| 137|| Ridley-C|| 30|| 5677.518|| 189.251
| 62 || someone8888-2 || 15 || 4100.731 || 273.382
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480 KB (21,341 words) - 16:46, 14 September 2025
<cmath>\begin{array}{c|c|c|c|c|c|c|c|c}
<cmath>\begin{array}{c|c|c|c|c|c|c|c}
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15 KB (2,351 words) - 07:28, 20 April 2023
<math> \mathrm{(A) \ -50 } \qquad \mathrm{(B) \ -49 } \qquad \mathrm{(C) \ 0 } \qquad \mathrm{(D) \ 49 } \qquad \mathrm{(E) \ 50 } </math>
<math>\text{(C) All equilateral triangles are equiangular.}</math>
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13 KB (1,945 words) - 20:04, 2 February 2025
<math>\textbf{(A)}\ -13 \qquad \textbf{(B)}\ -8 \qquad \textbf{(C)}\ -5 \qquad \textbf{(D)}\ -3 \qquad \textbf{(E)}\ 11</math>
<math>\textbf{(A)}\ 600 \qquad \textbf{(B)}\ 800 \qquad \textbf{(C)}\ 1000 \qquad \textbf{(D)}\ 1200 \qquad \textbf{(E)}\ 1400</math>
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15 KB (2,458 words) - 22:05, 4 July 2024
<math> \textbf{(A)}\ -1 \qquad\textbf{(B)}\ \frac{5}{36} \qquad\textbf{(C)}\ \frac{7}{12} \qquad\textbf{(D)}\ \frac{49}{20} \qquad\textbf{(E)}\ \frac
<math> \textbf{(A)}\ 600 \qquad\textbf{(B)}\ 800 \qquad\textbf{(C)}\ 1000 \qquad\textbf{(D)}\ 1200 \qquad\textbf{(E)}\ 1400 </math>
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12 KB (1,926 words) - 20:54, 6 October 2022
...\textbf{(A)}\ {-}\frac{2}{3}\qquad\textbf{(B)}\ \frac{7}{36}\qquad\textbf{(C)}\ \frac{7}{12}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\textbf{(E)}\ \frac{5}{
<math>\textbf{(A) } 0 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 30 \qquad\textbf{(D) } 45 \qquad\textbf{(E) } 60</math>
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13 KB (1,968 words) - 22:41, 12 October 2025
<math>\textbf{(A)}\ 10\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 20\%\qquad\textbf{(D)}\ 30\%\qquad\textbf{(E)}\ 35\%</math>
<math>\textbf{(A) } 17 \qquad \textbf{(B) } 15 + 2\sqrt{2} \qquad \textbf{(C) } 13 + 4\sqrt{2} \qquad \textbf{(D) } 11 + 6\sqrt{2} \qquad \textbf{(E) }
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14 KB (2,073 words) - 01:29, 7 October 2025
...89 \qquad \textbf{(C)}\ 104 \qquad \textbf{(D)}\ 144 \qquad \textbf{(E)}\ 273</math>
...y_{2} = 8</math>. Thus, <math>5y_{1} + 8y_{2} = 40 + 64 = \boxed{\textbf{(C) }104}</math>.
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3 KB (419 words) - 16:36, 10 March 2025
If <math>\texttt{a,b,}</math> and <math>\texttt{c}</math> are digits for which
&\ \texttt{c 7 3} \end{tabular}</math></center>
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1 KB (230 words) - 18:46, 10 March 2015
...117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306</math>
Thus, Our answer is <math>136+1=\boxed{\textbf{(C) } 137}</math>.
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7 KB (930 words) - 16:19, 19 October 2025
...th>. Let <math>a = \cot^{-1}(3)</math>, <math>b=\cot^{-1}(7)</math>, <math>
c=\cot^{-1}(13)</math>, and <math>d=\cot^{-1}(21)</math>. We have
<center><p><math>\tan(a)=\frac{1}{3},\quad\tan(b)=\frac{1}{7},\quad\tan(
c)=\frac{1}{13},\quad\tan(d)=\frac{1}{21}</math>,</p></center>
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3 KB (473 words) - 09:04, 16 August 2025
\textbf{(C) }42\qquad
(Solution by C-273)
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4 KB (702 words) - 00:19, 19 January 2025
...ac 1{15} \qquad \mathrm{(B) \ } \frac 1{91} \qquad \mathrm{(C) \ } \frac 1{273} \qquad \mathrm{(D) \ } \frac 1{455} \qquad \mathrm{(E) \ } \frac 1{1365}</
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1 KB (155 words) - 16:35, 2 February 2025
\text{(C)}\ 9 \qquad
...uarters, and half-dollars by <math>x</math>. Then <math>x+5x+10x+25x+50x = 273</math>, and <math>x = 3</math>. The total amount of coins is <math>5x</math
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602 bytes (98 words) - 11:25, 21 November 2016
\text{(C) }56
...ns place is not <math>3</math>, we get <math>203, 213, 223, 243, 253, 263, 273, 283,</math> and <math>293.</math> With cases in which the tens place is <
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2 KB (259 words) - 20:46, 15 October 2025
