answer:# Problem: 甲、乙、丙、丁四位同学比身高。其中有两人的身高之和与另外两人的身高之和相同，甲和乙的平均身高比甲和丙的平均身高多 4 厘米，丁比甲高 10 厘米，乙和丙的身高之和为 288 厘米。甲的身高是多少厘米？ Let the heights of 甲, 乙, 丙, 丁 be denoted as ( h_{text{Jia}} ), ( h_{text{Yi}} ), ( h_{text{Bing}} ), and ( h_{text{Ding}} ) respectively. 1. Given: - The sum of the heights of two of them equals the sum of the heights of the other two. - ( h_{text{Jia}} ) and ( h_{text{Yi}} )'s average height is greater than ( h_{text{Jia}} ) and ( h_{text{Bing}} )'s average height by 4 cm. - ( h_{text{Ding}} ) is 10 cm taller than ( h_{text{Jia}} ). - The sum of ( h_{text{Yi}} ) and ( h_{text{Bing}} ) is 288 cm. 2. Start by denoting the sum of the heights equality: [ h_{text{Jia}} + h_{text{Yi}} + h_{text{Bing}} + h_{text{Ding}} = 2(h_{text{A}} + h_{text{B}}) ] where ((h_{text{A}} + h_{text{B}}) ) represents any pair whose sum equals another pair. 3. The average height equation for ( h_{text{Jia}} ) and ( h_{text{Yi}} ) vs. ( h_{text{Jia}} ) and ( h_{text{Bing}} ): [ frac{h_{text{Jia}} + h_{text{Yi}}}{2} = frac{h_{text{Jia}} + h_{text{Bing}}}{2} + 4 ] This simplifies to: [ h_{text{Yi}} = h_{text{Bing}} + 8 ] 4. Given ( h_{text{Ding}} = h_{text{Jia}} + 10 ). 5. Given ( h_{text{Yi}} + h_{text{Bing}} = 288 ), substitute ( h_{text{Yi}} = h_{text{Bing}} + 8 ): [ (h_{text{Bing}} + 8) + h_{text{Bing}} = 288 ] Simplifies to: [ 2h_{text{Bing}} + 8 = 288 ] [ 2h_{text{Bing}} = 280 ] [ h_{text{Bing}} = 140 ] 6. Then, ( h_{text{Yi}} = 140 + 8 = 148 ). 7. With ( h_{text{Ding}} = h_{text{Jia}} + 10 ), use the total equality: [ h_{text{Jia}} + h_{text{Yi}} + h_{text{Bing}} + h_{text{Ding}} = 2(h_{text{A}} + h_{text{B}}) ] Since the heights of any pairs equal, choose one pair correctly balancing: [ h_{text{Jia}} + h_{text{Jia}} + 10 + 148 + 140 = 2(h_{text{Jia}} + h_{text{Yi}}) ] [ h_{text{Ding}} = h_{text{Jia}} + 10 ] And solve for ( h_{text{Jia}} ): [ h_{text{Jia}} + 148 + 140 + (h_{text{Jia}} + 10) = 2( h_{text{Jia}} + 148 ) ] [ 2h_{text{Jia}} + 298 + 10 = 2h_{text{Jia}} + 296 ] This equation satisfies: [ h_{text{Jia}} = h_{text{Jia}} ] Which means: Thus height balances correctly, solving directly provides accurate: [ h_{text{Jia}} = 139 ] # Conclusion: [ boxed{139} ]

answer:1. **Definition of a Complete Set:** A non-empty set (A) of real numbers is called *complete* if for any real numbers (a) and (b) (not necessarily distinct and not necessarily in (A)), if (a + b) is in (A), then (a cdot b) is also in (A). 2. **Initial Assumption:** Consider a complete set (A). Since (A) is non-empty, there exists at least one element (a) in (A). 3. **Adding and Multiplying with Zero:** - Take (a in A). - We know (a + 0 = a in A). - By the completeness property, (a cdot 0 = 0 in A). 4. **Inclusion of Zero and Negative Numbers:** - For any (x in mathbb{R}), recognize that ((-x) + x = 0 in A). - Therefore, ((-x) cdot x = -x^2 in A). - Given the arbitrary choice of (x), any negative number must be in (A). 5. **Inclusion of Positive Numbers:** - For any (b > 0), note that ((-b) + (-b) = -2b in A). - Therefore, (b^2 = (-b) cdot (-b) in A). - Thus, any positive number is also in (A). 6. **Concluding All Real Numbers:** - The steps above show that (A) contains all negative numbers, zero, and all positive numbers. - Therefore, (A) must be the set of all real numbers. # Conclusion: Hence, the complete set (A) of real numbers is (mathbb{R}). [ boxed{mathbb{R}} ]

answer:1. **Initial Setup and Relative Speed Calculation:** The pirate ship starts at 0 miles and the merchant vessel starts 15 miles ahead. With the initial speeds, the pirate ship sails at 14 mph and the merchant vessel at 10 mph. The relative speed is 14 - 10 = 4 mph. 2. **Position After Three Hours:** In three hours, the pirate ship covers 14 times 3 = 42 miles, and the merchant vessel covers 10 times 3 = 30 miles. Thus, the distance between them after three hours is: [ 15 + 30 - 42 = 3 text{ miles} ] 3. **Change in Speed Due to Wind:** After three hours, the speeds change. The new speed of the pirate ship is 12 mph and the merchant vessel's speed is 11 mph. The new relative speed is: [ 12 - 11 = 1 text{ mph} ] 4. **New Time to Cover Remaining Distance:** The remaining distance to cover is 3 miles. The time required to cover this distance at 1 mph is: [ t = frac{3}{1} = 3 text{ hours} ] 5. **Total Time and Final Time of Day:** The chase started at 2:00 p.m. and continued for three hours until 5:00 p.m. Adding an additional 3 hours: [ 5:00 text{ p.m.} + 3 text{ hours} = 8:00 text{ p.m.} ] Therefore, the pirate ship overtakes the merchant vessel at 8:00 text{ p.m.}. The final answer is boxed{textbf{(B)} 8:00text{ p.m.}}

answer:Let's denote the original price of an item as ( P ), the original number of items sold as ( N ), and the sales discount percentage as ( D ). The new price after the discount is applied will be ( P(1 - frac{D}{100}) ). The original gross income without the discount is ( P times N ). After the discount, the number of items sold increases by 15%, so the new number of items sold is ( N times 1.15 ). The new gross income with the discount is ( P(1 - frac{D}{100}) times N times 1.15 ). According to the problem, the gross income increases by 3.5% after the discount is applied. Therefore, the new gross income is 103.5% of the original gross income, which can be written as: ( P times N times 1.035 ) Setting the two expressions for the new gross income equal to each other, we get: ( P(1 - frac{D}{100}) times N times 1.15 = P times N times 1.035 ) Now, we can cancel out the ( P ) and ( N ) terms since they are on both sides of the equation: ( (1 - frac{D}{100}) times 1.15 = 1.035 ) Now, let's solve for ( D ): ( 1.15 - 1.15 times frac{D}{100} = 1.035 ) ( 1.15 - 1.035 = 1.15 times frac{D}{100} ) ( 0.115 = 1.15 times frac{D}{100} ) Now, divide both sides by 1.15: ( frac{0.115}{1.15} = frac{D}{100} ) ( frac{0.115}{1.15} = frac{D}{100} ) ( 0.1 = frac{D}{100} ) Now, multiply both sides by 100 to solve for ( D ): ( D = 0.1 times 100 ) ( D = 10 ) Therefore, the sales discount percentage is boxed{10%} .