💡Okay, so I have this problem here: I need to find the value of (cot(cot^{-1}4 - cot^{-1}9 + cot^{-1}14)). Hmm, that looks a bit complicated, but I think I can break it down step by step. Let me recall some trigonometric identities that might help me here.First, I remember that (cot^{-1}x) is the inverse cotangent function, which means that if (theta = cot^{-1}x), then (cottheta = x). So, in this problem, we're dealing with angles whose cotangents are 4, 9, and 14. I also recall that there are addition and subtraction formulas for cotangent. Specifically, for two angles (alpha) and (beta), we have:[cot(alpha + beta) = frac{cotalpha cotbeta - 1}{cotalpha + cotbeta}]and[cot(alpha - beta) = frac{cotalpha cotbeta + 1}{cotalpha - cotbeta}]These formulas might be useful here since the expression inside the cotangent function involves both subtraction and addition of inverse cotangent terms.Let me try to apply these formulas step by step. The expression is (cot(cot^{-1}4 - cot^{-1}9 + cot^{-1}14)). I can think of this as (cot((cot^{-1}4 - cot^{-1}9) + cot^{-1}14)). So, first, I can compute (cot^{-1}4 - cot^{-1}9), and then add (cot^{-1}14) to the result, and finally take the cotangent of the entire thing.Let me denote:[alpha = cot^{-1}4 quad text{and} quad beta = cot^{-1}9]So, (cotalpha = 4) and (cotbeta = 9). Using the subtraction formula for cotangent, I can find (cot(alpha - beta)):[cot(alpha - beta) = frac{cotalpha cotbeta + 1}{cotalpha - cotbeta} = frac{4 times 9 + 1}{4 - 9} = frac{36 + 1}{-5} = frac{37}{-5} = -frac{37}{5}]So, (cot(alpha - beta) = -frac{37}{5}), which means that (alpha - beta = cot^{-1}left(-frac{37}{5}right)). Wait, but the inverse cotangent function typically returns values in the range ((0, pi)), right? So, if the cotangent is negative, that would place the angle in the second quadrant where cotangent is negative. Hmm, that's something to keep in mind, but maybe it won't affect the calculation directly.Now, moving on, I need to compute (cot((alpha - beta) + gamma)), where (gamma = cot^{-1}14). So, let's denote:[gamma = cot^{-1}14 quad text{so} quad cotgamma = 14]Now, I need to find (cot((alpha - beta) + gamma)). Using the addition formula for cotangent:[cot((alpha - beta) + gamma) = frac{cot(alpha - beta) cotgamma - 1}{cot(alpha - beta) + cotgamma}]We already found that (cot(alpha - beta) = -frac{37}{5}), and (cotgamma = 14). Plugging these values in:[cot((alpha - beta) + gamma) = frac{left(-frac{37}{5}right) times 14 - 1}{left(-frac{37}{5}right) + 14}]Let me compute the numerator and the denominator separately.First, the numerator:[left(-frac{37}{5}right) times 14 = -frac{37 times 14}{5} = -frac{518}{5}]Then subtract 1:[-frac{518}{5} - 1 = -frac{518}{5} - frac{5}{5} = -frac{523}{5}]Now, the denominator:[-frac{37}{5} + 14 = -frac{37}{5} + frac{70}{5} = frac{33}{5}]So, putting it all together:[cot((alpha - beta) + gamma) = frac{-frac{523}{5}}{frac{33}{5}} = frac{-523}{33}]Simplifying this, we get:[cot((cot^{-1}4 - cot^{-1}9) + cot^{-1}14) = -frac{523}{33}]Therefore, the value of the given expression is (-frac{523}{33}).Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting from the beginning, I used the subtraction formula correctly for (cot(alpha - beta)), which gave me (-frac{37}{5}). Then, I used the addition formula with (gamma), plugging in the values correctly. The numerator was (-frac{518}{5} - 1), which is indeed (-frac{523}{5}), and the denominator was (-frac{37}{5} + 14 = frac{33}{5}). Dividing these gives (-frac{523}{33}), which seems correct.I think that's the right answer. So, the final value is (-frac{523}{33}).

