💡Okay, so I have this quadratic equation to solve: 3x² + 11x - 20 = 0. The problem specifies that x is greater than 0, so I need to find the positive solution. I remember that quadratic equations can be solved in a few different ways—factoring, completing the square, or using the quadratic formula. I'm not sure if this equation factors nicely, so maybe I should try factoring first because it's usually quicker if it works.Let me write down the equation again: 3x² + 11x - 20 = 0. To factor this, I need two numbers that multiply to 3 * (-20) = -60 and add up to 11. Hmm, let's think about the factors of 60. The pairs are 1 and 60, 2 and 30, 3 and 20, 4 and 15, 5 and 12, 6 and 10. Since the product is negative, one number has to be positive and the other negative. I need them to add up to 11, which is positive, so the larger number should be positive.Let me try 15 and -4. 15 * (-4) = -60, and 15 + (-4) = 11. Perfect! So I can rewrite the middle term using these numbers. That would give me:3x² + 15x - 4x - 20 = 0Now, I can factor by grouping. Let's group the first two terms and the last two terms:(3x² + 15x) + (-4x - 20) = 0Factor out the greatest common factor from each group:3x(x + 5) - 4(x + 5) = 0Now, I notice that both terms have a common factor of (x + 5), so I can factor that out:(3x - 4)(x + 5) = 0Now, set each factor equal to zero and solve for x:First factor: 3x - 4 = 03x = 4x = 4/3Second factor: x + 5 = 0x = -5But the problem says x > 0, so I discard the negative solution. Therefore, the solution is x = 4/3.Wait, let me double-check my factoring to make sure I didn't make a mistake. If I expand (3x - 4)(x + 5), I should get back the original equation.Multiply 3x by x: 3x²Multiply 3x by 5: 15xMultiply -4 by x: -4xMultiply -4 by 5: -20Combine like terms: 3x² + 15x - 4x - 20 = 3x² + 11x - 20. Yep, that's correct.Just to be thorough, I can also use the quadratic formula to verify. The quadratic formula is x = [-b ± sqrt(b² - 4ac)] / (2a). For the equation 3x² + 11x - 20 = 0, a = 3, b = 11, c = -20.Calculate the discriminant: b² - 4ac = 11² - 4*3*(-20) = 121 + 240 = 361. The square root of 361 is 19.So, x = [-11 ± 19] / (2*3) = (-11 ± 19)/6.First solution: (-11 + 19)/6 = 8/6 = 4/3.Second solution: (-11 - 19)/6 = -30/6 = -5.Again, the positive solution is 4/3. So both methods give me the same answer, which makes me confident that it's correct.I think I'm done here. The positive solution is 4/3.

💡First, I recognize that there are 36 positive integers less than or equal to 36.Next, I need to determine how many of these integers are factors of 36. To do this, I'll start by finding the prime factorization of 36, which is (36 = 2^2 cdot 3^2).Using the prime factorization, I can determine the number of divisors. Each divisor can be expressed in the form (2^a cdot 3^b), where (a) can be 0, 1, or 2, and (b) can also be 0, 1, or 2. This gives me 3 choices for (a) and 3 choices for (b).Multiplying the number of choices for (a) and (b) together, I find that there are (3 cdot 3 = 9) divisors of 36.Finally, to find the probability that a randomly selected positive integer less than or equal to 36 is a factor of 36, I divide the number of favorable outcomes (the number of divisors) by the total number of possible outcomes (36). This gives me (frac{9}{36} = frac{1}{4}).

