💡Okay, so I have this problem about Bernoulli trials and probability. Let me try to understand what's being asked here. First, it mentions a series of 7 Bernoulli trials. I remember that a Bernoulli trial is a random experiment with exactly two possible outcomes, often referred to as success and failure. In this case, a success is defined as a hit that topples one of the two posts to which a red section of the fence is attached. The probability of success is given as 2/7. The question is asking for the probability that the red section does not suffer, meaning it experiences 0 successes in these 7 trials. So, essentially, we need to find the probability that none of the 7 trials result in a success.Let me recall the formula for the probability of exactly k successes in n Bernoulli trials. It's given by the binomial distribution formula:P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)Where:- C(n, k) is the combination of n items taken k at a time.- p is the probability of success on a single trial.- (1 - p) is the probability of failure.In this problem:- n = 7 (since there are 7 trials)- k = 0 (we want 0 successes)- p = 2/7 (probability of success)Plugging these values into the formula:P(X = 0) = C(7, 0) * (2/7)^0 * (1 - 2/7)^(7 - 0)Let me compute each part step by step.First, C(7, 0). I know that C(n, 0) is always 1 because there's only one way to choose nothing from a set. So, C(7, 0) = 1.Next, (2/7)^0. Any number raised to the power of 0 is 1, so (2/7)^0 = 1.Now, (1 - 2/7) is 5/7. So, (5/7)^7. I need to calculate this value.Calculating (5/7)^7:First, 5/7 is approximately 0.7142857.Raising this to the 7th power:0.7142857^1 = 0.71428570.7142857^2 ≈ 0.7142857 * 0.7142857 ≈ 0.510204080.7142857^3 ≈ 0.51020408 * 0.7142857 ≈ 0.364047970.7142857^4 ≈ 0.36404797 * 0.7142857 ≈ 0.25989590.7142857^5 ≈ 0.2598959 * 0.7142857 ≈ 0.18570420.7142857^6 ≈ 0.1857042 * 0.7142857 ≈ 0.13264550.7142857^7 ≈ 0.1326455 * 0.7142857 ≈ 0.0946693So, approximately, (5/7)^7 ≈ 0.0946693.Putting it all together:P(X = 0) = 1 * 1 * 0.0946693 ≈ 0.0946693Rounding this to three decimal places, it's approximately 0.095.Therefore, the probability that the red section does not suffer (experiences 0 successes) is approximately 0.095, or 9.5%.Wait, let me double-check my calculations to make sure I didn't make any errors.First, confirming the binomial formula:Yes, for 0 successes, the formula simplifies to (1 - p)^n, which in this case is (5/7)^7.Calculating (5/7)^7 again:5/7 ≈ 0.71428570.7142857^2 ≈ 0.510204080.51020408 * 0.7142857 ≈ 0.36404797 (third power)0.36404797 * 0.7142857 ≈ 0.2598959 (fourth power)0.2598959 * 0.7142857 ≈ 0.1857042 (fifth power)0.1857042 * 0.7142857 ≈ 0.1326455 (sixth power)0.1326455 * 0.7142857 ≈ 0.0946693 (seventh power)Yes, that seems consistent.Alternatively, I can compute (5/7)^7 more accurately using logarithms or a calculator, but since I'm doing it manually, the approximation seems reasonable.So, I think my calculation is correct. The probability is approximately 0.095.

💡First, I need to determine the discounted price of the speakers by subtracting the amount saved from the initial price.The initial price of the speakers is 475.00.The amount saved due to the discount is 276.00.By subtracting the discount from the initial price, I can find the discounted price.So, 475.00 minus 276.00 equals 199.00.Therefore, the discounted price of the speakers is 199.00.

