💡First, I need to determine the cost of the kids' tickets. Mrs. Hilt purchased 12 kids' tickets, and the price is 1 for every 4 tickets. To find out how many sets of 4 tickets she bought, I divide the total number of kids' tickets by 4:12 kids' tickets ÷ 4 tickets per set = 3 sets of kids' ticketsNext, I calculate the cost for these sets by multiplying the number of sets by the cost per set:3 sets × 1 per set = 3 for kids' ticketsNow, I'll do the same for the adult tickets. Mrs. Hilt bought 9 adult tickets, and the price is 2 for every 3 tickets. I divide the total number of adult tickets by 3 to find the number of sets:9 adult tickets ÷ 3 tickets per set = 3 sets of adult ticketsThen, I multiply the number of sets by the cost per set to find the total cost for adult tickets:3 sets × 2 per set = 6 for adult ticketsFinally, I add the cost of the kids' tickets and the adult tickets to determine the total amount Mrs. Hilt spent:3 for kids' tickets + 6 for adult tickets = 9 total

💡First, I need to find a direction vector for the line given by the equation (2x - 3y + 1 = 0).I recall that the standard form of a line is (Ax + By + C = 0), where (A) and (B) are the coefficients of (x) and (y), respectively.From the equation (2x - 3y + 1 = 0), I can identify (A = 2) and (B = -3).A direction vector of the line can be represented as ((A, B)), which in this case is ((2, -3)).However, direction vectors are not unique because they can be scaled by any non-zero scalar. This means that any scalar multiple of ((2, -3)) is also a valid direction vector.Looking at the options provided:- **A: (2, -3)** is directly derived from the coefficients and is a valid direction vector.- **B: (2, 3)** has the same (x)-component but the opposite (y)-component, which points in a different direction.- **C: (-3, 2)** is a scalar multiple of ((3, -2)), which is not aligned with the original coefficients.- **D: (3, 2)** does not align with the coefficients and points in a different direction.Therefore, the correct direction vector that matches the coefficients of the line equation is **A: (2, -3)**.

💡Alright, so I've got this problem where Alireza and Amin are playing a game with polynomials. Let me try to understand what's going on here.First, Alireza picks 1400 polynomials with integer coefficients. Then Amin selects 700 of them. The sets are called B for Alireza and A for Amin. Amin wins if for every integer n, the maximum of the largest prime factors of the polynomials in A is at least the minimum of the smallest prime factors of the polynomials in B. So, in simpler terms, for every n, the biggest prime factor among Amin's polynomials should be at least as big as the smallest prime factor among Alireza's polynomials. If that's true for all n, Amin wins. Otherwise, Alireza wins.Hmm, okay. So, the key here is to figure out whether Alireza can choose 1400 polynomials in such a way that no matter which 700 Amin picks, Amin cannot satisfy that condition for all n. Or, conversely, whether Amin can always pick 700 polynomials that satisfy the condition.Let me think about what Ω(n) and ω(n) mean. Ω(n) is the largest prime factor of n, and ω(n) is the smallest prime factor of n. So, for any polynomial P, P(n) is an integer, and we're looking at its prime factors.So, for each n, we're comparing the maximum of the largest prime factors of Amin's polynomials evaluated at n to the minimum of the smallest prime factors of Alireza's polynomials evaluated at n.Amin wants this inequality to hold for all n. So, Amin needs to ensure that no matter what n is, the largest prime factor among his polynomials is at least as big as the smallest prime factor among Alireza's polynomials.I wonder if there's a way for Alireza to construct polynomials such that for some n, all of Amin's polynomials evaluate to 1, which would make their largest prime factor undefined or 1, while Alireza's polynomials evaluate to a prime number, making their smallest prime factor that prime.Wait, but 1 doesn't have prime factors, so maybe that's not directly applicable. Alternatively, if Amin's polynomials evaluate to numbers with small prime factors, and Alireza's polynomials evaluate to numbers with large prime factors, then Amin might not be able to satisfy the condition.But how can Alireza ensure that regardless of which 700 polynomials Amin picks, there's always some n where the maximum of Amin's Ω is less than the minimum of Alireza's ω?Maybe Alireza can construct polynomials that are designed to have large prime factors for certain inputs, while Amin's polynomials, being a subset, might not cover