answer:1. **Set up the equations for Jack and Jill's movements:** - Jack starts 8 minutes (or frac{2}{15} hours) earlier than Jill. - Jack's speed uphill is 12 km/hr and downhill is 18 km/hr. - Jill's speed uphill is 14 km/hr and downhill is 20 km/hr. 2. **Calculate the time Jack and Jill take to reach the top of the hill:** - Time taken by Jack: [ text{Time} = frac{7 text{ km}}{12 text{ km/hr}} = frac{7}{12} text{ hours} ] - Time taken by Jill: [ text{Time} = frac{7 text{ km}}{14 text{ km/hr}} = frac{1}{2} text{ hours} ] 3. **Equations for their movement:** - Jack’s downhill motion: [ y = 7 - 18(x - frac{7}{12}) quad text{(for } x geq frac{7}{12} text{)} ] - Jill’s uphill motion: [ y = 14(x - frac{2}{15}) quad text{(for } x geq frac{2}{15} text{)} ] 4. **Find the time when Jack and Jill meet:** - Set equations equal: [ 7 - 18(x - frac{7}{12}) = 14(x - frac{2}{15}) ] - Solve for x: [ 7 - 18x + frac{21}{2} = 14x - frac{28}{15} ] [ frac{35}{2} + frac{28}{15} = 32x ] [ frac{805}{30} + frac{56}{15} = 32x implies frac{937}{30} = 32x implies x = frac{937}{960} ] 5. **Calculate the position y where they meet:** - Using Jill's equation: [ y = 14left(frac{937}{960} - frac{2}{15}right) = 14left(frac{533}{960}right) approx frac{7462}{960} approx frac{3731}{480} ] 6. **Calculate the distance from the top of the hill where they meet:** - Distance from the top of the hill: [ 7 - frac{3731}{480} approx frac{960}{480} = 2 ] The distance from the top of the hill where Jack and Jill meet is approximately 2 km. boxed{- The final correct answer is (D) 2 km.}

answer:1. Given the function ( f(x) ) is odd, periodic with period 5, and continuous across the entire real number line. Also provided are the values ( f(-1) = f(2) = -1 ). 2. Since ( f(x) ) is odd, we have: [ f(-x) = -f(x) ] Thus, for ( x = 0 ): [ f(0) = -f(0) implies f(0) = 0 ] 3. Because ( f(x) ) is periodic with period 5, we have: [ f(x + 5) = f(x) ] Therefore: [ f(5) = f(0) = 0 ] 4. Using the odd property and periodicity again: [ f(1) = -f(-1) = 1, quad f(3) = f(3-5) = f(-2) implies f(3) = -f(2) = -(-1) = 1 ] Also: [ f(4) = f(-1+5) = f(-1) = -1 ] So the values of ( f(x) ) from ( x = 0 ) to ( x = 5 ): [ f(0) = 0, quad f(1) = 1, quad f(2) = -1, quad f(3) = 1, quad f(4) = -1, quad f(5) = 0 ] 5. To find the number of roots in each period, note the function must cross the x-axis at least once between each pair of intervals where it changes sign. Each period (length of 5 units) must contain at least 4 roots: [ f(x) = 0 ] must occur at least at the intervals ([0,1)), ([1,2)), ([2,3)), and ([3,4)). 6. To find the number of such periods within ([1755; 2017]), first find the range: [ 2017 - 1755 = 262 ] 7. The number of complete periods ( left [0, 5 right) ) that fit within 262 units: [ leftlfloor frac{262}{5} rightrfloor = 52 ] Each period contributes at least 4 roots: [ 52 times 4 = 208 ] 8. Additionally, we need to account for roots in the last partial period, within ([2015, 2017]). 9. Finally, adding these roots: [ 208 + 1 (text{for } 2015) + 1 (text{for } text{ interval from 2016 to 2017}) = 210 ] Conclusion: [ boxed{210} ]

answer:1. Simplify the inside of the first cube root: [ sqrt[3]{1+27} = sqrt[3]{28} ] 2. Calculate sqrt[3]{27} first then proceed to the second cube root calculation: [ sqrt[3]{27} = 3 ] [ sqrt[3]{1+3} = sqrt[3]{4} ] 3. Add both cube roots: [ sqrt[3]{28} + sqrt[3]{4} ] Simplifying cube root values exactly might involve approximate values or using tables or calculators. However, since the operation involves addition, the expression stays as: [ boxed{sqrt[3]{28} + sqrt[3]{4}} ] Conclusion: The problem's solution yields sqrt[3]{28} + sqrt[3]{4}, which retains its form as an exact value, not simplified further without the aid of numerical approximations or computational assistance.

answer:To prove the inequality (sum_{i=1}^n frac{1}{n-1+x_i} le 1) given that (x_1 x_2 cdots x_n = 1) and (x_i > 0) for all (i), we can use the AM-GM inequality and some algebraic manipulations. 1. **Apply the AM-GM Inequality:** The Arithmetic Mean-Geometric Mean (AM-GM) inequality states that for any non-negative real numbers (a_1, a_2, ldots, a_n), [ frac{a_1 + a_2 + cdots + a_n}{n} geq sqrt[n]{a_1 a_2 cdots a_n}. ] Given (x_1 x_2 cdots x_n = 1), we have: [ frac{x_1 + x_2 + cdots + x_n}{n} geq sqrt[n]{x_1 x_2 cdots x_n} = 1. ] Therefore, [ x_1 + x_2 + cdots + x_n geq n. ] 2. **Transform the Inequality:** Consider the function (f(x) = frac{1}{n-1+x}). We need to show that: [ sum_{i=1}^n frac{1}{n-1+x_i} le 1. ] 3. **Use the AM-HM Inequality:** The Harmonic Mean (HM) of (n) positive numbers (a_1, a_2, ldots, a_n) is given by: [ frac{n}{frac{1}{a_1} + frac{1}{a_2} + cdots + frac{1}{a_n}}. ] Applying the AM-HM inequality, we have: [ frac{a_1 + a_2 + cdots + a_n}{n} geq frac{n}{frac{1}{a_1} + frac{1}{a_2} + cdots + frac{1}{a_n}}. ] Let (a_i = n-1+x_i). Then, [ frac{(n-1+x_1) + (n-1+x_2) + cdots + (n-1+x_n)}{n} geq frac{n}{sum_{i=1}^n frac{1}{n-1+x_i}}. ] Simplifying the left-hand side, we get: [ frac{n(n-1) + (x_1 + x_2 + cdots + x_n)}{n} geq frac{n}{sum_{i=1}^n frac{1}{n-1+x_i}}. ] Using (x_1 + x_2 + cdots + x_n geq n), we have: [ frac{n(n-1) + n}{n} = n-1 + 1 = n. ] Therefore, [ n geq frac{n}{sum_{i=1}^n frac{1}{n-1+x_i}}. ] Dividing both sides by (n), we get: [ 1 geq sum_{i=1}^n frac{1}{n-1+x_i}. ] Thus, we have shown that: [ sum_{i=1}^n frac{1}{n-1+x_i} le 1. ] (blacksquare)