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question:A chocolate bar can be broken into 3 sections to make 3 s'mores. Ron is hosting a boy scout camp out in his backyard for 15 scouts. He wants to make sure that there are enough chocolate bars for everyone to have 2 s'mores each. He spent 15 on chocolate bars. How much does one chocolate bar cost?
answer:First, let's calculate the total number of s'mores needed for the 15 scouts if each scout is to have 2 s'mores: 15 scouts * 2 s'mores each = 30 s'mores Since one chocolate bar can be broken into 3 sections to make 3 s'mores, we need to find out how many chocolate bars are needed to make 30 s'mores: 30 s'mores / 3 s'mores per bar = 10 chocolate bars Now, we know that Ron spent 15 on chocolate bars, and he needs 10 chocolate bars to have enough for the s'mores. To find out the cost of one chocolate bar, we divide the total amount spent by the number of chocolate bars: 15 / 10 chocolate bars = 1.50 per chocolate bar Therefore, one chocolate bar costs boxed{1.50} .
question:The function f(x) is defined on the set of real numbers R, and it satisfies f(x+y)=f(x)+f(y). Also, when x>0, f(x)<0. Then f(x) is ( ) A: An odd function and monotonically increasing on R B: An odd function and monotonically decreasing on R C: An even function and monotonically decreasing on R D: An even function and not a monotonic function on R
answer:It is clear that the domain of f(x) is R, and it is symmetric about the origin. Since the function satisfies f(x+y)=f(x)+f(y) for all x, y, let x=y=0, we get f(0)=2f(0), thus f(0)=0. Then, let y=-x, we get f(0)=f(x)+f(-x), thus f(-x)=-f(x), which means f(x) is an odd function. For any x_1<x_2, x_2-x_1>0, then f(x_2-x_1)<0 Thus f(x_2)+f(-x_1)<0; For f(x+y)=f(x)+f(y), let x=y=0 we get: f(0)=0, then let y=-x we get f(x)+f(-x)=0 which means f(-x)=-f(x), thus we have f(x_2)-f(x_1)<0 Therefore, f(x_2)<f(x_1) Thus, f(x) is monotonically decreasing on R. Hence, the correct choice is boxed{text{B}}.
question:the Bamboo Nine Sections problem from The Nine Chapters on the Mathematical Art, there is a bamboo stick with 9 sections. The volume of each section from top to bottom forms an arithmetic sequence. The combined volume of the top 4 sections is 3 liters, and the combined volume of the bottom 3 sections is 4 liters. What is the volume of the topmost section of the bamboo stick? Choose the correct answer from the following options: A: frac{13}{22} B: frac{37}{33} C: frac{47}{44} D: frac{67}{66}
answer:Let the volumes of the sections of the bamboo from top to bottom be a_1, a_2, ldots, a_9, and assume they form an arithmetic sequence. From the problem, we know: a_1 + a_2 + a_3 + a_4 = 3 quad (1) a_7 + a_8 + a_9 = 4 quad (2) Since a_1, a_2, ldots, a_9 form an arithmetic sequence, we have the common difference d such that: a_n = a_1 + (n-1)d From equation (1), we have: 4 a_1 + 6d = 3 quad (3) Substituting n=7, 8, 9 into a_n, and adding them according to equation (2), we find: 3a_1 + 21d = 4 quad (4) Multiplying equation (3) by 4 and equation (4) by 3 and then subtracting one from the other, we get: 12 a_1 + 24d - 9 a_1 - 63d = 12 - 3 cdot 4 3 a_1 - 39d = -0 This simplifies to: 66d = 7 d = frac{7}{66} Substituting d = frac{7}{66} into equation (3): 4a_1 + 6 cdot frac{7}{66} = 3 4a_1 + frac{7}{11} = 3 4a_1 = 3 - frac{7}{11} 4a_1 = frac{33}{11} - frac{7}{11} 4a_1 = frac{26}{11} a_1 = frac{26}{11 cdot 4} a_1 = frac{13}{22} Now, to find the volume of the fifth section a_5, we use a_1 and the common difference d: a_5 = a_1 + 4d a_5 = frac{13}{22} + 4 cdot frac{7}{66} a_5 = frac{13}{22} + frac{28}{66} a_5 = frac{39}{66} + frac{28}{66} a_5 = frac{67}{66} Thus the volume of the fifth section of the bamboo stick is boxed{frac{67}{66}}.
question:Let ABC be a triangle. Let Gamma be its circumcircle and omega be its incircle. Suppose that P, Q, and R lie on Gamma such that (PQ) and (QR) are tangent to the incircle omega. Show that in this case the line (RP) is tangent to the incircle.
answer:1. Let ( triangle ABC ) be a triangle with ( Gamma ) as its circumcircle and ( omega ) as its incircle. We are given that points ( P, Q, ) and ( R ) lie on the circumcircle ( Gamma ) such that the lines ( PQ ) and ( QR ) are tangent to the incircle ( omega ). 2. We aim to show that the line ( RP ) is also tangent to ( omega ). 3. We start by considering a more generalized statement: Suppose ( P ) and ( Q ) lie on the circumcircle ( Gamma ), and all three lines ( PQ, QR, ) and ( RP ) are tangent to the incircle ( omega ). 4. We employ an inversion transformation with respect to the incircle ( omega ). Denote the points of tangency of the incircle with the sides of ( triangle ABC ) as ( D, E, ) and ( F ). 5. Under this inversion, ( A, B, ) and ( C ) map to points ( X, Y, ) and ( Z ) respectively. Importantly, ( X, Y, ) and ( Z ) are the midpoints of the sides of the medial triangle ( triangle DEF ). 6. The circumcircle of ( triangle XYZ ) (image of ( A, B, ) and ( C ) under the inversion) has a radius that is half that of ( omega ). This follows from the properties of inversion and the configuration of the medial triangle. 7. Similarly, the image of the circumcircle of ( triangle PQR ) under this inversion is another circle with radius half that of ( omega ). Since ( triangle PQR ) has two points in common with the nine-point circle (circumcircle of the medial triangle ( triangle DEF )), it must coincide with this nine-point circle. 8. Therefore, ( R ) also lies on the circumcircle of ( triangle ABC ), implying that the line ( RP ) must also be tangent to the incircle ( omega ). # Conclusion: Since we have completed all the steps and verified that ( R ) must also lie on the circumcircle ( Gamma ), it follows that if ( PQ ) and ( QR ) are tangent to the incircle ( omega ), then ( RP ) must also be tangent to the incircle ( omega ). Thus, we conclude that: [ boxed{text{In this configuration, the line } RP text{ is tangent to the incircle } omega.} ]