 is 2.5π, which is less than 4π, so 4θ can be in (0, π/2) or (2π, 5π/2). But 5π/2 is 2.5π, which is less than 4π, so 4θ can be in (0, π/2) or (2π, 5π/2).But 5π/2 is 2.5π, which is less than 4π, so 4θ can be in (0, π/2) or (2π, 5π/2).Wait, but 5π/2 is 2.5π, which is less than 4π, so 4θ can be in (0, π/2) or (2π, 5π/2).But 5π/2 is 2.5π, which is less than 4π, so 4θ can be in (0, π/2) or (2π, 5π/2).Wait, I think I'm repeating myself. Let me try to find the specific θ values.We have θ = π/12 or θ = 7π/12.Let's check for θ = π/12:4θ = 4*(π/12) = π/3, which is in the first quadrant (0, π/2). Wait, π/3 is approximately 1.047, which is less than π/2 (1.5708). So, 4θ = π/3 is in (0, π/2), so sin4θ > 0 and cos4θ > 0, which satisfies our condition.Now, for θ = 7π/12:4θ = 4*(7π/12) = 7π/3, which is equivalent to 7π/3 - 2π = π/3. Wait, 7π/3 is more than 2π, so subtracting 2π gives π/3, which is in the first quadrant. But 7π/3 is actually in the first quadrant as well, since 7π/3 = 2π + π/3, which is coterminal with π/3. So, sin7π/3 = sinπ/3 = √3/2 > 0, and cos7π/3 = cosπ/3 = 1/2 > 0. So, both sin4θ and cos4θ are positive, which satisfies our condition.Wait, but 4θ = 7π/3 is more than 2π, but when considering the unit circle, it's the same as π/3, so it's still in the first quadrant.So, both θ = π/12 and θ = 7π/12 satisfy the condition that sin4θ > 0 and cos4θ > 0, meaning arg ω is in the first quadrant, which is less than π/2.But wait, earlier when I calculated arg(ω) = π/2 - 4θ, but that might not be accurate because I didn't consider the periodicity.Let me think again. Since arg(ω) = arctan(Im/Re) = arctan[(sin8θ)/(2 sin²4θ)]But sin8θ = 2 sin4θ cos4θ, so Im/Re = (2 sin4θ cos4θ)/(2 sin²4θ) = cos4θ / sin4θ = cot4θSo, arg(ω) = arctan(cot4θ) = arctan(tan(π/2 - 4θ)) = π/2 - 4θ + kπBut since arg(ω) must be in (-π, π], and we know that arg(ω) < π/2, let's find the appropriate k.Given that 4θ is in (0, π/2) or (2π, 5π/2), as we found earlier.For θ = π/12, 4θ = π/3, so arg(ω) = π/2 - π/3 = π/6 < π/2, which is good.For θ = 7π/12, 4θ = 7π/3, which is equivalent to π/3 in terms of angle, so arg(ω) = π/2 - π/3 = π/6 < π/2, which is also good.Wait, but 4θ = 7π/3 is more than 2π, so when we subtract 2π, it becomes π/3, so arg(ω) = π/2 - π/3 = π/6.So, both θ = π/12 and θ = 7π/12 give arg(ω) = π/6, which is less than π/2.Therefore, both solutions satisfy the given conditions.But let me double-check by plugging in θ = π/12 and θ = 7π/12 into ω and see if the argument is indeed less than π/2.For θ = π/12:z = cos(π/12) + i sin(π/12)z^4 = cos(π/3) + i sin(π/3) = 0.5 + i (√3/2)ẑ^4 = cos(π/3) - i sin(π/3) = 0.5 - i (√3/2)So, ω = [1 - (0.5 - i √3/2)] / [1 + (0.5 + i √3/2)] = [0.5 + i √3/2] / [1.5 + i √3/2]Let me compute this:Multiply numerator and denominator by the conjugate of the denominator:[ (0.5 + i √3/2)(1.5 - i √3/2) ] / [ (1.5 + i √3/2)(1.5 - i √3/2) ]Denominator: (1.5)^2 + (√3/2)^2 = 2.25 + 0.75 = 3Numerator:0.5*1.5 + 0.5*(-i √3/2) + i √3/2*1.5 + i √3/2*(-i √3/2)= 0.75 - i (√3/4) + i (3√3/4) - i² (3/4)= 0.75 + i (2√3/4) + 3/4 (since i² = -1)= 0.75 + 0.5 + i (√3/2)= 1.25 + i (√3/2)So, ω = (1.25 + i √3/2) / 3 = (5/4 + i √3/2) / 3 = 5/12 + i √3/6So, the real part is 5/12 and the imaginary part is √3/6, both positive, so arg(ω) is in the first quadrant, which is less than π/2.Similarly, for θ = 7π/12:z = cos(7π/12) + i sin(7π/12)z^4 = cos(7π/3) + i sin(7π/3) = cos(π/3) + i sin(π/3) = 0.5 + i √3/2Similarly, ẑ^4 = 0.5 - i √3/2So, ω = [1 - (0.5 - i √3/2)] / [1 + (0.5 + i √3/2)] = same as before, which gives ω = 5/12 + i √3/6So, the argument is the same, π/6, which is less than π/2.Therefore, both θ = π/12 and θ = 7π/12 satisfy the given conditions.But wait, the problem states 0 < θ < π, so both solutions are within this range.Therefore, the solutions are θ = π/12 and θ = 7π/12.I think that's it. Both values of θ satisfy the given conditions.

