answer:The function f(x) is defined as follows: f(x) = begin{cases} 3x+2, & x<1 2^{x}, & xgeq 1 end{cases} To find f(0), we need to determine which part of the function to use. Since 0 is less than 1, we will use the first part of the function, which is 3x+2. Substituting x=0 into this expression, we get: f(0) = 3 times 0 + 2 = 2 Now, to find f(f(0)), we first note that f(0)=2. We now need to find f(2), which requires us to determine which part of the function to use again. Since 2 is greater than or equal to 1, we will use the second part of the function, which is 2^{x}. Substituting x=2 into this expression, we get: f(f(0)) = f(2) = 2^{2} = 4 Thus, the final answers are: boxed{f(0) = 2} boxed{f(f(0)) = 4}

answer:Given the problem statement, we begin by considering the polynomial ( P(x) = (x - x_0)(x - x_1) cdots (x - x_n)(x - x_{n+1}) ). We know the following: - ( x_0 = 0 ) - ( x_{n+1} = 1 ) - ( 0 < x_1 < x_2 < cdots < x_n < 1 ) - The numbers ( x_1, x_2, ldots, x_n ) satisfy the given system of equations: [ sum_{j=0, j neq i}^{n+1} frac{1}{x_i - x_j} = 0 quad text{for} quad i=1,2, ldots, n. ] Now, let's derive ( P'(x) ) and ( P''(x) ). 1. **Calculating the first derivative ( P'(x) )**: [ P'(x) = sum_{j=0}^{n+1} frac{P(x)}{x - x_j} ] Note that this follows from the product rule where each term in ( P(x) ) is considered. 2. **Calculating the second derivative ( P''(x) )**: [ P''(x) = sum_{j=0}^{n+1} sum_{k neq j} frac{P(x)}{(x - x_j)(x - x_k)} ] 3. **Evaluating ( P''(x_i) ) at ( x_i ) for ( i=1,2, ldots, n )**: Using the given condition: [ sum_{j=0, j neq i}^{n+1} frac{1}{x_i - x_j} = 0 ] we find that: [ P''(x_i) = 2 P'(x_i) sum_{j neq i} frac{1}{x_i - x_j} = 2 P'(x_i) cdot 0 = 0 ] Hence, [ P''(x_i) = 0 quad text{for} quad i=1,2, ldots, n. ] 4. **Formulating a differential equation**: Given ( P''(x_i) = 0 ), we know: [ x(x-1) P''(x) = (n+2)(n+1) P(x) ] 5. **Determining a unique monic polynomial**: Notice that there is a unique monic polynomial of degree ( n+2 ) that satisfies this differential equation. 6. **Considering polynomial ( Q(x) = (-1)^n P(1-x) )**: ( Q(x) ) shares the properties: - It is monic. - It has degree ( n+2 ). Given the condition ( (-1)^n P(1-x) = P(x) ), we can utilize the symmetry observed here to show that: [ x_{n+1-i} = 1 - x_i quad text{for} quad i=1,2,ldots, n. ] **Conclusion**: The solution is verified, hence: [ boxed{x_{n+1-i} = 1 - x_i quad text{for} quad i=1,2,ldots, n.} ]

answer:Let's start by calculating the time Paula takes to get to work by bus. Since she takes 3/5 of the time Luke takes, we can calculate her travel time as follows: Paula's travel time to work = 3/5 * Luke's travel time Paula's travel time to work = 3/5 * 70 minutes Paula's travel time to work = 42 minutes Now, let's calculate the total time they both spend traveling to work and back home each day. We know that Luke takes 70 minutes to get to work by bus and an unknown time to get back home by bike. Paula takes 42 minutes to get to work and the same amount of time to get back home by bus. The total time they spend traveling is 504 minutes. Let's denote the time Luke takes to get back home by bike as ( t ) minutes. So, we have: Luke's total travel time = 70 minutes (to work by bus) + ( t ) minutes (back home by bike) Paula's total travel time = 42 minutes (to work by bus) + 42 minutes (back home by bus) The sum of their travel times is 504 minutes: 70 + ( t ) + 42 + 42 = 504 Now, let's solve for ( t ): ( t ) + 154 = 504 ( t ) = 504 - 154 ( t ) = 350 minutes So, Luke takes 350 minutes to get back home by bike. Now, we need to find out how many times slower Luke's bike ride is compared to the bus. We know that the bus takes 70 minutes for the same distance. Let's denote the multiple slower as ( m ). We can set up the following equation: ( m ) * (time by bus) = time by bike ( m ) * 70 = 350 Now, let's solve for ( m ): ( m ) = 350 / 70 ( m ) = 5 Luke's bike ride is boxed{5} times slower than the bus.

answer:Given: - An arithmetic progression (AP) and a geometric progression (GP). - Each sequence starts with the term 1. - Both sequences have the same last term. - The GP has 4 terms, and the AP has 8 terms. - The sum of the GP is 21 more than the last term of the AP. We need to determine the terms in both sequences. 1. **Define the Progressions:** - Let the common ratio of the GP be ( q ). - Let the common difference of the AP be ( d ). - The 4-term GP can be represented as: ( 1, q, q^2, q^3 ). - The 8-term AP can be represented as: [ 1, 1+d, 1+2d, ldots, 1+7d ] 2. **Condition on Last Terms:** - Since the last terms of both sequences are equal, we have: [ q^3 = 1 + 7d quad text{(Equation 1)} ] 3. **Sum of the GP:** - The sum of the GP ((1, q, q^2, q^3)) is: [ 1 + q + q^2 + q^3 ] - According to the problem, this sum is 21 more than the last term of the AP: [ 1 + q + q^2 + q^3 = q^3 + 21 quad text{(Equation 2)} ] 4. **Simplify Equation 2:** - Substitute ( q^3 ) from Equation 1 into Equation 2: [ 1 + q + q^2 + q^3 = (1 + 7d) + 21 ] - This simplifies to: [ 1 + q + q^2 + 1 + 7d = 1 + 7d + 21 ] [ 1 + q + q^2 + q^3 = 1 + 7d + 21 ] [ q + q^2 = 21 ] 5. **Solve the Quadratic Equation:** - We rearrange the quadratic equation: [ q^2 + q - 21 = 0 ] - Solving this quadratic equation using the quadratic formula ( q = frac{-b pm sqrt{b^2 - 4ac}}{2a} ): [ q = frac{-1 pm sqrt{1 + 84}}{2} ] [ q = frac{-1 pm 9}{2} ] - So, we get: [ q_1 = frac{8}{2} = 4, quad q_2 = frac{-10}{2} = -5 ] 6. **Substitute ( q ) back into Equation 1:** - For ( q_1 = 4 ): [ 4^3 = 64, quad 64 = 1 + 7d ] [ 63 = 7d implies d_1 = 9 ] - For ( q_2 = -5 ): [ (-5)^3 = -125, quad -125 = 1 + 7d ] [ -126 = 7d implies d_2 = -18 ] # Conclusion: The terms of the progressions are: 1. For ( q = 4, d = 9 ): - GP: ( 1, 4, 16, 64 ) - AP: ( 1, 10, 19, 28, 37, 46, 55, 64 ) 2. For ( q = -5, d = -18 ): - GP: ( 1, -5, 25, -125 ) - AP: ( 1, -17, -35, -53, -71, -89, -107, -125 ) boxed{text{The values of } q text{ and } d text{ are } q = 4 text{ and } d = 9 text{ or } q = -5 text{ and } d = -18.}