answer:To find the number of paths that spell the word "WATERLOO" in the given diagram, we start from the initial 'W' at the top, and account for every possible path that follows the arrows to the final 'O's at the bottom. 1. **Initial 'W' to the first layer 'A's**: - There is only **1 path** from 'W' to each of the two 'A's directly below it. - Hence, the number of paths from 'W' to each 'A' is: [ text{Paths to A (first layer)} = 1 text{ for each A} ] 2. **Paths from 'A's to 'T's**: - Each 'A' leads to two 'T's below it. - Every path to the first 'T' at the second layer can come from either of the 'A's directly above it: [ text{Paths to leftmost T} = 1 + 1 = 2 ] - Results for each 'T' node from left to right: [ text{Paths} = 1 + 1 = 2, quad 1 + 1 + 1 + 1 = 2 + 2 = 4 ] - 3. **Paths from 'T's to 'E's**: - Each 'T' leads to two 'E's below it: [ text{Paths to leftmost E} = 2 ] [ text{Next E} = 2 + 2 = 4 ] [ text{Paths to rightmost E} = 4 + 4 = 8 ] 4. **Paths from 'E's to 'R's**: - Following similar steps from 'E's to 'R's: [ text{Paths to leftmost R} = 2 ] [ text{Next R} = 2 + 4 = 6 ] [ text{To next R} = 4 + 8 = 12 ] [ text{To rightmost R} = 8 ] 5. **Paths from 'R's to 'L's**: - Here each 'R' leads to two 'L's below: [ left ( begin{array}{l} 2 6 + 2 = 8 6 + 12 - 18 8 end{array} right ) ] 6. **Paths from 'L's to 'O's (second layer to third)**: - Every connections: [ left (begin{array}{l) 'L' gets 2 (18+8) 28 20 end{array} right ) ] In the final path diagram, the value reached under last step leading "OO" displayed 20 Therefore, all the correct count paths followed be spelling out destinated ``WATERLOO`` are available outcomes. Conclusion: boxed{20}

answer:To find the sum of these fractions, we need to add them together. However, before we can add them, we need to find a common denominator. The denominators are 3, 2, 6, 5, 4, 20, and 15. The least common multiple (LCM) of these numbers is the smallest number that all of these denominators can divide into without leaving a remainder. The LCM of 3, 2, 6, 5, 4, 20, and 15 is 60. Now, we convert each fraction to an equivalent fraction with a denominator of 60: 1/3 = 20/60 1/2 = 30/60 -5/6 = -50/60 1/5 = 12/60 1/4 = 15/60 -9/20 = -27/60 -2/15 = -8/60 Now we can add them together: 20/60 + 30/60 - 50/60 + 12/60 + 15/60 - 27/60 - 8/60 Combine the numerators: (20 + 30 - 50 + 12 + 15 - 27 - 8)/60 Now, calculate the sum of the numerators: 20 + 30 = 50 50 - 50 = 0 0 + 12 = 12 12 + 15 = 27 27 - 27 = 0 0 - 8 = -8 So the sum of the numerators is -8. The sum of the fractions is: -8/60 We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 4: -8 ÷ 4 = -2 60 ÷ 4 = 15 So the simplified sum of the fractions is: boxed{-2/15}

answer:First, note that sqrt[4]{625} = sqrt[4]{5^4} = 5. Therefore, the expression simplifies to sqrt[4]{5} + 5. Let's denote u = sqrt[4]{5}. Then, u^4 = 5, and u + 5 is the root we are considering. To find a polynomial with rational coefficients for which u + 5 is a root, we start by considering the polynomial for u, which is x^4 - 5 = 0. Now, u + 5 will be a root of the polynomial obtained by substituting x - 5 in place of x in x^4 - 5, yielding: [ (x - 5)^4 - 5 = 0. ] Expanding (x - 5)^4 gives: [ x^4 - 20x^3 + 150x^2 - 500x + 625 - 5 = x^4 - 20x^3 + 150x^2 - 500x + 620 = 0. ] By Vieta's formulas, the product of the roots of x^4 - 20x^3 + 150x^2 - 500x + 620 = 0 is the constant term, 620, divided by the leading coefficient, which is 1. Therefore, the product of all the roots is boxed{620}.

answer:To determine the number of people from each school, let ( x ) be the number of people from School One and ( y ) be the number of people from School Two. According to the problem, we have two types of buses with capacities 14 seats and 19 seats, and two scenarios: 1. Both schools rent 14-seater buses, and together they need 72 buses. 2. Both schools rent 19-seater buses, and School Two rents 7 more buses than School One. We need to express these conditions mathematically. Step 1: Express the number of buses required for each type When using 14-seater buses: [ frac{x}{14} + frac{y}{14} = 72 implies frac{x + y}{14} = 72 x + y = 72 times 14 = 1008 ] When using 19-seater buses: [ frac{y}{19} = frac{x}{19} + 7 ] Rewriting this into simpler equations: [ frac{x + y}{19} = frac{x}{19} + frac{y}{19} = frac{x}{19} + left( frac{x}{19} + 7 right) implies frac{y - x}{19} = 7 implies y - x = 133 + frac{ x}{19} ] Thus, we have two linear equations: [ x + y = 1008 quad text{(Equation 1)} ] [ y - x = 133 quad text{(Equation 2)} ] Step 2: Solve the system of linear equations Adding and subtracting Equation 1 and Equation 2: [ (x + y) + (y - x) = 1008 + 133 implies 2y = 1141 implies y = 570 ] [ y - x = 133 implies 570 - x = 133 implies x = 437 ] Thus, the number of people from School One and School Two are: boxed{437 text{ people from School One,} 570 text{ people from School Two}} Conclusion Thus, School One has 437 people, and School Two has 570 people, validating our calculations with the given conditions. boxed{437 text{ people from School One,} 570 text{ people from School Two}}