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  • ...72 174 175 176 177 178 180 182 183 184 185 186 187 188 189 190 192 194 195 196 198 200 201 202 203 204 205 206 207 208 209 210 212 213 214 215 216 217 218
    6 KB (350 words) - 11:58, 26 September 2023
  • pair C=(0,0), D=(0,-14),A=(-(961-196)^.5,0),B=IP(circle(C,21),circle(A,18));
    2 KB (217 words) - 20:43, 2 February 2014
  • \qquad\mathrm{(C)}\ 196
    13 KB (2,049 words) - 12:03, 19 February 2020
  • ...s that <math>n = 131 + 65 = 196</math>; the total number of cards is <math>196 \cdot 2 = \boxed{392}</math>.
    2 KB (384 words) - 23:31, 25 July 2018
  • <cmath>169 = 196 + 225 - 2 \cdot 14 \cdot 15 \cdot \cos{C} \Rightarrow \cos{C} = \frac{3}{5} ...lines. Hence, we have <cmath>\frac{BE}{15-BE}\cdot \frac{9}{6}=\frac{169}{196}\implies BE=\frac{2535}{463}</cmath> so our desired answer is <math>\boxed{
    14 KB (2,340 words) - 15:38, 21 August 2024
  • <math>EG=\sqrt{EB^2+BG^2}=\sqrt{(\frac{17x}{14})^2+17^2}=17\sqrt{\frac{x^2}{196}+1}</math> <cmath>[FCBE]=\frac{1}{2} \cdot 17 \cdot \sqrt{\frac{x^2}{196}+1} \cdot \frac{5\sqrt{34}}{2}-\frac{9x}{28}</cmath>
    9 KB (1,500 words) - 19:06, 8 October 2024
  • <cmath>AB^2 = 36 + 100 + 60 = 196, BC^2 = 36 + x^2 + 6x, CA^2 = 100 + x^2 + 10x</cmath> <cmath>x^2 + 10x + 100 = x^2 + 6x + 36 + 196</cmath>
    1 KB (212 words) - 12:37, 11 December 2024
  • \hline & & 206 - 2x & 196 - x & 186 \
    5 KB (877 words) - 19:57, 27 December 2024
  • ...th>. Computing this and dividing by 100 gives us an answer of <math>\boxed{196}</math>. ...\frac{50 * 49 * 48}{3 * 2} = 19600</math>, and <math>\frac n{100} = \boxed{196}</math>.
    5 KB (684 words) - 17:52, 19 June 2024
  • c^2 &= b^2 + 196 - 28b \cos x \
    7 KB (1,184 words) - 12:25, 22 December 2022
  • &=\sqrt{196}\
    8 KB (1,241 words) - 12:47, 1 October 2024
  • ...we must have <math>E(2,0,12)</math> so <math>CE=\sqrt{4^2+6^2+12^2}=\sqrt{196}=14</math> (the other solution, <math>E(10,0,12)</math> does not lie over t
    7 KB (1,181 words) - 14:56, 1 July 2024
  • ...>450</math> and the other factor <math>2</math>, but solving gives <math>b=196</math>, which is already a perfect square, so we have to keep going. In ord
    1 KB (218 words) - 13:14, 25 June 2021
  • ...</math>, so <math>r=14</math>. Thus the area of the circle is <math>\boxed{196}\pi</math>.
    795 bytes (129 words) - 09:22, 4 April 2012
  • 1. 196
    124 bytes (0 words) - 14:02, 3 April 2012
  • Squaring both sides of <math>a^2+a^{-2}=14</math> gives <math>a^4+a^{-4}+2=196,</math> from which <math>a^4+a^{-4}=\boxed{\textbf{(D)}\ 194}.</math> a^4+2a^2a^{-2}+a^{-4}&=196 \
    4 KB (663 words) - 06:32, 4 November 2022
  • <math>\mathrm{(A)}\ 194 \qquad \mathrm{(B)}\ 195 \qquad \mathrm{(C)}\ 196 \qquad \mathrm{(D)}\ 197 \qquad \mathrm{(E)}\ 198</math>
    3 KB (383 words) - 19:12, 15 October 2016
  • <math>\mathrm{(A)}\ 195\qquad\mathrm{(B)}\ 196\qquad\mathrm{(C)}\ 224\qquad\mathrm{(D)}\ 378\qquad\mathrm{(E)}\ 405</math>
    13 KB (2,025 words) - 12:56, 2 February 2021
  • <math>\mathrm{(A)}\ 195\qquad\mathrm{(B)}\ 196\qquad\mathrm{(C)}\ 224\qquad\mathrm{(D)}\ 378\qquad\mathrm{(E)}\ 405</math>
    5 KB (895 words) - 21:54, 9 January 2021
  • ...common difference <math>27</math> are <math>13^2=169</math> and <math>14^2=196</math>. One third of the way between them is <math>178</math> and two third ...er <math>abc</math> is one-third of the way from <math>169</math> to <math>196</math>, which is <math>178</math>. <math>1+7+8 = 16.</math>
    5 KB (759 words) - 01:38, 2 November 2024

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