answer:To find the integer closest to sqrt[3]{250}, we first determine which integers sqrt[3]{250} falls between. We know 6^3 = 216 and 7^3 = 343. Since 250 is between 216 and 343, the possible integers are 6 or 7. Next, to find out which is closer, calculate the midpoint between 6^3 and 7^3: [ text{Midpoint} = frac{216 + 343}{2} = frac{559}{2} = 279.5 ] The number 250 is closer to 216 than to 279.5. Therefore, sqrt[3]{250} is closer to 6 than to 7. Hence, the closest integer to the cube root of 250 is boxed{6}.

answer:1. Since a_{n+1}b_{n}=S_{n}+1, a_{1}=1, b_{n}= frac {n}{2}, we have: a_{2}= frac {a_{1}+1}{b_{1}}= frac {1+1}{ frac {1}{2}}=4 a_{3}= frac {S_{2}+1}{b_{2}}= frac {1+4+1}{1}=6 a_{4}= frac {S_{3}+1}{b_{3}}= frac {1+4+6+1}{ frac {3}{2}}=8 Therefore, the value of a₄ is boxed{8}. 2. Let's assume that a_{n}=a_{1}q^{n-1} (q≠1), then S_{n}= frac {a_{1}(1-q^{n})}{1-q}. Since a_{n+1}b_{n}=S_{n}+1, we have: b_{n}= frac {S_{n}+1}{a_{n+1}}= frac { frac {a_{1}(1-q^{n})}{1-q}+1}{a_{1}q^{n}}= frac {a_{1}-a_{1}q^{n}+1-q}{(1-q)a_{1}q^{n}} Therefore, b_{n}+ frac {1}{1-q}= frac {a_{1}-a_{1}q^{n}+1-q}{(1-q)a_{1}q^{n}}+ frac {1}{1-q}= frac {a_{1}+1-q}{(1-q)a_{1}q^{n}} And, b_{n+1}+ frac {1}{1-q}= frac {a_{1}+1-q}{(1-q)a_{1}q^{n+1}} Hence, frac {b_{n+1}}{b_{n}}=q Since q is a constant, the sequence {b_{n}+ frac {1}{1-q}} is a geometric sequence. 3. Since {b_n} is an arithmetic sequence with common difference d, when n≥2, we have: (a_{n+1}-a_{n})b_{n}=(1-d)a_{n} Since all items of {a_n} are non-zero, we have: a_{n+1}-a_{n}≠0, 1-d≠0 When n≥2, frac {b_{n}}{1-d}= frac {a_{n}}{a_{n+1}-a_{n}} When n≥3, frac {b_{n-1}}{1-d}= frac {a_{n-1}}{a_{n}-a_{n-1}} Subtracting the two equations, when n≥3, we get: frac {a_{n}}{a_{n+1}-a_{n}}- frac {a_{n-1}}{a_{n}-a_{n-1}}= frac {b_{n}-b_{n-1}}{1-d}= frac {d}{1-d} First, we prove sufficiency: If d= frac {1}{2}, then frac {a_{n}}{a_{n+1}-a_{n}}- frac {a_{n-1}}{a_{n}-a_{n-1}}=1 When n≥3, frac {a_{n-1}}{a_{n}-a_{n-1}}+1= frac {a_{n}}{a_{n+1}-a_{n}} Since a_n≠0, a_{n+1}-a_{n}=a_{n}-a_{n-1} Hence, a₂, a₃, …, a_n, … is an arithmetic sequence. Next, we prove necessity: If a₂, a₃, …, a_n, … is an arithmetic sequence, then when n≥3, a_{n+1}-a_{n}=a_{n}-a_{n-1} Therefore, frac {a_{n}}{a_{n+1}-a_{n}}- frac {a_{n-1}}{a_{n}-a_{n-1}}= frac {a_{n-1}}{a_{n}-a_{n-1}}- frac {a_{n-1}}{a_{n}-a_{n-1}}=1= frac {d}{1-d} Hence, d= frac {1}{2}. In conclusion, the necessary and sufficient condition for a₂, a₃, …, a_n, … to be an arithmetic sequence is boxed{d= frac {1}{2}}.

answer:Let's denote the initial amount of milk in the can as M liters and the initial amount of water as W liters. According to the problem, the capacity of the can is 60 liters. When 20 liters of milk are added, the total amount of milk becomes M + 20 liters, and the total volume of the mixture becomes 60 liters (since the can is then full). At this point, the ratio of milk to water is 3:1, which means that for every 3 parts of milk, there is 1 part of water. Let's set up the ratio equation for the situation after adding the 20 liters of milk: (M + 20) / W = 3 / 1 This simplifies to: M + 20 = 3W (Equation 1) We also know that initially, the sum of milk and water is less than the capacity of the can by 20 liters (since adding 20 liters of milk fills the can): M + W = 60 - 20 M + W = 40 (Equation 2) Now we have two equations with two unknowns (M and W). We can solve this system of equations to find the values of M and W. From Equation 2, we can express W in terms of M: W = 40 - M (Equation 3) Now, we can substitute Equation 3 into Equation 1: M + 20 = 3(40 - M) M + 20 = 120 - 3M 4M = 100 M = 25 Now that we have the value of M, we can find W using Equation 3: W = 40 - M W = 40 - 25 W = 15 So, the initial amount of milk is 25 liters, and the initial amount of water is 15 liters. The initial ratio of milk to water is therefore: Milk : Water = 25 : 15 To simplify the ratio, we can divide both numbers by their greatest common divisor, which is 5: Milk : Water = (25/5) : (15/5) Milk : Water = 5 : 3 The initial ratio of milk to water in the can is boxed{5:3} .

answer:1. **Radius of semicircles**: Each side of the original 6 by 6 square is the diameter of a semicircle. So, the radius of each semicircle is half of this measurement: [ text{Radius} = frac{6}{2} = 3 ] 2. **Position and side length of square EFGH**: - The sides of square EFGH are tangent to the semicircles. The distance from the center of any side of the original square to the tangent point on EFGH is equal to the radius of the semicircle. - Thus, starting from the center of a side of the original square, moving outward by the radius of 3 units, crossing the side of EFGH, and moving outward by another radius of 3, leads to the next side’s center of the original square. - Therefore, the total side length of square EFGH considering all segments is: [ text{Side length of } EFGH = 6 + 3 + 3 = 12 ] 3. **Area of square EFGH**: - The area of square EFGH is calculated by squaring its side: [ text{Area} = 12^2 = 144 ] - Therefore, the area of square EFGH is 144. The final answer is boxed{D) 144}