💡Alright, so I'm trying to figure out this problem where Kolya and Leva each have a segment of length ( k ) and ( l ) respectively. They each divide their segments into three parts, and then we check if we can form two triangles from all six segments. If we can, Leva wins; otherwise, Kolya wins. The goal is to determine who can ensure victory based on the ratio ( k/l ).First, I need to recall the triangle inequality theorem, which states that for any three lengths to form a triangle, the sum of any two sides must be greater than the third side. So, if any of the three parts from either Kolya or Leva is too long compared to the others, it might prevent forming a triangle.Let me think about how Kolya can prevent Leva from winning. If Kolya can make sure that at least one of his segments is so long that it can't form a triangle with any other two segments, then Leva can't form two triangles. Similarly, Leva wants to divide his segment in such a way that regardless of how Kolya divided his, Leva can still form two triangles.Maybe I should consider the maximum possible length of a segment after division. For Kolya, if he divides his segment ( k ) into three parts, the longest part he can have is just under ( k ), but realistically, if he wants to prevent triangle formation, he might make one part very long. Similarly, Leva would want to balance his segments to ensure they can form triangles.Let me consider the ratio ( k/l ). If ( k ) is much larger than ( l ), Kolya might have an advantage because he can create a very long segment that can't form a triangle. On the other hand, if ( l ) is large enough, Leva can balance his segments to form triangles regardless of Kolya's division.Suppose ( k > l ). Then, Kolya can divide his segment such that one part is longer than ( l ). For example, if Kolya divides ( k ) into ( l + epsilon ), ( frac{k - l - epsilon}{2} ), and ( frac{k - l - epsilon}{2} ) where ( epsilon ) is a very small positive number. Then, the segment ( l + epsilon ) is longer than ( l ), and since Leva's segments are all parts of ( l ), none of Leva's segments can be longer than ( l ). Therefore, the segment ( l + epsilon ) cannot form a triangle with any two of Leva's segments because ( l + epsilon ) is longer than the sum of any two segments from Leva (since each of Leva's segments is at most ( l ), and ( l + epsilon > l )).Wait, is that correct? If Kolya has a segment longer than ( l ), and Leva's segments are all parts of ( l ), then indeed, Kolya's long segment can't form a triangle with any two of Leva's segments. But what about forming triangles among Kolya's segments? If Kolya makes one segment very long, the other two segments might still be able to form a triangle with some of Leva's segments.Hmm, maybe I need to think more carefully. If Kolya divides his segment into three parts, one of which is longer than ( l ), then that part can't form a triangle with any two of Leva's segments. But the other two parts of Kolya's segment might still be able to form triangles with Leva's segments. So, Leva might still be able to form two triangles using Kolya's smaller segments and his own.Wait, no. If Kolya has one very long segment, say ( l + epsilon ), and the other two segments are very small, then Leva can only use his segments to form triangles. But if Kolya's long segment is too long, it might interfere with forming triangles using the other segments.Actually, maybe it's better to think in terms of the maximum possible length of a segment. If Kolya can ensure that one of his segments is longer than the sum of any two segments from Leva, then that segment can't form a triangle. But Leva can try to balance his segments to avoid this.Alternatively, maybe the key is to consider the total lengths. If ( k ) is much larger than ( l ), Kolya can make one segment so long that it can't form a triangle with any of Leva's segments. Conversely, if ( l ) is sufficiently large, Leva can ensure that his segments are balanced enough to form triangles regardless of Kolya's division.Let me try to formalize this. Suppose Kolya divides ( k ) into ( a ), ( b ), and ( c ) such that ( a geq b geq c ). To prevent forming a triangle, Kolya needs ( a geq b + c ). If he can do this, then the segment ( a ) can't form a triangle with ( b ) and ( c ). But Leva will divide ( l ) into ( d ), ( e ), and ( f ). Leva wants to ensure that from the six segments ( a, b, c, d, e, f ), two triangles can be formed.So, if Kolya makes ( a geq b + c ), then ( a ) can't form a triangle with ( b ) and ( c ). But Leva can try to form triangles using his segments and Kolya's segments. For example, Leva might pair ( d ) with ( e ) and ( f ), and Kolya might pair ( b ) and ( c ) with some of Leva's segments.Wait, but if ( a geq b + c ), then ( a ) is too long to form a triangle