answer:: 1. **Determine the relative speeds of Petya and Vasya:** - According to the problem, Petya finishes a 100-meter race while Vasya is 10 meters behind him. Therefore, Vasya covers 90 meters in the time it takes Petya to cover 100 meters. - Let ( v_P ) be Petya's speed and ( v_V ) be Vasya's speed. Then: [ v_P cdot t = 100 quad text{and} quad v_V cdot t = 90 ] - Here ( t ) is the time taken by Petya to complete the race. [ v_V = frac{90}{100} v_P ] [ v_V = 0.9 v_P ] 2. **Analyze the second race scenario:** - Petya starts 10 meters behind Vasya. Both run with their constant respective speeds, ( v_P ) and ( v_V ). - Let ( d ) be the time it takes Petya to catch up to Vasya. During this time, the distance Petya will cover is ( d cdot v_P ) and the distance Vasya will cover is ( d cdot v_V ). - Petya needs to cover Vasya's lead plus an additional 10 meters, i.e., the total distance to catch up will be: [ d cdot v_P = 10 + d cdot v_V ] - Substitute ( v_V = 0.9 v_P ): [ d cdot v_P = 10 + d cdot (0.9 v_P) ] [ d cdot v_P = 10 + 0.9 d cdot v_P ] - Isolate ( d cdot v_P ): [ d cdot v_P - 0.9 d cdot v_P = 10 ] [ 0.1 d cdot v_P = 10 ] [ d cdot v_P = 100 ] - Therefore, [ d = frac{100}{v_P} ] 3. **Calculate the remaining part of the race:** - The total length of the race is 100 meters. By the time Petya has run 100 meters, Vasya has run: [ 100 - 10 = 90 quad text{meters} ] 4. **Evaluate the final distances when Petya finishes:** - Petya runs the full 100 meters. Calculate how far Vasya runs in the same time: [ text{Time taken by Petya to complete 100 meters} = frac{100}{v_P} quad text{seconds} ] [ text{Distance Vasya runs in this time} = v_V times frac{100}{v_P} = 0.9 v_P times frac{100}{v_P} = 90 quad text{meters} ] 5. **Determine who finishes first and by how much:** - When Petya finishes the 100 meters race, Vasya is at 90 meters. - Hence, Petya finishes 10 meters ahead of Vasya. - Because they are running at a constant speed, the result repeats proportionally. In the second race, Petya is initially 10 meters behind, consolidates his lead, and finishes 1 meter ahead. # Conclusion: Therefore, Petya finishes first in the second race, and he finishes 1 meter ahead of Vasya. [ boxed{text{Petya outpaced Vasya by 1 meter.}} ]

answer:Solution: (1) According to the problem, there are 36 basic events for the event (a,b), The events where the sum of the numbers is greater than 7 are (2,6), (3,5), (3,6), (4,4), (4,5), (4,6), (5,3), (5,4), (5,5), (5,6), (6,2), (6,3), (6,4), (6,5), (6,6), totaling 15, The probability that the sum of the numbers is greater than 7 is dfrac{15}{36}= boxed{dfrac{5}{12}}; (2) From the system of equations begin{cases} & ax+by=3 & x+2y=2 end{cases}, we can derive begin{cases}left(2a-bright)x=6-2b left(2a-bright)y=2a-3end{cases}, For the system to have only one solution, we need 2a-bneq 0, i.e., bneq 2a, The events where b=2a are (1,2), (2,4), (3,6), totaling 3, The probability that the system has only one solution is 1- dfrac{3}{36}= boxed{dfrac{11}{12}}; (3) For the system of equations begin{cases} & ax+by=3 & x+2y=2 end{cases} to have only positive solutions, we need bneq 2a, and begin{cases}x= dfrac{6-2a}{2a-b} > 0 y= dfrac{2a-3}{2a-b} > 0end{cases}, This implies left{begin{matrix}2a > b a > dfrac{3}{2} b < 3end{matrix}begin{matrix} end{matrix}right. or left{begin{matrix}2a < b a < dfrac{3}{2} b > 3end{matrix}begin{matrix} end{matrix}right., The events included are (2,1), (3,1), (4,1), (5,1), (6,1), (2,2), (3,2), (4,2), (5,2), (6,2), (1,4), (1,5), (1,6), totaling 13, Therefore, the probability is boxed{dfrac{13}{36}}.

answer:1. **Identify the Shape and its Components**: - The given diagram consists of four identical squares. - Each square has sides of length (3 text{ cm}). - The squares are arranged in such a way that they form a larger geometric shape. 2. **Calculate the Area of One Square**: - The area (A) of a square is given by the formula: [ A = text{side length}^2 ] - Given the side length is (3 text{ cm}): [ A = 3 text{ cm} times 3 text{ cm} = 9 text{ cm}^2 ] 3. **Calculate the Total Area**: - Since there are four identical squares, the total area (A_{text{total}}) is: [ A_{text{total}} = 4 times 9 text{ cm}^2 = 36 text{ cm}^2 ] 4. **Conclusion**: - Therefore, the area of the shape is: [ boxed{36} ]

answer:Let's examine each proposition: - Proposition ① states that the length of the zero vector is zero, and its direction is arbitrary. This is true according to the definition of the zero vector. A zero vector has no specific direction because it has no magnitude. - Proposition ② says that if overrightarrow{a} and overrightarrow{b} are both unit vectors, then overrightarrow{a} = overrightarrow{b}. This is false because even though unit vectors have the same magnitude (length), they may not necessarily have the same direction. Therefore, two unit vectors are not equal unless they have both the same magnitude and direction. - Proposition ③ asserts that vector overrightarrow{AB} is equal to vector overrightarrow{BA}. This is false because overrightarrow{AB} and overrightarrow{BA} are opposite in direction. In vector notation, overrightarrow{AB} = -overrightarrow{BA}. - Proposition ④ claims that if non-zero vector overrightarrow{AB} and overrightarrow{CD} are collinear, then points A, B, C, D are collinear. This is false because even though the vectors overrightarrow{AB} and overrightarrow{CD} can be collinear (parallel or antiparallel), this does not necessarily mean that all four points lie on the same straight line; they can be part of separate lines that are parallel or antiparallel to each other. Based on the analysis above: ① is correct, ② is incorrect, ③ is incorrect, ④ is incorrect. Therefore, the correct option is: boxed{A}