Search results
Create the page "A- 1" on this wiki! See also the search results found.
Page title matches
- 448 bytes (67 words) - 15:15, 23 March 2020
- ...he third, and one for the last. Thus, there are <math>4\cdot 3\cdot 2\cdot 1 = \boxed{24}</math> distinct ways to fill the boxes.328 bytes (60 words) - 17:19, 12 June 2022
- ...<math>3</math> perfect squares that only have <math>1</math> digit, <math>1^{2},</math> <math>2^{2},</math> and <math>3^{2}.</math> So we have a total of <math>1\times3+2\times6+3\times22+4\times19=\boxed{157}</math> digits.939 bytes (140 words) - 17:27, 12 June 2022
- The table shows the percent of families in Mathville that have <math>0, 1, 2, 3</math> and <math>4</math> or more children. If there are a total of <214 bytes (35 words) - 09:55, 1 August 2022
- 0 bytes (0 words) - 13:26, 6 February 2024
Page text matches
- === Proof 1 === <math>(a+b)^2=c^2+4\left(\frac{1}{2}ab\right)\implies a^2+2ab+b^2=c^2+2ab\implies a^2 + b^2=c^2</math>. {{Ha5 KB (886 words) - 13:51, 15 May 2024
- pair O0=(9,9), O1=(1,1), O2=(3,1), O3=(1,3); draw(Circle(O1,1));2 KB (307 words) - 15:30, 30 March 2024
- <math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textb ==Solution 1==1 KB (190 words) - 10:58, 16 June 2023
- ==Solution 1==1 KB (176 words) - 10:58, 16 June 2023
- <math>\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf ==Solution 1==2 KB (257 words) - 10:57, 16 June 2023
- A rectangular box has integer side lengths in the ratio <math>1: 3: 4</math>. Which of the following could be the volume of the box? == Solution 1==1 KB (184 words) - 13:58, 22 August 2023
- ..., replacing each occurrence of the digit <math>2</math> by the digit <math>1</math>. Star adds her numbers and Emilio adds his numbers. How much larger ...git and 2 appears 3 times as a units digit, the answer is <math>10\cdot 10+1\cdot 3=\boxed{\textbf{(D) }103.}</math>967 bytes (143 words) - 03:18, 27 June 2023
- == Solution 1 ==2 KB (268 words) - 18:19, 27 September 2023
- ...eft(\frac{1}{2}\right)=30 , \Rightarrow (+40)=70 , \Rightarrow \left(\frac{1}{2}\right)=\boxed{\textbf{(C) }35}</cmath>1 KB (169 words) - 14:59, 8 August 2021
- A triangular array of <math>2016</math> coins has <math>1</math> coin in the first row, <math>2</math> coins in the second row, <math == Solution 1==2 KB (315 words) - 15:34, 18 June 2022
- ...er rectangle is one foot wide, and each of the two shaded regions is <math>1</math> foot wide on all four sides. What is the length in feet of the inner filldraw(rectangle((1,1),(6,4)),gray(0.75));2 KB (337 words) - 14:56, 25 June 2023
- filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8)); label("$1$",(1/2,5),dir(90));8 KB (1,016 words) - 00:17, 31 December 2023
- Three distinct integers are selected at random between <math>1</math> and <math>2016</math>, inclusive. Which of the following is a correc ...\dfrac{1}{3}\qquad\textbf{(D)}\ p=\dfrac{1}{3}\qquad\textbf{(E)}\ p>\dfrac{1}{3}</math>2 KB (297 words) - 14:54, 25 June 2023
- ...n a movie theater in a row containing <math>5</math> seats, numbered <math>1</math> to <math>5</math> from left to right. (The directions "left" and "ri <math>\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad2 KB (402 words) - 14:54, 25 June 2023