💡Alright, so I'm trying to figure out the eccentricity of this "Golden Ellipse" by analogy with the "Golden Hyperbola." Let me start by recalling what I know about hyperbolas and ellipses.First, for a hyperbola centered at the origin, the standard equation is (frac{x^2}{a^2} - frac{y^2}{b^2} = 1). The foci are located at ((pm c, 0)), where (c^2 = a^2 + b^2). The eccentricity (e) is given by (e = frac{c}{a}), which is always greater than 1 for hyperbolas.In the problem statement, it's mentioned that the left focus (F), the right vertex (A), and the vertex (B) on the imaginary axis form a right-angled triangle. For a hyperbola, the vertices on the imaginary axis are at ((0, pm b)). So, in this case, point (B) would be ((0, b)).Given that these three points form a right-angled triangle, we can use the Pythagorean theorem. Let's denote the coordinates of the points:- Left focus (F = (-c, 0))- Right vertex (A = (a, 0))- Vertex on the imaginary axis (B = (0, b))Now, we need to determine which angle is the right angle. Since the triangle is formed by (F), (A), and (B), the right angle could be at any of these points. Let's consider each possibility.1. **Right angle at (F):** The vectors (FA) and (FB) would be perpendicular. The vector (FA) is from (F) to (A), which is ((a - (-c), 0 - 0) = (a + c, 0)). The vector (FB) is from (F) to (B), which is ((0 - (-c), b - 0) = (c, b)). The dot product of these vectors should be zero for them to be perpendicular: [ (a + c) cdot c + 0 cdot b = c(a + c) = 0 ] This simplifies to (c(a + c) = 0). Since (c) is positive, this would imply (a + c = 0), which isn't possible because both (a) and (c) are positive. So, the right angle isn't at (F).2. **Right angle at (A):** The vectors (AF) and (AB) would be perpendicular. Vector (AF) is from (A) to (F), which is ((-c - a, 0 - 0) = (-c - a, 0)). Vector (AB) is from (A) to (B), which is ((0 - a, b - 0) = (-a, b)). The dot product should be zero: [ (-c - a)(-a) + 0 cdot b = a(c + a) = 0 ] Again, since (a) and (c) are positive, this implies (a(c + a) = 0), which isn't possible. So, the right angle isn't at (A).3. **Right angle at (B):** The vectors (BA) and (BF) would be perpendicular. Vector (BA) is from (B) to (A), which is ((a - 0, 0 - b) = (a, -b)). Vector (BF) is from (B) to (F), which is ((-c - 0, 0 - b) = (-c, -b)). The dot product should be zero: [ a(-c) + (-b)(-b) = -ac + b^2 = 0 ] So, we have: [ -ac + b^2 = 0 implies b^2 = ac ] This is a valid equation, so the right angle is at (B).Now, for hyperbolas, we know that (c^2 = a^2 + b^2). Substituting (b^2 = ac) into this equation:[c^2 = a^2 + ac]Let's divide both sides by (a^2) to express this in terms of eccentricity (e = frac{c}{a}):[left(frac{c}{a}right)^2 = 1 + frac{c}{a}][e^2 = 1 + e]Rearranging the equation:[e^2 - e - 1 = 0]This is a quadratic equation in terms of (e). Using the quadratic formula:[e = frac{1 pm sqrt{1 + 4}}{2} = frac{1 pm sqrt{5}}{2}]Since eccentricity is positive, we take the positive root:[e = frac{1 + sqrt{5}}{2}]Which is approximately 1.618, known as the golden ratio. Hence, the eccentricity of the "Golden Hyperbola" is (frac{1 + sqrt{5}}{2}).Now, moving on to the "Golden Ellipse." An ellipse centered at the origin has the standard equation (frac{x^2}{a^2} + frac{y^2}{b^2} = 1), where (a > b). The foci are located at ((pm c, 0)), where (c^2 = a^2 - b^2). The eccentricity (e) is given by (e = frac{c}{a}), which is always less than 1 for ellipses.By analogy, we need to form a right-angled triangle with the left focus (F), the right vertex (A), and the vertex (B) on the minor axis (which for an ellipse is the