💡Okay, so I need to find the smallest prime number that comes after a sequence of seven consecutive nonprime numbers. Hmm, let me think about how to approach this. First, I know that prime numbers are numbers greater than 1 that have no positive divisors other than 1 and themselves. Nonprime numbers, on the other hand, are either composite numbers or 1. Since 1 is not considered a prime, it can be part of the sequence of nonprimes.The problem is asking for seven consecutive nonprime numbers, and then the next number after that should be a prime. So, essentially, I'm looking for a prime number such that the seven numbers before it are all nonprime.I guess the first step is to list out prime numbers and look for gaps between them. If there's a gap of at least eight numbers between two primes, that means there are seven nonprime numbers in between. So, I need to find the first occurrence where two primes are eight numbers apart, and then the prime after that gap will be my answer.Let me start listing the prime numbers and their positions:2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, etc.Now, let's look at the differences between consecutive primes:3 - 2 = 15 - 3 = 27 - 5 = 211 - 7 = 413 - 11 = 217 - 13 = 419 - 17 = 223 - 19 = 429 - 23 = 631 - 29 = 237 - 31 = 641 - 37 = 443 - 41 = 247 - 43 = 453 - 47 = 659 - 53 = 661 - 59 = 267 - 61 = 671 - 67 = 473 - 71 = 279 - 73 = 683 - 79 = 489 - 83 = 697 - 89 = 8Okay, so the difference between 89 and 97 is 8. That means there are seven numbers between them, right? Let me check what those numbers are.After 89, the next number is 90, then 91, 92, 93, 94, 95, 96, and then 97. So, 90 to 96 are seven consecutive numbers. Are all of these nonprime?Let's verify each one:90: Divisible by 2, 3, 5, etc. Definitely composite.91: Hmm, 91 divided by 7 is 13. So, 7 × 13 = 91. Composite.92: Even number, divisible by 2. Composite.93: Divisible by 3 (9 + 3 = 12, which is divisible by 3). Composite.94: Even number, divisible by 2. Composite.95: Ends with 5, divisible by 5. Composite.96: Even number, divisible by 2. Composite.So, all seven numbers from 90 to 96 are composite. That means 97 is the next prime after seven consecutive nonprimes.Is there a smaller prime that fits this condition? Let me check the differences before 89.Looking back at the differences:After 2, the differences are 1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2, 6, 4, 6, 8.The first difference of 8 is between 89 and 97. Before that, the largest difference was 6. So, there wasn't a gap of seven nonprimes before 89.Therefore, 97 is indeed the smallest prime that occurs after seven consecutive nonprimes.I think that's the answer. To be thorough, maybe I should check the numbers just before 89 to ensure there isn't a smaller gap I missed.Looking at primes before 89:..., 83, 89.Difference is 6, which is less than 8. So, between 83 and 89, there are five numbers: 84, 85, 86, 87, 88. That's only five numbers, so not seven.Similarly, the difference before that was between 79 and 83, which is 4. So, only three numbers in between.Continuing back, the differences are smaller or equal to 6, so no seven-number gaps before 89.Hence, 97 is the correct answer.

💡Okay, so I've got this problem here with three sets: A, B, and C. Let me try to figure out what each part is asking and how to solve it step by step.First, let me write down what each set is. Set A is defined as all integers x such that the absolute value of x is less than or equal to 6. So, that means A includes all integers from -6 up to 6. Let me list them out to make sure I have it right: A = {-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6}. Yeah, that looks right.Set B is given as {1, 2, 3}, and set C is {3, 4, 5, 6}. Okay, so B has the numbers 1, 2, 3 and C has 3, 4, 5, 6.Now, the problem has two parts:1. Find A ∩ (B ∩ C)2. Find A ∩ C_A(B ∪ C)Hmm, I need to figure out what these operations mean. Let's take them one at a time.Starting with part (1): A ∩ (B ∩ C). I remember that the intersection of two sets, denoted by ∩, is the set of elements that are common to both sets. So, B ∩ C would be the elements that are in both B and C. Let me find that first.Looking at B = {1, 2, 3} and C = {3, 4, 5, 6}, the only element common to both is 3. So, B ∩ C = {3}.Now, I need to find A ∩ {3}. Since A includes all integers from -6 to 6, and {3} is just the single element 3, the intersection would just be {3}, because 3 is in both A and {3}.So, for part (1), the answer is {3}. That seems straightforward.Moving on to part (2): A ∩ C_A(B ∪ C). Hmm, I need to parse this notation. I think C_A might mean the complement of set A, but wait, usually the complement is denoted with a superscript c or something. Maybe it's a typo or notation difference. Alternatively, it could mean the complement of (B ∪ C) with respect to A. That is, all elements in A that are not in (B ∪ C). Let me think about that.First, let's figure out what B ∪ C is. The union of B and C, denoted by ∪, is the set of all elements that are in either B or C or both. So, B = {1, 2, 3} and C = {3, 4, 5, 6}. Combining these, we get {1, 2, 3, 4, 5, 6}. So, B ∪ C = {1, 2, 3, 4, 5, 6}.Now, if C_A(B ∪ C) is the complement of (B ∪ C) with respect to A, that would mean all elements in A that are not in (B ∪ C). So, A is {-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6}, and (B ∪ C) is {1, 2, 3, 4, 5, 6}. Therefore, the complement would be all elements in A that are not in {1, 2, 3, 4, 5, 6}. So, that would be the negative numbers and zero: {-6, -5, -4, -3, -2, -1, 0}.So, C_A(B ∪ C) = {-6, -5, -4, -3, -2, -1, 0}.Now, we need to find A ∩ C_A(B ∪ C). But wait, A is the universal set here, right? Because C_A(B ∪ C) is defined with respect to A. So, intersecting A with its own subset would just give us that subset. So, A ∩ C_A(B ∪ C) is just C_A(B ∪ C), which is {-6, -5, -4, -3, -2, -1, 0}.Let me double-check that. If I take A, which is all integers from -6 to 6, and intersect it with the complement of (B ∪ C) in A, which is the negative numbers and zero, then yes, the intersection is just those negative numbers and zero because they are already in A.So, part (2) is {-6, -5, -4, -3, -2, -1, 0}.Wait, let me make sure I didn't make a mistake in interpreting C_A(B ∪ C). If C_A meant something else, like the Cartesian product or something, but I don't think so. Given the context, it's more likely the complement. Also, since the problem is about set operations, complement makes more sense here.Alternatively, if C_A was a typo and they meant C^A, but that usually denotes the set of functions from A to C, which doesn't fit here. So, I think my initial interpretation is correct.Just to recap:1. A ∩ (B ∩ C) = A ∩ {3} = {3}2. A ∩ C_A(B ∪ C) = A ∩ {-6, -5, -4, -3, -2, -1, 0} = {-6, -5, -4, -3, -2, -1, 0}I think that's it. I don't see any mistakes in my reasoning.