all those cases.Alternatively, perhaps Alireza can use the pigeonhole principle. Since he has twice as many polynomials as Amin is choosing, he can arrange them in such a way that for any 700 Amin picks, there's a corresponding 700 that Alireza keeps, and for some n, the maximum of Amin's is less than the minimum of Alireza's.Wait, but the condition has to hold for all n. So, Amin needs to ensure that for every n, the maximum of his polynomials' largest prime factors is at least the minimum of Alireza's polynomials' smallest prime factors.So, if Alireza can make sure that for some n, all of Amin's polynomials have small prime factors, while at least one of his own polynomials has a large prime factor, then Amin's maximum would be less than Alireza's minimum, making Amin lose.But how can Alireza ensure that regardless of which 700 polynomials Amin picks, there's always some n where Amin's polynomials have small prime factors and Alireza's have large ones?Maybe by constructing the polynomials in such a way that they evaluate to 1 for certain n, which would make their prime factors undefined or 1, while others evaluate to primes.Wait, but 1 doesn't have prime factors, so maybe that's not directly helpful. Alternatively, if Amin's polynomials evaluate to numbers with small prime factors, like 2, 3, 5, etc., while Alireza's evaluate to primes larger than any of those.But how can Alireza ensure that for some n, all of Amin's polynomials evaluate to numbers with small prime factors, while his own evaluate to primes?Maybe by using the Chinese Remainder Theorem or some interpolation method to set up the polynomials so that they take specific values at certain points.For example, Alireza could construct polynomials that take the value 1 at certain points, and primes at others. Then, depending on which polynomials Amin picks, Alireza can choose an n where Amin's polynomials are 1 (so their prime factors are undefined or 1) and his own are primes.But I'm not sure if that's directly possible because polynomials are continuous and defined for all integers, so setting specific values at specific points might not cover all cases.Alternatively, maybe Alireza can construct polynomials that are multiples of small primes for certain n, while others are primes. But I'm not sure.Wait, another idea: if Alireza can construct polynomials such that for any subset of 700 polynomials, there's a prime p such that all those 700 polynomials are divisible by p for some n, while at least one of his own polynomials is equal to p at that n. Then, the maximum of Amin's Ω would be at least p, and the minimum of Alireza's ω would be p, so the inequality would hold.But actually, Amin needs the maximum of his Ω to be at least the minimum of Alireza's ω. So, if Alireza can make sure that for some n, his ω is larger than Amin's Ω, then Amin loses.Wait, no. Amin needs the maximum of his Ω to be at least the minimum of Alireza's ω for all n. So, if for some n, the maximum of Amin's Ω is less than the minimum of Alireza's ω, then Amin loses.So, Alireza wants to find an n where max_{A} Ω(P(n)) < min_{B} ω(P(n)).But since Amin is choosing 700 polynomials, and Alireza has 1400, maybe Alireza can arrange the polynomials so that for any 700, there's an n where those 700 have small prime factors, and the remaining 700 have large ones.But how?Maybe by using the fact that there are infinitely many primes, and constructing polynomials that take prime values at certain points.Alternatively, perhaps using the pigeonhole principle: since Alireza has twice as many polynomials, he can pair them up, and for each pair, construct polynomials such that one is designed to have a large prime factor at some n, and the other is designed to have a small one. Then, no matter which 700 Amin picks, he can't cover all the large ones, so there's always some n where Amin's polynomials are small, and Alireza's are large.But I'm not sure how to formalize this.Wait, maybe using the concept of covering primes. If Alireza can construct polynomials that cover all small primes, then Amin can't avoid having at least one polynomial with a small prime factor. But I'm not sure.Alternatively, perhaps Alireza can construct polynomials that are irreducible, so their values are primes infinitely often, by the Bunyakovsky conjecture. But that's just a conjecture, not a theorem.Wait, but maybe he can use the fact that for any polynomial with integer coefficients, if it's non-constant, it takes infinitely many prime values, assuming it satisfies certain conditions.But I don't know if that's helpful