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Home » This Mysterious Mole Man Finally Revealed: The Secret Behind His Underground Life! - Get link 4share

This Mysterious Mole Man Finally Revealed: The Secret Behind His Underground Life! - Get link 4share

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 Solution: Let $ h(x) = ax^2 + bx + c $. From $ h(1) = a + b + c = 5 $ and $ h(-1) = a - b + c = 3 $, adding gives $ 2a + 2c = 8 $, so $ a + c = 4 $. The sum of roots is $ -rac{b}{a} = 4 $, so $ b = -4a $. Substituting $ b = -4a $ into $ a + b + c = 5 $: $ a - 4a + c = 5 $ â $ -3a + c = 5 $. Since $ a + c = 4 $, subtracting gives $ -4a = 1 $, so $ a = -rac{1}{4} $. Then $ c = 4 - a = 4 + rac{1}{4} = rac{17}{4} $, and $ b = -4a = 1 $. Thus, $ h(x) = -rac{1}{4}x^2 + x + rac{17}{4} $. Multiplying by 4 to eliminate fractions: $ h(x) = -x^2 + 4x + 17 $. Verifying $ h(1) = -1 + 4 + 17 = 20 $? Wait, inconsistency. Rechecking: $ a = -1/4 $, $ c = 17/4 $, $ b = 1 $. Then $ h(1) = -1/4 + 1 + 17/4 = (-1 + 4 + 17)/4 = 20/4 = 5 $, correct. $ h(-1) = -1/4 -1 + 17/4 = ( -1 -4 + 17 )/4 = 12/4 = 3 $, correct. Sum of roots $ -b/a = -1 / (-1/4) = 4 $, correct. Final answer: $ oxed{-x^2 + 4x + \dfrac{17}{4}} $ or $ oxed{-\dfrac{1}{4}x^2 + x + \dfrac{17}{4}} $.

📌 Question: A science communicator observes that the number of views $ V(t) $ on a video grows quadratically over time $ t $ (in days). If $ V(1) = 120 $, $ V(2) = 200 $, and $ V(3) = 300 $, find $ V(4) $.

📌 Solution: Assume $ V(t) = at^2 + bt + c $. From $ V(1) = a + b + c = 120 $, $ V(2) = 4a + 2b + c = 200 $, $ V(3) = 9a + 3b + c = 300 $. Subtract first equation from the second: $ 3a + b = 80 $. Subtract second from the third: $ 5a + b = 100 $. Subtract these: $ 2a = 20 $ â $ a = 10 $. Then $ 3(10) + b = 80 $ â $ b = 50 $. From $ a + b + c = 120 $: $ 10 + 50 + c = 120 $ â $ c = 60 $. Thus, $ V(t) = 10t^2 + 50t + 60 $. For $ t = 4 $: $ V(4) = 10(16) + 50(4) + 60 = 160 + 200 + 60 = 420 $. Final answer: $ oxed{420} $.

📌 Question: An underwater robotâs depth $ d(t) $ (in meters) satisfies $ d(t) = pt^3 + qt^2 + rt + s $. Given $ d(1) = 10 $, $ d'(1) = 12 $, $ d(2) = 28 $, and $ d'(2) = 30 $, find $ d(0) $.

📌 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 $ â $ q = -18 $. From $ d'(1) $: $ 3(6) + 2(-18) + r = 12 $ â $ 18 - 36 + r = 12 $ â $ r = 30 $. From $ d(1) $: $ 6 - 18 + 30 + s = 10 $ â $ 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) $ â $ k(4) + 2k(4) = 0 + 0 $ â $ 3k(4) = 0 $ â $ k(4) = 0 $. Assume $ k(x) = 0 $ for all $ x $, but $ k(1) = 1

📌 eq 0 $. Contradiction? Wait, from $ k(2) = 0 $, check $ x = 1, y = -1 $: $ k(0) = k(1) + k(-1) - 2k(-1) = 1 + k(-1) - 2k(-1) = 1 - k(-1) $. Also $ k(0) = k(0 + 0) = 2k(0) - 2k(0) = 0 $? No: $ k(0) = k(0 + 0) = 2k(0) - 2k(0) = 0 $. So $ k(0) = 0 $. Then $ 0 = 1 - k(-1) $ â $ k(-1) = 1 $. Then $ x = -1, y = -1 $: $ k(-2) = 2k(-1) - 2k(1) = 2(1) - 2(1) = 0 $. $ x = 1, y = -1 $: $ k(0) = k(1) + k(-1) - 2k(-1) = 1 + 1 - 2(1) = 0 $, consistent. Now $ x = 2, y = -1 $: $ k(1) = k(2) + k(-1) - 2k(-2) = 0 + 1 - 0 = 1 $, matches. No contradiction. Thus $ k(2) = 0 $. Final answer: $ oxed{0} $.

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