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• 2002 AMC 10A Problems/Problem 23
  ...ath>, we find <math>AE=\sqrt{x^2+12x+100}</math>. From symmetry, <math>DE=\sqrt{x^2+12x+100}</math> as well. Now, we use the fact that the perimeter of <ma We have <math>2\sqrt{x^2+12x+100}+2x+12=2(32)</math> so <math>\sqrt{x^2+12x+100}=26-x</math>. Squaring both sides, we have <math>x^2+12x+100=67
  2 KB (378 words) - 21:38, 19 July 2023
• 2002 AMC 10B Problems/Problem 14
  Taking the root, we get <math>N=\sqrt{25^{64}\cdot 64^{25}}=5^{64}\cdot 8^{25}</math>.
  1 KB (148 words) - 08:08, 27 November 2020
• 1972 USAMO Problems/Problem 4
  <center><math>\left|\frac{aR^2+bR+c}{dR^2+eR+f}-\sqrt[3]{2}\right|<|R-\sqrt[3]{2}|</math></center> .../math>, <math>\frac{aR^2+bR+c}{dR^2+eR+f}</math> must also approach <math>\sqrt[3]{2}</math> for the given inequality to hold. Therefore
  1 KB (245 words) - 15:09, 2 June 2018
• 1962 IMO Problems/Problem 2
  <math>\sqrt{\sqrt{3-x}-\sqrt{x+1}}>\dfrac{1}{2}</math> Obviously we need <math>\sqrt{3-x} \geq \sqrt{x+1}</math> for the outer square root to be defined, <math>x\leq 3</math> for t
  3 KB (445 words) - 14:43, 2 February 2009
• 1998 AHSME Problems/Problem 23
  ...th>, which is a circle centered at <math>(2,6)</math> with radius <math>r=\sqrt{40+k}</math>. ...>\sqrt{40+k} \geq 2</math> we get that <math>k\geq -36</math>. From <math>\sqrt{40+k}\leq 12</math> we get <math>k\leq 104</math>.
  1 KB (193 words) - 09:12, 2 December 2018
• 1999 AHSME Problems/Problem 16
  Using [[Pythagorean theorem]] we can compute <math>a=\sqrt{ (d_1/2)^2 + (d_2/2)^2 }=13</math>. pair A=(-3,0), B=(0,-1.25), C=(3,0), D=(0,1.25);
  2 KB (308 words) - 14:35, 5 July 2013
• 2009 Zhautykov International Olympiad Problems
  ==Day 1== ===Problem 1===
  2 KB (402 words) - 13:52, 3 February 2009
• 2009 AMC 12A Problems/Problem 21
  *<math>x^4 - 2009 = 0</math>: The real roots are <math>\pm \sqrt [4]{2009}</math>, and there are two complex roots. *<math>x^4 - 9002 = 0</math>: The real roots are <math>\pm \sqrt [4]{9002}</math>, and there are two complex roots.
  2 KB (322 words) - 10:25, 29 July 2020
• 2009 AMC 12A Problems/Problem 13
  ...^\circ</math>, hence <math>BD_2 = BC_2 / 2</math> and <math>D_2C_2 = BC_2 \sqrt 3 / 2</math>. label("$20/\sqrt 2$",midpoint(B--D1),S);
  5 KB (739 words) - 10:24, 9 February 2015
• 2009 AMC 12B Problems/Problem 13
  ...BC</math>. Then <math>BD = \sqrt {13^2 - 12^2} = 5</math> and <math>DC = \sqrt {15^2 - 12^2} = 9</math>. Thus <math>BC = BD + BC = 5 + 9 = 14</math> or
  883 bytes (137 words) - 18:42, 23 February 2017
• 1966 AHSME Problems/Problem 4
  ...\quad \text{(C) } \sqrt{3} \quad \text{(D) } 2\sqrt{2} \quad \text{(E) } 2\sqrt{3}</math> ...onal to be <math>2\sqrt{2}x</math>. Half of that is the radius, or <math>x\sqrt{2}</math>. Using the same equation as before, you get the area of the large
  1 KB (191 words) - 22:09, 14 January 2018
• 2006 Alabama ARML TST Problems/Problem 14
  Thus, <math>|x+yi|^2 = \sqrt[3]{2^3 \cdot 37^3} = 2\cdot37 = 74</math>. <math>-1 < \tan{(3\theta)} = -\dfrac{182}{610} < 0</math>.
  2 KB (318 words) - 22:55, 29 March 2009
• 2007 Alabama ARML TST Problems/Problem 3
  ...2 cents is <math>\frac{144}{a+2}</math> cents a dozen. Their difference is 1, so we have <math>\frac{144}{a+2}+1=\frac{144}{a}</math>
  1,020 bytes (167 words) - 13:14, 6 May 2009
• 2010 AIME II Problems/Problem 5
  ...{10}x)(\log_{10} yz) + (\log_{10}y) (\log_{10}z) = 468</math>. Find <math>\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}</math>. ...>\ a+b+c</math> and <math>\ 2ab+2ac+2bc</math>, and we want to find <math>\sqrt {a^2 + b^2 + c^2}</math>.
  2 KB (392 words) - 23:49, 5 February 2022
• 1961 AHSME Problems/Problem 7
  ...ified, the third term in the expansion of <math>(\frac{a}{\sqrt{x}}-\frac{\sqrt{x}}{a^2})^6</math> is: <cmath>\binom{6}{2}(\frac{a}{\sqrt{x}})^{4}(\frac{-\sqrt{x}}{a^2})^2</cmath>
  723 bytes (112 words) - 20:09, 17 May 2018
• 1961 AHSME Problems/Problem 10
  <math>\textbf{(A)}\ \sqrt{18} \qquad \textbf{(B)}\ \sqrt{28} \qquad
  1 KB (176 words) - 23:27, 26 May 2018
• Euclidean space
  ...he [[distance formula|distance]] [[metric]], <math>d(\bold{x},\bold{y}) = \sqrt{(x_1-y_1)^2 + \cdots + (x_n - y_n)^2}</math>. Similarly, the Euclidean spa
  1,018 bytes (164 words) - 12:46, 13 March 2010
• 2010 AMC 12B Problems/Problem 22
  ...{389}{2}} \qquad \textbf{(D)}\ \sqrt{\dfrac{425}{2}} \qquad \textbf{(E)}\ \sqrt{\dfrac{533}{2}}</math> ...+ 49 + 36 + 144 = 425</math>, so our answer is <math>\boxed{\textbf{(D)} \sqrt{\frac{425}{2}}}</math>.
  2 KB (262 words) - 16:43, 15 February 2021
• 2011 AMC 10A Problems/Problem 11
  ...\,\frac78 \qquad\text{(D)}\,\frac{5\sqrt{2}}{8} \qquad\text{(E)}\,\frac{\sqrt{14}}{4} </math> ...s^2}=\sqrt{50s^2}</math>, and hence the ratio in the areas is <math>\frac{\sqrt{50s^2}^2}{(8s)^2}=\frac{50}{64} = \boxed{\frac{25}{32} \ \mathbf{(B)}}</mat
  991 bytes (155 words) - 03:42, 23 January 2023
• 2012 IMO Problems/Problem 1
  ...dot \frac{\sqrt{\frac{(s-a)(s-b)}{ab}}}{\sqrt{\frac{(s-b)(s-c)}{bc}}} = c \sqrt{\frac{(s-a)(s-c)}{ac}} = c \sin \frac{B}{2}.</cmath>
  7 KB (1,189 words) - 01:22, 19 November 2023

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