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- ...17 418 420 422 423 424 425 426 427 428 429 430 432 434 435 436 437 438 440 441 442 444 445 446 447 448 450 451 452 453 454 455 456 458 459 460 462 464 4656 KB (350 words) - 11:58, 26 September 2023
- \qquad\mathrm{(E)}\ 441</math>10 KB (1,547 words) - 03:20, 9 October 2022
- \frac{56}{441}1 KB (188 words) - 21:10, 9 June 2016
- and so the answer is <math>118 + 323 = \boxed{441}</math>.2 KB (330 words) - 12:42, 1 January 2015
- ...hown in the figures below. Find <math>AC + CB</math> if area <math>(S_1) = 441</math> and area <math>(S_2) = 440</math>.6 KB (869 words) - 14:34, 22 August 2023
- ...hown in the figures below. Find <math>AC + CB</math> if area <math>(S_1) = 441</math> and area <math>(S_2) = 440</math>. ..., the area of triangle <math>ABC</math> is equal to both <math>T_1 + T_2 + 441</math> and <math>T_3 + T_4 + T_5 + 440.</math>5 KB (838 words) - 17:05, 19 February 2022
- &= 441(r+1) +546r + 169 \10 KB (1,595 words) - 15:30, 24 August 2024
- <cmath>9(a + b + c) + 54 = abc=\boxed{441}</cmath> ...{A^2B+A^2C+B^2A+B^2C+C^2A+C^2B+2ABC}{ABC}=\frac{49}{3} \implies abc=\boxed{441}</math>4 KB (743 words) - 20:02, 8 December 2024
- ...or a <math>6</math>): <math>\frac {1}{21} \cdot \frac {6}{21} = \frac {6}{441}</math>. ...or a <math>5</math>): <math>\frac {2}{21} \cdot \frac {5}{21} = \frac {10}{441}</math>.4 KB (512 words) - 03:05, 21 October 2024
- ...math>, or <math>\frac{1}{6}\pi (42)^2 - \frac{42^2 \sqrt{3}}{4} = 294\pi - 441\sqrt{3}</math>. Thus, the desired area is <math>360\sqrt{3} + 294\pi - 441\sqrt{3} = 294\pi - 81\sqrt{3}</math>, and <math>m+n+d = \boxed{378}</math>.3 KB (484 words) - 12:11, 14 January 2023
- 441(50+x) &=& 621x\ 180x = 441 \cdot 50 &\Longrightarrow & x = \frac{245}{2}3 KB (472 words) - 18:03, 21 June 2024
- \qquad\mathrm{(E)}\ 441</math>2 KB (261 words) - 22:34, 18 March 2023
- ...s must be odd. So we generate <math>N = 1, 21, 45, 73, 105, 141, 181, 381, 441, 721, 801</math>. <math>N</math> cannot exceed <math>1000</math> since it i4 KB (628 words) - 15:23, 2 January 2024
- If <math>n = 2</math>, consider the triangle formed by <math>A(21, 441), B(3, 9), C(-3, 9)</math>. It is parabolic and has area <math>1/2 \cdot 66 KB (1,001 words) - 19:21, 11 September 2021
- \textbf{(B)}\ 441 \qquad13 KB (1,994 words) - 12:52, 3 July 2021
- \textbf{(B)}\ 441 \qquad ...e get a total of <math>\frac{1}{2}\times 18 \times (25+24) = 9 \times 49 = 441 \rightarrow \boxed{\textbf{B}}</math>1 KB (196 words) - 22:47, 15 July 2020
- <cmath>441, 484, 529,576,625,676,729,784,841,900,961</cmath>We see that <math>484</mat3 KB (547 words) - 14:39, 1 December 2024
- ...to <math>m^{2}c + nc + 40mc + 441c = 1261 \Rightarrow c(m^{2} + n + 40m + 441) = 1261</math> ...<math>P(x) = x^2 - 20x + (400+a)</math>, and <math>Q(x) = x^2 + (c-21)x + (441 - 21c)</math>.8 KB (1,347 words) - 20:14, 9 November 2024
- 112 441 455 [16 - 63 - 65] 420 441 609 [20 - 21 - 29]55 KB (3,566 words) - 10:28, 29 September 2024
- ...(2\cdot 6\cdot 7) = 19/21</math>, so <math>\sin (\angle BCD) = \sqrt{1-361/441} = 4\sqrt{5}/21</math>. Therefore the area of this trapezoid is <math>\frac4 KB (683 words) - 07:03, 6 September 2024