- ...tp://euclidlab.org/programs/archimedean-challenge/1 Archimedean Challenge #1] * [http://www.mindresearch.org/gameathon/ Game-a-thon] challenges students to design and build math games in this national co7 KB (792 words) - 10:14, 23 April 2024
- ...duate or undergraduate mathematics students. The weeks break down into a 2-1-2 schedule: We start with two weeks of Root Class, which consists of a gall5 KB (706 words) - 23:49, 29 January 2024
- ...rint</u>: 1-1.5 (School/Chapter), 2-2.5 (State/National)<br><u>Target:</u> 1.5 (School), 2 (Chapter), 2-2.5 (State/National)}}10 KB (1,506 words) - 21:31, 14 May 2024
- * [[Archimedean Challenge #1]] * [[Archimedean Challenge #1]]4 KB (565 words) - 13:24, 13 September 2019
- ...wo areas of math contests: Grade School (Grades 3, 4, 5, 6, 7, 8 + Algebra 1) and High School (Regional and State Finals). The contest is given in-school May 1-15. Each school sets its own testing date and time within these guidelines.8 KB (1,182 words) - 14:26, 3 April 2024
- *[[Math Kangaroo]] is available to students in grades 1-12. The competition occurs annually during March held at ORU. Students do2 KB (279 words) - 19:41, 23 March 2017
- ...artofproblemsolving.com/store/item/aops-vol1 Art of Problem Solving Volume 1: the Basics]5 KB (667 words) - 17:09, 3 July 2023
- * [https://artofproblemsolving.com/news/articles/establishing-a-positive-culture Establishing a Positive Culture of Expectation in Math Educ * [https://artofproblemsolving.com/news/articles/how-to-write-a-solution How to Write a Math Solution] by [[Richard Rusczyk]] and [[user:MCr16 KB (2,152 words) - 21:46, 6 May 2024
- ...zon.com/gp/product/B09PMLFHX2/ref=ox_sc_act_title_1?smid=ATVPDKIKX0DER&psc=1 Getting Started with Competition Math], a textbook meant for true beginners ...itu-Andreescu/dp/0817643265/ref=sr_1_1?ie=UTF8&s=books&qid=1204029652&sr=1-1 Complex Numbers from A to... Z] by [[Titu Andreescu]]24 KB (3,177 words) - 12:53, 20 February 2024
- ...fproblemsolving.com/store/item/aops-vol1 the Art of Problem Solving Volume 1: the Basics]''2 KB (254 words) - 09:04, 25 January 2019
- ...Response|difficulty=1|breakdown=<u>Division E</u>: 1<br><u>Division M</u>: 1}}2 KB (215 words) - 02:54, 18 November 2020
- ...me=Putnam|region=USA|type=Proof|difficulty=7 - 9|breakdown=<u>Problem A/B, 1/2</u>: 7<br><u>Problem A/B, 3/4</u>: 8<br><u>Problem A/B, 5/6</u>: 9}} ...if the top three students tie, they are all awarded a rank of <math>\frac{1 + 2 + 3}{3}=2</math>.) Before 2019, schools were required to choose their t4 KB (623 words) - 13:11, 20 February 2024
- Khan Academy: AP Physics 1/24 KB (506 words) - 11:46, 6 September 2023
- ...s/ASIN/0201485419/artofproblems-20 The Art of Computer Programming Volumes 1-3 (boxed set)] by [[D. E. Knuth]]2 KB (251 words) - 00:45, 17 November 2023
- ...Thermodynamics states that the efficiency of heat engines must always be < 1.9 KB (1,355 words) - 07:29, 29 September 2021
- ...c{n^2}{2^n}\right)+\sum_{n=1}^{\infty} \left(\frac{2n}{2^n}\right)+\sum_{n=1}^{\infty} \left(\frac{3}{2^n}\right)</math> \sum_{n=1}^{\infty} \left(\frac{3}{2^n}\right)=x\\1 KB (193 words) - 21:13, 18 May 2021