with ( b ) and ( c ). But Leva can still try to form triangles using his own segments. If Leva divides ( l ) into three equal parts, ( d = e = f = l/3 ), then each of these can form a triangle with any two segments from Kolya, provided Kolya's segments are not too long.But if Kolya makes one segment very long, say ( a = k - 2epsilon ), and ( b = c = epsilon ), then ( a ) is almost ( k ), and ( b ) and ( c ) are very small. Then, Leva's segments are ( l/3 ) each. If ( k > l ), then ( a = k - 2epsilon ) is greater than ( l ), so ( a > l ). Therefore, ( a ) is longer than any of Leva's segments, which are ( l/3 ). So, ( a ) can't form a triangle with any two of Leva's segments because ( a > l/3 + l/3 = 2l/3 ), but ( a ) is greater than ( l ), which is greater than ( 2l/3 ).Wait, that might not necessarily prevent forming two triangles. Leva can still form triangles using his own segments. For example, if Leva divides ( l ) into three equal parts, each ( l/3 ), then he can form a triangle with any two of his segments and one of Kolya's segments, provided Kolya's segments are not too long.But if Kolya's segment ( a ) is too long, say ( a > l ), then ( a ) can't form a triangle with any two of Leva's segments because ( a > l/3 + l/3 = 2l/3 ), but ( a > l ), which is greater than ( 2l/3 ). However, Leva can still form triangles using his own segments. For example, ( l/3, l/3, l/3 ) can form an equilateral triangle, but that's only one triangle. To form two triangles, Leva needs to use all six segments, but if Kolya has one segment that can't form a triangle, then maybe Leva can only form one triangle.Wait, no. Leva needs to form two triangles from all six segments. So, if Kolya has one segment that can't form a triangle, but the other two segments can, then Leva might still be able to form two triangles by combining segments appropriately.This is getting a bit confusing. Maybe I should consider specific cases.Case 1: ( k > l )Kolya can divide his segment into ( k = a + b + c ), where ( a = l + epsilon ), ( b = c = frac{k - l - epsilon}{2} ). Then, ( a > l ), and since Leva's segments are all parts of ( l ), none of Leva's segments can be longer than ( l ). Therefore, ( a ) can't form a triangle with any two of Leva's segments because ( a > l geq d + e ) for any ( d, e ) from Leva's segments.But what about forming triangles with Kolya's other segments? If ( b ) and ( c ) are small enough, maybe they can form triangles with Leva's segments. For example, if ( b ) and ( c ) are each less than ( l/3 ), then ( b + c < 2l/3 ), and since ( l/3 + l/3 = 2l/3 ), it's possible that ( b + c < 2l/3 ), but ( b ) and ( c ) are each less than ( l/3 ), so ( b + c < 2l/3 ). Therefore, ( b ) and ( c ) can form triangles with Leva's segments.Wait, but if ( b ) and ( c ) are each less than ( l/3 ), then ( b + c < 2l/3 ), but Leva's segments are ( l/3 ) each. So, ( b + c < 2l/3 ), but ( b + c ) could be greater than ( l/3 ), depending on ( k ) and ( l ).This is getting too tangled. Maybe I need to think about the maximum length of a segment after division. If Kolya can make one segment longer than ( l ), then that segment can't form a triangle with any two of Leva's segments. But Leva can still try to form triangles using his own segments.Wait, but Leva needs to form two triangles from all six segments. If Kolya has one segment that can't form a triangle, but the other two segments can, then Leva might still be able to form two triangles by combining segments appropriately.Alternatively, maybe Kolya can make two segments that are too long to form triangles with any of Leva's segments. But that might not be possible because Kolya only has one segment to divide into three parts.Wait, no. Kolya divides his segment into three parts, so he can have one very long segment and two very short ones. Then, the long segment can't form a triangle with any two of Leva's segments, but the short segments can.But Leva needs to form two triangles. If Kolya has one long segment and two short ones, Leva can use his own segments to form two triangles, each using one of Kolya's short segments and two of his own.Wait, but if Kolya's long segment is too long, it might interfere with forming the triangles. For example, if Kolya's long segment is longer than the sum of any two of Leva's segments, then it can't be used in any triangle. But Leva can still form two triangles using his own segments and Kolya's short segments.Wait, no. If Kolya's long segment is too long, it can't form a triangle, but Leva can still form two triangles using his own segments and Kolya's short segments. For example, Leva can form one triangle with two of his segments and one of Kolya's short segments, and another triangle with the remaining segments.But I'm not sure. Maybe I need to think about specific values.Suppose ( k = 4 ) and ( l = 3 ). Then, Kolya can divide his segment into ( 3 + epsilon ), ( epsilon ), ( epsilon ). Then, the long segment ( 3 + epsilon ) is longer than ( l = 3 ). Leva divides his segment into three parts, say ( 1 ), ( 1 ), ( 1 ). Then, Kolya's long segment ( 3 + epsilon ) can't form a triangle with any two of Leva's segments because ( 3 + epsilon > 1 + 1 = 2 ). But Leva can form two triangles: one with ( 1 ), ( 1 ), and ( epsilon ), and another with ( 1 ), ( epsilon ), and ( epsilon ). Wait, but ( epsilon ) is very small, so ( 1 + epsilon > epsilon ), but ( 1 + epsilon > 1 ) is true, so the triangle inequality holds. Therefore, Leva can form two triangles.Hmm, so even if Kolya makes one segment longer than ( l ), Leva can still form two triangles using his own segments and Kolya's short segments. Therefore, maybe Kolya can't prevent Leva from winning if ( k leq l ).Wait, but in this example, ( k = 4 > l = 3 ), but Leva still can form two triangles. So maybe my initial thought was wrong.Alternatively, maybe Kolya needs to make one segment so long that it can't form a triangle with any two segments, including Kolya's own segments. But if Kolya makes one segment longer than the sum of the other two, then he can't form a triangle with his own segments, but Leva can still form triangles with his own segments.Wait, but Leva needs to form two triangles from all six segments. If Kolya can't form a triangle with his own segments, but Leva can form two triangles using his own segments and Kolya's segments, then Leva still wins.This is confusing. Maybe I need to think about the total number of segments. There are six segments in total. To form two triangles, we need to partition these six segments into two sets of three segments each, where each set satisfies the triangle inequality.So, if Kolya makes one segment too long to form a triangle with any two others, but the other two segments are small enough, then Leva can still form two triangles by combining his segments with Kolya's small segments.Wait, but if Kolya's long segment is too long, it can't be used in any triangle, so it has to be excluded. But then, we only have five segments left, which is not enough to form two triangles (since each triangle requires three segments). Therefore, if Kolya can make one segment that can't be used in any triangle, then we only have five usable segments, which is insufficient for two triangles. Therefore, Kolya can win by making one segment too long.But wait, in my earlier example, ( k = 4 ), ( l = 3 ), Kolya divides into ( 3 + epsilon ), ( epsilon ), ( epsilon ). Then, the long segment ( 3 + epsilon ) can't form a triangle with any two of Leva's segments, but Leva can still form two triangles using his own segments and Kolya's small segments. Specifically, Leva can form one triangle with two of his segments and one of Kolya's small segments, and another triangle with the remaining segments.But wait, if Kolya's long segment is excluded, we have five segments: ( epsilon ), ( epsilon ), ( 1 ), ( 1 ), ( 1 ). Can we form two triangles from these? Let's see:First triangle: ( 1 ), ( 1 ), ( epsilon ). This satisfies the triangle inequality because ( 1 + 1 > epsilon ), ( 1 + epsilon > 1 ), and ( 1 + epsilon > 1 ).Second triangle: ( 1 ), ( epsilon ), ( epsilon ). This also satisfies the triangle inequality because ( 1 + epsilon > epsilon ), ( epsilon + epsilon > 1 ) (but ( epsilon ) is very small, so this might not hold). Wait, ( epsilon + epsilon = 2epsilon ), which is less than 1, so ( 2epsilon < 1 ), which means ( epsilon + epsilon < 1 ). Therefore, the triangle inequality fails for the second triangle because ( epsilon + epsilon ) is not greater than 1.Therefore, we can only form one triangle from the five segments, which means Leva can't form two triangles. Therefore, Kolya wins.Wait, so in this case, Kolya can make one segment too long, and the remaining segments can only form one triangle, so Leva can't form two triangles. Therefore, Kolya wins.But in this example, ( k = 4 > l = 3 ). So, if ( k > l ), Kolya can win by making one segment longer than ( l ), which can't form a triangle with any two of Leva's segments, and the remaining segments can only form one triangle.On the other hand, if ( k leq l ), then Kolya can't make a segment longer than ( l ), so Leva can divide his segment in such a way that all segments are balanced, allowing two triangles to be formed.Therefore, the conclusion is:- If ( k > l ), Kolya can ensure victory by making one segment longer than ( l ), preventing Leva from forming two triangles.- If ( k leq l ), Leva can ensure victory by balancing his segments to allow two triangles to be formed regardless of Kolya's division.So, the ratio ( k/l ) determines the winner. If ( k/l > 1 ), Kolya wins; otherwise, Leva wins.

