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question:A box with a volume of 16 text{cm}^3 can hold X paperclips. A box with a volume of 48 text{cm}^3 could hold 150 paperclips. What is the value of unknown variable X?
answer:To solve this problem, we need to determine the value of x, which represents the number of paperclips that can be held in the box with a volume of 16 text{cm}^3. We know that the volume of the box is directly proportional to the number of paperclips it can hold. Let's set up a proportion using the given information: frac{text{Volume of box 1}}{text{Number of paperclips in box 1}} = frac{text{Volume of box 2}}{text{Number of paperclips in box 2}} frac{16}{x} = frac{48}{150} Let's simplify and solve for x: 16 * 150 = 48 * x 2400 = 48x To isolate x, we divide both sides of the equation by 48: x = frac{2400}{48} x = 50 The value of x is 50. The answer is: 50
question:Define the function f(x) = 2x - 5. For what value of x is f(x) equal to f^{-1}(x)?
answer:We know that f(x) = 2x - 5. To find f^{-1}(x), we need to solve the equation f(f^{-1}(x)) = x. Substituting f^{-1}(x) into the equation, we have f(f^{-1}(x)) = 2(f^{-1}(x)) - 5. Setting this equal to x gives 2(f^{-1}(x)) - 5 = x. Solving for f^{-1}(x), we find that f^{-1}(x) = frac{x + 5}{2}. To find the value of x for which f(x) = f^{-1}(x), we set 2x - 5 = frac{x + 5}{2}. Simplifying, we get 4x - 10 = x + 5. Subtracting x from both sides and adding 10 to both sides, we have 3x = 15. Dividing both sides by 3, we find that x = boxed{5}. The answer is: 5
question:today's field day challenge, the 4th graders were competing against the 5th graders. Each grade had 2 different classes. The first 4th grade class had 12 girls and 13 boys. The second 4th grade class had 15 girls and 11 boys. The first 5th grade class had 9 girls and 13 boys while the second 5th grade class had 10 girls and 11 boys. In total, how many more boys were competing than girls?
answer:In the first 4th grade class, there were 12 girls + 13 boys = 25 students. In the second 4th grade class, there were 15 girls + 11 boys = 26 students. In the first 5th grade class, there were 9 girls + 13 boys = 22 students. In the second 5th grade class, there were 10 girls + 11 boys = 21 students. In total, there were 25 + 26 + 22 + 21 = 94 students competing. The total number of girls competing was 12 + 15 + 9 + 10 = 46 girls. The total number of boys competing was 13 + 11 + 13 + 11 = 48 boys. Therefore, there were 48 - 46 = 2 more boys competing than girls. 2 The answer is: 2
question:The solution of 8x+1equiv 5 pmod{12} is xequiv apmod{m} for some positive integers mgeq 2 and a<m. Find a+m.
answer:To solve the congruence 8x+1equiv 5 pmod{12}, we first subtract 1 from both sides to get 8xequiv 4 pmod{12}. Then we can divide both sides by 4 (since 4 is relatively prime to 12) to get 2xequiv 1 pmod{3}. Multiplying both sides by 2 (since 2 is relatively prime to 3) gives 4xequiv 2 pmod{3}. Finally, dividing both sides by 2 (since 2 is relatively prime to 3) gives 2xequiv 1 pmod{3}. The solutions to 2xequiv 1 pmod{3} are xequiv 2 pmod{3}, so the solution to 8x+1equiv 5 pmod{12} is xequiv 2 pmod{3}. Therefore, a=2 and m=3, so a+m=2+3=boxed{5}. The answer is: 5