answer:Let's assume the one-way distance from the starting point to the service center is D miles. Therefore, the round-trip distance would be 2D miles (going to the service center and then returning). The technician completes the drive to the center, which is D miles, and then completes 30 percent of the drive from the center. Thirty percent of D is 0.30D. So, the total distance the technician has traveled is D (to the center) + 0.30D (30 percent of the way back), which equals 1.30D. To find out what percent of the round-trip this distance represents, we divide the distance traveled by the total round-trip distance and then multiply by 100 to get the percentage: (1.30D / 2D) * 100 = (1.30 / 2) * 100 = 0.65 * 100 = 65% The technician has completed boxed{65} percent of the round-trip.

answer:Let the roots of the quadratic equation x^2 - 16x + 15 = 0 be denoted as r_1 and r_2. According to Vieta's formulas, the sum of the roots (r_1 + r_2) and the product of the roots (r_1 cdot r_2) can be derived from the coefficients of the polynomial: 1. The sum of the roots is given by r_1 + r_2 = 16. 2. The product of the roots is given by r_1 cdot r_2 = 15. To find the sum of the reciprocals of the roots, note that: [ frac{1}{r_1} + frac{1}{r_2} = frac{r_1 + r_2}{r_1 cdot r_2} = frac{16}{15} ] Thus, the sum of the reciprocals of the roots of the equation x^2 - 16x + 15 = 0 is boxed{frac{16}{15}}.

answer:The first few prime numbers greater than 3 are: 5, 7, 11, 13, 17, 19, 23, ldots. Since the triangle is scalene with consecutive prime sides, let the sides be p, p+2, and p+4 where all are prime. For a triangle with sides p, p+2, and p+4, the perimeter is: [ p + (p+2) + (p+4) = 3p + 6. ] We need this perimeter to be prime. Since p is prime and greater than 3, it must be odd. Thus, 3p+6 is odd only if p = 5. This results in a perimeter of: [ 3 times 5 + 6 = 21, ] which is not prime. Next, try p = 7: [ 3 times 7 + 6 = 27, ] which is also not prime. Next, try p = 11: [ 3 times 11 + 6 = 39, ] which is not prime. Continue this process until p = 13: [ 3 times 13 + 6 = 45, ] which is not prime. Finally, test p = 17: [ 3 times 17 + 6 = 57, ] which is not prime. Thus, we need to find a set of three consecutive primes such that their sum is prime. Testing p=5, p+2=7, and p+4=11 yields: [ 5 + 7 + 11 = 23, ] which is prime. Therefore, the smallest possible perimeter is boxed{23}.

answer:Given the dimensions of the field and the information about the area, we can establish the equation for the area: [ (3m + 8)(m - 3) = 100 ] Expanding this expression gives: [ 3m^2 - 9m + 8m - 24 = 100 ] [ 3m^2 - m - 24 = 100 ] [ 3m^2 - m - 124 = 0 ] Solving this quadratic equation, we use the quadratic formula where (a = 3), (b = -1), and (c = -124): [ m = frac{-(-1) pm sqrt{(-1)^2 - 4 times 3 times (-124)}}{2 times 3} ] [ m = frac{1 pm sqrt{1 + 1488}}{6} ] [ m = frac{1 pm sqrt{1489}}{6} ] Since square root of 1489 is 38.583, the values of (m) become: [ m = frac{1 + 38.583}{6} = frac{39.583}{6} approx 6.597 text{ and } m = frac{1 - 38.583}{6} = frac{-37.583}{6} approx -6.264 ] However, dimensions should be positive, so we consider only (m approx 6.597). Conclusion: The valid solution for (m) that satisfies the given problem condition is boxed{6.597} (approximately).