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question:Given the vertices of a triangle are A(3,4), B(0,0), and C(c,2c-6), if angle BAC is an obtuse angle, then the range of values for c is ______.
answer:From the given information, we know: overrightarrow{AB}=(-3,-4), overrightarrow{AC}=(c-3,2c-10).If angle BAC is an obtuse angle, then cos angle A 0.Thus, -3(c-3)+(-4)(2c-10) frac{49}{11}, and c neq 9.Hence, the range of values for c is boxed{left( frac{49}{11},+inftyright)} and c neq 9.If angle A is an obtuse angle, then cos angle A < 0 and cos angle A neq -1. The condition cos angle A < 0 translates to overrightarrow{AB} cdot overrightarrow{AC} < 0, which allows us to derive the relationship for c and thus find the answer.This problem mainly examines the application of plane vector operations in solving problems related to triangles. It's easy to overlook that when two vectors are collinear and in opposite directions, the angle between them is 180^circ. The necessary and sufficient condition for the angle between two non-zero vectors to be obtuse is that their dot product is less than 0 and they are not parallel, making it a medium-difficulty problem.
question:3. The present ages in years of two brothers A and B, and their father C are three distinct positive integers a, b, and c respectively. Suppose frac{b-1}{a-1} and frac{b+1}{a+1} are two consecutive integers, and frac{c-1}{b-1} and frac{c+1}{b+1} are two consecutive integers. If a+b+c leq 150 determine a, b and c.
answer:Solution: We havefrac{b-1}{a-1}=l, quad frac{b+1}{a+1}=l-1, quad frac{c-1}{b-1}=m, quad frac{c+1}{b+1}=m-1(If we take a leq b, we see that frac{b-1}{a-1} geq frac{b+1}{a+1}.)Consider the first two relations: b-1=l(a-1), b+1=(l-1)(a+1). Solving for a, we get a=2 l-3 and hence b=2 l^{2}-4 l+1. Using the second set of relations, we obtain b=2 m-3 and c=2 m^{2}-4 m+1. Thius we have 2 m-3=2 l^{2}-4 l+1 or m=(l-1)^{2}+1. Obviously l>1. If l=2, we get a=1 which forces b=1 and c=1, which is impossible. If l=3, we get a=3, b=7 and c=31. If l geq 4, then m geq 10 and c geq 161. But then a+b+c>150. Thus the only choice is a=3, b=7 and c=31.
question:The number 0.2012 is between(A) 0 and frac{1}{10}(B) frac{1}{10} and frac{1}{5}(C) frac{1}{5} and frac{1}{4}(D) frac{1}{4} and frac{1}{3}(E) frac{1}{3} and frac{1}{2}
answer:Since the number in question (0.2012) is in decimal form, it is easiest to determine into which of the 5 given ranges it falls by converting the ranges into decimal form also.Converting, frac{1}{10} is 0.1, frac{1}{5} is 0.2, frac{1}{4} is 0.25, frac{1}{3} is 0 . overline{3}, and frac{1}{2} is 0.5 .Since 0.2012 is greater than 0.2 but less than 0.25 , it is between frac{1}{5} and frac{1}{4}.ANsWER: (C)
question:# Task № 4.2## Condition:Six friends live in a small town in southern France. One day they decided to go to Paris together for one day and see the sights - the Eiffel Tower, the Louvre, Notre-Dame Cathedral, the Rodin Museum, the Bois de Boulogne, and the Champs-Élysées. The friends agreed to split up and see them separately, and since time was limited, each chose three attractions and visited only those. In the evening, they met and began discussing which places each had visited. Each named their three attractions, but two of the friends accidentally got everything mixed up and instead of the places they visited, they named the three places they did not visit. Here is the list of places each friend mentioned:- Friend 1: Eiffel Tower, Notre-Dame Cathedral, Rodin Museum- Friend 2: Eiffel Tower, Louvre, Notre-Dame Cathedral- Friend 3: Eiffel Tower, Louvre, Rodin Museum- Friend 4: Louvre, Notre-Dame Cathedral, Rodin Museum- Friend 5: Eiffel Tower, Rodin Museum, Bois de Boulogne- Friend 6: Eiffel Tower, Rodin Museum, Champs-ÉlyséesWhen the friends later began to sort things out, comparing photos, museum tickets, and sharing what they enjoyed, it turned out that each place was visited by exactly three of them. Help to restore who among the friends got everything mixed up and named the places they did NOT visit. In each field, write separately the numbers of the friends who meet the condition.
answer:# Answer:circ 1circ 3## Exact match of the answer -1 pointSolution by analogy with task №4.1#
question:Define the set operation A⊙B as {z | z=xy(x+y), x in A, y in B}. Given the sets A={0, 1} and B={2, 3}, find the sum of all elements in the set A⊙B.
answer:According to the definition of the set operation A⊙B, we have z such that z=xy(x+y), where x in A and y in B.Since A = {0, 1} and B = {2, 3}, we can calculate each element in A⊙B by pairing each element from A with each element from B and applying the operation:1. For x=0, regardless of the value of y, we will have z = 0 cdot y cdot (0 + y) = 0.2. For x=1 and y=2, we get z = 1 cdot 2 cdot (1 + 2) = 6.3. For x=1 and y=3, we get z = 1 cdot 3 cdot (1 + 3) = 12.Therefore, the set A⊙B consists of the elements {0, 6, 12}.Finally, the sum of all elements in the set A⊙B is 0 + 6 + 12 = 18.[boxed{18}]
question:Ex. 94. The diagonals of the inscribed quadrilateral A B C D intersect at point E, and angle A D B=frac{pi}{8}, B D=6 and A D cdot C E=D C cdot A E. Find the area of the quadrilateral A B C D.
answer:Ex. 94. 9 sqrt{2}.