answer:Let's break down the 50 cents into ten 5-cent blocks. We can use nickels or pennies to fill a 5-cent block, and a dime can fill two 5-cent blocks. We will consider different cases based on the number of quarters used. **Two Quarters:** Using two quarters fills the entire amount. Since we cannot use a single 50-cent coin, this case does not apply. **One Quarter:** Using one quarter fills five out of the ten blocks. We must fill the remaining five blocks using dimes, nickels, and pennies. We can fill these blocks in the following ways: - 0 dimes: 6 ways (from 0 to 5 nickels, rest in pennies) - 1 dime: 4 ways (from 0 to 3 nickels, rest in pennies) - 2 dimes: 3 ways (from 0 to 2 nickels, rest in pennies) - 3 dimes: 2 ways (from 0 to 1 nickel, rest in pennies) - 4 dimes: 1 way (all dimes, no nickels or pennies) Total for this case: 6 + 4 + 3 + 2 + 1 = 16 ways. **No Quarters:** Using no quarters, we fill ten blocks using only dimes, nickels, and pennies: - 0 dimes: 11 ways (from 0 to 10 nickels, rest in pennies) - 1 dime: 9 ways (from 0 to 8 nickels, rest in pennies) - 2 dimes: 7 ways (from 0 to 6 nickels, rest in pennies) - 3 dimes: 5 ways (from 0 to 4 nickels, rest in pennies) - 4 dimes: 3 ways (from 0 to 2 nickels, rest in pennies) - 5 dimes: 1 way (all dimes, no nickels or pennies) Total for this case: 11 + 9 + 7 + 5 + 3 + 1 = 36 ways. Thus, the total number of ways to make change for 50 cents, excluding a single 50-cent coin, is 16 + 36 = boxed{52} ways.

answer:First, we need to convert the speed of the train from kilometers per hour (kmph) to meters per second (m/s) because the lengths are given in meters. 1 kilometer = 1000 meters 1 hour = 3600 seconds So, to convert 54 kmph to m/s: 54 kmph * (1000 meters / 1 kilometer) * (1 hour / 3600 seconds) = 54 * 1000 / 3600 m/s 54 * 1000 / 3600 = 15 m/s Now, we need to calculate the total distance the train needs to cover to completely cross the bridge. This is the length of the train plus the length of the bridge: 100 meters (length of the train) + 150 meters (length of the bridge) = 250 meters Now, we can calculate the time it will take for the train to cross the bridge by dividing the total distance by the speed of the train: Time = Distance / Speed Time = 250 meters / 15 m/s Time = 16.67 seconds So, it will take the train approximately boxed{16.67} seconds to cross the bridge.

answer:According to the problem, the sequence {a_n} satisfies S_n = 3a_n - 2. (1) When n geq 2, we have S_{n-1} = 3a_{n-1} - 2. (2) Subtracting (2) from (1), we get a_n = 3a_n - 3a_{n-1}, which simplifies to 2a_n = 3a_{n-1}. When n=1, we have S_1 = a_1 = 3a_1 - 2, which gives a_1 = 1. Thus, the sequence {a_n} is a geometric sequence with the first term a_1 = 1 and common ratio frac{3}{2}. Therefore, a_n = (frac{3}{2})^{n-1}. The sum of the first n terms of the sequence {na_n} is T_n, so T_n = 1 + 2 times frac{3}{2} + 3 times (frac{3}{2})^2 + dots + n times (frac{3}{2})^{n-1}. (3) Then, we have frac{3}{2}T_n = frac{3}{2} + 2 times (frac{3}{2})^2 + 3 times (frac{3}{2})^3 + dots + n times (frac{3}{2})^n. (4) Subtracting (3) from (4), we get -frac{1}{2}T_n = 1 + (frac{3}{2}) + (frac{3}{2})^2 + dots + (frac{3}{2})^{n-1} - n times (frac{3}{2})^n = -2(1 - frac{3^n}{2^n}) - n times (frac{3}{2})^n. Simplifying, we get T_n = 4 + (2n - 4) times (frac{3}{2})^n. If T_n > 100, then 4 + (2n - 4) times (frac{3}{2})^n > 100. Upon analysis, we find that n geq 7. Therefore, the smallest value of n such that T_n > 100 is 7. Hence, the answer is boxed{7}. This problem tests the understanding of recursive formulas in sequences. The key is to analyze the general term formula of the sequence {a_n}. This is a basic problem.

answer:Firstly, using the relationship that f(f^{-1}(x)) = x, we begin by substituting f(x) in the given equation for f^{-1}(x): [ x = fleft(frac{2 - 3x}{3x}right) ] Substituting f(x) = frac{2}{3x + c}: [ x = frac{2}{3left(frac{2 - 3x}{3x}right) + c} ] Simplifying the denominator: [ x = frac{2}{frac{6 - 9x}{3x} + c} = frac{2}{frac{6 - 9x + 3cx}{3x}} = frac{2 cdot 3x}{6 - 9x + 3cx} ] Setting the denominators equal to resolve for x: [ 2 cdot 3x = x(6 - 9x + 3cx) ] [ 6x = 6x - 9x^2 + 3cx^2 ] [ 0 = -9x^2 + 3cx^2 ] [ 0 = x^2(3c - 9) ] For this to hold true for all x, we must have 3c - 9 = 0: [ 3c = 9 ] [ c = boxed{3} ]