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Create the page "Problem 18" on this wiki! See also the search results found.
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- == Problem ==953 bytes (152 words) - 01:40, 16 August 2023
- ==Problem== {{AHSME 40p box|year=1962|before=Problem 17|num-a=19}}902 bytes (147 words) - 22:17, 3 October 2014
- ==Problem==3 KB (412 words) - 22:30, 18 December 2023
- == Problem==485 bytes (86 words) - 01:57, 3 January 2014
- ==Problem==5 KB (791 words) - 03:18, 20 June 2022
- ==Problem== ...f{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 18 </math>1 KB (170 words) - 03:08, 28 May 2021
- ==Problem==2 KB (410 words) - 21:18, 31 May 2020
- ==Problem== ...roblem is quite similar to [[2004_AMC_12A_Problems/Problem_16|2004 AMC 12A Problem 16]].4 KB (703 words) - 19:04, 10 July 2021
- ==Problem==2 KB (369 words) - 17:44, 30 January 2021
- {{duplicate|[[2014 AMC 12B Problems|2014 AMC 12B #18]] and [[2014 AMC 10B Problems|2014 AMC 10B #24]]}} ==Problem==2 KB (331 words) - 04:43, 12 January 2021
- == Problem == Day 18: Al works; Barb rests2 KB (294 words) - 16:49, 9 September 2020
- == Problem ==1,008 bytes (157 words) - 03:02, 20 February 2018
- == Problem ==1 KB (181 words) - 01:52, 16 August 2023
- == Problem ==987 bytes (146 words) - 11:52, 4 February 2016
- == Problem ==1 KB (180 words) - 20:08, 23 February 2024
- == Problem ==608 bytes (84 words) - 16:23, 2 July 2016
- == Problem ==830 bytes (142 words) - 20:45, 18 June 2021
- == Problem ==880 bytes (138 words) - 01:40, 22 December 2015
- == Problem == <math>\text{(F) }18\qquad2 KB (393 words) - 17:01, 10 June 2018
- ==Problem==1 KB (232 words) - 14:03, 27 February 2018
Page text matches
- == Problem == ...than 1, so <math>3n^2 > 999</math> and <math>n > \sqrt{333}</math>. <math>18^2 = 324 < 333 < 361 = 19^2</math>, so we must have <math>n \geq 19</math>.4 KB (673 words) - 19:48, 28 December 2023
- == Problem == ...to 3, it would overlap with case 1). Thus, there are <math>2(3 \cdot 3) = 18</math> cases.3 KB (547 words) - 22:54, 4 April 2016
- == Problem == ...s: <math>\sqrt{(12 - (-2))^2 + (8 - (-10))^2 + (-16 - 5)^2} = \sqrt{14^2 + 18^2 + 21^2} = 31</math>. The largest possible distance would be the sum of th697 bytes (99 words) - 18:46, 14 February 2014
- == Problem == <cmath>3(ab + bc + ca) + 18(a + b + c) + 81 = abc + 3(ab + bc + ca) + 9(a + b + c) + 27</cmath>4 KB (727 words) - 23:37, 7 March 2024
- == Problem == <math>(-3+i)^1=-3+i ; (-3+i)^2=8-6i ; (-3+i)^3=-18+26i</math>2 KB (408 words) - 17:28, 16 September 2023
- == Problem == ...be a [[tetrahedron]] with <math>AB=41</math>, <math>AC=7</math>, <math>AD=18</math>, <math>BC=36</math>, <math>BD=27</math>, and <math>CD=13</math>, as2 KB (376 words) - 13:49, 1 August 2022
- == Problem == ...e [[Pythagorean Theorem]]. Thus <math>A = \frac 12(12\sqrt{3})\sqrt{399} = 18\sqrt{133}</math>.7 KB (1,086 words) - 08:16, 29 July 2023
- == Problem == The sets <math>A = \{z : z^{18} = 1\}</math> and <math>B = \{w : w^{48} = 1\}</math> are both sets of comp3 KB (564 words) - 04:47, 4 August 2023
- == Problem == <!-- replaced: Image:1990 AIME Problem 7.png by I_like_pie -->8 KB (1,319 words) - 11:34, 22 November 2023
- == Problem == The area of a 2x3 rectangle and a 3x4 rectangle combined is 18, so a 4x4 square is impossible without overlapping. Thus, the next smallest1 KB (242 words) - 18:35, 15 August 2023
- == Problem == ...=expi(-pi/2+acos(475/486)+3*acos(7/18)), F=expi(-pi/2-acos(475/486)-acos(7/18));2 KB (284 words) - 03:56, 23 January 2023
- == Problem == {{AMC10 box|year=2006|ab=B|num-b=16|num-a=18}}1 KB (211 words) - 04:32, 4 November 2022
- == Problem == ...2, \pm 72),( \pm 3, \pm 48),( \pm 4, \pm 36),( \pm 6, \pm 24),( \pm 8, \pm 18),( \pm 9, \pm 16),( \pm 12, \pm 12)</math>; since <math>c</math> is the [[m2 KB (310 words) - 11:25, 13 June 2023
- == Problem == ...s take a value of 7. So, <math>\lfloor r\rfloor \le ...\le \lfloor r+\frac{18}{100}\rfloor \le 7</math> and <math>\lfloor r+\frac{92}{100}\rfloor \ge ...3 KB (447 words) - 17:02, 24 November 2023
- == Problem == <math>a_{19}=0=a_1+18(a_2-a_1)+\binom{18}{2}=18a_2-17a_1+153</math>5 KB (778 words) - 21:36, 3 December 2022
- == Problem == {{AMC10 box|year=2006|ab=B|num-b=18|num-a=20}}5 KB (873 words) - 15:39, 29 May 2023
- == Problem == ...mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } \frac{55}{3} </math>5 KB (861 words) - 00:53, 25 November 2023
- == Problem == <math> \mathrm{(A) \ } 18\sqrt{3}\qquad \mathrm{(B) \ } 24\sqrt{2}\qquad \mathrm{(C) \ } 36\qquad \ma4 KB (558 words) - 14:38, 6 April 2024
- == Problem == Another way to do the problem is by the process of elimination. The only possible correct choices are the5 KB (878 words) - 14:39, 3 December 2023
- == Problem == ...19)\\ &= P_{17}(((x - 20) - 19) - 18)\\ &= \cdots\\ &= P_0(x - (20 + 19 + 18 + \ldots + 2 + 1)).\end{align*}</cmath>2 KB (355 words) - 13:25, 31 December 2018
