answer:First, let's find out the total distance from their house to the water park. Since Jake can bike at 11 miles per hour and it takes him 2 hours to get to the water park, the total distance is: Distance = Speed × Time Distance = 11 miles/hour × 2 hours Distance = 22 miles Now, let's find out how far Jake's dad drives at 60 miles per hour during the second half of the journey. Since the total journey takes 30 minutes (or 0.5 hours), the second half of the journey also takes 0.5 hours / 2 = 0.25 hours. The distance covered at 60 miles per hour is: Distance = Speed × Time Distance = 60 miles/hour × 0.25 hours Distance = 15 miles Since the total distance to the water park is 22 miles, and Jake's dad covers 15 miles at 60 miles per hour, the remaining distance for the first half of the journey is: Remaining distance = Total distance - Distance covered at 60 mph Remaining distance = 22 miles - 15 miles Remaining distance = 7 miles Jake's dad also takes 0.25 hours to cover this distance during the first half of the journey. Therefore, his speed during the first half is: Speed = Distance / Time Speed = 7 miles / 0.25 hours Speed = 28 miles per hour So, Jake's dad drives at a speed of boxed{28} miles per hour during the first half of the journey.

answer:The given problem requires finding all bijective functions f:left{x_1, x_2, cdots, x_nright} rightarrow left{x_1, x_2, cdots, x_nright} such that left|f(x_i) - x_iright| = a quad text{for all } i = 1,2,cdots,n, where ( n ) is an odd number and ( x_1, x_2, cdots, x_n ) are real numbers. 1. **Identify Function Property**: [ left|f(x_i) - x_iright| = a quad text{implies that} quad f(x_i) = x_i pm a quad text{for each } i. ] 2. **Sum Study**: Note that ( f ) maps the set ({x_1, x_2, dots, x_n}) to itself. Therefore, it must hold that the sum of the values under ( f ) equals the sum of the original values: [ sum_{i=1}^{n} f(x_i) = sum_{i=1}^{n} x_i. ] 3. **Difference in Sums**: Compute the difference of the sums: [ 0 = sum_{i=1}^{n} f(x_i) - sum_{i=1}^{n} x_i = sum_{i=1}^{n} left[f(x_i) - x_iright]. ] 4. **Form Expression**: Using ( f(x_i) = x_i pm a ): [ sum_{i=1}^{n} left[f(x_i) - x_iright] = (x_1 pm a - x_1) + (x_2 pm a - x_2) + ldots + (x_n pm a - x_n). ] 5. **Simplify**: Each term simplifies to ( pm a ): [ sum_{i=1}^{n} left[f(x_i) - x_iright] = pm a pm a pm a cdots pm a, ] where there are ( n ) terms. Let ( m ) be the number of ( +a ) terms and ( k ) be the number of ( -a ) terms such that ( m + k = n ). 6. **Total Sum**: This results in: [ sum_{i=1}^{n} left[f(x_i) - x_iright] = ma - ka = (m-k)a. ] 7. **Condition for Zero Sum**: Since the sum must be zero: [ (m - k)a = 0. ] 8. **Odd Number Implication**: Since ( n ) is odd, ( m ) and ( k ) cannot both be equal. Therefore, ( a ) must be zero for the equation to hold: [ (m - k)a = 0 implies a = 0 quad text{since } m neq k. ] 9. **Final Function**: Thus, ( f(x_i) = x_i ) for all ( i ). Conclusion: [ boxed{f(x) = x text{ for all } x in {x_1, x_2, cdots, x_n}} ] This is the unique solution satisfying the given condition.

answer:**Step-by-Step Solution:** **(1)** To express the quantities in algebraic expressions involving x, we follow these steps: ① **The profit per item** is calculated by subtracting the cost price and the decrease in price from the original price. The original price is 60, the cost price is 40, and the price decreases by x dollars. Therefore, the profit per item is: [60 - 40 - x = 20 - x] So, the profit per item in dollars is boxed{20-x}. ② **The number of items sold per week** increases by 30 items for every 1 decrease in price. If the price decreases by x dollars, the increase in items sold is 30x. The original sales volume is 300 items per week, so the new sales volume is: [300 + 30x] Thus, the number of items sold per week is boxed{300+30x}. ③ **The functional relationship of y with respect to x** combines the profit per item and the number of items sold. The profit per item is (20-x), and the number of items sold is (300+30x). Multiplying these gives the total profit, y: [y = (20-x)(300+30x) = -30x^2 + 300x + 6000] Therefore, the functional relationship is boxed{y=-30x^2+300x+6000}. **(2)** To find the pricing that maximizes the weekly profit and determine the maximum value, we analyze the functional relationship: Given y = -30x^2 + 300x + 6000, we can complete the square to find the vertex of the parabola, which represents the maximum point since the coefficient of x^2 is negative: [y = -30(x^2 - 10x) + 6000] [y = -30[(x - 5)^2 - 25] + 6000] [y = -30(x - 5)^2 + 6750] This shows that the maximum value of y occurs when x = 5, leading to a maximum profit of 6750 dollars. The price decrease is 5 dollars, so the new price is: [60 - 5 = 55] Therefore, the pricing should be set at boxed{55} dollars per item to maximize the weekly profit, with a maximum value of boxed{6750} dollars.

answer:First, let's simplify each part of the expression: 1. **Simplify (-5^{4})^{2}**: Recall that (-a)^{n} = a^n if n is even, so (-5^4)^2 = (5^4)^2 = 5^{8}. However, since the negative sign is outside the initial exponentiation, the result is -5^8. 2. **Simplify frac{1}{(-5^{8})}**: Since frac{1}{a^n} = a^{-n}, frac{1}{-5^8} = -5^{-8}. 3. **Multiply -5^{-8} and -5^9**: Using the rule a^m cdot a^n = a^{m+n}, [ -5^{-8}cdot(-5)^9 = (-1)cdot5^{-8}cdot(-1)cdot5^9 = 5^{-8+9} = 5^1 = 5. ] Thus, the solution simplifies to boxed{5}.