💡Alright, so I have this problem here: Given n points on a plane, with no three points collinear (and n is greater than 4), any two points can be connected to form C(n,4) quadrilaterals. I need to prove that at least 20% of these quadrilaterals are convex. Hmm, okay, let me try to wrap my head around this.First off, I know that a quadrilateral is convex if all its interior angles are less than 180 degrees, and none of its sides cross each other. In other words, a convex quadrilateral is one where the four points form a shape without any indentations. On the other hand, a concave quadrilateral has one interior angle greater than 180 degrees, giving it a "dent."Now, the problem states that no three points are collinear, which is important because if three points were on the same line, that could complicate things, especially when trying to form quadrilaterals. But since that's not the case here, I don't have to worry about that.The total number of quadrilaterals that can be formed from n points is C(n,4), which is the number of ways to choose 4 points out of n. So, that's straightforward. Now, I need to figure out how many of these quadrilaterals are convex and show that at least 20% of them are convex.I remember that in any set of points, the convex hull is the smallest convex polygon that contains all the points. The convex hull can be a triangle, a quadrilateral, a pentagon, etc., depending on the arrangement of the points. If the convex hull is a quadrilateral, then that quadrilateral is convex. If it's a pentagon or higher, then there are multiple convex quadrilaterals within it.Wait, but how does this relate to the problem? Maybe I need to consider the convex hull of the entire set of points. If the convex hull has k points, then any subset of 4 points that includes at least 3 points from the convex hull will form a convex quadrilateral. Hmm, is that right?Let me think. Suppose the convex hull has k points. If I choose 4 points where at least 3 are on the convex hull, then those 3 points will form a triangle, and the fourth point will be inside or on the hull. If it's on the hull, then the quadrilateral is convex. If it's inside, then the quadrilateral might still be convex or concave, depending on the position.Wait, no. If three points are on the convex hull, and the fourth is inside, then the quadrilateral formed by these four points is actually convex because the three hull points form a triangle, and the fourth point is inside, but connecting them in order would still result in a convex shape. Is that correct?Actually, no. If three points are on the convex hull and the fourth is inside, the quadrilateral can be concave if the internal angle at the inside point is greater than 180 degrees. Hmm, so maybe that's not a guaranteed convex quadrilateral.Maybe I need a different approach. I recall something about the Erdős–Szekeres theorem, which deals with convex polygons and points in general position. But I'm not sure if that's directly applicable here.Alternatively, maybe I can use an averaging argument. If I consider all possible quadrilaterals, and figure out the probability that a random quadrilateral is convex, then show that this probability is at least 20%.But how would I calculate that probability? Maybe by considering the number of convex quadrilaterals divided by the total number of quadrilaterals, which is C(n,4).Wait, but I need to find a lower bound for the number of convex quadrilaterals. Maybe I can use some combinatorial geometry principles here.I remember that in a set of points in general position (no three collinear), the number of convex quadrilaterals is related to the number of empty convex quadrilaterals, but I'm not sure if that's helpful here.Alternatively, maybe I can use the fact that any five points in general position contain at least one convex quadrilateral. Is that a theorem? Let me think.Yes, I think that's a result from combinatorial geometry. Any five points in general position (no three collinear) contain at least one convex quadrilateral. So, if I have five points, no three on a line, then among the C(5,4)=5 quadrilaterals, at least one is convex.If that's the case, then maybe I can use this to bound the number of convex quadrilaterals in the entire set.So, if I consider all subsets of five points from the n points, each subset contains at least one convex quadrilateral. The total number of such subsets is C(n,5). Each convex quadrilateral is counted in multiple subsets, specifically in C(n-4,1) subsets, because once you fix four points forming a convex quadrilateral, the fifth point can be any of the remaining n-4 points.Therefore, the total number of convex quadrilaterals is at least C(n,5)/(n-4). So, the number of convex quadrilaterals is at least C(n,5)/(n-4).Now, let's compute this. C(n,5) is n!