💡Okay, so I have this geometry problem here with four statements, labeled A through D. Each statement involves lines and planes in space, and I need to figure out which one has a converse that isn't necessarily true. Hmm, okay. Let me try to break this down step by step.First, let me recall what a converse of a statement is. If I have a statement like "If P, then Q," the converse would be "If Q, then P." So, for each of these statements, I need to not only understand what the original statement is saying but also what its converse would be and whether that converse is always true or not.Alright, let's go through each option one by one.**Option A:**The statement is: If line c is perpendicular to plane α, and line c is also perpendicular to plane β, then planes α and β are parallel.Okay, so if a single line is perpendicular to two different planes, those planes must be parallel. That makes sense because if two planes are both perpendicular to the same line, they can't intersect each other; otherwise, the line would have to be perpendicular to both planes along their line of intersection, which isn't possible unless they're parallel.Now, what's the converse of this statement? The converse would be: If planes α and β are parallel, then any line c that is perpendicular to α is also perpendicular to β. Hmm, does that hold true?Well, if two planes are parallel, then a line perpendicular to one should indeed be perpendicular to the other. Because the direction perpendicular to one plane is the same as the direction perpendicular to the other plane since they're parallel. So, the converse seems to be true here. So, A's converse is true.**Option B:**The statement is: If line b is contained within plane α, and line b is perpendicular to plane β, then plane α is perpendicular to plane β.Alright, so if a line lying in plane α is perpendicular to plane β, then the two planes must be perpendicular. That makes sense because for two planes to be perpendicular, it's sufficient that one plane contains a line perpendicular to the other plane.Now, the converse of this statement would be: If plane α is perpendicular to plane β, then any line b contained within plane α is perpendicular to plane β. Hmm, is that necessarily true?Wait, no. If two planes are perpendicular, it doesn't mean every line in one plane is perpendicular to the other plane. Only lines that are specifically aligned in a certain way (like the line of intersection) would be perpendicular. Most lines in plane α would just intersect plane β at some angle, not necessarily 90 degrees. So, the converse isn't true. That seems like a strong candidate for the answer, but let me check the others just to be sure.**Option C:**The statement is: If line b is contained within plane α, and line c is the projection of line a onto plane α, and line b is perpendicular to line c, then line a is perpendicular to line b.Okay, so if I have line a somewhere in space, and I project it onto plane α, getting line c. If line b in plane α is perpendicular to this projection c, then the original line a is perpendicular to line b.I think this is related to the three perpendiculars theorem or something like that. Let me recall. If you have a line and its projection onto a plane, then any line in the plane that's perpendicular to the projection is also perpendicular to the original line. Yeah, that sounds right.So, the converse would be: If line a is perpendicular to line b, then line c, the projection of a onto α, is perpendicular to line b. Is that true?Well, if line a is perpendicular to line b, then projecting a onto plane α should result in a projection c that's also perpendicular to b. Because the projection would retain the perpendicularity in the plane. So, the converse seems to hold here as well. So, C's converse is true.**Option D:**The statement is: If line b is contained within plane α, and line c is not contained within plane α, and line c is parallel to plane α, then line b is parallel to line c.Wait, so if line c is parallel to plane α, and it's not lying on α, then any line b in α must be parallel to c? Hmm, is that correct?Wait, no. If line c is parallel to plane α, that means it doesn't intersect α and lies in a direction that's parallel to α. But lines within plane α can have various directions. So, line c could be parallel to some lines in α but not necessarily all of them. So, the original statement seems a bit off. Let me think again.Wait, the original statement says: If b is in α, c is not in α, and c is parallel to α, then b is parallel to c. That doesn't sound right. Because c being parallel to α just means it's parallel to some