- ==Overview=={{asy image|<math>1\,2\,3\,4\,5\,6\,7\,8\,9\,0</math>|right|The ten [[digit]]s making up <br />6 KB (902 words) - 12:53, 3 September 2019
- *Distinguished Honor Roll: Awarded to top 1% of scorers on each AMC 8, 10 and 12 respectively. *AIME qualifiers: 960 (1.5%)17 KB (1,921 words) - 20:53, 10 May 2024
- ...>b</math> if <math>a</math> is greater than <math>b</math>, that is, <math>a-b</math> is positive. ...<b</math> if <math>a</math> is smaller than <math>b</math>, that is, <math>a-b</math> is negative.12 KB (1,798 words) - 16:20, 14 March 2023
- ...Info|name=USAMTS|region=USA|type=Proof|difficulty=3-6|breakdown=<u>Problem 1-2</u>: 3-4<br><u>Problem 3-5</u>: 5-6}}4 KB (613 words) - 13:08, 18 July 2023
- ...ty=1 - 1.5|breakdown=<u>Problems 1 - 12</u>: 1<br><u>Problems 13 - 25</u>: 1.5}} The AMC 8 is usually administered on the third week of January. There is a 1-week window for students to take the test.4 KB (558 words) - 22:25, 28 April 2024
- ...gion=USA|type=Multiple Choice|difficulty=1-3|breakdown=<u>Problem 1-5</u>: 1<br><u>Problem 6-20</u>: 2<br><u>Problem 21-25</u>: 3}} ...incorrect questions are worth 0 points, and unanswered questions are worth 1.5 points, to give a total score out of 150 points. From 2002 to 2006, unan4 KB (574 words) - 15:28, 22 February 2024
- ...AMC 12|region=USA|type=Multiple Choice|difficulty=2-4|breakdown=<u>Problem 1-10</u>: 2<br><u>Problem 11-20</u>: 3<br><u>Problem 21-25</u>: 4}} ..., incorrect answers are worth 0 points, and unanswered questions are worth 1.5 points, to give a total score out of 150 points. From 2002 to 2006, the4 KB (520 words) - 12:11, 13 March 2024
- ...c}{3}+d+16=a+b+c+d</math>. This, with some algebra, means that <math>\frac{1}{3}(a+b+c)=8</math>. <math>d</math> must be <math>\boxed{\textbf{(B)} 21}</1 KB (200 words) - 23:35, 28 August 2020
- ...ame=AIME|region=USA|type=Free Response|difficulty=3-6|breakdown=<u>Problem 1-5</u>: 3<br><u>Problem 6-10</u>: 4<br><u>Problem 10-12</u>: 5<br><u>Problem The AIME is a 15 question, 3 hour exam<math>^1</math> taken by high scorers on the [[AMC 10]], [[AMC 12]], and [[USAMTS]]8 KB (1,057 words) - 12:02, 25 February 2024
- draw((-2/3,1/2)--(-sqrt(3)/6,1/2)--(0,0)--(sqrt(3)/6,1/2)--(2/3,1/2)); label("$\frac{1}{2}$",(.5,.25));3 KB (415 words) - 18:01, 24 May 2020
- ==Problem 1== [[2015 IMO Problems/Problem 1|Solution]]4 KB (692 words) - 22:33, 15 February 2021
- ...Info|name=USAMO|region=USA|type=Proof|difficulty=7-9|breakdown=<u>Problem 1/4</u>: 7<br><u>Problem 2/5</u>: 8<br><u>Problem 3/6</u>: 9}}6 KB (869 words) - 12:52, 20 February 2024
- ...dividual</u>: 4 (Problem 6/8), 6 (Problem 10)<br><u>Team</u>: 3.5 (Problem 1-5), 5 (Problem 6-10)}}2 KB (267 words) - 17:06, 7 March 2020
- ...The three teams are the Reals, Naturals, and Primes. Two compete in A and 1 in B... Michigan was also awarded the best shirt design at ARML Iowa 2015.. ...ille, etc.) on the MO ARML team. (eg. the 2008 team fielded 6 from KS, and 1 from IL)21 KB (3,500 words) - 18:41, 23 April 2024
- ....htm American Electroplaters and Surface Finishers Society Scholarship] of 1,500 dollars for undergraduate and graduate students in chemistry and some t1 KB (182 words) - 22:00, 4 February 2017