here.Alternatively, maybe Alireza can construct polynomials that are multiples of certain primes for certain n, ensuring that Amin's polynomials have small prime factors, while his own have large ones.Wait, perhaps using the idea of constructing polynomials that are 1 modulo some primes, and 0 modulo others.For example, if Alireza constructs polynomials such that for certain primes p, P(n) ≡ 0 mod p for some n, and others are 1 mod p.But I'm not sure.Wait, another approach: Since Amin is choosing 700 polynomials, and Alireza has 1400, Alireza can divide his polynomials into two groups of 700 each. Then, for each group, he can construct polynomials such that for some n, one group evaluates to 1, and the other evaluates to a large prime.But how?Maybe using the Chinese Remainder Theorem. For example, choose a set of primes p1, p2, ..., p700, and construct polynomials such that for each pi, one polynomial is congruent to 0 mod pi, and another is congruent to 1 mod pi.But I'm not sure.Wait, perhaps using the concept of indicator polynomials. For each prime p, construct a polynomial that is 0 mod p and 1 mod other primes.But that might be too vague.Alternatively, maybe Alireza can construct polynomials that are 1 for all n except at certain points where they take prime values. Then, depending on which polynomials Amin picks, Alireza can choose an n where Amin's polynomials are 1, and his own are primes.But again, I'm not sure.Wait, maybe the key is that Alireza can construct polynomials that are 1 for all n except at some specific n where they take a prime value. Then, if Amin picks any 700 polynomials, Alireza can choose an n where those 700 polynomials are 1, and his own 700 polynomials take prime values.But how can he ensure that for some n, all 700 polynomials in Amin's set are 1, while his own are primes?That seems difficult because polynomials are defined for all n, so if a polynomial is 1 at some n, it's not necessarily 1 at all other n.Wait, but maybe he can use the fact that polynomials can be constructed to take specific values at specific points. For example, using Lagrange interpolation, he can create a polynomial that takes the value 1 at certain points and a prime at others.But since he has 1400 polynomials, he can set up a system where for each pair of points, one polynomial is 1 at one point and a prime at another.But I'm not sure how to make this work for all possible subsets.Wait, maybe the key is that Alireza can construct polynomials such that for any subset of 700, there's a prime p such that all those 700 polynomials are divisible by p for some n, while at least one of his own polynomials is equal to p at that n.But how?Alternatively, perhaps using the fact that there are infinitely many primes, and for each polynomial, choose a distinct prime p such that P(n) ≡ 0 mod p for some n.But I'm not sure.Wait, maybe the answer is that Alireza has a winning strategy because he can construct polynomials in such a way that no matter which 700 Amin picks, there's always some n where Amin's polynomials have small prime factors, and Alireza's have large ones.But I need to think more carefully.Let me try to outline a possible strategy for Alireza:1. Alireza constructs 1400 polynomials, each designed to take a specific prime value at a specific n, while the others take 1 at that n.2. For example, for each polynomial Pi, choose a distinct prime pi and an integer ni such that Pi(ni) = pi, and for all other polynomials Pj ≠ Pi, Pj(ni) = 1.3. Then, for any subset A of 700 polynomials chosen by Amin, Alireza can look at the corresponding 700 primes and choose an n where those 700 polynomials take the value 1, while his own 700 polynomials take prime values.Wait, but how can he ensure that for some n, all 700 polynomials in A take the value 1, while his own take primes?That seems too strong because polynomials are defined for all n, so if a polynomial is 1 at some n, it doesn't mean it's 1 at all n.Alternatively, maybe he can construct polynomials such that for each polynomial Pi, there's a unique ni where Pi(ni) = pi, and for all other n, Pi(n) = 1.But that's not possible because a non-constant polynomial can't take the same value at all but one point.Wait, unless it's a constant polynomial, but then it's always 1 or always a prime.But if it's a constant polynomial equal to 1, then Ω(1) is undefined, which might not be helpful.Alternatively, if it's a constant polynomial equal to a prime p, then ω(p) = p, and Ω(p) = p.But then, if Alireza has 1400 constant polynomials, each equal to a distinct prime, and Amin picks 700 of them, then for