answer:First, find the area of triangle DEF using Heron's formula. The semiperimeter s is: [ s = frac{12 + 15 + 17}{2} = 22. ] The area K is: [ K = sqrt{s(s-12)(s-15)(s-17)} = sqrt{22 cdot 10 cdot 7 cdot 5} = 70. ] The inradius r is: [ r = frac{K}{s} = frac{70}{22} = frac{35}{11}. ] Let the points of tangency of the incircle with overline{DF} be U and the excircle be V. Since ED = s - 17 = 5 and EV = s = 22, and using the fact that tangents from a point to a circle are equal, we have DU = DU = 5 and DV = DV = 22. The angle bisector of angle E passes through the centers of the incircle (I) and excircle (E). Triangles triangle EDI and triangle EDV are right triangles. By the Pythagorean Theorem: [ EI = sqrt{ED^2 + DI^2} = sqrt{5^2 + left(frac{35}{11}right)^2} = sqrt{25 + frac{1225}{121}} = sqrt{frac{4225}{121}} = frac{65}{11}. ] Using the similarity of triangles triangle EDI and triangle EDV, we find: [ frac{EI}{EV} = frac{ED}{EV} = frac{5}{22}. ] Hence, EV = 22 cdot frac{65}{11} = 130. Finally, the distance between the centers IE is: [ IE = EV - EI = 130 - frac{65}{11} = frac{1300}{11} - frac{65}{11} = frac{1235}{11} = 112.2727. ] Thus, the distance between the centers of the two circles is boxed{112.2727}.

answer:C Key point: Intersection and its operations. Topic: Sets. Analysis: The solution can be found by applying the basic operations of sets. Solution: Since A={-1, 0, 2} and B={x|-1<xleq4}, Therefore, Acap B={0, 2}, Hence, the correct choice is: boxed{C} Review: This question mainly examines the basic operations of sets, which is quite fundamental.

answer:To prove that if (2^p + 3^p = a^n) for a prime number (p) and positive nonzero integers (a) and (n), then (n=1), we will use the Lifting The Exponent (LTE) lemma. 1. **Check small prime cases:** - For (p = 2): [ 2^2 + 3^2 = 4 + 9 = 13 ] Since 13 is a prime number, it cannot be expressed as (a^n) for (n > 1). Thus, (n = 1). - For (p = 5): [ 2^5 + 3^5 = 32 + 243 = 275 ] We need to check if 275 can be expressed as (a^n) for (n > 1). The prime factorization of 275 is: [ 275 = 5^2 times 11 ] Since 275 is not a perfect power (i.e., it cannot be written as (a^n) for (n > 1)), (n = 1). 2. **General case for (p neq 2) and (p neq 5):** - By the Lifting The Exponent (LTE) lemma, we have: [ v_5(2^p + 3^p) = v_5(2 + 3) + v_5(p) ] where (v_5) denotes the 5-adic valuation (the exponent of 5 in the prime factorization of the number). - Since (2 + 3 = 5), we have: [ v_5(2 + 3) = v_5(5) = 1 ] - For any prime (p neq 5), (v_5(p) = 0) because (p) is not divisible by 5. Therefore: [ v_5(2^p + 3^p) = 1 + 0 = 1 ] - This means that the exponent of 5 in the number (2^p + 3^p) is 1. Hence, (2^p + 3^p) has exactly one factor of 5, implying that it cannot be a perfect power (a^n) for (n > 1). 3. **Conclusion:** - Since (2^p + 3^p) cannot be a perfect power for (n > 1), we conclude that (n = 1). (blacksquare)

answer:Let's call the number of tickets Connie spent on the earbuds E. We know that Connie redeemed 50 tickets in total. She spent half of them on a stuffed koala bear, which means she spent 50 / 2 = 25 tickets on the koala bear. She also spent 15 tickets on glow bracelets. So the total tickets spent on the koala bear and the glow bracelets is 25 (koala) + 15 (bracelets) = 40 tickets. Since Connie redeemed 50 tickets in total, the remaining tickets she spent on the earbuds would be 50 (total tickets) - 40 (koala + bracelets) = 10 tickets. Therefore, Connie spent boxed{10} tickets on the pair of earbuds.