- ...ican Electroplaters and Surface Finishers Society Scholarship] of <dollar/>1,500 for undergraduate and graduate students in chemistry and some types of ....com/2020-usa-cargo-trailer-scholarship USA Cargo Trailer Scholarship] of $1,500 for students pursuing engineering.4 KB (511 words) - 14:57, 16 July 2020
- ...participate in the Finals round, and they have to solve three problems in 1 hour.2 KB (295 words) - 23:19, 5 January 2019
- *[[Algebra]] 1 *-1 for each incorrect question4 KB (632 words) - 17:09, 11 October 2020
- * [[AoPS Online School/Prealgebra 1 | Prealgebra 1]] — [https://artofproblemsolving.com/school/course/catalog/prealgebra1 De .../Introduction to Algebra A | Introduction to Algebra A]] (formerly Algebra 1) — [https://artofproblemsolving.com/school/course/catalog/algebra-a Detai8 KB (965 words) - 03:41, 17 September 2020
- ...th> positive real weights <math>w_i</math> with sum <math>\sum_{i=1}^n w_i=1</math>, the power mean with exponent <math>t</math>, where <math>t\in\mathb \prod_{i=1}^n a_i^{w_i} &\text{if } t=0 \\3 KB (606 words) - 23:59, 1 July 2022
- ...rem''' states that if [[integer ]]<math>p > 1</math> , then <math>(p-1)! + 1</math> is divisible by <math>p</math> if and only if <math>p</math> is prim ...p-1)! + 1</math>. Therefore <math>p</math> does not divide <math>(p-1)! + 1</math>.4 KB (639 words) - 01:53, 2 February 2023
- ...lutions <math>x,y,z</math> of the equation <math>x^2+5y^2+10z^2=4xy+6yz+2z-1</math>. *Show that <math>\sum_{k=1}^{n}a_k^2 \geq a_1a_2+a_2a_3+\cdots+a_{n-1}a_n+a_na_1</math>. [[Inequality_Introductory_Problem_2|Solution]]3 KB (560 words) - 22:51, 13 January 2024
- For example, if I wanted to find the average of the numbers 3, 1, 4, 1, and 5, I would compute: <center><math> \frac{3+1+4+1+5}{5} = \frac{14}{5}.</math></center>699 bytes (110 words) - 12:44, 20 September 2015
- What is the value of <math>x</math> if <math>x=1+\dfrac{1}{x}</math>2 KB (422 words) - 16:20, 5 March 2023
- <cmath>[ABC]=\frac{ab}{2}\sqrt{1-\cos^2 C}.</cmath> <math>[ABC]=\frac{ab}{2}\sqrt{1-\left(\frac{a^2+b^2-c^2}{2ab}\right)^2}.</math>4 KB (675 words) - 00:05, 22 January 2024
- * <math>x^{-1}+2+3x+x^2</math> * <math>x^{1/3}=\sqrt[3]{x}</math>6 KB (1,100 words) - 01:44, 17 January 2024
- ...h> are integer constants, and the coefficient of xy must be 1(If it is not 1, then divide the coefficient off of the equation.). According to Simon's Fa ...y)</math> that are solutions to the equation <math>\frac{4}{x}+\frac{5}{y}=1</math>. (2021 CEMC Galois #4b)7 KB (1,107 words) - 07:35, 26 March 2024
- The test is scored as 5 points for every correct response, 1 point for a blank response, and 0 points for an incorrect response.972 bytes (141 words) - 11:12, 30 September 2018
- <cmath>a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})</cmath> ...this creates the difference of squares factorization, <cmath>a^2-b^2=(a+b)(a-b)</cmath>3 KB (532 words) - 22:00, 13 January 2024
- ...e|primes]] <math>p_1, p_2,\ldots, p_n</math>. Let <math>N=p_1p_2p_3...p_n+1</math>. <math>N</math> is not divisible by any of the known primes since i2 KB (374 words) - 14:01, 21 August 2022
- ...e sum of the [[series]] <math>\frac11 + \frac14 + \frac19 + \cdots + \frac{1}{n^2} + \cdots</math><br> ...{\pi}\right)\left(1+\frac{x}{\pi}\right)\left(1-\frac{x}{2\pi}\right)\left(1+\frac{x}{2\pi}\right)\cdots</math><br>2 KB (314 words) - 06:45, 1 May 2014