y-axis). So, the points are:- Left focus (F = (-c, 0))- Right vertex (A = (a, 0))- Vertex on the minor axis (B = (0, b))Again, we need to determine where the right angle is. Let's go through the same process.1. **Right angle at (F):** Vectors (FA) and (FB) should be perpendicular. Vector (FA = (a + c, 0)), vector (FB = (c, b)). Dot product: [ (a + c)c + 0 cdot b = c(a + c) = 0 ] As before, this implies (c(a + c) = 0), which isn't possible since (c) and (a) are positive.2. **Right angle at (A):** Vectors (AF) and (AB) should be perpendicular. Vector (AF = (-a - c, 0)), vector (AB = (-a, b)). Dot product: [ (-a - c)(-a) + 0 cdot b = a(a + c) = 0 ] Again, this implies (a(a + c) = 0), which isn't possible.3. **Right angle at (B):** Vectors (BA) and (BF) should be perpendicular. Vector (BA = (a, -b)), vector (BF = (-c, -b)). Dot product: [ a(-c) + (-b)(-b) = -ac + b^2 = 0 ] So, we have: [ -ac + b^2 = 0 implies b^2 = ac ] This is similar to the hyperbola case.For an ellipse, we know that (c^2 = a^2 - b^2). Substituting (b^2 = ac) into this equation:[c^2 = a^2 - ac]Let's divide both sides by (a^2) to express in terms of eccentricity (e = frac{c}{a}):[left(frac{c}{a}right)^2 = 1 - frac{c}{a}][e^2 = 1 - e]Rearranging the equation:[e^2 + e - 1 = 0]This is another quadratic equation in terms of (e). Using the quadratic formula:[e = frac{-1 pm sqrt{1 + 4}}{2} = frac{-1 pm sqrt{5}}{2}]Since eccentricity is positive, we take the positive root:[e = frac{-1 + sqrt{5}}{2}]Which simplifies to:[e = frac{sqrt{5} - 1}{2}]This is approximately 0.618, which is the reciprocal of the golden ratio. Therefore, the eccentricity of the "Golden Ellipse" is (frac{sqrt{5} - 1}{2}).I should double-check my steps to ensure there are no mistakes. Starting from the right-angled triangle condition, I correctly identified the coordinates and vectors. The dot product condition led me to (b^2 = ac), which I substituted into the ellipse relationship (c^2 = a^2 - b^2). Dividing by (a^2) to express in terms of (e) was the right move, leading to the quadratic equation. Solving that gave me the eccentricity, and since it's an ellipse, the eccentricity is less than 1, which matches the result.So, I'm confident that the eccentricity of the "Golden Ellipse" is (frac{sqrt{5} - 1}{2}).

💡First, I need to determine the total amount in Ram's account at the end of two years. Since the account balance increased by 32% from the initial deposit of 1000, I calculate the total increase over two years by finding 32% of 1000.Next, I add this total increase to the initial deposit to find the account balance at the end of two years.I already know that at the end of the first year, the account balance was 1100. To find the increase in the second year, I subtract the balance at the end of the first year from the balance at the end of the two years.Finally, to find the percentage increase for the second year, I divide the increase in the second year by the account balance at the end of the first year and then multiply by 100 to convert it to a percentage.

💡Alright, so I have this geometry problem here. Let me try to parse it step by step. We have triangle ABC where angle C is twice the size of angle B. Okay, so if I let angle B be x degrees, then angle C would be 2x degrees. Since the sum of angles in a triangle is 180 degrees, angle A would be 180 - x - 2x, which is 180 - 3x degrees. Got that.Next, segment CD is the angle bisector of angle C. So, CD splits angle C into two equal parts. Since angle C is 2x, each of the angles at C, namely angle BCD and angle ACD, would be x degrees each. That makes sense.Then, M is the