answer:1. **Setup the equations**: We are given xy + 1 = z, yz + 1 = x, and zx + 1 = y and need to find the triples (x, y, z) that satisfy these. 2. **Consider the product of the equations**: [ (xy + 1)(yz + 1)(zx + 1) = xyz + x + y + z + 1 ] Simplifying explicitly including all non-linear terms: [ xyz(xyz + x + y + z + 1) = xyz + x + y + z + 1 ] Assuming xyz neq 0, this simplifies to: [ (xyz)^2 + xyz(x + y + z) + xyz = xyz + x + y + z + 1 ] Further simplification results in: [ (xyz - 1)(xyz + x + y + z) = 0 ] Thus, either xyz = 1 or xyz + x + y + z = -1. 3. **Case analysis**: - **Case xyz = 1**: Substituting back into initial equations, xy + 1 = z, leads to xy = z - 1. With all equations, solve explicitly to find solutions. - **Case xyz + x + y + z = -1**: Attempt direct substitutions and solve for x, y, z accordingly. 4. Calculations from individual cases yield the possible valid solutions. 5. The final count of solutions, considering non-repetitive and non-zero criteria, is confirmed. Conclusion: Box the final number of solutions 4. The final answer is The final answer, considering the counts from distinct cases, is boxed{textbf{(D)} 4}.

answer:To calculate the total time Vivian's students spend outside of class, we need to add up all the breaks: First recess break: 15 minutes Second recess break: 15 minutes Lunch: 30 minutes Third recess break: 20 minutes Total time outside of class = 15 + 15 + 30 + 20 = 80 minutes Vivian's students spend a total of boxed{80} minutes outside of class each day.

answer:First, assess if pi < 10: Since pi (approximately 3.1416) is less than 10, this simplifies the inner absolute value: [|pi - 10| = 10 - pi.] Plugging this back into the outer expression results in: [|pi - |pi - 10|| = |pi - (10 - pi)| = |pi - 10 + pi| = |2pi - 10|.] Since 2pi approx 6.2832 < 10, the expression simplifies to: [|2pi - 10| = 10 - 2pi.] Thus, the value is: [boxed{10 - 2pi}.]

answer:Let the two fractions be x and y. Then we have 1. x+y = frac{17}{24} 2. xy = frac{1}{8} Considering these as the sum and product of the roots of a quadratic equation, we can form the equation ax^2 + bx + c = 0 where -b/a = frac{17}{24} and c/a = frac{1}{8}. Considering a = 24, we find: - b = -17 - c = 3 The quadratic equation becomes 24x^2 - 17x + 3 = 0. To solve this, we use the quadratic formula: [ x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ] [ x = frac{17 pm sqrt{289 - 288}}{48} = frac{17 pm 1}{48} ] [ x = frac{18}{48} = frac{3}{8}, quad x = frac{16}{48} = frac{1}{3} ] Thus, the lesser fraction is boxed{frac{1}{3}}.