answer:Firstly, we restate the conditions for an integer ( a ) being a primitive root modulo 11: ( a ) must have powers that yield every residue class ( 1 ) through ( 10 ) (since 11-1 = 10) exactly once before reaching ( a^{10} equiv 1 pmod{11} ). We start by quickly eliminating: - ( 1^k equiv 1 pmod{11} ) for any ( k ), so 1 is not a primitive root. Next, we check higher candidates up to 8: - ( 2 )'s powers modulo 11: ( 2, 4, 8, 5, 10, 9, 7, 3, 6, 1 ) thus, 2 is a primitive root. - ( 3 )'s powers modulo 11: ( 3, 9, 5, 4, 1 ), then repeats, so 3 is not a primitive root. - ( 4 )'s powers modulo 11: ( 4, 5, 9, 3, 1 ), then repeats, so 4 is not a primitive root. - ( 5 )'s powers modulo 11: ( 5, 3, 4, 9, 1 ), then repeats, so 5 is not a primitive root. - ( 6 )'s powers modulo 11: ( 6, 3, 7, 9, 10, 5, 8, 4, 2, 1 ) thus, 6 is a primitive root. - ( 7 )'s powers modulo 11: ( 7, 5, 2, 3, 10, 4, 6, 9, 8, 1 ) thus, 7 is a primitive root. - ( 8 )'s powers modulo 11: ( 8, 9, 6, 4, 10, 3, 2, 5, 7, 1 ) thus, 8 is a primitive root. The integers that are primitive roots from the set {1,2,3,4,5,6,7,8} modulo 11 are (2, 6, 7, 8). Hence, their sum: 2 + 6 + 7 + 8 = boxed{23}

answer:We first find the minimum value m. Observe that for any a and b in the given range, we have frac{3}{a}+b geq frac{3}{a}+a. By the AM-GM inequality, we know that frac{3}{a}+a geq 2sqrt{3}. Thus, m=2sqrt{3}. Next, we find the maximum value M. When a=1 and b=2, we have frac{3}{a}+b=5. Hence, M=5. Therefore, the answer is boxed{(5, 2sqrt{3})}. This problem involves finding the minimum and maximum values of a set by utilizing the properties of arithmetic and geometric means. It is a basic problem that tests one's understanding of set theory and inequalities.

answer:Let X and Y be the points where the folded portion of the triangle crosses AB, and Z be the location of the original vertex C after folding. Assume the fold preserves parallel lines and thus maintains the angles, making triangle ACB similar to triangle XZY. Since triangle XZY is folded over such that its area is 25% = (0.5)^2 of triangle ABC, the sides of triangle XZY must be 0.5 times the sides of triangle ACB due to the properties of similar triangles and their areas scaling with the square of side lengths. Draw the altitude of triangle ACB from C down to P on AB, crossing DE at Q and extending through AB to Z. 1. CP = CQ + QP = ZQ + QP = ZP + 2PQ. 2. Since the sides of triangle XZY are 0.5 times as long as those of triangle ACB, ZP = 0.5 times CP. 3. With CP = ZP + 2PQ, solving for PQ gives PQ = 0.25 times CP, and so CQ = CP - PQ = 0.75 times CP. Using the proportions established by the similar triangles, DE is 0.75 times the length of AB, so DE = 0.75 times 20 = boxed{15} text{ cm}.

answer:Let's denote Eddy's average speed as ( V_e ) and Freddy's average speed as ( V_f ). According to the given ratio, we have: [ frac{V_e}{V_f} = 1.7391304347826086 ] We know that Freddy takes 4 hours to complete his journey to city C, which is 460 km away. Therefore, we can calculate Freddy's average speed as: [ V_f = frac{Distance_{AC}}{Time_{Freddy}} ] [ V_f = frac{460 text{ km}}{4 text{ hours}} ] [ V_f = 115 text{ km/h} ] Now, using the ratio, we can find Eddy's average speed: [ V_e = 1.7391304347826086 times V_f ] [ V_e = 1.7391304347826086 times 115 text{ km/h} ] [ V_e = 200 text{ km/h} ] Now that we have Eddy's average speed, we can calculate the time it takes for him to travel from city A to city B, which is 600 km away: [ Time_{Eddy} = frac{Distance_{AB}}{V_e} ] [ Time_{Eddy} = frac{600 text{ km}}{200 text{ km/h}} ] [ Time_{Eddy} = 3 text{ hours} ] So, Eddy takes boxed{3} hours to complete his journey from city A to city B.