midpoint of side BC. So, BM equals MC. From M, a perpendicular MN is drawn to segment CD. So, MN is perpendicular to CD, meaning that angle MND is 90 degrees. On side AB, there exists a point K such that triangle KMN is equilateral. So, triangle KMN has all sides equal and all angles equal to 60 degrees. We need to prove that points M, N, and A are collinear. Hmm, okay. So, I need to show that M, N, and A lie on the same straight line. Maybe I can find some properties or use some theorems related to angle bisectors, midpoints, and equilateral triangles.Let me try to sketch this triangle to visualize it better. Let me draw triangle ABC with angle C at the top, angle B at the bottom left, and angle A at the bottom right. CD is the angle bisector from C to side AB, splitting angle C into two equal angles of x each. M is the midpoint of BC, so BM equals MC. From M, I draw a perpendicular to CD, which is MN. Then, on AB, there's a point K such that triangle KMN is equilateral.Since triangle KMN is equilateral, all its sides are equal, so KM equals MN equals NK, and all its angles are 60 degrees. That might be useful.I think I need to find some relationships between the sides or angles in the triangle. Maybe using the Angle Bisector Theorem since CD is an angle bisector. The Angle Bisector Theorem states that the angle bisector divides the opposite side into segments proportional to the adjacent sides. So, in this case, CD is the angle bisector of angle C, so it divides AB into segments proportional to AC and BC.So, if I let AD be the segment from A to D on AB, and DB be the segment from D to B, then AD/DB = AC/BC. Hmm, but I don't know the lengths of AC or BC yet.Wait, maybe I can express AC and BC in terms of the angles. Since in triangle ABC, the sides are proportional to the sines of their opposite angles. So, by the Law of Sines, AC/sin B = BC/sin A = AB/sin C.Given that angle B is x, angle C is 2x, and angle A is 180 - 3x. So, AC/sin x = BC/sin(180 - 3x) = AB/sin(2x). Since sin(180 - 3x) is sin(3x), so AC/sin x = BC/sin(3x). Therefore, AC = BC * (sin x / sin 3x). That might be useful later.Now, since M is the midpoint of BC, BM = MC. So, BM = MC = BC/2.From M, we draw a perpendicular to CD, which is MN. So, MN is perpendicular to CD, meaning that triangle MND is a right triangle with right angle at N.Since triangle KMN is equilateral, angle KMN is 60 degrees. So, angle between KM and MN is 60 degrees. Hmm, maybe I can relate this angle to other angles in the triangle.Wait, since MN is perpendicular to CD, and triangle KMN is equilateral, perhaps there is some relationship between the angles at point M.Let me think about the coordinates. Maybe assigning coordinates to the points could help. Let me place point B at (0,0), point C at (2c,0), so that M, the midpoint, is at (c,0). Then, point A can be somewhere in the plane. Let me try to assign coordinates.But before I get into coordinate geometry, maybe there's a synthetic approach. Let me see.Since CD is the angle bisector, and M is the midpoint of BC, maybe there's some symmetry or properties related to medians and angle bisectors.Also, since triangle KMN is equilateral, point K must be such that it's at a 60-degree rotation from point N around point M. Maybe I can use some rotation properties.Alternatively, perhaps using vectors or coordinate geometry would make this more straightforward.Let me try coordinate geometry. Let me place point B at (0,0), point C at (2,0), so M is at (1,0). Let me assume BC is of length 2 units for simplicity, so BM = MC = 1.Now, I need to find coordinates of point A. Since angle