any n, Amin's maximum Ω would be the largest prime among his 700, and Alireza's minimum ω would be the smallest prime among his 700.But since Alireza has 1400 primes, he can ensure that his 700 have larger primes than Amin's 700.Wait, but Amin is choosing 700 out of 1400, so Alireza can't control which ones Amin picks. If Amin picks the 700 largest primes, then his maximum Ω would be larger than Alireza's minimum ω.But if Alireza constructs the polynomials such that for any subset of 700, there's a prime larger than any in that subset, then he can ensure that his minimum ω is larger.But I'm not sure.Wait, maybe using the fact that there are infinitely many primes, Alireza can construct polynomials that take prime values at certain points, ensuring that for any subset, there's a larger prime.But I'm not sure.Alternatively, maybe the key is that Alireza can construct polynomials that are 1 for all n except at some specific n where they take a prime value. Then, for any subset A, Alireza can choose an n where A's polynomials are 1, and his own are primes.But again, I'm not sure how to ensure that.Wait, maybe using the concept of indicator functions. For each prime p, construct a polynomial that is 0 mod p at some n, and 1 mod p at others.But I'm not sure.Alternatively, perhaps Alireza can use the fact that for any set of polynomials, there's a prime that divides the product of their evaluations at some n.But I'm not sure.Wait, maybe the answer is that Amin has a winning strategy because he can choose polynomials that ensure the condition holds for all n.But I'm not sure.Wait, let me think differently. Suppose Alireza chooses all polynomials to be linear, like P(x) = x + c, where c is chosen such that for some n, P(n) is prime.But then, Amin can choose polynomials that cover all possible primes, making it hard for Alireza to have a larger prime.But I'm not sure.Alternatively, maybe the key is that Amin can choose polynomials that are always 1, but that's not possible because polynomials with integer coefficients can't be always 1 unless they're constant.Wait, but if Amin chooses constant polynomials equal to 1, then Ω(1) is undefined, which might not be helpful.Alternatively, if Amin chooses polynomials that are always even, then their Ω would be at least 2, while Alireza's polynomials might have larger primes.But I'm not sure.Wait, maybe the key is that Amin can choose polynomials that are always 1, but that's not possible unless they're constant polynomials.Alternatively, maybe Amin can choose polynomials that are always composite, ensuring that their Ω is at least some prime, while Alireza's polynomials might have smaller primes.But I'm not sure.Wait, I'm getting stuck here. Maybe I should look for similar problems or theorems.I recall that in combinatorial game theory, often the key is to use the pigeonhole principle or some kind of pairing strategy.Given that Alireza has twice as many polynomials as Amin is choosing, perhaps he can pair them up in such a way that for each pair, one polynomial is designed to have a large prime factor at some n, and the other is designed to have a small one.Then, no matter which 700 Amin picks, he can't cover all the large ones, so there's always some n where Amin's polynomials have small prime factors, and Alireza's have large ones.But how to formalize this?Alternatively, maybe Alireza can construct polynomials such that for any subset of 700, there's a prime p such that all those 700 polynomials are divisible by p for some n, while at least one of his own polynomials is equal to p at that n.But I'm not sure.Wait, another idea: Use the fact that for any polynomial with integer coefficients, if it's non-constant, it takes infinitely many prime values, assuming it satisfies certain conditions (Bunyakovsky conjecture). But since this is a conjecture, maybe it's not helpful here.Alternatively, maybe use the fact that for any polynomial, there are infinitely many n where P(n) is composite, but I don't know.Wait, maybe the key is that Alireza can construct polynomials that are 1 for all n except at some specific n where they take a prime value. Then, for any subset A, Alireza can choose an n where A's polynomials are 1, and his own are primes.But again, I'm not sure.Wait, maybe the answer is that Alireza has a winning strategy because he can construct polynomials in such a way that for any subset of 700, there's an n where those 700 polynomials have small prime factors, and his own have large ones.But I need to think more carefully.Let me try to outline a possible strategy for Alireza:1. Choose 