WebLet A = {1,2,3,4,5,6}. Given the equivalence relation R = { (1,1), (2,2), (3,3), (4,4), (5,5), (1,3), (3,1), (6,6), (2,4), (4,2), (4,5), (2,5), (5,4), (5,2)} on A. Find the equivalence class of each element of A. How many distinct equivalence classes are there on … Web3 jul. 2024 · 2 Answers Sorted by: 3 On one hand, there's a constant-folding optimization that'll precompute constant tuples, so in (1, 2, 3) doesn't go through the expense of building the whole tuple each time. Without needing to build the tuple every time, the in approach actually has less bytecode interpretation overhead and runs quicker than the == approach.
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If x = 1 + 5^(1/3) + 5^(2/3), then the value of x^3 - 3x^2 - 12x + 6
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