C is twice angle B, and angles at B and C are x and 2x respectively, with angle A being 180 - 3x.Using the Law of Sines, AC/sin x = BC/sin(180 - 3x) = AB/sin(2x). Since BC is 2, AC = 2 * sin x / sin(3x). Similarly, AB = 2 * sin(2x) / sin(3x).Let me compute sin(3x). Using the identity, sin(3x) = 3 sin x - 4 sin^3 x. Hmm, but maybe it's better to express AC and AB in terms of x.Alternatively, maybe I can assign coordinates to point A. Let me suppose that point A is at (a,b). Then, I can express the angles at B and C in terms of coordinates.But this might get complicated. Maybe another approach.Since CD is the angle bisector, by the Angle Bisector Theorem, AD/DB = AC/BC. We have AC = 2 sin x / sin(3x), and BC = 2. So, AD/DB = (2 sin x / sin(3x)) / 2 = sin x / sin(3x).So, AD = (sin x / sin(3x)) * DB. Since AD + DB = AB, which is 2 sin(2x)/ sin(3x). So, AD = (sin x / sin(3x)) * DB, and AD + DB = 2 sin(2x)/ sin(3x).Let me denote DB = y, then AD = (sin x / sin(3x)) y. So, (sin x / sin(3x)) y + y = 2 sin(2x)/ sin(3x). Factor out y: y (sin x / sin(3x) + 1) = 2 sin(2x)/ sin(3x).So, y = [2 sin(2x)/ sin(3x)] / [1 + sin x / sin(3x)] = [2 sin(2x)] / [sin(3x) + sin x].Using the identity sin(3x) + sin x = 2 sin(2x) cos x. So, y = [2 sin(2x)] / [2 sin(2x) cos x] = 1 / cos x.Therefore, DB = 1 / cos x, and AD = (sin x / sin(3x)) * (1 / cos x).But sin(3x) = 3 sin x - 4 sin^3 x, so AD = (sin x / (3 sin x - 4 sin^3 x)) * (1 / cos x) = [1 / (3 - 4 sin^2 x)] * (1 / cos x).Hmm, this is getting a bit messy. Maybe I should assign specific values to x to make it easier. Let me assume x is 30 degrees, so angle B is 30 degrees, angle C is 60 degrees, and angle A is 90 degrees. So, triangle ABC is a 30-60-90 triangle.In that case, sides are in the ratio 1 : sqrt(3) : 2. So, if BC is 2 units, then AB is 1 unit, and AC is sqrt(3) units.Wait, but in a 30-60-90 triangle, the sides opposite 30, 60, 90 degrees are 1, sqrt(3), 2 respectively. So, if angle B is 30 degrees, side AC is opposite angle B, so AC = 1. Angle C is 60 degrees, side AB is opposite angle C, so AB = sqrt(3). Angle A is 90 degrees, side BC is opposite angle A, so BC = 2.Wait, but in that case, AC = 1, AB = sqrt(3), BC = 2.So, point D is on AB such that CD bisects angle C. So, using the Angle Bisector Theorem, AD/DB = AC/BC = 1/2. So, AD = (1/3) AB, DB = (2/3) AB.Since AB is sqrt(3), AD = sqrt(3)/3, DB = 2 sqrt(3)/3.So, coordinates: Let me place point B at (0,0), point C at (2,0). Since triangle ABC is right-angled at A, point A must be at (0, sqrt(3)), because AB is sqrt(3), and AC is 1.Wait, hold on. If angle A is 90 degrees, then sides AB and AC are the legs, and BC is the hypotenuse.So, AB is opposite angle C (60 degrees), so AB = sqrt(3), AC is opposite angle B (30 degrees), so AC = 1, and BC is 2.So, coordinates: Let me place point A at (0,0). Then, since AC is 1 unit along the x-axis, point C is at (1,0). Point B is at (0, sqrt(3)), because AB is sqrt(3). Then, BC is from (1,0) to (0, sqrt(3)), which has length sqrt(1 + 3) = 2, which matches.So, points:- A: (0,0)- B: (0, sqrt(3))- C: (1,0)Midpoint M of BC: Coordinates of B (0, sqrt(3)) and C (1,0). So, midpoint M is ((0+1)/2, (sqrt(3)+0)/2) = (0.5, sqrt(3)/2).Now, CD is the angle bisector from C to AB. Point D is on AB. Since AB is from (0,0) to (0, sqrt(3)). Wait, AB is along the y-axis from (0,0) to (0, sqrt(3)). So, point D is somewhere on AB.Using the Angle Bisector Theorem, AD/DB = AC/BC = 1/2. So, AD = (1/3) AB, DB = (2/3) AB.Since AB is sqrt(3), AD = sqrt(3)/3, DB = 2 sqrt(3)/3.So, point D is sqrt(3)/3 units from A along AB. Since AB is along the y-axis, point