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  • ...39 940 942 943 944 945 946 948 949 950 951 952 954 955 956 957 958 959 960 961 962 963 964 965 966 968 969 970 972 973 974 975 976 978 979 980 981 982 984
    6 KB (350 words) - 11:58, 26 September 2023
  • ...0</math>, we have <math>n+1 < 1001</math>. We first try <math>n+1 = 31^2 = 961</math>(since it is the smallest square below <math>1000</math>, which gives
    10 KB (1,702 words) - 21:23, 25 July 2024
  • pair C=(0,0), D=(0,-14),A=(-(961-196)^.5,0),B=IP(circle(C,21),circle(A,18)); <div style="text-align:center"><math> (AD)^2 = 961 </math></div>
    2 KB (217 words) - 20:43, 2 February 2014
  • ...ath> is less than <math>abc</math>, <math>k^2</math> must be at most <math>961</math>, or <math>31^2</math>. Therefore <math>k \le 31</math>. Plugging thi
    5 KB (759 words) - 01:38, 2 November 2024
  • .... Circles <math>\omega_1</math> and <math>\omega_2</math> with radii <math>961</math> and <math>625</math>, respectively, intersect at distinct points <ma
    12 KB (2,125 words) - 07:38, 23 May 2024
  • <cmath>441, 484, 529,576,625,676,729,784,841,900,961</cmath>We see that <math>484</math> and <math>784</math> differ by <math>30
    3 KB (547 words) - 14:39, 1 December 2024
  • (a + b)^2 &= 961\
    1 KB (186 words) - 10:05, 13 July 2021
  • <math>31^2 = 961</math>, and <math>32^2 = 1024</math>. Thus, <math>\sqrt {1000}</math> must
    2 KB (293 words) - 19:22, 3 September 2021
  • <cmath>625 Q^2 + 1250 QR + 625 R^2 = 961 Q^2 + 961 R^2</cmath>
    5 KB (796 words) - 14:22, 31 July 2024
  • #961
    252 bytes (26 words) - 19:17, 31 October 2020
  • ...ac{5^8-336^2-17^2\cdot 5^4}{5^8-336^2}</math>, which gives <math>q= \boxed{961}</math>. ...^2}{31^2}=\frac{90 \cdot 336}{961}</math>, yielding the answer <math>\fbox{961}</math>.
    7 KB (1,112 words) - 23:19, 2 January 2025
  • .../math> digits are close to <math>1000</math>, but not there, such as <math>961</math> or <math>974</math>. Now, the best we can do is to keep on listing s ...is over <math>1000</math>. We find that <math>k=31</math> gives <math>k^2=961</math>, and so <math>(k+1)^2=32^2=1024</math>. We can be sure that this ski
    5 KB (842 words) - 19:08, 26 December 2024
  • ...ath>\sqrt{96+961} = \sqrt{1057}</math>. Since <math>\sqrt{96+529}<\sqrt{96+961}</math>, <math>25</math> is the shorter length*, so the answer is <math>\bo
    4 KB (652 words) - 08:18, 23 September 2021
  • <cmath>(a+b)^2=a^2+2ab+b^2=31^2=961</cmath> <cmath>(a^2+2ab+b^2)-(a^2+b^2)=961-625</cmath>
    1 KB (211 words) - 20:26, 16 April 2017
  • Circles <math>\omega_1</math> and <math>\omega_2</math> with radii <math>961</math> and <math>625</math>, respectively, intersect at distinct points <ma ...s well. It follows that <cmath>O_{1}O_{2}=O_{1}G-O_{2}G=2(O_{1}D-O_{2}E)=2(961-625)=\textbf{672}.</cmath>
    17 KB (2,852 words) - 02:59, 7 February 2024
  • Circles <math>\omega_1</math> and <math>\omega_2</math> with radii <math>961</math> and <math>625</math>, respectively, intersect at distinct points <ma
    7 KB (1,182 words) - 13:54, 13 March 2023
  • 2^{-4}&\equiv(-31)^2=961\equiv-39, \
    16 KB (2,240 words) - 22:16, 26 January 2024
  • ...math> numbers. Because of our pattern, we know that the numbers from <math>961</math> thru <math>1000</math> will have the same remainders as <math>1</mat
    8 KB (1,127 words) - 15:40, 1 January 2025
  • ...5\cdot 3\sqrt7)^2}}{16}=\frac{15\sqrt{31^2+(3\sqrt7)^2}}{16}=\frac{15\sqrt{961+63}}{16}=\frac{15\sqrt{1024}}{16}=\frac{15}{16}\cdot 32=30.</cmath>
    11 KB (1,733 words) - 10:11, 23 November 2023
  • ...to the ratio of the sines. Therefore it is <cmath> \frac{\sqrt{\frac{936}{961}}}{\sqrt{\frac{416}{441}}} = \sqrt{\frac{936}{416}} \cdot \frac{21}{31} = \
    16 KB (2,614 words) - 18:35, 28 December 2024

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