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  • ...78 879 880 882 884 885 886 888 889 890 891 892 893 894 895 896 897 898 899 900 901 902 903 904 905 906 908 909 910 912 913 914 915 916 917 918 920 921 922
    6 KB (350 words) - 11:58, 26 September 2023
  • ...largest even square <math>< 1000</math> is <math>900</math> so <math>n+1= 900 => n= \boxed{899}</math> Next, we try the second smallest value, which is <math>n = 30^2 = 900</math>, which tells us that <math>n=899</math>. <math>n-1</math> is indeed
    10 KB (1,702 words) - 21:23, 25 July 2024
  • <math> n < 900 </math> <cmath> 6.25 < n < 900 </cmath>
    1 KB (164 words) - 13:58, 14 April 2020
  • Square <math> ABCD </math> has center <math> O, AB=900, E </math> and <math> F </math> are on <math> AB </math> with <math> AE<BF
    7 KB (1,119 words) - 20:12, 28 February 2020
  • [[Square]] <math>ABCD </math> has [[center]] <math> O,\ AB=900,\ E </math> and <math> F </math> are on <math> AB </math> with <math> AE<BF ...)); label("x",(B+F)/2,(0,1)); label("400",(E+F)/2,(0,1)); label("\(900\)",(C+D)/2,(0,-1));
    13 KB (2,080 words) - 12:14, 23 July 2024
  • <cmath>9c^2 - 180c + 900 = 4c^2 - 400</cmath>
    5 KB (906 words) - 22:15, 6 January 2024
  • ...h> miles it slows to <math>900</math> mph, and thus takes <math>\frac{250}{900}(60)=\frac{50}{3}</math> minutes to travel the next <math>250</math> miles. ...rter takes to complete the same amount of work, which is <math>\frac{1000}{900}\cdot\frac{1}{4}\cdot t=\frac{5}{18}t</math>.
    4 KB (592 words) - 18:02, 26 September 2020
  • ...9 \cdot 8</math> numbers that have all distinct digits so there are <math>900 - 9 \cdot 9 \cdot 8</math> total three digit numbers that work.
    3 KB (508 words) - 00:16, 19 January 2024
  • ...are <math>A=(900,300), B=(1800,600), C=(600,1800),</math> and <math>D=(300,900).</math> Let <math>k_{}</math> be the area of the region enclosed by the i
    7 KB (1,094 words) - 12:39, 16 August 2020
  • ...{h^2w^2}{h^2 + w^2} = \frac {1}{\frac {1}{h^2} + \frac {1}{w^2}} = \frac {900}{13}</cmath> <cmath>\frac {1}{l^2} + \frac {1}{w^2} = \frac {45}{900}</cmath>
    2 KB (346 words) - 12:13, 22 July 2020
  • ...dot450=\frac{18}{17}d</math>. Since <math>FD'=BC-EE'</math>, we have <math>900-\frac{33}{17}d=d</math>, so <math>d=\boxed{306}</math>.
    11 KB (1,879 words) - 20:04, 8 December 2024
  • ...ed{890}</math>; we can double-check manually and we find that indeed <math>900\mid 890^3+100</math>. ...</math> must be divisible by <math>k</math>, which is largest when <math>k=900</math> and <math>n=\boxed{890}</math>
    2 KB (338 words) - 18:56, 15 October 2023
  • ...cdot 46=47\cdot 45\cdot 46</math>. But we didn't count 100, 200, 300, ..., 900. We add another 45 to get <math>45\cdot 2163</math>. The largest prime fact
    2 KB (275 words) - 09:27, 14 June 2024
  • ...are <math>A=(900,300), B=(1800,600), C=(600,1800),</math> and <math>D=(300,900).</math> Let <math>k_{}</math> be the area of the region enclosed by the i <cmath>\begin{eqnarray*}A' = & (\sqrt {900}, \sqrt {300})\
    2 KB (354 words) - 15:42, 20 July 2021
  • ...a square, so we have <math>b^2+60b=n^2</math>, and <math>(b+n+30)(b-n+30)=900</math>. The sum of these two factors is <math>2b+60</math>, so they must bo
    1 KB (218 words) - 13:14, 25 June 2021
  • So, <math>m+n+p+q = \boxed{900}</math>.
    2 KB (336 words) - 18:30, 24 June 2020
  • Thus there are <math>900</math> three-digit numbers in base-10 Therefore the desired probability is <math>\frac{608}{900}\approx 0.7 \Rightarrow\boxed{\mathrm{(E)}\ 0.7}</math>.
    1 KB (200 words) - 23:04, 31 July 2023
  • For each of the <math>9\cdot10\cdot10=900</math> possible values of <math>q</math>, there are at least <math>\left\lf ...congruent to <math>0\pmod{11}</math>, each of the <math>\left\lfloor \frac{900}{11} \right\rfloor = 81</math> values of <math>q</math> that are congruent
    5 KB (831 words) - 16:04, 5 January 2025
  • For each of the <math>9\cdot10\cdot10=900</math> possible values of <math>q</math>, there are at least <math>\left\lf ...congruent to <math>0\pmod{11}</math>, each of the <math>\left\lfloor \frac{900}{11} \right\rfloor = 81</math> values of <math>q</math> that are congruent
    3 KB (426 words) - 17:20, 18 July 2022
  • <math>\textbf{(A) } 9,800 \qquad \textbf{(B) } 9,900 \qquad \textbf{(C) } 10,000 \qquad \textbf{(D) } 10,100 \qquad \textbf{(E)
    13 KB (1,968 words) - 17:05, 23 November 2024

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