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  • ...65 266 267 268 270 272 273 274 275 276 278 279 280 282 284 285 286 287 288 289 290 291 292 294 295 296 297 298 299 300 301 302 303 304 305 306 308 309 310
    6 KB (350 words) - 11:58, 26 September 2023
  • \qquad\mathrm{(C)}\ 289
    10 KB (1,547 words) - 03:20, 9 October 2022
  • trying all the terms from the third condition, it is clear that <math>a_9 = 289</math> is the only solution.
    3 KB (547 words) - 10:41, 7 August 2024
  • pair Da=IP(Circle(A,289),A--B),E=IP(Circle(C,324),B--C),Ea=IP(Circle(B,270),B--C); pair Da=IP(Circle(A,289),A--B),E=IP(Circle(C,324),B--C),Ea=IP(Circle(B,270),B--C);
    11 KB (1,879 words) - 20:04, 8 December 2024
  • ...</math>. Thus, <math>(m - n^2)(m + n^2) = 289\cdot 1 = 17\cdot 17 = 1\cdot 289.</math> The only possible value, then, for <math>m</math> is <math>145</mat ...c {S_{n} + 17}{n^{2}}\ge \frac {n^{2}}{S_{n} - 17}\rightarrow S_{n}^{2}\ge 289 + n^{4}
    4 KB (658 words) - 15:58, 10 November 2023
  • ...D is a median of triangle <math>BF'F</math> . Therefore, <math>CB/FB=\frac{289}{240}</math>. ...<math>MEB</math>. For simplicity, put masses of <math>240</math> and <math>289</math> at <math>C</math> and <math>F</math> respectively. To find the mass
    8 KB (1,382 words) - 23:37, 11 July 2024
  • ...nteger]]s, so we must have <math>y - x = 1</math> and <math>y + x = 17^2 = 289</math> from which we get the solution <math>x = 144, y= 145</math>.
    2 KB (329 words) - 14:53, 3 April 2012
  • <math>4 , 9 , 25 , 49 , 121, 169 , 289 , 361 , 529</math>
    8 KB (1,255 words) - 21:56, 23 October 2024
  • ...ative integers <math>a_1 < a_2 < … < a_k</math> such that<cmath>\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.</cmath>What is <math>k?<
    13 KB (1,968 words) - 17:05, 23 November 2024
  • ...13}</math>. By the Pythagorean Theorem, we have <math>EF=\sqrt{\left(\frac{289}{13}\right)^2+\left(\frac{119}{13} \right)^2}=\frac{\sqrt{(17^2)(17^2+7^2)}
    6 KB (933 words) - 23:05, 7 July 2023
  • \qquad\mathrm{(C)}\ 289
    2 KB (261 words) - 22:34, 18 March 2023
  • ..., the minimum value of <math>PQ</math> is <math>\sqrt{\dfrac{225 \cdot 64}{289}} = \dfrac{120}{17}</math>, which is closest to <math>7 \Rightarrow \boxed{
    4 KB (551 words) - 13:17, 23 June 2022
  • * <math>x_1=289</math> gives <math>y_1=\sqrt{289\cdot 144}=17\cdot 12 = 204</math>, hence <math>(x,y)=(577,408)</math>, and
    3 KB (478 words) - 22:41, 5 January 2014
  • ...AB}}{2})^2} = \sqrt {17^2 - (\frac{16}{2})^2} = \sqrt {289 - 8^2} = \sqrt {289 - 64} = \sqrt {225} = 15</math>
    2 KB (256 words) - 00:45, 26 June 2016
  • \mathrm{(D)}\ 289
    14 KB (2,126 words) - 16:46, 13 June 2024
  • \mathrm{(D)}\ 289
    990 bytes (151 words) - 11:20, 25 October 2024
  • ...th>d</math>. We square the length and the width and multiply both sides by 289 (the denominator of the LHS) to reach the equation <math>(12+d^2)^2+100d^2=
    5 KB (873 words) - 03:33, 17 December 2023
  • 136 255 289 [8 - 15 - 17] 161 240 289
    55 KB (3,566 words) - 10:28, 29 September 2024
  • There are 9: <math>9,25,49,81,121,169,225,289,</math>and <math>361</math>. Therefore, the answer is <math>\boxed{D}</math
    2 KB (290 words) - 13:24, 27 June 2021
  • ...s <math>(\frac{336}{625})^2</math>, after seeing that <math>C_1C_0 = \frac{289}{625}</math>, . Now it suffices to find 90 times ratio <math>[B_0B_1C_1]</m ...h>, which simplifies to <math> \dfrac{90 \cdot 17^2}{25^2} \dfrac{625^2}{(289)(961)}</math>. Cancelling all common factors, we get the reduced fraction
    7 KB (1,112 words) - 23:19, 2 January 2025

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