answer:We need to find the expressions for f(g(x)) and g(f(x)) first and then their difference: begin{aligned} f(g(x)) &= fleft(frac{x}{4} + 1right) &= 4left(frac{x}{4} + 1right) - 5 &= x + 4 - 5 &= x - 1, end{aligned} and begin{aligned} g(f(x)) &= g(4x - 5) &= frac{4x - 5}{4} + 1 &= x - frac{5}{4} + 1 &= x - frac{5}{4} + frac{4}{4} &= x - frac{1}{4}. end{aligned} Now, calculate f(g(x)) - g(f(x)): begin{aligned} f(g(x)) - g(f(x)) &= (x - 1) - left(x - frac{1}{4}right) &= x - 1 - x + frac{1}{4} &= boxed{frac{1}{4}}. end{aligned} Conclusion: The result frac{1}{4} makes sense as the functions have been altered, and their compositions have still led to a valid and consistent result.

answer:# Problem: Compute the volume of the solid formed by the rotation of the region bounded by the curves [ y = (x-1)^2, quad y = 1 ] about the y-axis. The volume ( V ) of a solid of revolution about the ( y )-axis can be found using the integral formula [ V = pi cdot int_{a}^{b} x^{2} , dy. ] First, express ( x ) as a function of ( y ) from the given curve ( y = (x-1)^2 ): [ y = (x-1)^2 implies x = 1 pm sqrt{y}. ] Since we are rotating the region bounded by ( y = (x-1)^2 ) and ( y = 1 ) about the ( y )-axis, we need to determine the limits of integration. # Step 1: Determine the limits of integration From the given bounds: [ y = (x-1)^2 ] Evaluate for ( x = 1 ): [ y_1 = (1-1)^2 = 0. ] So, the lower limit ( y_1 ) is ( 0 ). The upper bound is given directly as ( y_2 = 1 ). # Step 2: Set up the integral for the volume Now, considering the region between the curves ( y = 0 ) and ( y = 1). We use: [ x = 1 pm sqrt{y}, ] we consider the outer radius ( 1 + sqrt{y} ) and inner radius ( 1 - sqrt{y} ): [ begin{align*} V &= pi cdot int_{0}^{1} (x_{text{outer}}^2 - x_{text{inner}}^2) , dy &= pi cdot int_{0}^{1} left((1 + sqrt{y})^2 - (1 - sqrt{y})^2right) , dy. end{align*} ] # Step 3: Simplify the integrand Simplify ( (1 + sqrt{y})^2 - (1 - sqrt{y})^2 ): [ begin{align*} (1 + sqrt{y})^2 - (1 - sqrt{y})^2 &= (1 + 2sqrt{y} + y) - (1 - 2sqrt{y} + y) &= 1 + 2sqrt{y} + y - 1 + 2sqrt{y} - y &= 4sqrt{y}. end{align*} ] # Step 4: Integrate Thus, the integral becomes: [ begin{align*} V &= pi cdot int_{0}^{1} 4sqrt{y} , dy &= 4pi cdot int_{0}^{1} y^{1/2} , dy. end{align*} ] Use the power rule for integration: [ begin{align*} int y^{1/2} , dy &= frac{2}{3} y^{3/2}. end{align*} ] # Step 5: Evaluate the definite integral Now evaluate the definite integral from ( 0 ) to ( 1 ): [ begin{align*} V &= 4pi cdot left. frac{2}{3} y^{3/2} right|_{0}^{1} &= 4pi cdot left( frac{2}{3} cdot 1^{3/2} - frac{2}{3} cdot 0^{3/2} right) &= 4pi cdot frac{2}{3} &= frac{8pi}{3}. end{align*} ] # Conclusion The volume of the solid formed by rotating the region bounded by ( y = (x-1)^2 ) and ( y = 1 ) about the ( y )-axis is: [ boxed{frac{8pi}{3}}. ]

answer:1. Let ( n ) be the number of marbles. According to the problem, we have the following congruences: [ n equiv 1 pmod{3} ] [ n equiv 2 pmod{7} ] [ n equiv 0 pmod{5} ] 2. We need to solve this system of congruences using the Chinese Remainder Theorem (CRT). First, we solve the congruences ( n equiv 1 pmod{3} ) and ( n equiv 2 pmod{7} ). 3. Since ( n equiv 0 pmod{5} ), we can write ( n = 5k ) for some integer ( k ). Substituting ( n = 5k ) into the other congruences, we get: [ 5k equiv 1 pmod{3} ] [ 5k equiv 2 pmod{7} ] 4. Simplify the first congruence: [ 5k equiv 1 pmod{3} implies 2k equiv 1 pmod{3} ] The multiplicative inverse of 2 modulo 3 is 2, since ( 2 times 2 equiv 1 pmod{3} ). Therefore: [ k equiv 2 pmod{3} implies k = 3m + 2 text{ for some integer } m ] 5. Substitute ( k = 3m + 2 ) into the second congruence: [ 5(3m + 2) equiv 2 pmod{7} ] Simplify: [ 15m + 10 equiv 2 pmod{7} implies m + 3 equiv 2 pmod{7} implies m equiv -1 pmod{7} implies m equiv 6 pmod{7} ] Therefore: [ m = 7p + 6 text{ for some integer } p ] 6. Substitute ( m = 7p + 6 ) back into ( k = 3m + 2 ): [ k = 3(7p + 6) + 2 = 21p + 20 ] 7. Substitute ( k = 21p + 20 ) back into ( n = 5k ): [ n = 5(21p + 20) = 105p + 100 ] 8. To find the largest ( n ) less than 400, we solve: [ 105p + 100 < 400 implies 105p < 300 implies p < frac{300}{105} approx 2.857 ] The largest integer ( p ) is 2. 9. Substitute ( p = 2 ) back into ( n = 105p + 100 ): [ n = 105 times 2 + 100 = 310 ] Conclusion: [ boxed{310} ]

answer:The path of the light beam reflects geometrically analogous to the original problem setup with increased complexity: 1. The changes to X and Y are 8 and 3, respectively, which are relatively prime to 14. To match the face again, light must travel such that changes in XYZ are divisible by 14. 2. Because both 8 and 3 are relatively prime to 14, total reflections needed for X and Y values to reach a vertex again are 14 times 1 = 14 reflections. 3. The distance traveled per reflection is now sqrt{ (14^2) + (8^2) + (3^2) } = sqrt{285}. 4. Thus, the total distance covered is 14 cdot sqrt{285}. The final answer, with m=14 and n=285, is boxed{299}.