direction in α, but it doesn't have to be parallel to every line in α.Wait, maybe I misread. Let me check: "If b is contained in α, and c is not contained in α, and c is parallel to α, then b is parallel to c." Hmm, that seems incorrect because c could be parallel to α but not necessarily parallel to every line in α. So, perhaps the original statement is false? But the question is about the converse.Wait, no, the question is about which statement's converse is not true. So, first, let's figure out the original statement.Wait, the original statement in D is: If b is in α, c is not in α, and c is parallel to α, then b is parallel to c. Hmm, that seems incorrect because c being parallel to α doesn't imply it's parallel to every line in α. So, maybe the original statement is false? But the question is about the converse.Wait, maybe I need to think differently. Let me clarify.The original statement is: If b is in α, c is not in α, and c is parallel to α, then b is parallel to c.Is this statement true? Well, if c is parallel to α, then c is parallel to some line in α, but not necessarily to every line in α. So, unless b is specifically aligned with the direction of c, b and c might not be parallel. So, the original statement seems to be false.But the question is about the converse. So, the converse of D would be: If b is in α, c is not in α, and b is parallel to c, then c is parallel to α.Is that true? Well, if b is in α and c is parallel to b, then c must be parallel to α because if a line is parallel to a line in a plane, it's either parallel to the plane or lies within the plane. But since c is not contained in α, it must be parallel to α. So, the converse is true.Wait, but the original statement is false, but the question is about the converse. So, the original statement is false, but its converse is true. Hmm, but the question is asking for the statement whose converse is not true. So, if the original statement is false, does that affect the converse?Wait, no. The question is about the converse of each statement, regardless of whether the original is true or not. So, even if the original statement is false, we still need to consider its converse.But in this case, the converse is true, so D is not the answer.Wait, but going back to the original statement of D: If b is in α, c is not in α, and c is parallel to α, then b is parallel to c. That seems false because c could be parallel to α but not to b. So, the original statement is false, but the converse is true.So, the original statement is false, but the converse is true. So, the converse is true, so D is not the answer.Wait, but the question is asking for the statement whose converse is not true. So, among A, B, C, D, which one has a converse that is not true.From earlier, A's converse is true, B's converse is not true, C's converse is true, D's converse is true. So, the answer should be B.Wait, but let me double-check.For B: Original statement: If b is in α and b is perpendicular to β, then α is perpendicular to β. That's true because if a line in α is perpendicular to β, then α must be perpendicular to β.Converse: If α is perpendicular to β, then any line b in α is perpendicular to β. That's not true because only specific lines in α (like the line of intersection) are perpendicular to β, not all lines.So, yes, B's converse is not true, making B the correct answer.But just to make sure, let me think about D again. The original statement is: If b is in α, c is not in α, and c is parallel to α, then b is parallel to c. That's false because c could be parallel to α but not to b. So, the original statement is false, but the converse is: If b is in α, c is not in α, and b is parallel to c, then c is parallel to α. That's true because if c is parallel to b, which is in α, and c is not in α, then c must be parallel to α.So, D's converse is true, so D is not the answer.Similarly, for C: Original statement is true, converse is also true.For A: Original statement is true, converse is also true.So, only B's converse is not true. Therefore, the answer is B.**Final Answer**boxed{B}

