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📰 Solution: $ d(t) = pt^3 + qt^2 + rt + s $. Compute $ d'(t) = 3pt^2 + 2qt + r $. From $ d(1) = p + q + r + s = 10 $, $ d'(1) = 3p + 2q + r = 12 $, $ d(2) = 8p + 4q + 2r + s = 28 $, $ d'(2) = 12p + 4q + r = 30 $. Subtract first equation from third: $ 7p + 3q + r = 18 $. Subtract $ d'(1) $ from this: $ (7p + 3q + r) - (3p + 2q + r) = 4p + q = 6 $. From $ d'(2) $: $ 12p + 4q + r = 30 $, and $ d'(1) $: $ 3p + 2q + r = 12 $. Subtract: $ 9p + 2q = 18 $. Now solve $ 4p + q = 6 $ and $ 9p + 2q = 18 $. Multiply first by 2: $ 8p + 2q = 12 $. Subtract: $ p = 6 $. Then $ 4(6) + q = 6 $ → $ q = -18 $. From $ d'(1) $: $ 3(6) + 2(-18) + r = 12 $ → $ 18 - 36 + r = 12 $ → $ r = 30 $. From $ d(1) $: $ 6 - 18 + 30 + s = 10 $ → $ s = -8 $. Thus, $ d(0) = s = -8 $. Final answer: $ oxed{-8} $. 📰 Question: A philosopher of science considers a function $ k(x) $ modeling the "distance" from theory to observation, satisfying $ k(x + y) = k(x) + k(y) - 2k(xy) $. If $ k(1) = 1 $, find $ k(2) $. 📰 Solution: Set $ x = y = 1 $: $ k(2) = k(1) + k(1) - 2k(1 \cdot 1) = 1 + 1 - 2(1) = 0 $. Verify consistency: Try $ x = 2, y = 1 $: $ k(3) = k(2) + k(1) - 2k(2) = 0 + 1 - 0 = 1 $. Try $ x = y = 2 $: $ k(4) = k(2) + k(2) - 2k(4) $ → $ k(4) + 2k(4) = 0 + 0 $ → $ 3k(4) = 0 $ → $ k(4) = 0 $. Assume $ k(x) = 0 $ for all $ x $, but $ k(1) = 1 📰 Kpop Demon Hunters Saja Boys 584521 📰 The Last Digit Is Not 0 Or 5 So 77 Is Not Divisible By 5 1983413 📰 Your Hair Will Look Professional Without Ever Pulling A Single Strandmaster This Braid Today 3973570 📰 Verizon Youtubetv 7578985 📰 Top Ten Free To Play Games 9930777 📰 Cool Roblox Names 1266155 📰 Mp Materials Is This The Stock That Will Definitively Rise This Week 8682532 📰 First Time Meme 2938078 📰 Verizon Wireless Riverside Plaza 7686095 📰 How To Disable Hardware Acceleration In Chrome 3073050 📰 Whats A Swinger 6007794 📰 Skin Color Hex 8166795 📰 You Wont Let This Hairstyle Down Slick Back Secrets That Steal The Spotlight 1889576 📰 Saint Clares Hospital Nj 2784603 📰 Superhero Villains 8161973