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• Furman University Wylie Mathematics Tournament/1996 Senior Exam/Solutions
  == Problem 1 == Multiplying the denominator by <math> 1-2i </math> gives us that the expression is <math> 2+3i, </math> the coeffic
  9 KB (1,364 words) - 15:59, 21 July 2006
• University of South Carolina High School Math Contest/1993 Exam/Problem 27
  ...mathrm{(C) \ }36 \qquad \mathrm{(D) \ }12\sqrt{2} \qquad \mathrm{(E) \ }12\sqrt{3} </math></center>
  1 KB (149 words) - 16:27, 18 August 2006
• University of South Carolina High School Math Contest/1993 Exam/Problem 26
  ...Then the first nonzero digit in the decimal expansion of <math>\sqrt{n^2 + 1} - n</math> is <center><math> \mathrm{(A) \ }1 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }3 \qquad \mathrm{(D) \ }4 \qq
  2 KB (258 words) - 16:34, 18 August 2006
• University of South Carolina High School Math Contest/1993 Exam/Problem 16
  ...r><math> \mathrm{(A) \ }1 \qquad \mathrm{(B) \ }4/3 \qquad \mathrm{(C) \ }\sqrt{2} \qquad \mathrm{(D) \ }3/2 \qquad \mathrm{(E) \ }2 </math></center> ...P</math> is [[parallel]] to <math>AB</math> and thus <math>\frac{CQ}{QN} = 1</math> and <math>CQ = QN = 4</math>. Then <math>CN = CQ + QN = 8</math>.
  1 KB (221 words) - 11:45, 17 August 2006
• Distance formula
  ...math>(a_1,a_2,...,a_n)</math> and <math>(b_1,b_2,...,b_n)</math> is <math>\sqrt{(a_1-b_1)^2+(a_2-b_2)^2+\cdots+(a_n-b_n)^2}</math>. <cmath>\dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}</cmath>
  2 KB (340 words) - 18:34, 8 September 2018
• 2006 Romanian NMO Problems/Grade 8/Problem 1
  ...e and a point <math>Q</math> on the circle, <math>PQ=\sqrt{PO_1^2+O_1Q^2}=\sqrt{PO_1^2+r_1^2}</math>, which does not depend on the point <math>Q</math> cho
  3 KB (509 words) - 23:22, 15 August 2012
• 1969 Canadian MO Problems/Problem 3
  ...have lengths <math>a</math> and <math>b</math>. Prove that <math>a+b\le c\sqrt{2}</math>. When does the equality hold?
  790 bytes (129 words) - 22:39, 17 November 2007
• Equilateral triangle
  ...f an equilateral triangle can be found in terms of a side: <math>\frac{s^2\sqrt{3}}{4}</math>.
  1 KB (186 words) - 19:57, 15 September 2022
• Notation
  ...easier to express using notation: <math>\displaystyle x+x^2=\frac{x}{2}+x\sqrt{x}</math>. The latter is easier to understand, and we can immediately jump
  692 bytes (109 words) - 10:28, 4 August 2006
• Galois theory
  ...ath>. Then the map <math>f:K\to K</math> given by <math>f(a+b\sqrt{2})=a-b\sqrt{2}</math> is a field automorphism; that is, <math>f(\alpha\beta)=f(\alpha)f ...he case. For example, if <math>K=\mathbb{Q}</math> and <math>L=\mathbb{Q}(\sqrt[3]{2})</math>, then <math>Gal(L/K)</math> is the trivial group, so every el
  2 KB (385 words) - 14:09, 5 May 2008
• 1969 Canadian MO Problems
  == Problem 1 == [[1969 Canadian MO Problems/Problem 1 | Solution]]
  3 KB (536 words) - 12:46, 8 October 2007
• Mock USAMO by probability1.01 dropped problems
  == Problem 1 == Let <math>n>1</math> be a fixed positive integer, and let <math>a_1,a_2,\ldots,a_n</math>
  3 KB (572 words) - 02:46, 16 May 2009
• Finite Fourier Analysis
  ...th>f(0) = f(1)</math> and <math>\frac{d^m f}{dx^m}(0) = \frac{d^m f}{dx^m}(1)</math> for all <math>m \in \mathbb{Z}^+</math> . <math>S</math> is a vect ...rac1N</math>, <math>x=\frac2N</math>, <math>\ldots</math>, <math>x=\frac{N-1}N </math>. For each <math>n \in \mathbb{Z}</math>, let <math>f_n</math> be
  4 KB (724 words) - 19:15, 9 September 2006
• Real part
  ...can be written in the form <math>z = a + bi</math> where <math>i = \sqrt{-1}</math> is the [[imaginary unit]] and <math>a</math> and <math>b</math> ar ...athrm{Re}(4(\cos \frac \pi6 + i \sin \frac\pi 6)) = 4 \cos \frac \pi 6 = 2\sqrt 3</math>
  2 KB (281 words) - 15:56, 5 September 2008
• Imaginary part
  ...can be written in the form <math>z = a + bi</math> where <math>i = \sqrt{-1}</math> is the [[imaginary unit]] and <math>a</math> and <math>b</math> ar * <math>\mathrm{Im}((1 + i)\cdot(2 + i)) = \mathrm{Im}(1 + 3i) = 3</math>. Note in particular that <math>\mathrm Im</math> is ''not
  2 KB (269 words) - 15:56, 5 September 2008
• Euclidean metric
  ...{y} = (y_1, y_2, \ldots, y_n)</math> is given by <math>d(\mathbf{x, y}) = \sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2 + \ldots + (x_n - y_n)^2}</math>. It is stra
  813 bytes (132 words) - 17:49, 28 March 2009
• Mock AIME 5 2005-2006 Problems
  == Problem 1 == ...5</math> and <math>f(101) = 0</math>). Evaluate the remainder when <math>f(1)+f(2)+\cdots+f(99)</math> is divided by <math>1000</math>.
  8 KB (1,370 words) - 21:52, 27 February 2007
• Hexagon
  [[Area]]: <math>\frac{3s^2\sqrt{3}}{2}</math> Where <math>s</math> is the side length of the hexagon. [[Apothem]], or [[inradius]]: <math>\dfrac{s\sqrt{3}}{2}</math>
  688 bytes (94 words) - 21:00, 14 December 2018
• Coefficient
  ...</math>, and <math>c</math> in the [[Quadratic Formula]], <math>\frac{b\pm\sqrt{b^2-4ac}}{2a}</math> are all coefficients of the polynomial <math>ax^2+bx+c
  1 KB (227 words) - 22:38, 6 October 2020
• Rectangular prism
  ...e interior [[diagonal]]s can be determined by using the formula <math>d = \sqrt{l^2 + w^2 + h^2}</math>. Proof: To get a base diagonal, we use the [[pythagorean theorem]]: <math> \sqrt{l^2+w^2}</math>. We call that v. Then we use the pythagorean theorem again
  883 bytes (132 words) - 09:04, 12 September 2007

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