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  • ...07 708 710 711 712 713 714 715 716 717 718 720 721 722 723 724 725 726 728 729 730 731 732 734 735 736 737 738 740 741 742 744 745 746 747 748 749 750 752
    6 KB (350 words) - 11:58, 26 September 2023
  • <math> \mathrm{(A) \ } \frac{1}{2187}\qquad \mathrm{(B) \ } \frac{1}{729}\qquad \mathrm{(C) \ } \frac{2}{243}\qquad \mathrm{(D) \ } \frac{1}{81}\qqu
    15 KB (2,223 words) - 12:43, 28 December 2020
  • ...any positive integers <math>b</math> have the property that <math>\log_{b} 729</math> is a positive integer?
    13 KB (1,948 words) - 09:35, 16 June 2024
  • <math> \textbf{(A) } \frac{1}{2187}\qquad \textbf{(B) } \frac{1}{729}\qquad \textbf{(C) } \frac{2}{243}\qquad \textbf{(D) } \frac{1}{81} \qquad
    6 KB (1,083 words) - 13:05, 25 November 2024
  • <math>\mathrm{(A)}\ \frac{1}{2187}\qquad\mathrm{(B)}\ \frac{1}{729}\qquad\mathrm{(C)}\ \frac{2}{243}\qquad\mathrm{(D)}\ \frac{1}{81}\qquad\mat
    13 KB (2,028 words) - 15:32, 22 March 2022
  • ...ation into three consecutive [[integer]]s. Since 720 is close to <math>9^3=729</math>, we try 8, 9, and 10, which works, so <math>n - 3 = 10</math> and <m
    1 KB (239 words) - 10:54, 31 July 2023
  • ...along that edge to the vertex at its opposite end. Let <math>p = \frac{n}{729}</math> be the probability that the bug is at vertex <math>A</math> when it P(7)&=\frac13(1-P(6))&&=\frac{182}{729}.
    19 KB (3,128 words) - 20:38, 23 July 2024
  • ...change it back to base 10 for the answer, which is <math>3^6 + 3^5 + 3^2 = 729 + 243 + 9 = \boxed {981}</math>. ...h>729</math>. Now out of <math>64</math> terms which are of the form <math>729</math> + <math>'''S'''</math>, <math>32</math> of them include <math>243</m
    5 KB (866 words) - 23:00, 21 December 2022
  • <math>\left(4 \cos^2 \theta - \dfrac{3}{2}\right)^2 = \dfrac{729}{324} - \dfrac{200}{324} = \left(\dfrac{23}{18}\right)^2</math>
    4 KB (547 words) - 03:46, 1 December 2024
  • ...ac{8\cdot 26 + 4 \cdot 24 \ldots 2 \cdot 8 + 1 \cdot 0}{27^2} = \frac{532}{729}</math>, and the solution is <math>\boxed{532}</math>. So we have <math>p=(4/9)+(4/9)^2+(4/9)^3=4/9+16/81+64/729=532/729</math>. Therefore, the answer is <math>\boxed{532}</math>.
    5 KB (813 words) - 05:10, 25 February 2024
  • <math>(a_6,b_6)=\left(\frac{364}{729}, \frac{365}{729}\right)</math> Since <math>a_6=\frac{364}{729}</math>, <math>m+n = 1093 \equiv \boxed{093} \pmod{1000}</math>.
    7 KB (1,058 words) - 19:57, 22 December 2020
  • ...Theorem. But the order of the sets doesn't matter, so we get <math>\dfrac{729 - 1}{2} + 1 = \boxed{365}.</math>
    9 KB (1,400 words) - 13:09, 12 January 2024
  • ...mes by treating the set as ordered. The final solution is then <math>\frac{729-27}{6}=\boxed{117}</math>
    3 KB (585 words) - 18:37, 25 April 2022
  • ...math>141/729</math>. Then the desired answer is <math>\frac{1 - \frac{141}{729}}{2} = \frac{98}{243}</math>, so the answer is <math>m+n = \boxed{341}</mat
    3 KB (415 words) - 22:25, 20 February 2023
  • <cmath>f(2001) = 729\left[1 - \left| \frac{2001}{729} - 2\right|\right] = 186.</cmath> ...ct that the slope of the line is <math>-1</math>, we compute <math>f(2001)=729-543=186</math>. However, we want the minimum value such that <math>f(x)=186
    5 KB (779 words) - 20:42, 28 December 2024
  • ...25\qquad \mathrm{(C) \ } 27\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 729 </math>
    13 KB (1,900 words) - 21:27, 6 January 2021
  • ...25\qquad \mathrm{(C) \ } 27\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 729 </math>
    7 KB (993 words) - 20:06, 29 September 2024
  • ...along that edge to the vertex at its opposite end. Let <math>p = \frac{n}{729}</math> be the probability that the bug is at vertex <math>A</math> when it
    7 KB (1,071 words) - 18:24, 23 February 2024
  • ...positive [[integer]]s <math>b</math> have the property that <math>\log_{b} 729</math> is a positive integer? ...be <math>3</math> to some [[factor]] of 6. Thus, there are four (3, 9, 27, 729) possible values of <math>b \Longrightarrow \boxed{\mathrm{E}}</math>.
    616 bytes (86 words) - 22:49, 29 December 2023
  • ...ath>2004_9</math> miles. By base conversion, <math>2004_9=9^3(2)+9^0(4)=729(2)+1(4)=1458+4=\boxed{1462}</math>
    4 KB (536 words) - 17:50, 26 November 2024

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