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• 2010 USAMO Problems/Problem 3
  \frac{4i^3+24i^2+41i+21}{4i^3+24i^2+41i+15} \\ \left(1+\frac{6}{4i^3+24i^2+41i+15}\right) \\
  11 KB (1,889 words) - 12:45, 4 July 2013
• 2018 AIME II Problems
  ...t <math>xy = -80 - 320i</math>, <math>yz = 60</math>, and <math>zx = -96 + 24i</math>, where <math>i</math> <math>=</math> <math>\sqrt{-1}</math>. Then th
  9 KB (1,385 words) - 23:26, 20 January 2024
• 2018 AIME II Problems/Problem 5
  ...t <math>xy = -80 - 320i</math>, <math>yz = 60</math>, and <math>zx = -96 + 24i</math>, where <math>i</math> <math>=</math> <math>\sqrt{-1}</math>. Then th ...th>. Substituting this into the third equation, we get <math>z^2=\frac{-96+24i}{-\frac{4}{3}-\frac{16}{3}i}=3\cdot \frac{-24+6i}{-1-4i}=3\cdot \frac{(-24+
  11 KB (2,077 words) - 19:15, 12 January 2024
• Arcticturn Prep
  ...t <math>xy = -80 - 320i</math>, <math>yz = 60</math>, and <math>zx = -96 + 24i</math>, where <math>i</math> <math>=</math> <math>\sqrt{-1}</math>. Then th
  18 KB (3,089 words) - 10:53, 27 August 2024
• 2024 AMC 12A Problems/Problem 10
  <math> 25e^{i2A} = (4+3i)^2 = 16 + 9 * i^2 + 2*12 * i= 16- 9 + 24i = 7+ 24i </math>
  5 KB (833 words) - 02:54, 11 November 2024