answer:Answer: 18.5 | 37 / 2## Examples of answer notation:171 / 71.7

answer:If the base edge of the pyramid is x, then according to the problemfrac{x^{2}(x-4)}{3}+18=frac{(x+2)^{2}(x-6)}{3}orx^{2}-10 x-39=0from which the positive value of x is 13. Therefore, the volume of the pyramidK=frac{x^{2}(x-4)}{3}=507 mathrm{~cm}^{3}

answer:To solve for a given that x=5 is a solution to the equation ax-8=10+4a, we substitute x=5 into the equation:[5a - 8 = 10 + 4a]To isolate a, we first subtract 4a from both sides of the equation:[5a - 4a - 8 = 10 + 4a - 4a]This simplifies to:[a - 8 = 10]Next, we add 8 to both sides to solve for a:[a - 8 + 8 = 10 + 8]Which simplifies to:[a = 18]Therefore, the value of a is boxed{18}.

answer:-、1.(D).It is easy to get -1<f(x+1)<3.because f(0)=3, f(3)=-1, and f(x) is a decreasing function,therefore the solution to the inequality -1<f(x)<3 is 0<x<3.Therefore, when -1<f(x+1)<3, we have 0<x+1<3.

answer:Analysis According to the problem, we classify the set using the remainder of division by 17.Solution (1) Divide 1,2, cdots, 366 into 17 categories based on the remainder when divided by 17: [0],[1], cdots,[16] (where [i] represents the category of numbers that leave a remainder of i when divided by 17). Since 366=17 times 21+9, the categories [1], [2], cdots, [9] each contain 22 numbers; the categories [10],[11], cdots,[16] and [0] each contain 21 numbers.(1) When a, b in[0], the number of subsets with property P is C_{21}^{2}=210;(2) When a in[k], b in[17-k], k=1,2, cdots, 7, the number of subsets with property P is C_{22}^{1} cdot C_{21}^{1}=462;(3) When a in[8], b in[9], the number of subsets with property P is C_{22}^{1} cdot C_{22}^{1}=484.Therefore, the number of subsets of A with property P is 210+462 times 7+484=3928.(2) To ensure that the binary subsets do not intersect, when a, b in[0], 10 subsets can be paired; when a in[k], b in[17-k], k=1,2, cdots, 7, 21 subsets can be paired for each; when a in[8], b in[9], 22 subsets can be paired.Thus, the number of pairwise disjoint subsets with property P is 10+21 times 7+22=179.

answer:1. Let the coordinates of points ( A, B, C ) be ( (a, 0, 0), (0, b, 0), (0, 0, c) ) respectively. These points lie on the coordinate axes ( O_x, O_y, O_z ).2. The sphere passing through ( A, B, C ) intersects the coordinate axes again at points ( A', B', C' ) with coordinates ( (a', 0, 0), (0, b', 0), (0, 0, c') ) respectively.3. The centroid ( M ) of triangle ( ABC ) is given by: [ M = left( frac{a}{3}, frac{b}{3}, frac{c}{3} right) ]4. The centroid ( M' ) of triangle ( A'B'C' ) is given by: [ M' = left( frac{a'}{3}, frac{b'}{3}, frac{c'}{3} right) ]5. The midpoint ( S ) of segment ( MM' ) is given by: [ S = left( frac{frac{a}{3} + frac{a'}{3}}{2}, frac{frac{b}{3} + frac{b'}{3}}{2}, frac{frac{c}{3} + frac{c'}{3}}{2} right) ] Simplifying, we get: [ S = left( frac{a + a'}{6}, frac{b + b'}{6}, frac{c + c'}{6} right) ]6. Let ( A'', B'', C'' ) be the midpoints of ( AA', BB', CC' ) respectively. Then: [ A'' = left( frac{a + a'}{2}, 0, 0 right), quad B'' = left( 0, frac{b + b'}{2}, 0 right), quad C'' = left( 0, 0, frac{c + c'}{2} right) ]7. The coordinates of ( S ) can be rewritten as: [ S = left( frac{a''}{3}, frac{b''}{3}, frac{c''}{3} right) ]8. Since ( A'', B'', C'' ) are the midpoints of ( AA', BB', CC' ), they lie on the coordinate axes. Therefore, the coordinates of ( S ) are: [ S = left( frac{x}{3}, frac{y}{3}, frac{z}{3} right) ] where ( x, y, z ) are the coordinates of the midpoints of ( AA', BB', CC' ) respectively.Conclusion:The locus of the midpoint of ( MM' ) is a line passing through the origin and is given by:[S = left( frac{x}{3}, frac{y}{3}, frac{z}{3} right)]The final answer is ( boxed{ left( frac{x}{3}, frac{y}{3}, frac{z}{3} right) } ).