answer:1. **Volume of Spheres**: The volume (V) of a sphere with radius (r) is given by ( V = frac{4}{3} pi r^3 ). - For the first sphere (radius = 4 inches): ( V_1 = frac{4}{3} pi (4)^3 = frac{256}{3} pi ). - For the second sphere (radius = 6 inches): ( V_2 = frac{4}{3} pi (6)^3 = 288 pi ). - For the third sphere (radius = 8 inches): ( V_3 = frac{4}{3} pi (8)^3 = frac{2048}{3} pi ). 2. **Volume of Cylinder**: The volume (V) of a cylinder with radius (r) and height (h) is given by ( V = pi r^2 h ). - For the cylinder (radius = 5 inches, height = 10 inches): ( V_4 = pi (5)^2 (10) = 250 pi ). 3. **Total Volume**: The total volume ( V_{text{total}} ) is the sum of the volumes of the three spheres and the cylindrical base: [ V_{text{total}} = left(frac{256}{3} pi + 288 pi + frac{2048}{3} pi + 250 pi right) = left(frac{256+2048}{3} pi + (288 + 250) pi right) = left(frac{2304}{3} pi + 538 pi right) = boxed{1250 pi} ]

answer:Given the problem, we need to find all pairs ((x, y)) of positive integers such that [ xy mid x^2 + 2y - 1. ] 1. **Initial Divisibility Condition:** Since (xy mid x^2 + 2y - 1), it implies that (x mid x^2 + 2y - 1). Therefore, we can write: [ x mid x^2 + 2y - 1 implies x mid 2y - 1. ] This means there exists an integer (k) such that: [ 2y - 1 = xk. ] Solving for (y), we get: [ y = frac{xk + 1}{2}. ] 2. **Substituting (y) into the Original Condition:** We substitute (y = frac{xk + 1}{2}) into the condition (xy mid x^2 + 2y - 1): [ xy = x cdot frac{xk + 1}{2} = frac{x^2 k + x}{2}. ] We need: [ frac{x^2 k + x}{2} mid x^2 + 2 left(frac{xk + 1}{2}right) - 1. ] Simplifying the right-hand side: [ x^2 + 2 left(frac{xk + 1}{2}right) - 1 = x^2 + xk + 1 - 1 = x^2 + xk. ] Therefore, we need: [ frac{x^2 k + x}{2} mid x^2 + xk. ] 3. **Simplifying the Divisibility Condition:** Since (frac{x^2 k + x}{2} mid x^2 + xk), we can multiply both sides by 2 to clear the fraction: [ x^2 k + x mid 2(x^2 + xk). ] Simplifying the right-hand side: [ 2(x^2 + xk) = 2x^2 + 2xk. ] Therefore, we need: [ x^2 k + x mid 2x^2 + 2xk. ] 4. **Analyzing the Divisibility:** We can rewrite the condition as: [ x^2 k + x mid 2x^2 + 2xk - (x^2 k + x) = 2x^2 + 2xk - x^2 k - x = 2x^2 + xk - x. ] Therefore, we need: [ x^2 k + x mid 2x^2 + xk - x. ] 5. **Considering Different Values of (t):** We have the equation: [ x + k = yt. ] Substituting (y = frac{xk + 1}{2}), we get: [ x + k = t left(frac{xk + 1}{2}right). ] Simplifying, we get: [ 2(x + k) = t(xk + 1). ] [ 2x + 2k = txk + t. ] - **Case (t = 1):** [ 2x + 2k = xk + 1. ] [ xk - 2x - 2k = -1. ] [ (x - 2)(k - 2) = 3. ] The integer solutions are: [ (x, k) = (3, 5), (5, 3). ] Substituting back, we get: [ (x, y) = (3, 8), (5, 8). ] - **Case (t = 2):** [ 2x + 2k = 2xk + 2. ] [ xk - x - k = 1. ] [ (x - 1)(k - 1) = 2. ] The integer solutions are: [ (x, k) = (2, 2), (3, 1), (1, 3). ] Substituting back, we get: [ (x, y) = (1, n), (2n - 1, n). ] - **Case (t = 3):** [ 2x + 2k = 3xk + 3. ] [ 3xk - 2x - 2k = 3. ] [ frac{2}{k} + frac{2}{x} = 3 + frac{3}{xk}. ] If (x, k geq 2), then (LHS < RHS). Thus, at least one of them must be 1. After checking, no new solutions are found. Hence, the solutions are: [ (x, y) = (3, 8), (5, 8), (1, n), (2n - 1, n). ] The final answer is ( boxed{ (3, 8), (5, 8), (1, n), (2n - 1, n) } ).

answer:To solve this, note that gcd(x, y) divides both x and y, and hence must also divide x + y = 780. The largest gcd(x, y) corresponds to the largest divisor of 780. 1. List the divisors of 780: - 1, 2, 3, 4, 5, 6, 10, 12, 13, 15, 20, 26, 30, 39, 52, 60, 65, 78, 130, 156, 195, 260, 390, 780. 2. The largest divisor less than 780 is 390. 3. Assign x = 390 and y = 780 - 390 = 390. Therefore, we calculate gcd(390, 390) = 390. Hence, the largest possible value of gcd(x, y) when their sum is 780 is boxed{390}.

answer:1. Convert the logarithmic equation to its exponential form: [ 3x - 5 = 8^2 ] Since ( log_8(3x - 5) = 2 ) implies ( 3x - 5 = 8^2 ). 2. Calculate ( 8^2 ): [ 8^2 = 64 ] So, the equation becomes: [ 3x - 5 = 64 ] 3. Solve for ( x ): [ 3x = 64 + 5 = 69 ] [ x = frac{69}{3} = 23 ] Conclusion: [ boxed{x = 23} ]