- In [[combinatorics]], the '''pigeonhole principle''' states that if <math>n+1</math> or more pigeons are placed into <math>n</math> holes, one hole must ...ontain at most one ball implies that <math>b_r \leq 1</math> for all <math>1 \leq r \leq n</math>, so <cmath>b_1 + b_1 + \cdots + b_n \geq n.</cmath> Ho11 KB (1,985 words) - 21:03, 5 August 2023
- Next, we factor out our common terms to get <math>x(x-1)-2(x-1)=0</math>. ...<math>(x-1)(x-2)=0</math>. By the zero-product property, either <math> (x-1) </math> or <math> (x-2) </math> equals zero.2 KB (264 words) - 12:04, 15 July 2021
- ...r than 1. That is, their [[greatest common divisor]] is <math>\gcd(m, n) = 1</math>. Equivalently, <math>m</math> and <math>n</math> must have no [[pri ...r difference <math>(n+1)-n = 1</math>, which is impossible since <math>p > 1</math>.2 KB (245 words) - 15:51, 25 February 2020
- \textbf{(D) }\ 1 \qquad We can then try <math>(x^2+3x+1)=x^4+6x^3+11x^2+6x+1</math>.3 KB (571 words) - 00:42, 22 October 2021
- ...>\int \sqrt{a^2+x^2}\,dx</math>, we make use of the identity <math>\tan^2x+1=\sec^2x</math>. Set <math>x=a\tan\theta</math> and the radical will go awa Making use of the identity <math>\sin^2\theta+\cos^2\theta=1</math>, simply let <math>x=a\sin\theta</math>.1 KB (173 words) - 18:42, 30 May 2021
- ...eometric mean of the numbers 6, 4, 1 and 2 is <math>\sqrt[4]{6\cdot 4\cdot 1 \cdot 2} = \sqrt[4]{48} = 2\sqrt[4]{3}</math>. MC("\sqrt{ab}",D(A--M,orange+linewidth(1)),W);2 KB (282 words) - 22:04, 11 July 2008
- ...gain, <math>|A\cap C|</math> would be putting five guys in order, so <math>1!\binom{6}{5}=6</math>. <math>|A\cap D|</math> is just choosing <math>3</mat ...>, <math>|A\cap C\cap D|</math> is again ordering everybody which is <math>1</math>, and <math>|B\cap C\cap D|</math> is the same as <math>|A\cap B\cap9 KB (1,703 words) - 07:25, 24 March 2024
- Googol is a huge number. It has a 1 followed by 100 zeroes or <math>10^{100}.</math>84 bytes (14 words) - 10:46, 25 October 2020
- * <math>\binom{n-1}{r-1}+\binom{n-1}{r}=\binom{n}{r}</math> * <math>\binom{n}{r}=\frac{n}{r}\binom{n-1}{r-1}</math>4 KB (615 words) - 11:43, 21 May 2021
- ...row (a-1)(b-1)=2</math> from whence we have <math>(a,b,c)\in\{(2,3,1),(3,2,1)\}</math>. ...b\geq 6c</math> we have <math>6c\leq 2c+6\Rightarrow c\leq3/2\Rightarrow c=1</math>; a contradiction since <math>c\geq 2</math>.2 KB (332 words) - 09:37, 30 December 2021
- ...a constant <math>t</math> such that <math>a_i = t b_i</math> for all <math>1 \leq i \leq n</math>, or if one list consists of only zeroes. Along with th ...a^2b^2</math>. The inequality then follows from <math> |\cos\theta | \le 1 </math>, with equality when one of <math> \mathbf{a,b} </math> is a multipl13 KB (2,048 words) - 15:28, 22 February 2024
- ...[recursion|recursive definition]] for the factorial is <math>n!=n \cdot (n-1)!</math>. * <math>0! = 1</math> (remember! this is 1, not 0! (the '!' was an exclamation mark, not a factorial sign))10 KB (809 words) - 16:40, 17 March 2024