D is at (0, sqrt(3)/3).So, coordinates of D: (0, sqrt(3)/3).Now, CD is the segment from C (1,0) to D (0, sqrt(3)/3). Let me find the equation of CD.The slope of CD is (sqrt(3)/3 - 0)/(0 - 1) = (sqrt(3)/3)/(-1) = -sqrt(3)/3.So, equation of CD: y - 0 = (-sqrt(3)/3)(x - 1), so y = (-sqrt(3)/3)x + sqrt(3)/3.Now, from midpoint M (0.5, sqrt(3)/2), we need to draw a perpendicular to CD, which is MN.The slope of CD is -sqrt(3)/3, so the slope of MN, being perpendicular, is the negative reciprocal, which is 3/sqrt(3) = sqrt(3).So, equation of MN: y - sqrt(3)/2 = sqrt(3)(x - 0.5).We need to find the intersection point N of MN and CD.So, set the two equations equal:(-sqrt(3)/3)x + sqrt(3)/3 = sqrt(3)(x - 0.5) + sqrt(3)/2.Let me simplify this equation.First, multiply both sides by 3 to eliminate denominators:-sqrt(3) x + sqrt(3) = 3 sqrt(3)(x - 0.5) + (3 sqrt(3))/2.Simplify the right side:3 sqrt(3) x - (3 sqrt(3))/2 + (3 sqrt(3))/2 = 3 sqrt(3) x.So, left side: -sqrt(3) x + sqrt(3).Right side: 3 sqrt(3) x.Bring all terms to left:-sqrt(3) x + sqrt(3) - 3 sqrt(3) x = 0.Combine like terms:(-sqrt(3) - 3 sqrt(3))x + sqrt(3) = 0 => (-4 sqrt(3))x + sqrt(3) = 0.So, -4 sqrt(3) x = -sqrt(3) => x = (-sqrt(3))/(-4 sqrt(3)) = 1/4.So, x = 1/4. Then, plug back into equation of CD to find y:y = (-sqrt(3)/3)(1/4) + sqrt(3)/3 = (-sqrt(3)/12) + (4 sqrt(3)/12) = (3 sqrt(3))/12 = sqrt(3)/4.So, point N is at (1/4, sqrt(3)/4).Now, we have point M at (0.5, sqrt(3)/2), point N at (1/4, sqrt(3)/4), and point K on AB such that triangle KMN is equilateral.Let me find point K on AB such that triangle KMN is equilateral. Since AB is the y-axis from (0,0) to (0, sqrt(3)), point K must be at (0, k) for some k between 0 and sqrt(3).We need to find k such that distances KM, KN, and MN are equal.First, compute MN:Distance between M (0.5, sqrt(3)/2) and N (1/4, sqrt(3)/4):MN = sqrt[(0.5 - 1/4)^2 + (sqrt(3)/2 - sqrt(3)/4)^2] = sqrt[(1/4)^2 + (sqrt(3)/4)^2] = sqrt[1/16 + 3/16] = sqrt[4/16] = sqrt[1/4] = 1/2.So, MN = 1/2. Therefore, KM and KN must also be 1/2.Compute KM: distance from K (0, k) to M (0.5, sqrt(3)/2):KM = sqrt[(0.5 - 0)^2 + (sqrt(3)/2 - k)^2] = sqrt[0.25 + (sqrt(3)/2 - k)^2].Set this equal to 1/2:sqrt[0.25 + (sqrt(3)/2 - k)^2] = 1/2.Square both sides:0.25 + (sqrt(3)/2 - k)^2 = 0.25.Subtract 0.25:(sqrt(3)/2 - k)^2 = 0.Take square root:sqrt(3)/2 - k = 0 => k = sqrt(3)/2.So, point K is at (0, sqrt(3)/2).Wait, but point M is at (0.5, sqrt(3)/2). So, point K is at (0, sqrt(3)/2). So, the distance between K and M is 0.5, which is correct.Similarly, compute KN: distance from K (0, sqrt(3)/2) to N (1/4, sqrt(3)/4):KN = sqrt[(1/4 - 0)^2 + (sqrt(3)/4 - sqrt(3)/2)^2] = sqrt[(1/16) + ( - sqrt(3)/4)^2] = sqrt[1/16 + 3/16] = sqrt[4/16] = 1/2. So, that's correct.So, triangle KMN is equilateral with all sides equal to 1/2.Now, we need to prove that points M, N, and A are collinear.Point A is at (0,0), point M is at (0.5, sqrt(3)/2), and point N is at (1/4, sqrt(3)/4).Let me check if these three points lie on the same straight line.Compute the slope from A to M: (sqrt(3)/2 - 0)/(0.5 - 0) = (sqrt(3)/2)/0.5 = sqrt(3).Compute the slope from M to N: (sqrt(3)/4 - sqrt(3)/2)/(1/4 - 0.5) = (-sqrt(3)/4)/(-1/4) = sqrt(3).Compute the slope from N to A: (sqrt(3)/4 - 0)/(1/4 - 0) = (sqrt(3)/4)/(1/4) = sqrt(3).So, all slopes are equal to sqrt(3), meaning points A, N, and M lie on a straight line with slope sqrt(3). Therefore, they are collinear.Hence, proved that points M, N, and A are collinear.**Final Answer**boxed{M}, boxed{N}, and boxed{A} are collinear.**Final Answer**boxed{M}, boxed{N}, and boxed{A} are collinear.