💡Alright, so I've got this multiple-choice question here, and I need to figure out which statement is correct. The options are A, B, C, and D. Let me take them one by one and think through each.Starting with option A: ∅ ∈ {0}. Okay, so ∅ is the empty set, which means it has no elements. The set {0} contains one element, which is 0. Now, the symbol ∈ means "is an element of." So, is the empty set an element of the set {0}? Hmm, I don't think so. The set {0} only has 0 as its element, not the empty set. So, I think A is incorrect.Moving on to option B: ∅ ⊆ {0}. The symbol ⊆ means "is a subset of." A subset is a set where all its elements are also in the other set. Since the empty set has no elements, it doesn't have any elements that aren't in {0}. In fact, the empty set is a subset of every set because there are no elements in it that could possibly not be in another set. So, I think B is correct. But let me check the other options just to be sure.Option C: 0 ⊊ ∅. The symbol ⊊ means "is a proper subset of." Wait, 0 is an element, not a set, right? So, can an element be a subset of something? I don't think so. Subsets are relationships between sets, not between elements and sets. Also, the empty set has no elements, so 0 can't be a subset of it. Therefore, C is definitely incorrect.Lastly, option D: ∅ ∉ ℂℝ∅. Hmm, this one looks a bit confusing. I'm not entirely sure what ℂℝ∅ stands for. Maybe it's a typo or some notation I'm not familiar with. If I interpret ℂℝ as the set of complex numbers or something like that, but ∅ is the empty set. So, is the empty set not an element of the complex numbers? Well, complex numbers are numbers like a + bi, where a and b are real numbers. The empty set isn't a number, so it doesn't belong to the set of complex numbers. But I'm not entirely sure about this notation. Maybe ℂℝ∅ refers to something else? Without more context, it's hard to say, but I think D is also incorrect because the empty set isn't a complex number.So, after going through all the options, I'm pretty confident that option B is the correct one because the empty set is always a subset of any set, including {0}. The other options either involve incorrect relationships or are based on misunderstandings of set theory basics.

💡First, I need to determine the total number of strawberries and tomatoes Nathan harvested. He planted 5 strawberry plants and each plant yielded 14 strawberries, so the total strawberries are 5 multiplied by 14, which equals 70 strawberries. Similarly, he planted 7 tomato plants and each plant produced 16 tomatoes, so the total tomatoes are 7 multiplied by 16, resulting in 112 tomatoes.Next, I'll calculate how many baskets of each fruit Nathan made. Since each basket holds 7 strawberries, the number of strawberry baskets is 70 strawberries divided by 7, which equals 10 baskets. For tomatoes, the number of baskets is 112 tomatoes divided by 7, resulting in 16 baskets.Now, I know that Nathan sold each basket of tomatoes for 6. Therefore, the total revenue from tomato baskets is 16 baskets multiplied by 6 per basket, which equals 96.Given that Nathan's total revenue from both strawberries and tomatoes was 186, I can find out the revenue from strawberry baskets by subtracting the revenue from tomatoes. So, 186 minus 96 equals 90, which is the total revenue from strawberry baskets.Finally, to determine the price per basket of strawberries, I'll divide the total revenue from strawberries by the number of strawberry baskets. That is, 90 divided by 10 baskets equals 9 per basket.Therefore, Nathan sold each basket of strawberries for 9.