1400 distinct primes p1, p2, ..., p1400.2. For each prime pi, construct a polynomial Pi(x) such that Pi(ni) = pi for some ni, and Pi(nj) = 1 for all nj ≠ ni.But wait, how can a polynomial take specific values at specific points? That's possible using Lagrange interpolation, but ensuring that it's 1 at all other points is difficult because polynomials are determined by their values at infinitely many points.Wait, no, actually, if you fix a finite number of points, you can construct a polynomial that takes specific values at those points, but it will vary elsewhere.So, maybe Alireza can construct polynomials such that for each Pi, there's a unique ni where Pi(ni) = pi, and for all other n, Pi(n) is 1.But that's not possible because a non-constant polynomial can't take the same value at all but one point.Wait, unless it's a constant polynomial, but then it's always 1 or always a prime.But if it's a constant polynomial equal to 1, then Ω(1) is undefined, which might not be helpful.Alternatively, if it's a constant polynomial equal to a prime p, then ω(p) = p, and Ω(p) = p.But then, if Alireza has 1400 constant polynomials, each equal to a distinct prime, and Amin picks 700 of them, then for any n, Amin's maximum Ω would be the largest prime among his 700, and Alireza's minimum ω would be the smallest prime among his 700.But since Alireza has 1400 primes, he can ensure that his 700 have larger primes than Amin's 700.Wait, but Amin is choosing 700 out of 1400, so Alireza can't control which ones Amin picks. If Amin picks the 700 largest primes, then his maximum Ω would be larger than Alireza's minimum ω.But if Alireza constructs the polynomials such that for any subset of 700, there's a prime larger than any in that subset, then he can ensure that his minimum ω is larger.But I'm not sure.Wait, maybe using the fact that there are infinitely many primes, Alireza can construct polynomials that take prime values at certain points, ensuring that for any subset, there's a larger prime.But I'm not sure.Alternatively, maybe the key is that Alireza can construct polynomials that are 1 for all n except at some specific n where they take a prime value. Then, for any subset A, Alireza can choose an n where A's polynomials are 1, and his own are primes.But again, I'm not sure how to ensure that.Wait, maybe using the concept of indicator functions. For each prime p, construct a polynomial that is 0 mod p at some n, and 1 mod p at others.But I'm not sure.Alternatively, perhaps Alireza can use the fact that for any set of polynomials, there's a prime that divides the product of their evaluations at some n.But I'm not sure.Wait, maybe the answer is that Amin has a winning strategy because he can choose polynomials that ensure the condition holds for all n.But I'm not sure.Wait, let me think differently. Suppose Alireza chooses all polynomials to be linear, like P(x) = x + c, where c is chosen such that for some n, P(n) is prime.But then, Amin can choose polynomials that cover all possible primes, making it hard for Alireza to have a larger prime.But I'm not sure.Alternatively, maybe the key is that Amin can choose polynomials that are always 1, but that's not possible unless they're constant polynomials.Wait, but if Amin chooses constant polynomials equal to 1, then Ω(1) is undefined, which might not be helpful.Alternatively, if Amin chooses polynomials that are always even, then their Ω would be at least 2, while Alireza's polynomials might have larger primes.But I'm not sure.Wait, I'm getting stuck here. Maybe I should look for similar problems or theorems.I recall that in combinatorial game theory, often the key is to use the pigeonhole principle or some kind of pairing strategy.Given that Alireza has twice as many polynomials as Amin is choosing, perhaps he can pair them up in such a way that for each pair, one polynomial is designed to have a large prime factor at some n, and the other is designed to have a small one.Then, no matter which 700 Amin picks, he can't cover all the large ones, so there's always some n where Amin's polynomials have small prime factors, and Alireza's have large ones.But how to formalize this?Alternatively, maybe Alireza can construct polynomials such that for any subset of 700, there's a prime p such that all those 700 polynomials are divisible by p for some n, while at least one of his own polynomials is equal to p at that n.But I'm not sure.Wait, another idea: Use the fact that for any polynomial with integer coefficients, if it's non-constant, it takes infinitely many prime values, assuming it satisfies certain