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- == Problem ==2 KB (302 words) - 00:57, 18 October 2024
- {{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #1]] and [[2001 AMC 10 Problems|2001 AMC 10 #3]]}} == Problem ==904 bytes (136 words) - 15:10, 15 July 2024
- {{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #2]] and [[2001 AMC 10 Problems|2001 AMC 10 #6]]}} == Problem ==1,007 bytes (165 words) - 19:31, 27 October 2024
- {{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #3]] and [[2001 AMC 10 Problems|2001 AMC 10 #9]]}} ==Problem==2 KB (221 words) - 20:40, 15 July 2024
- {{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #1]] and [[2000 AMC 10 Problems|2000 AMC 10 #1]]}} ==Problem==2 KB (320 words) - 18:37, 30 November 2024
- == Problem ==911 bytes (147 words) - 17:54, 13 December 2017
- == Problem ==1 KB (239 words) - 12:29, 17 December 2024
- {{duplicate|[[2005 AMC 12B Problems|2005 AMC 12B #12]] and [[2005 AMC 10B Problems|2005 AMC 10B #16]]}} == Problem ==2 KB (322 words) - 01:57, 11 November 2024
- == Problem == <!-- [[Image:2006 AMC 10A-12.GIF]] -->3 KB (424 words) - 09:14, 17 December 2021
- ...ate|[[2006 AMC 12A Problems|2006 AMC 12A #12]] and [[2006 AMC 10A Problems/Problem 14|2006 AMC 10A #14]]}} == Problem ==2 KB (292 words) - 10:56, 17 December 2021
- == Problem == ...^2</math> equal and solving for <math>x</math> (it is helpful to scale the problem down by a factor of 50 first), we get <math>x = 250\pm 50\sqrt{7}</math>. S13 KB (2,080 words) - 12:14, 23 July 2024
- == Problem ==4 KB (647 words) - 01:29, 4 May 2021
- == Problem ==2 KB (303 words) - 17:43, 16 October 2024
- == Problem ==3 KB (431 words) - 22:21, 4 July 2013
- == Problem ==2 KB (412 words) - 17:23, 1 January 2024
- == Problem == Since this is a recursive problem, list out the functions f(2) and f(7) and figure out what is equivalent wit3 KB (588 words) - 13:37, 22 July 2020
- == Problem == ...math>(x+x^2+x^3)^n</math>, so the generating function of interest for this problem is <math>(x+x^2+x^3)^7</math>. Our goal is to find the coefficients of ever19 KB (3,128 words) - 20:38, 23 July 2024
- == Problem == ...ve more than 4 elements, otherwise its sum would be at most <math>15+14+13+12=54</math>.2 KB (382 words) - 18:52, 20 January 2025
- == Problem ==4 KB (673 words) - 18:48, 28 December 2023
- == Problem == [[Image:1988_AIME-12.png]]4 KB (743 words) - 20:02, 8 December 2024
Page text matches
- <cmath>5-12-13</cmath> <cmath>12-35-37</cmath>6 KB (943 words) - 09:44, 17 January 2025
- == Problem == ...p with <math>12x^3</math>. Taking the answer choices and dividing by <math>12</math>, we get <math>(A) 4</math>,1 KB (216 words) - 23:23, 5 September 2024
- == Problem == ...+ 2 + 3 \cdots + 10 = 55.</math> Knowing this, we can say that <math>11 + 12 \cdots + 20 = 155</math> and <math>21 + \cdots +30 =255</math> and so on. T2 KB (395 words) - 22:29, 3 December 2024
- == Problem == ...the shaded area is <math>(8 \cdot 5) - 2(\frac{15}{2}) - 2(6) = 40 - 15 - 12 = 13</math>. Since the desired area is half the shaded region, our area is8 KB (1,016 words) - 23:17, 30 December 2023
- == Problem == {{AMC10 box|year=2016|ab=A|num-b=12|num-a=14}}2 KB (402 words) - 13:54, 25 June 2023
- * [[Noetic Learning Math Contest]] - semiannual problem solving contest for elementary and middle school students. [Grades 2-8] ...athcon.org MathCON] hosts annual math competition for students in grades 4-12.5 KB (512 words) - 22:26, 13 January 2025
- * [https://www.hardestmathproblem.org Hardest Math Problem] math contest for grades 5-8 with great prizes. * [[Noetic Learning Math Contest]]: a popular problem-solving contest for students in grades 2-8.7 KB (805 words) - 22:27, 13 January 2025