💡Okay, so I need to add these four polynomials together: f(x), g(x), h(x), and j(x). Each of them is a quadratic polynomial, which means they all have an x squared term, an x term, and a constant term. My goal is to combine them into a single polynomial, making sure the terms are ordered from the highest degree to the lowest. First, I should write down each polynomial so I can see them clearly:f(x) = -2x² + 2x - 5g(x) = -6x² + 4x - 9h(x) = 4x² + 6x + 3j(x) = 3x² - x + 2Now, I need to add all these together. To do this, I should combine like terms. Like terms are terms that have the same variable raised to the same power. So, I'll group the x² terms together, the x terms together, and the constant terms together.Starting with the x² terms:f(x) has -2x²g(x) has -6x²h(x) has 4x²j(x) has 3x²So, adding these coefficients together: -2 + (-6) + 4 + 3Let me calculate that step by step:-2 + (-6) = -8-8 + 4 = -4-4 + 3 = -1So, the combined x² term is -1x², which is just -x².Next, the x terms:f(x) has 2xg(x) has 4xh(x) has 6xj(x) has -1xAdding these coefficients together: 2 + 4 + 6 + (-1)Calculating step by step:2 + 4 = 66 + 6 = 1212 + (-1) = 11So, the combined x term is 11x.Now, the constant terms:f(x) has -5g(x) has -9h(x) has 3j(x) has 2Adding these together: -5 + (-9) + 3 + 2Calculating step by step:-5 + (-9) = -14-14 + 3 = -11-11 + 2 = -9So, the combined constant term is -9.Putting it all together, the polynomial after adding all four is:- x² + 11x - 9I should double-check my work to make sure I didn't make any mistakes in adding the coefficients.For the x² terms: -2 -6 +4 +3-2 -6 is -8, plus 4 is -4, plus 3 is -1. That seems correct.For the x terms: 2 +4 +6 -12 +4 is 6, plus 6 is 12, minus 1 is 11. That looks good.For the constants: -5 -9 +3 +2-5 -9 is -14, plus 3 is -11, plus 2 is -9. Correct.So, combining all these, the final polynomial is -x² + 11x -9.I think that's the correct answer. It's always good to verify each step, especially when dealing with negative numbers, to ensure that the signs are handled properly. But in this case, everything seems to add up correctly.

💡Okay, so I have this problem where a bucket weighs a certain amount when it's three-fourths full of water, and a different amount when it's one-third full. I need to figure out how much it weighs when it's completely full. Hmm, let's see.First, I think I need to set up some variables. Let me call the weight of the empty bucket "x" and the weight of the water when the bucket is full "y". That makes sense because the total weight when the bucket is full would be the weight of the bucket plus the weight of the water.So, when the bucket is three-fourths full, the total weight is "a" kilograms. That means the weight of the bucket plus three-fourths of the water's weight equals "a". In equation form, that would be:x + (3/4)y = aSimilarly, when the bucket is one-third full, the total weight is "b" kilograms. So, the weight of the bucket plus one-third of the water's weight equals "b". That equation would be:x + (1/3)y = bNow, I have two equations with two variables, x and y. I need to solve this system of equations to find x and y, and then add them together to get the total weight when the bucket is full.Let me write down the equations again:1. x + (3/4)y = a2. x + (1/3)y = bI think the best way to solve this is by elimination. If I subtract the second equation from the first, I can eliminate x and solve for y.So, subtracting equation 2 from equation 1:(x + (3/4)y) - (x + (1/3)y) = a - bSimplify the left side:x - x + (3/4)y - (1/3)y = a - bThat simplifies to:(3/4 - 1/3)y = a - bNow, I need to compute 3/4 minus 1/3. To subtract these fractions, they need a common denominator. The least common denominator for 4 and 3 is 12.Convert 3/4 to 9/12 and 1/3 to 4/12.So, 9/12 - 4/12 = 5/12Therefore, (5/12)y = a - bTo solve for y, multiply both sides by 12/5:y = (12/5)(a - b)Okay, so y is equal to (12/5)(a - b). That's the weight of the water when the bucket is full.Now, I need to find x, the weight of the empty bucket. I can use either of the original equations. Let me use the second one because the numbers seem smaller.From equation 2:x + (1/3)y = bWe already know y, so substitute y with (12/5)(a - b):x + (1/3)*(12/5)(a - b) = bSimplify (1/3)*(12/5):(1/3)*(12/5) = (12/15) = (4/5)So, the equation becomes:x + (4/5)(a - b) = bNow, solve for x:x = b - (4/5)(a - b)Let me distribute the (4/5):x = b - (4/5)a + (4/5)bCombine like terms:x = (-4/5)a + (1 + 4/5)bConvert 1 to 5/5 to add the fractions:x = (-4/5)a + (5/5 + 4/5)bx = (-4/5)a + (9/5)bSo, x is equal to (-4/5)a + (9/5)b.Now, I need to find the total weight when the bucket is full, which is x + y.We have x = (-4/5)a + (9/5)b and y = (12/5)(a - b)Let me compute x + y:x + y = (-4/5)a + (9/5)b + (12/5)(a - b)First, expand (12/5)(a - b):(12/5)a - (12/5)bNow, substitute back into x + y:x + y = (-4/5)a + (9/5)b + (12/5)a - (12/5)bCombine like terms:For a terms: (-4/5 + 12/5)a = (8/5)aFor b terms: (9/5 - 12/5)b = (-3/5)bSo, x + y = (8/5)a - (3/5)bTherefore, the total weight of the bucket when it's completely full of water is (8/5)a - (3/5)b.Let me check the answer choices:A) (3/5)a - (8/5)bB) (8/5)a - (3/5)bC) (2/5)a + (1/5)bD) (10/5)a - (5/5)bE) (6/5)a - (6/5)bLooks like option B matches my result. So, the correct answer should be B.**Final Answer**boxed{B}