💡First, I need to understand the original equation provided by the user: 7 3/4 ã · 2 1/2 = 3231. The symbol "ã" is not a recognized mathematical operator, so I'll assume it represents multiplication.Next, I'll convert the mixed numbers to improper fractions for easier calculation. 7 3/4 becomes 31/4 and 2 1/2 becomes 5/2. Multiplying these fractions: (31/4) × (5/2) = 155/8, which simplifies to 19 3/8. This shows that the original equation is incorrect since 19 3/8 does not equal 3231.Moving on to the second part of the problem: 77/4 × 5/2. I'll multiply the numerators and denominators: (77 × 5) / (4 × 2) = 385/8. Converting this to a mixed number gives 48 1/8.Finally, I'll present the solution in a clear and organized manner, ensuring that all steps are understandable and properly formatted.

💡Okay, so I have this geometry problem here. Let me read it again: A moving point P draws two tangents PA and PB to the circle x² + y² = 1. The points where the tangents touch the circle are A and B, respectively. The angle APB is 60 degrees. I need to find the equation of the trajectory of point P, and the options are given as A: x² + y² = 4, B: x² + y² = 3, C: x² + y² = 2, D: x² + y² = 1.Hmm, okay. So, point P is moving such that it always draws two tangents to the unit circle, and the angle between these two tangents at point P is 60 degrees. I need to find the path that point P follows.First, I remember that the set of all points from which tangents can be drawn to a circle lie on another circle called the director circle. But wait, is that right? Or is it something else? Let me think.Actually, the director circle of a circle is the locus of points from which the two tangents to the circle are perpendicular. So, if the angle between the tangents is 90 degrees, then the point lies on the director circle. But in this case, the angle is 60 degrees, not 90. So, maybe it's a different circle.Let me recall the formula for the angle between two tangents from an external point to a circle. I think the angle depends on the distance from the external point to the center of the circle. If the circle has radius r, and the external point is at a distance d from the center, then the angle θ between the two tangents can be found using the formula:θ = 2 arcsin(r / d)Is that correct? Let me verify. So, if I have a circle with radius r, and an external point P at distance d from the center, then the length of the tangent from P to the circle is sqrt(d² - r²). The angle between the two tangents can be found using trigonometry in the triangle formed by the center, point P, and one of the points of tangency.Yes, that triangle is a right triangle because the radius is perpendicular to the tangent at the point of contact. So, in triangle OPA, where O is the center, A is the point of tangency, and P is the external point, we have a right angle at A. The angle at P is half of the angle between the two tangents, so θ/2.So, in triangle OPA, sin(θ/2) = OA / OP. Since OA is the radius r, and OP is the distance d from the center to P, we have:sin(θ/2) = r / dTherefore, d = r / sin(θ/2)In this problem, the circle is x² + y² = 1, so the radius r is 1. The angle θ is 60 degrees, so θ/2 is 30 degrees. Therefore,d = 1 / sin(30°)Sin(30°) is 0.5, so d = 1 / 0.5 = 2.So, the distance from the center to point P is 2. Therefore, the locus of point P is a circle with radius 2 centered at the origin. So, the equation is x² + y² = 4.Looking back at the options, option A is x² + y² = 4. So, that should be the correct answer.Wait, just to make sure I didn't make a mistake. Let me go through the steps again.1. The circle is x² + y² = 1, so radius r = 1.2. Point P is external, drawing two tangents PA and PB, with angle APB = 60°.3. The angle between the two tangents is 60°, so θ = 60°, which means θ/2 = 30°.4. In triangle OPA, sin(θ/2) = OA / OP => sin(30°) = 1 / d.5. Sin(30°) is 0.5, so 0.5 = 1 / d => d = 2.6. Therefore, the distance from the center to P is 2, so the locus is x² + y² = 4.Yes, that seems correct. So, the answer should be A.But just to double-check, let me think about the director circle. The director circle is the locus of points where the angle between the two tangents is 90°, right? For a unit circle, the director circle would have radius sqrt(2), because the formula is sqrt(2)r. So, for r=1, it's sqrt(2). But in this case, the angle is 60°, which is less than 90°, so the radius should be larger than sqrt(2). Since 2 is larger than sqrt(2) (~1.414), that makes sense.Alternatively, if the angle were larger, say 120°, the radius would be smaller. So, as the angle decreases, the radius increases, which is consistent with our result.Therefore, I'm confident that the correct answer is A: x² + y² = 4.**Final Answer**boxed{A}