/(5!(n-5)!), and dividing by (n-4) gives us n!/(5!(n-5)!*(n-4)) = n!/(5!(n-4)!*(n-4)) = n!/(5!(n-4)!*(n-4)).Wait, that seems a bit messy. Maybe I can simplify it differently. Let's see:C(n,5) = n(n-1)(n-2)(n-3)(n-4)/120Divided by (n-4) gives:n(n-1)(n-2)(n-3)/120Which is equal to C(n,4)/5.Because C(n,4) is n(n-1)(n-2)(n-3)/24, so dividing by 5 gives n(n-1)(n-2)(n-3)/120, which is exactly what we have.So, the number of convex quadrilaterals is at least C(n,4)/5.Therefore, the proportion of convex quadrilaterals is at least (C(n,4)/5)/C(n,4) = 1/5 = 20%.So, that seems to do it. By considering all subsets of five points, each contributing at least one convex quadrilateral, and accounting for the overcounting, we find that at least 20% of all quadrilaterals must be convex.Wait, but is this tight? Is there a case where exactly 20% are convex? Or is this just a lower bound?I think it's a lower bound. There might be configurations where more than 20% are convex, but this shows that you can't have fewer than 20%.Let me see if I can think of a specific example where exactly 20% are convex. Maybe arranging the points in a convex position, but that would make all quadrilaterals convex, which is way more than 20%.Alternatively, arranging points in a way that many quadrilaterals are concave. But I don't know if it's possible to get exactly 20%.Anyway, the problem only asks to prove that at least 20% are convex, so this seems sufficient.So, to recap:1. Any five points in general position contain at least one convex quadrilateral.2. The total number of subsets of five points is C(n,5).3. Each convex quadrilateral is counted in C(n-4,1) subsets.4. Therefore, the number of convex quadrilaterals is at least C(n,5)/(n-4) = C(n,4)/5.5. Hence, the proportion is at least 20%.Yeah, that makes sense. I think I've got it.**Final Answer**boxed{20%}

💡Okay, so I've got this problem here: Given that (4^{a} = 9^{b} = k) and (frac{1}{a} + frac{1}{b} = 2), I need to find the value of (k). Hmm, let me try to figure this out step by step.First, I know that if (4^{a} = k), then (a) must be the exponent that you raise 4 to in order to get (k). Similarly, (9^{b} = k) means that (b) is the exponent you raise 9 to in order to get (k). So, I can express (a) and (b) in terms of logarithms. Let me write that down:(a = log_{4}k) and (b = log_{9}k).Okay, so now I have expressions for both (a) and (b). The next part of the problem says that (frac{1}{a} + frac{1}{b} = 2). I need to substitute the expressions for (a) and (b) into this equation.So, substituting, I get:(frac{1}{log_{4}k} + frac{1}{log_{9}k} = 2).Hmm, this looks a bit complicated. Maybe I can simplify it using logarithmic identities. I remember that (frac{1}{log_{c}d} = log_{d}c). Let me apply that here.So, (frac{1}{log_{4}k} = log_{k}4) and (frac{1}{log_{9}k} = log_{k}9). Therefore, the equation becomes:(log_{k}4 + log_{k}9 = 2).Alright, now I have the sum of two logarithms with the same base. I recall that (log_{c}d + log_{c}e = log_{c}(d times e)). Applying that property here, I can combine the two logarithms:(log_{k}(4 times 9) = 2).Calculating (4 times 9), that's 36. So, the equation simplifies to:(log_{k}36 = 2).Now, what does this mean? A logarithm (log_{k}36 = 2) means that (k) raised to the power of 2 equals 36. In other words:(k^{2} = 36).To solve for (k), I take the square root of both sides:(k = sqrt{36}).Which gives me:(k = 6) or (k = -6).But wait, (k) is defined as (4^{a}) and (9^{b}). Since 4 and 9 are positive numbers, any real exponent of them will also be positive. Therefore, (k) must be positive. So, (k = 6) is the only valid solution.Let me just recap to make sure I didn't make a mistake:1. Expressed (a) and (b) in terms of logarithms with base 4 and 9 respectively.2. Substituted these into the given equation involving reciprocals.3. Used the logarithmic identity to convert the reciprocals into logarithms with base (k).4. Combined the logarithms into a single logarithm using the product rule.5. Converted the logarithmic equation back into its exponential form.6. Solved for (k) and considered the domain to discard the negative solution.Everything seems to check out. So, I'm confident that (k = 6) is the correct answer.**Final Answer**The value of (k) is boxed{6}.