- **[[AMC 12]] ...athcon.org MathCON] hosts annual math competition for students in grades 4-12, with more than 200,000 participants since 2008.6 KB (625 words) - 22:27, 13 January 2025
- * [[Art of Problem Solving Academy]] (Locations Nationwide) [https://aopsacademy.org/courses/s * [[MehtA+ AI in Visual Arts Camp]] is a mini-bootcamp for grades 5-12 [https://mehtaplustutoring.com/ai-in-visual-arts/ Website]14 KB (1,894 words) - 09:54, 3 January 2025
- ...cluding Art of Problem Solving, the focus of MATHCOUNTS is on mathematical problem solving. Students are eligible for up to three years, but cannot compete be ...ics]]. The focus of MATHCOUNTS curriculum is in developing [[mathematical problem solving]] skills.10 KB (1,504 words) - 13:10, 1 December 2024
- *[[Doane triMATHlon Math Contest]]. Takes place at Doane University for 9-12 graders, includes material up to Algebra II. [https://www.doane.edu/trimath ...]. Consists of two problem solving rounds and a speed round - for grades 9-12.2 KB (200 words) - 09:11, 29 December 2020
- ...idual articles often have sample problems and solutions for many levels of problem solvers. Many also have links to books, websites, and other resources rele ...ttps://competifyhub.com/resources/ Free Competition Resources for Grades 1-12]17 KB (2,291 words) - 22:33, 13 January 2025
- These '''math books''' are recommended by [[Art of Problem Solving]] administrators and members of the [http://aops.com/community AoPS * Intermediate is recommended for students who can expect to pass the AMC 10/12.24 KB (3,202 words) - 14:33, 13 January 2025
- '''David Patrick''' is a teacher and curriculum developer for [[Art of Problem Solving]] (AoPS). Dr. Patrick joined AoPS in 2004. ...erican High School Mathematics Examination]] (AHSME), now called the [[AMC 12]]. That same year, Patrick was a winner of the [[USA Mathematical Olympiad3 KB (379 words) - 10:18, 27 September 2024
- ...ks''' page is for compiling a list of [[textbook]]s for mathematics -- not problem books, contest books, or general interest books. See [[math books]] for mo * Intermediate is recommended for students grades 9 to 12.7 KB (902 words) - 14:34, 13 January 2025
- ...n mathematics and also to help middle school students learn [[mathematical problem solving]]. ...gion=USA|type=Multiple Choice|difficulty=1 - 1.5|breakdown=<u>Problems 1 - 12</u>: 1<br><u>Problems 13 - 25</u>: 1.5}}4 KB (584 words) - 23:33, 15 January 2025
- ...'''American Mathematics Contest 10''' ('''AMC 10'''), along with the [[AMC 12]], is one of the first exams in the series of exams used to challenge brigh High scoring AMC 10 and AMC 12 students are invited to take the [[American Invitational Mathematics Examin4 KB (636 words) - 21:50, 17 January 2025
- ...irst exam in the series of exams used to challenge bright students, grades 12 and below, on the path towards choosing the team that represents the United High scoring AMC 12 students are invited to take the more challenging [[American Invitational M4 KB (529 words) - 08:01, 24 July 2024
- ...erica Mathematics Olympiad]] (USAMO) for qualification from taking the AMC 12 or United States of America Junior Mathematics Olympiad (USAJMO) for qualif ...ministered by the [[Mathematical Association of America]] (MAA). [[Art of Problem Solving]] (AoPS) is a proud sponsor of the AMC!8 KB (1,062 words) - 18:04, 17 January 2025
- ...th Jam''' is a free online class or information session hosted by [[Art of Problem Solving]] (AoPS) in the [[AoPS Schoolhouse|classroom]]. ...sessions for discussion of the problems from each year's [[AMC 10]], [[AMC 12]], and [[AIME]] exams.989 bytes (130 words) - 15:20, 20 August 2020