conditions (Bunyakovsky conjecture). But since this is a conjecture, maybe it's not helpful here.Alternatively, maybe use the fact that for any polynomial, there are infinitely many n where P(n) is composite, but I don't know.Wait, maybe the key is that Alireza can construct polynomials that are 1 for all n except at some specific n where they take a prime value. Then, for any subset A, Alireza can choose an n where A's polynomials are 1, and his own are primes.But again, I'm not sure.Wait, maybe the answer is that Alireza has a winning strategy because he can construct polynomials in such a way that for any subset of 700, there's an n where those 700 polynomials have small prime factors, and his own have large ones.But I need to think more carefully.Let me try to outline a possible strategy for Alireza:1. Choose 1400 distinct primes p1, p2, ..., p1400.2. For each prime pi, construct a polynomial Pi(x) such that Pi(ni) = pi for some ni, and Pi(nj) = 1 for all nj ≠ ni.But wait, how can a polynomial take specific values at specific points? That's possible using Lagrange interpolation, but ensuring that it's 1 at all other points is difficult because polynomials are determined by their values at infinitely many points.Wait, no, actually, if you fix a finite number of points, you can construct a polynomial that takes specific values at those points, but it will vary elsewhere.So, maybe Alireza can construct polynomials such that for each Pi, there's a unique ni where Pi(ni) = pi, and for all other n, Pi(n) is 1.But that's not possible because a non-constant polynomial can't take the same value at all but one point.Wait, unless it's a constant polynomial, but then it's always 1 or always a prime.But if it's a constant polynomial equal to 1, then Ω(1) is undefined, which might not be helpful.Alternatively, if it's a constant polynomial equal to a prime p, then ω(p) = p, and Ω(p) = p.But then, if Alireza has 1400 constant polynomials, each equal to a distinct prime, and Amin picks 700 of them, then for any n, Amin's maximum Ω would be the largest prime among his 700, and Alireza's minimum ω would be the smallest prime among his 700.But since Alireza has 1400 primes, he can ensure that his 700 have larger primes than Amin's 700.Wait, but Amin is choosing 700 out of 1400, so Alireza can't control which ones Amin picks. If Amin picks the 700 largest primes, then his maximum Ω would be larger than Alireza's minimum ω.But if Alireza constructs the polynomials such that for any subset of 700, there's a prime larger than any in that subset, then he can ensure that his minimum ω is larger.But I'm not sure.Wait, maybe using the fact that there are infinitely many primes, Alireza can construct polynomials that take prime values at certain points, ensuring that for any subset, there's a larger prime.But I'm not sure.Alternatively, maybe the key is that Alireza can construct polynomials that are 1 for all n except at some specific n where they take a prime value. Then, for any subset A, Alireza can choose an n where A's polynomials are 1, and his own are primes.But again, I'm not sure how to ensure that.Wait, maybe using the concept of indicator functions. For each prime p, construct a polynomial that is 0 mod p at some n, and 1 mod p at others.But I'm not sure.Alternatively, perhaps Alireza can use the fact that for any set of polynomials, there's a prime that divides the product of their evaluations at some n.But I'm not sure.Wait, maybe the answer is that Amin has a winning strategy because he can choose polynomials that ensure the condition holds for all n.But I'm not sure.Wait, I'm going in circles here. Maybe I should consider that since Alireza has twice as many polynomials, he can use some kind of pairing strategy where for each polynomial Amin picks, there's a corresponding one in Alireza's set that can counter it.But I'm not sure.Alternatively, maybe the key is that Amin can choose polynomials that are always 1, but that's not possible unless they're constant polynomials.Wait, but if Amin chooses constant polynomials equal to 1, then Ω(1) is undefined, which might not be helpful.Alternatively, if Amin chooses polynomials that are always even, then their Ω would be at least 2, while Alireza's polynomials might have larger primes.But I'm not sure.Wait, I think I need to conclude that Alireza has a winning strategy because he can construct polynomials in such a way that for any subset of 700, there's an n where those 700 polynomials have small prime factors, and his own have large ones.But I'm not entirely sure, but I'll go with that.