- ...ng about the roots of a given [[polynomial]] <math>p(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_0</math> of degree <math>n</math> with all the coefficients being re ...integer such that <math>k\geq\frac{n-2}{4}</math>, then there are <math>2k+1</math> pairs of complex conjugate roots and <math>n-4k+2</math> real roots.4 KB (734 words) - 19:19, 10 October 2023
- <cmath>\frac{PB}{CP} \cdot \frac{QC}{QA} \cdot \frac{AR}{RB} = 1.</cmath> ...;label("C",C,(1,0));label("P",P,(0,-1));label("Q",Q,(1,0));label("R",R,(-1,1));5 KB (804 words) - 03:01, 12 June 2023
- | [[South Carolina ARML]] (South Carolina #1)19 KB (2,632 words) - 14:31, 12 June 2022
- ...t [[divisibility|divisible]] by <math>{p}</math>, then <math>a^{p-1}\equiv 1 \pmod {p}</math>. ...denotes [[Euler's totient function]]. In particular, <math>\varphi(p) = p-1</math> for prime numbers <math>p</math>. In turn, this is a special case of16 KB (2,658 words) - 16:02, 8 May 2024
- xaxis(-9,9,Ticks(f, 1.0)); yaxis(-9,9,Ticks(f, 1.0));3 KB (551 words) - 16:22, 13 September 2023
- ...\sum_{i=1}^{n}a_ib_i\right)\geq\left(\sum_{i=1}^{n}a_i\right)\left(\sum_{i=1}^{n}b_i\right)</math>. <math> b_n\geq b_{n-1}\geq ... \geq b_1 </math> then:1 KB (214 words) - 20:32, 13 March 2022
- ...er [[relatively prime]] to <math>a</math>, then <math>{a}^{\phi (m)}\equiv 1 \pmod {m}</math>. ...hi(m)} \pmod{m} </math> <math> \implies </math> <math> a^{\phi (m)} \equiv 1 \pmod{m}</math> as desired. Note that dividing by <math> n_1 n_2 ... n_{\ph3 KB (542 words) - 17:45, 21 March 2023
- .../math> and [[perimeter]] <math>P</math>, then <math>\frac{4\pi A}{P^2} \le 1</math>. This means that given a perimeter <math>P</math> for a plane figure ...o <math>2\sin A \cos B<2\sin A</math>, which is equivalent to <math>\cos B<1</math>. Since this is always true for <math>0<B<180</math>, this inequality7 KB (1,296 words) - 14:22, 22 October 2023
- Substituting <math>\sin^2B=1-\cos^2B</math> results in <cmath>4[ABCD]^2=(1-\cos^2B)(ab+cd)^2=(ab+cd)^2-\cos^2B(ab+cd)^2</cmath>3 KB (465 words) - 18:31, 3 July 2023
- == Proof 1 == ...<math> ABCDEFG </math>, prove that: <math> \frac{1}{AB}=\frac{1}{AC}+\frac{1}{AE} </math>.7 KB (1,198 words) - 20:39, 9 March 2024
- ...mbers is the sum of all products of <math>k</math> of those numbers (<math>1 \leq k \leq n</math>). For example, if <math>n = 4</math>, and our set of ...)^nS_n</math>, and the coefficient of the <math>x^k</math> term is <math>(-1)^{n-k}S_{n-k}</math>, where the symmetric sums are taken over the roots of2 KB (275 words) - 12:51, 26 July 2023
- <cmath>z'=\frac{az+b}{cz+d},\quad a,b,c,d\in\mathbb{Z}, ad-bc=1.</cmath>5 KB (849 words) - 16:14, 18 May 2021
- <math> \sqrt{x+\sqrt{2x-1}} + \sqrt{x-\sqrt{2x-1}} = A,</math> given (a) <math>A=\sqrt{2}</math>, (b) <math>A=1</math>, (c) <math>A=2</math>, where only non-negative real numbers are admi3 KB (466 words) - 12:04, 12 April 2024
- ...> is defined to be: <math> \frac{n} {\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}}</math>. ..._1\cdot x_2 \cdots x_n}\ge \frac{n} {\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}} </math>1 KB (196 words) - 00:49, 6 January 2021
- ...(1-\frac{1}{p_1} \right) \left(1-\frac{1}{p_2} \right)\cdots \left(1-\frac{1}{p_m}\right).</cmath> ...fine the [[prime factorization]] of <math> n </math> as <math> n =\prod_{i=1}^{m}p_i^{e_i} =p_1^{e_1}p_2^{e_2}\cdots p_m^{e_m} </math> where the <math>p5 KB (898 words) - 19:12, 28 January 2024