💡Okay, so I have this problem where real numbers ( x ) and ( y ) satisfy the equation ( 4x^2 - 5xy + 4y^2 = 5 ). I need to find the value of ( frac{1}{s_{max}} ) where ( s = x^2 + y^2 ). Hmm, let me try to figure this out step by step.First, I know that ( s = x^2 + y^2 ) is the sum of the squares of ( x ) and ( y ). The given equation is a quadratic form in ( x ) and ( y ). Maybe I can express this quadratic form in terms of ( s ) and some other terms.Let me write down the given equation again:[ 4x^2 - 5xy + 4y^2 = 5 ]I notice that this equation is symmetric in ( x ) and ( y ), which might help. Maybe I can rewrite it in terms of ( s ) and ( xy ). Let me see.I know that ( (x + y)^2 = x^2 + 2xy + y^2 ) and ( (x - y)^2 = x^2 - 2xy + y^2 ). But I'm not sure if that's directly helpful here. Alternatively, maybe I can express the given quadratic form in terms of ( x^2 + y^2 ) and ( xy ).Let me try that. The given equation is:[ 4x^2 - 5xy + 4y^2 = 5 ]I can factor this as:[ 4(x^2 + y^2) - 5xy = 5 ]Yes, that's correct because ( 4x^2 + 4y^2 = 4(x^2 + y^2) ). So now, substituting ( s = x^2 + y^2 ), the equation becomes:[ 4s - 5xy = 5 ]So, ( 4s - 5xy = 5 ). I need to find ( s_{max} ), the maximum value of ( s ). To do this, I think I need another equation involving ( s ) and ( xy ) so that I can solve for ( s ).I recall that ( (x + y)^2 = x^2 + 2xy + y^2 = s + 2xy ) and ( (x - y)^2 = x^2 - 2xy + y^2 = s - 2xy ). But I'm not sure if that helps directly. Maybe I can use the Cauchy-Schwarz inequality or something related to quadratic forms.Alternatively, I remember that for quadratic forms, we can use eigenvalues to find the maximum and minimum values. Maybe that's a way to go.Let me think about this. The quadratic form ( 4x^2 - 5xy + 4y^2 ) can be represented as a matrix:[ begin{bmatrix} x & y end{bmatrix} begin{bmatrix} 4 & -frac{5}{2} -frac{5}{2} & 4 end{bmatrix} begin{bmatrix} x y end{bmatrix} = 5 ]Yes, because the coefficient of ( x^2 ) is 4, the coefficient of ( y^2 ) is 4, and the coefficient of ( xy ) is -5, which is split equally between the off-diagonal terms.Now, to find the maximum value of ( s = x^2 + y^2 ) subject to the constraint ( 4x^2 - 5xy + 4y^2 = 5 ), I can use the method of Lagrange multipliers or relate it to eigenvalues.I think using eigenvalues might be more straightforward here. The maximum value of ( s ) under the given constraint is related to the smallest eigenvalue of the matrix associated with the quadratic form.Wait, actually, the maximum value of ( s ) would be related to the inverse of the smallest eigenvalue. Let me recall.If we have a quadratic form ( mathbf{x}^T A mathbf{x} = c ), then the extrema of ( mathbf{x}^T B mathbf{x} ) subject to this constraint can be found using the generalized eigenvalues. In this case, ( B ) is the identity matrix since ( s = x^2 + y^2 ).So, the maximum value of ( s ) is the reciprocal of the smallest eigenvalue of matrix ( A ). Let me verify this.The quadratic form ( 4x^2 - 5xy + 4y^2 = 5 ) can be written as ( mathbf{x}^T A mathbf{x} = 5 ) where ( A ) is the matrix:[ A = begin{bmatrix} 4 & -frac{5}{2} -frac{5}{2} & 4 end{bmatrix} ]We need to find the maximum of ( s = x^2 + y^2 = mathbf{x}^T I mathbf{x} ) subject to ( mathbf{x}^T A mathbf{x} = 5 ).This is a constrained optimization problem, and the extrema occur where the gradient of ( s ) is proportional to the gradient of the constraint. Using Lagrange multipliers, we can set up the equation:[ nabla s = lambda nabla ( mathbf{x}^T A mathbf{x} - 5 ) ]Calculating the gradients:[ nabla s = 2 begin{bmatrix} x y end{bmatrix} ][ nabla ( mathbf{x}^T A mathbf{x} ) = 2 A begin{bmatrix} x y end{bmatrix} ]So, setting them proportional:[ 2 begin{bmatrix} x y end{bmatrix} = lambda 2 A begin{bmatrix} x y end{bmatrix} ][ begin{bmatrix} x y end{bmatrix} = lambda A begin{bmatrix} x y end{bmatrix} ]This implies that ( begin{bmatrix} x y end{bmatrix} ) is an eigenvector of ( A ) with eigenvalue ( frac{1}{lambda} ).Let me find the eigenvalues of matrix ( A ). The characteristic equation is:[ det(A - mu I) = 0 ][ det begin{bmatrix} 4 - mu & -frac{5}{2} -frac{5}{2} & 4 - mu end{bmatrix} = 0 ][ (4 - mu)^2 - left( -frac{5}{2} right)^2 = 0 ][ (4 - mu)^2 - frac{25}{4} = 0 ][ 16 - 8mu + mu^2 - frac{25}{4} = 0 ][ mu^2 - 8mu + 16 - 6.25 = 0 ][ mu^2 - 8mu + 9.75 = 0 ]Using the quadratic formula:[ mu = frac{8 pm sqrt{64 - 39}}{2} ][ mu = frac{8 pm sqrt{25}}{2} ][ mu = frac{8 pm 5}{2} ]So, the eigenvalues are:[ mu_1 = frac{8 + 5}{2} = frac{13}{2} = 6.5 ][ mu_2 = frac{8 - 