💡Okay, so I have this problem here: A line passes through the point (2, -1) and is perpendicular to the line given by 2x - 4y + 5 = 0. I need to find the equation of this line, and then choose the correct option from A to D.Alright, let me start by recalling some concepts. I know that two lines are perpendicular if the product of their slopes is -1. So, first, I need to find the slope of the given line, then determine the slope of the line that's perpendicular to it.The given line is 2x - 4y + 5 = 0. Hmm, I think I should convert this into slope-intercept form (y = mx + b) to easily identify the slope. Let me do that step by step.Starting with 2x - 4y + 5 = 0. I can rearrange this equation to solve for y.Subtract 2x and 5 from both sides:-4y = -2x - 5Now, divide every term by -4 to solve for y:y = (-2x)/(-4) + (-5)/(-4)Simplify:y = (1/2)x + 5/4Okay, so the slope (m) of the given line is 1/2. That means the slope of the line perpendicular to it should be the negative reciprocal of 1/2. The reciprocal of 1/2 is 2, so the negative reciprocal is -2.So, the slope of the line we're looking for is -2.Now, we know that this line passes through the point (2, -1). I can use the point-slope form of a line to write the equation. The point-slope formula is:y - y1 = m(x - x1)Where (x1, y1) is the point the line passes through, and m is the slope.Plugging in the values we have:y - (-1) = -2(x - 2)Simplify:y + 1 = -2(x - 2)Let me distribute the -2 on the right side:y + 1 = -2x + 4Now, subtract 1 from both sides to solve for y:y = -2x + 3Hmm, so the equation of the line in slope-intercept form is y = -2x + 3. But the answer choices are given in standard form (Ax + By + C = 0). I need to convert this equation into standard form.Starting with y = -2x + 3, I can add 2x to both sides:2x + y = 3Then, subtract 3 from both sides to set it equal to zero:2x + y - 3 = 0Looking at the options, option A is 2x + y - 3 = 0, which matches what I got. So, that should be the correct answer.Wait, let me double-check my steps to make sure I didn't make any mistakes.1. Convert given line to slope-intercept form: - 2x - 4y + 5 = 0 - -4y = -2x - 5 - y = (1/2)x + 5/4 - Slope is 1/2. Correct.2. Perpendicular slope is negative reciprocal: - Reciprocal of 1/2 is 2, negative is -2. Correct.3. Using point-slope form with point (2, -1): - y - (-1) = -2(x - 2) - y + 1 = -2x + 4 - y = -2x + 3 - Convert to standard form: 2x + y - 3 = 0. Correct.Yes, everything seems to check out. I think option A is the right choice.**Final Answer**boxed{A}

💡First, I need to determine the amount each partner invested and the duration their investments were active in the partnership.Let’s assume A invested an amount of Rs. ( x ) at the beginning of the year.B invested double the amount A invested, which is ( 2x ), but B joined after 6 months. Therefore, B's investment was active for 6 months.C invested thrice the amount A invested, which is ( 3x ), and C joined after 8 months. Therefore, C's investment was active for 4 months.Next, I calculate the product of the investment amount and the duration for each partner to determine their share of the profit.For A:[text{Investment} times text{Time} = x times 12 = 12x]For B:[text{Investment} times text{Time} = 2x times 6 = 12x]For C:[text{Investment} times text{Time} = 3x times 4 = 12x]Now, I sum up the total investment-time products to find the total ratio:[12x + 12x + 12x = 36x]To find A's share of the annual gain, I calculate the proportion of A's investment-time product to the total investment-time product and multiply it by the total gain:[text{A's Share} = left( frac{12x}{36x} right) times 27000 = frac{1}{3} times 27000 = 9000]Therefore, A's share of the annual gain is Rs. 9000.

💡Okay, so I need to figure out how many paper clips Lexie's younger brother collected. He wants to distribute them into 9 boxes, and each box will have 9 paper clips. Hmm, let's break this down step by step.First, I know that distribution means putting items into different containers or groups. In this case, the containers are boxes, and each box will have the same number of paper clips. So, if there are 9 boxes and each box has 9 paper clips, I need to find out the total number of paper clips.I think the way to approach this is by using multiplication. Multiplication is basically adding the same number multiple times. So, if each box has 9 paper clips and there are 9 boxes, I can multiply the number of boxes by the number of paper clips per box to find the total.Let me write that out: 9 boxes * 9 paper clips per box. That should give me the total number of paper clips. So, 9 times 9 is... hmm, what's 9 times 9? I remember that 9 times 9 is 81. Yeah, that's right. So, 9 multiplied by 9 equals 81.Wait, let me double-check that. If I have 9 boxes and each has 9 paper clips, then adding them up would be 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9 + 9. That's adding 9 nine times. So, 9 added nine times is 81. Yep, that confirms it.Is there another way to think about this? Maybe using division? If I had the total number of paper clips and wanted to find out how many per box, I would divide by the number of boxes. But in this case, I have the number per box and the number of boxes, so multiplication is the right approach.I guess another way to visualize it is by thinking of it as an array. If I imagine 9 rows of paper clips, with each row having 9 paper clips, the total number would be the area of the array, which is length times width, or 9 times 9, which again is 81.So, after considering different ways to approach the problem—using multiplication, repeated addition, and even visualizing it as an array—it all points to the same answer: 81 paper clips in total.I don't think I'm missing anything here. The problem seems straightforward, and all methods lead to the same conclusion. So, I'm pretty confident that Lexie's younger brother collected 81 paper clips in total.