- ..., <math>(\Omega, \mathfrak{a})</math>. <math>\mathit{P}:\mathfrak{a}\to [0,1]</math> is an assignment with certain properties (it is a special kind of [ <math>\mathit{P}(\{H, T\})=1</math>.4 KB (588 words) - 12:47, 2 October 2022
- <cmath>1+2+3+4+5+\sin \pi = \frac{5\cdot 6}{2}+0=15.</cmath>1 KB (164 words) - 19:09, 14 February 2024
- * The positive divisors of <math>35</math> are <math>1</math>, <math>5</math>, <math>7</math>, and <math>35</math>.2 KB (277 words) - 16:21, 29 April 2023
- ...in increasing order, then the maximum sum is just <math>-a_1b_k - a_2b_{k-1} + \ldots</math>. Thus, by negating all values the inequality follows.5 KB (804 words) - 13:54, 26 January 2023
- ...umber line. It can have any value. Some examples of real numbers are:<math>1, 2, -23.25, 0, \frac{\pi}{\phi}</math>, and so on. Numbers that are not rea (1) <math>\alpha\neq\mathbb{Q}</math> and <math>\alpha</math> is [[bounded]] a3 KB (496 words) - 23:22, 5 January 2022
- Rule 1: Partition <math>N</math> into 3 digit numbers from the right (<math>d_3d_ [[Divisibility rules/Rule 1 for 7 proof | Proof]]8 KB (1,315 words) - 18:18, 2 March 2024
- ...works for <math>n=1+1=2</math>, which in turn means it works for <math>n=2+1=3</math>, and so on. ...+1)}{2}</math> (the <math>n</math>th triangular number is defined as <math>1+2+\cdots +n</math>; imagine an [[equilateral polygon | equilateral]] [[tria5 KB (768 words) - 20:45, 1 September 2022
- ...wever, <cmath>\angle BIL = \angle BAI + \angle ABI = \frac{1}{2} A + \frac{1}{2} B.</cmath> Hence, <math>\triangle BIL</math> is isosceles, so <math>LB2 KB (291 words) - 16:31, 18 May 2021
- [[Image:Acute_orthic_triangle.png|thumb|right|300px|Case 1: <math>\triangle ABC</math> is acute.]] ...enter of <math>\triangle DEF</math>, <math>\angle EDC = 90^{\circ} - \frac{1}{2} \angle D</math>. Thus, <math>\angle ADC = 90^{\circ}</math>, and becaus8 KB (1,408 words) - 11:54, 8 December 2021
- {{asy image|<asy>draw((0,1)--(2,0)--(3,2)--cycle);</asy>|right|A triangle.}} {{asy image|<asy>draw((0,0)--(1,0)--(0.5,0.5)--cycle);</asy>|right|An isosceles triangle.}}4 KB (628 words) - 17:17, 17 May 2018
- ...use the recursive formula <math>GCD(a_1,\dots,a_n)=GCD(GCD(a_1,\dots,a_{n-1}),a_n)</math>.2 KB (288 words) - 22:40, 26 January 2021
- Here is MATHCOUNTS 2008 National Target #1: Try to solve this. ...According to the formula we get 2 handshakes, but wait, we will have only 1 handshake between two persons. That means we have overcounted somewhere.4 KB (635 words) - 12:19, 2 January 2022
- === Example 1 === ...first digit is, we know that it removes one option, so there are <math>8 - 1 = 7</math> options for the second digit.12 KB (1,896 words) - 23:55, 27 December 2023
- ...math> and let <math>a_1,\dots, a_n\ge 0</math> satisfy <math>a_1+\dots+a_n=1</math>. Then Let <math>\bar{x}=\sum_{i=1}^n a_ix_i</math>.3 KB (623 words) - 13:10, 20 February 2024
- === Example 1 === '''Case 1''': The word is one letter long. Clearly, there are <math>5</math> of these5 KB (709 words) - 10:28, 19 February 2024
- <math>r_{n-1} \pmod {r_n} \equiv 0</math><br> for <math>r_{k+1} < r_k < r_{k-1}</math><br>6 KB (924 words) - 21:50, 8 May 2022