5}{2} = frac{3}{2} = 1.5 ]So, the eigenvalues of ( A ) are 6.5 and 1.5. Therefore, the corresponding ( lambda ) values are ( frac{1}{mu_1} = frac{2}{13} ) and ( frac{1}{mu_2} = frac{2}{3} ).Wait, but I'm trying to find the maximum value of ( s ). How does this relate?I think the maximum value of ( s ) occurs when ( mathbf{x} ) is in the direction of the eigenvector corresponding to the smallest eigenvalue of ( A ). Because the quadratic form ( mathbf{x}^T A mathbf{x} ) is minimized when ( mathbf{x} ) is in the direction of the smallest eigenvalue, and since we have a constraint ( mathbf{x}^T A mathbf{x} = 5 ), the maximum ( s ) would correspond to the smallest eigenvalue.Let me think again. If ( A ) has eigenvalues ( mu_1 = 6.5 ) and ( mu_2 = 1.5 ), then the quadratic form ( mathbf{x}^T A mathbf{x} ) can be written as ( mu_1 a^2 + mu_2 b^2 ) where ( a ) and ( b ) are coefficients in the eigenvector basis.Given the constraint ( mu_1 a^2 + mu_2 b^2 = 5 ), we want to maximize ( s = a^2 + b^2 ). To maximize ( s ), we need to allocate as much as possible to the variable with the smaller coefficient, which is ( mu_2 ). So, to maximize ( a^2 + b^2 ), we set ( a = 0 ) and solve for ( b ).Wait, but if ( a = 0 ), then ( mu_2 b^2 = 5 ), so ( b^2 = frac{5}{mu_2} = frac{5}{1.5} = frac{10}{3} ). Therefore, ( s = 0 + frac{10}{3} = frac{10}{3} ).Alternatively, if I use the Lagrange multiplier method, I can set up the equations:Let ( f(x, y) = x^2 + y^2 ) and ( g(x, y) = 4x^2 - 5xy + 4y^2 - 5 = 0 ).The gradients are:[ nabla f = begin{bmatrix} 2x 2y end{bmatrix} ][ nabla g = begin{bmatrix} 8x - 5y -5x + 8y end{bmatrix} ]Setting ( nabla f = lambda nabla g ):[ 2x = lambda (8x - 5y) ][ 2y = lambda (-5x + 8y) ]So, we have the system:1. ( 2x = 8lambda x - 5lambda y )2. ( 2y = -5lambda x + 8lambda y )Let me rearrange these equations:1. ( 2x - 8lambda x + 5lambda y = 0 )2. ( 5lambda x + 2y - 8lambda y = 0 )Factor out terms:1. ( x(2 - 8lambda) + y(5lambda) = 0 )2. ( x(5lambda) + y(2 - 8lambda) = 0 )This is a homogeneous system, so for non-trivial solutions, the determinant of the coefficients must be zero:[ begin{vmatrix} 2 - 8lambda & 5lambda 5lambda & 2 - 8lambda end{vmatrix} = 0 ]Calculating the determinant:[ (2 - 8lambda)^2 - (5lambda)^2 = 0 ][ 4 - 32lambda + 64lambda^2 - 25lambda^2 = 0 ][ 4 - 32lambda + 39lambda^2 = 0 ]This is a quadratic equation in ( lambda ):[ 39lambda^2 - 32lambda + 4 = 0 ]Using the quadratic formula:[ lambda = frac{32 pm sqrt{(-32)^2 - 4 cdot 39 cdot 4}}{2 cdot 39} ][ lambda = frac{32 pm sqrt{1024 - 624}}{78} ][ lambda = frac{32 pm sqrt{400}}{78} ][ lambda = frac{32 pm 20}{78} ]So, the solutions are:1. ( lambda = frac{32 + 20}{78} = frac{52}{78} = frac{2}{3} )2. ( lambda = frac{32 - 20}{78} = frac{12}{78} = frac{2}{13} )So, the Lagrange multipliers are ( lambda = frac{2}{3} ) and ( lambda = frac{2}{13} ).Now, let's find the corresponding ( x ) and ( y ) for each ( lambda ).First, for ( lambda = frac{2}{3} ):Substitute into equation 1:[ x(2 - 8 cdot frac{2}{3}) + y(5 cdot frac{2}{3}) = 0 ][ x(2 - frac{16}{3}) + y(frac{10}{3}) = 0 ][ x(-frac{10}{3}) + y(frac{10}{3}) = 0 ][ -10x + 10y = 0 ][ -x + y = 0 ][ y = x ]Now, substitute ( y = x ) into the constraint equation ( 4x^2 - 5xy + 4y^2 = 5 ):[ 4x^2 - 5x^2 + 4x^2 = 5 ][ (4 - 5 + 4)x^2 = 5 ][ 3x^2 = 5 ][ x^2 = frac{5}{3} ][ x = pm sqrt{frac{5}{3}} ]So, ( y = pm sqrt{frac{5}{3}} ) as well.Then, ( s = x^2 + y^2 = frac{5}{3} + frac{5}{3} = frac{10}{3} ).Next, for ( lambda = frac{2}{13} ):Substitute into equation 1:[ x(2 - 8 cdot frac{2}{13}) + y(5 cdot frac{2}{13}) = 0 ][ x(2 - frac{16}{13}) + y(frac{10}{13}) = 0 ][ x(frac{26 - 16}{13}) + y(frac{10}{13}) = 0 ][ x(frac{10}{13}) + y(frac{10}{13}) = 0 ][ 10x + 10y = 0 ][ x + y = 0 ][ y = -x ]Substitute ( y = -x ) into the constraint equation:[ 4x^2 - 5x(-x) + 4x^2 = 5 ][ 4x^2 + 5x^2 + 4x^2 = 5 ][ 13x^2 = 5 ][ x^2 = frac{5}{13} ][ x = pm sqrt{frac{5}{13}} ]So, ( y = mp sqrt{frac{5}{13}} ).Then, ( s = x^2 + y^2 = frac{5}{13} + frac{5}{13} = frac{10}{13} ).So, we have two possible values for ( s ): ( frac{10}{3} ) and ( frac{10}{13} ). Since we are looking for ( s_{max} ), the maximum value is ( frac{10}{3} ).Therefore, ( frac{1}{s_{max}} = frac{3}{10} ).I think that makes sense. The maximum value of ( s ) is ( frac{10}{3} ), so its reciprocal is ( frac{3}{10} ).**Final Answer**boxed{dfrac{3}{10}}

💡First, I need to determine how many push-ups David did. The problem states that David did 3 times more push-ups than Zachary.Zachary did 19 push-ups, so I multiply this number by 3 to find out how many push-ups David did.19 push-ups multiplied by 3 equals 57 push-ups.Therefore, David did 57 push-ups.