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question:There were 27 boys and 35 girls playing soccer, 15 boys and 20 girls on the swings, and 10 boys and 5 girls eating their snacks on the playground at recess. There were _____ children on the playground at recess.
answer:To find the total number of children on the playground at recess, we need to add the number of boys and girls in each activity. Playing soccer: 27 boys + 35 girls = 62 children On the swings: 15 boys + 20 girls = 35 children Eating snacks: 10 boys + 5 girls = 15 children Now, we add the totals from each activity to find the overall total: 62 children (soccer) + 35 children (swings) + 15 children (snacks) = 112 children Therefore, there were boxed{112} children on the playground at recess.
question:Let (S) be the set of all nonzero real numbers. Let (f : S to S) be a function such that [ f(x) + f(y) = fleft(frac{xy}{x+y}right) ] for all (x, y in S) such that (x + y neq 0). Let (n) be the number of possible values of (f(2)), and let (s) be the sum of all possible values of (f(2)). Find (n times s).
answer:1. Consider (x = y) in the given functional equation: [ f(x) + f(x) = fleft(frac{x^2}{2x}right) = fleft(frac{x}{2}right). ] This simplifies to: [ 2f(x) = fleft(frac{x}{2}right). ] 2. Check the behavior for a potential function. Let's try (f(x) = frac{1}{x}): [ 2left(frac{1}{x}right) = frac{1}{x/2} = frac{2}{x}, ] which holds true; thus, (f(x) = frac{1}{x}) is a solution. 3. To ensure uniqueness, assume another (f(x)) satisfies the equation and derive a contradiction or confirm it leads back to (f(x) = frac{1}{x}). Given (f) as (f(x) = frac{1}{x}), calculate (f(2)): [ f(2) = frac{1}{2}. ] Hence, (n = 1) (the number of solutions) and (s = frac{1}{2}) (the sum of possible values). Therefore, [ n times s = 1 times frac{1}{2} = boxed{frac{1}{2}}. ]
question:Compute (5+7)^3+(5^3+7^3).
answer:Following the order of operations: 1. Compute the sum inside the cube (5+7)^3. 2. Compute the cube of the sum (5+7)^3. 3. Compute the cubes of 5 and 7 individually, and then sum these cubes. Start by solving (5+7)^3. First, compute the sum inside the parentheses: [ 5 + 7 = 12. ] Next, compute the cube of 12: [ 12^3 = 1728. ] Now, compute 5^3 and 7^3: [ 5^3 = 125, quad 7^3 = 343. ] Add these two values: [ 125 + 343 = 468. ] Finally, add the results of 12^3 and (5^3 + 7^3): [ 1728 + 468 = 2196. ] Therefore, the final answer is: [ boxed{2196}. ]
question:how many ways can 10 people be seated in a row of chairs if four of the people, John, Wilma, Paul, and Alice, refuse to sit in four consecutive seats?
answer:1. Calculate the total number of arrangements for 10 people: [ 10! ] 2. Calculate the number of ways the restricted group (John, Wilma, Paul, and Alice) can sit consecutively: - Treat John, Wilma, Paul, and Alice as a single unit. This unit together with the other 6 people makes 7 units. - The units can be arranged in: [ 7! ] - Within the unit, John, Wilma, Paul, and Alice can be arranged among themselves in: [ 4! ] The total number of restricted arrangements is: [ 7! times 4! ] 3. Subtract the restricted arrangements from the total arrangements: [ 10! - 7!times 4! ] Performing the calculation: [ 3628800 - (5040 times 24) = 3628800 - 120960 = 3507840 ] The number of acceptable seating arrangements where John, Wilma, Paul, and Alice do not sit in four consecutive seats is: [ boxed{3507840} ]