💡First, I recognize that in a rectangle, all interior angles are right angles, which means the sides form a right triangle with the diagonal.Given that AB is 4 units and AC is 5 units, I can use the Pythagorean theorem to find the length of BC.Applying the theorem:AB² + BC² = AC²4² + BC² = 5²16 + BC² = 25BC² = 25 - 16BC² = 9BC = √9BC = 3Now that I have the lengths of both sides AB and BC, I can calculate the area of the rectangle by multiplying these two lengths.Area = AB × BCArea = 4 × 3Area = 12Therefore, the area of rectangle ABCD is 12 square units.

💡Alright, so I've got this problem here: Given points A(2,4), B(-1,-5), and C(3,-2), I need to find the vector AC plus one-third of vector BA. Hmm, okay, let's break this down step by step. First, I remember that to find a vector between two points, say from point X to point Y, you subtract the coordinates of X from Y. So, vector XY is (Yx - Xx, Yy - Xx). Wait, no, that's not quite right. It should be (Yx - Xx, Yy - Xy). Yeah, that makes sense because you're moving from X to Y, so you subtract the starting point's coordinates from the ending point's.So, let's start with vector AC. Point A is (2,4) and point C is (3,-2). So, vector AC should be (3 - 2, -2 - 4). Let me calculate that: 3 minus 2 is 1, and -2 minus 4 is -6. So, vector AC is (1, -6). Okay, that seems straightforward.Next, I need to find vector BA. Point B is (-1,-5) and point A is (2,4). So, vector BA would be (2 - (-1), 4 - (-5)). Let me compute that: 2 minus (-1) is 3, and 4 minus (-5) is 9. So, vector BA is (3,9). Got it.Now, the problem asks for one-third of vector BA. So, I need to multiply each component of BA by 1/3. That would be (3*(1/3), 9*(1/3)) which simplifies to (1,3). Okay, so one-third of BA is (1,3).Now, I need to add vector AC and one-third of BA together. Vector AC is (1, -6) and one-third BA is (1,3). Adding these component-wise: 1 + 1 is 2, and -6 + 3 is -3. So, the resulting vector is (2, -3).Let me double-check my calculations to make sure I didn't make any mistakes. Starting with vector AC: (3-2, -2-4) = (1, -6). That seems correct. Vector BA: (2 - (-1), 4 - (-5)) = (3,9). That also looks right. One-third of BA: (1,3). Adding (1, -6) and (1,3) gives (2, -3). Yeah, that all checks out.So, the final answer should be (2, -3). Looking back at the options provided: A is (2,-3), B is (-2,-3), C is (-2,3), D is (2,3). So, the correct choice is A.Wait a second, just to make sure I didn't mix up any points. Vector AC is from A to C, which is (3-2, -2-4). Yep, that's (1, -6). Vector BA is from B to A, which is (2 - (-1), 4 - (-5)) = (3,9). Then, one-third of BA is (1,3). Adding those vectors: (1+1, -6+3) = (2, -3). Yep, that's definitely correct.I think I'm confident with this answer. It's always good to go through the steps again to ensure there are no calculation errors, especially with negative signs and fractions involved. But everything seems to add up correctly here.