💡Okay, so I have this problem about distributing fertilizer from two factories, C and D, to two planting bases, A and B. The goal is to figure out how much fertilizer each factory donates, set up a transportation cost function, and then adjust for a change in transportation cost. Let me try to break this down step by step.First, part (1) asks for the quantities of fertilizer donated by factories C and D. It says that the total donation is 100 tons, and the amount donated by C is 20 tons less than twice the amount donated by D. Hmm, okay, so let me denote the amount donated by D as 'd' tons. Then, the amount donated by C would be '2d - 20' tons. Since the total is 100 tons, I can write the equation:d + (2d - 20) = 100Let me solve this:d + 2d - 20 = 100 3d - 20 = 100 3d = 120 d = 40So, factory D donated 40 tons. Then, factory C donated 2*40 - 20 = 80 - 20 = 60 tons. That seems straightforward.Moving on to part (2). It says to let x tons be transported from factory C to location A. Then, y is the total transportation cost. I need to find the relationship between y and x and determine the minimum total transportation cost.First, let me figure out how much fertilizer each factory is sending to each location. Since the total fertilizer is 100 tons, and it's distributed evenly to A and B, each location gets 50 tons.So, from factory C, x tons go to A, and the rest, which is 60 - x tons, go to B. From factory D, since A needs 50 tons in total, and x tons are coming from C, the remaining 50 - x tons must come from D. Similarly, for location B, it needs 50 tons, and since 60 - x tons are coming from C, the remaining 50 - (60 - x) = x - 10 tons must come from D.Wait, hold on. If x is the amount from C to A, then D must send 50 - x to A. But for B, since C sends 60 - x to B, D must send 50 - (60 - x) = x - 10 to B. But x - 10 has to be non-negative because you can't send negative fertilizer. So, x - 10 ≥ 0 ⇒ x ≥ 10. Also, since D can only send up to 40 tons, x - 10 ≤ 40 ⇒ x ≤ 50. So, x is between 10 and 50.Now, let's calculate the transportation cost. The cost from C to A is 22元 per ton, so that's 22x. From C to B, it's 20元 per ton, so 20*(60 - x). From D to A, it's 20元 per ton, so 20*(50 - x). From D to B, it's 22元 per ton, so 22*(x - 10). Adding all these up:y = 22x + 20*(60 - x) + 20*(50 - x) + 22*(x - 10)Let me expand this:y = 22x + 1200 - 20x + 1000 - 20x + 22x - 220Combine like terms:22x - 20x - 20x + 22x = (22 - 20 - 20 + 22)x = 4x1200 + 1000 - 220 = 1980So, y = 4x + 1980Now, to find the minimum total transportation cost, since y is a linear function of x, and the coefficient of x is positive (4), the minimum occurs at the smallest x, which is x = 10.Plugging x = 10 into y:y = 4*10 + 1980 = 40 + 1980 = 2020元So, the minimum total transportation cost is 2020元.Now, part (3) says that the transportation cost from D to B is reduced by 'a' yuan per ton, where 0 < a < 6. How should the transportation be adjusted to minimize the total transportation cost?Let me adjust the cost function. The new cost from D to B is 22 - a元 per ton. So, the new total cost y becomes:y = 22x + 20*(60 - x) + 20*(50 - x) + (22 - a)*(x - 10)Let me expand this:y = 22x + 1200 - 20x + 1000 - 20x + (22 - a)x - (22 - a)*10Simplify:22x - 20x - 20x + (22 - a)x = (22 - 20 - 20 + 22 - a)x = (4 - a)x1200 + 1000 - (22 - a)*10 = 2200 - 220 + 10a = 1980 + 10aSo, y = (4 - a)x + 1980 + 10aNow, depending on the value of 'a', the coefficient of x changes:- If 4 - a > 0 ⇒ a < 4, then y increases as x increases, so to minimize y, set x as small as possible, which is x = 10.- If 4 - a = 0 ⇒ a = 4, then y is constant, doesn't depend on x.- If 4 - a < 0 ⇒ a > 4, then y decreases as x increases, so to minimize y, set x as large as possible, which is x = 50.So, the optimal x depends on the value of 'a':- For 0 < a < 4, set x = 10.- For a = 4, any x between 10 and 50 is fine.- For 4 < a < 6, set x = 50.Let me verify this with some example values.If a = 2 (which is less than 4), then y = (4 - 2)x + 1980 + 20 = 2x + 2000. To minimize, set x = 10, y = 20 + 2000 = 2020.If a = 4, y = 0x + 1980 + 40 = 2020, which is constant.If a = 5 (which is greater than 4), y = (4 - 5)x + 1980 + 50 = -x + 2030. To minimize, set x = 50, y = -50 + 2030 = 1980.Wait, that seems contradictory. When a = 5, y at x=50 is 1980, which is less than 2020. But when a=4, it's 2020, and when a=5, it's 1980. That makes sense because as a increases beyond 4, the cost saving from sending more from D to B outweighs the increased cost from sending more from C to A.So, summarizing:- If 0 < a < 4, send x=10 tons from C to A.- If a=4, any x is fine.- If 4 < a <6, send x=50 tons from C to A.This way, the total transportation cost is minimized based on the value of 'a'.