💡Okay, so I have this problem: I need to find the remainder when ( s^{10} + 1 ) is divided by ( s - 2 ). Hmm, I remember something about the Remainder Theorem from algebra class. Let me try to recall how that works.Alright, the Remainder Theorem states that if you divide a polynomial ( f(s) ) by ( s - a ), the remainder is just ( f(a) ). So in this case, my polynomial is ( s^{10} + 1 ) and I'm dividing by ( s - 2 ). That means ( a ) is 2 here. So, according to the theorem, the remainder should be ( f(2) ).Let me write that down to make sure I'm on the right track. So, ( f(s) = s^{10} + 1 ). Therefore, ( f(2) = 2^{10} + 1 ). Now, I need to calculate ( 2^{10} ). I know that ( 2^1 = 2 ), ( 2^2 = 4 ), ( 2^3 = 8 ), ( 2^4 = 16 ), ( 2^5 = 32 ), ( 2^6 = 64 ), ( 2^7 = 128 ), ( 2^8 = 256 ), ( 2^9 = 512 ), and ( 2^{10} = 1024 ). So, ( 2^{10} ) is 1024.Adding 1 to that gives me ( 1024 + 1 = 1025 ). So, the remainder should be 1025. Wait, is that all? It seems straightforward, but let me double-check to make sure I didn't make a mistake.Another way to think about this is polynomial long division. If I were to divide ( s^{10} + 1 ) by ( s - 2 ), the remainder would indeed be a constant term because the divisor is linear. So, after performing the division, the remainder would just be a number, which we've found to be 1025 using the Remainder Theorem.I guess I could also plug in ( s = 2 ) into the polynomial and see what I get. Let's do that: ( 2^{10} + 1 = 1024 + 1 = 1025 ). Yep, that matches what I found earlier.Is there any other way to verify this? Maybe synthetic division? Let's try that. For synthetic division, I set up the coefficients of the polynomial ( s^{10} + 1 ). Since it's ( s^{10} + 0s^9 + 0s^8 + dots + 0s + 1 ), the coefficients are 1 followed by nine zeros and then 1.I'll use 2 as the root for synthetic division because we're dividing by ( s - 2 ). So, I bring down the 1, multiply by 2, add to the next coefficient, and so on. But this might take a while since it's a 10th-degree polynomial. Let me see:1. Bring down the 1.2. Multiply by 2: 1 * 2 = 2. Add to next coefficient: 0 + 2 = 2.3. Multiply by 2: 2 * 2 = 4. Add to next coefficient: 0 + 4 = 4.4. Multiply by 2: 4 * 2 = 8. Add to next coefficient: 0 + 8 = 8.5. Multiply by 2: 8 * 2 = 16. Add to next coefficient: 0 + 16 = 16.6. Multiply by 2: 16 * 2 = 32. Add to next coefficient: 0 + 32 = 32.7. Multiply by 2: 32 * 2 = 64. Add to next coefficient: 0 + 64 = 64.8. Multiply by 2: 64 * 2 = 128. Add to next coefficient: 0 + 128 = 128.9. Multiply by 2: 128 * 2 = 256. Add to next coefficient: 0 + 256 = 256.10. Multiply by 2: 256 * 2 = 512. Add to next coefficient: 0 + 512 = 512.11. Multiply by 2: 512 * 2 = 1024. Add to last coefficient: 1 + 1024 = 1025.So, after performing synthetic division, the remainder is indeed 1025. That confirms my earlier result. I think I'm confident now that the remainder is 1025.Wait, just to make sure, let me think about what the Remainder Theorem really means. It's saying that when you divide a polynomial by a linear divisor ( s - a ), the remainder is the value of the polynomial evaluated at ( s = a ). So, in this case, evaluating ( s^{10} + 1 ) at ( s = 2 ) gives the remainder. That makes sense because if you write the division as ( f(s) = (s - 2)Q(s) + R ), where ( Q(s) ) is the quotient and ( R ) is the remainder, then plugging in ( s = 2 ) gives ( f(2) = R ). So, yeah, that's consistent.I guess another way to see it is that when ( s = 2 ), the term ( (s - 2)Q(s) ) becomes zero, leaving just the remainder ( R ). So, ( f(2) = R ). That's a neat theorem because it simplifies finding the remainder without having to perform the entire division.Alright, I think I've thought this through enough. The remainder is definitely 1025.