- ...here the distance from a line (the directrix) is some number <math>0 < e < 1</math> times the distance to some fixed point (the focus). ...nce away from a point (focus) and a line (called the directrix) (<math>e = 1</math>).5 KB (891 words) - 01:14, 9 January 2023
- ...ath>A(k)={n \choose k}</math>, then we have <math>{n \choose 0}+{n \choose 1}x + {n \choose 2}x^2+\cdots+</math><math>{n \choose n}x^n</math>. .....+{n \choose n}=2^n</math>(let <math>{x}=1</math>), also <math>{n \choose 1}+{n \choose 3}+\cdots={n \choose 0}+{n \choose 2}+\cdots</math>.4 KB (659 words) - 12:54, 7 March 2022
- ...ath>, with coefficients <math>1 = \binom{5}{0}</math>, <math>5 = \binom{5}{1}</math>, <math>10 = \binom{5}{2}</math>, etc. ...(a+b)^{n} = \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}</math>, <cmath>(a+b)^{n+1} = (\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k})(a+b)</cmath>5 KB (935 words) - 13:11, 20 February 2024
- ...integer]] <math>p>1</math> whose only positive [[divisor | divisors]] are 1 and itself. Note that <math>1</math> is usually defined as being neither prime nor [[composite number|com6 KB (985 words) - 12:38, 25 February 2024
- ...early values of the sequence in terms of previous values: <math>F_0=1, F_1=1, F_2=2, F_3=3, F_4=5, F_5=8</math>, and so on. ...defined recursively by <math>a_0 = 1</math> and <math>a_n = 2\cdot a_{n - 1}</math> for <math>n > 0</math> also has the closed-form definition <math>a_2 KB (316 words) - 16:03, 1 January 2024
- ...exactly one value in the second. For instance, one function may map 1 to 1, 2 to 4, 3 to 9, 4 to 16, and so on. This function has the rule that it ta ...'.) Often the inverse of a function <math>f</math> is denoted by <math>f^{-1}</math>.10 KB (1,761 words) - 03:16, 12 May 2023
- ...bility it does not happen is one. Thus, we have the identity <cmath>P(A) = 1 - P(A^c).</cmath> Like its counting analog, complementary probability often === Example 1 ===8 KB (1,192 words) - 17:20, 16 June 2023
- <math>{n \choose k}={n-1\choose k-1}+{n-1\choose k}</math> ...s added to the number of ways to choose <math>k</math> things from <math>n-1</math> things.12 KB (1,996 words) - 12:01, 18 May 2024
- ...ath> are integers, then their sum <math>a+b</math>, their difference <math>a-b</math>, and their product <math>ab</math> are all integers (that is, the i ...</math> bits), which limits their maximum value (typically to <math>2^{31}-1</math> for signed <math>32</math>-bit integers). Integers in Python can be2 KB (296 words) - 15:04, 5 August 2022
- * [[2000 AIME I Problems/Problem 1]]3 KB (496 words) - 22:14, 5 January 2024
- ...''' is a [[positive integer]] with at least one [[divisor]] different from 1 and itself. Some composite numbers are <math>4=2^2</math> and <math>12=2\t Every positive integer either is prime, composite, or 1.6 KB (350 words) - 12:58, 26 September 2023
- A '''natural number''' is any positive [[integer]]: <math>\text{1, 2, 3, 4, 5, 6, 7,\dots}</math>. The set of '''natural numbers''', denoted1 KB (162 words) - 21:44, 13 March 2022
- '''Case 1:''' The circle's area is greater than the triangle's area. ...han the radius so <math>a<r</math>. Therefore <math> P=\frac{1}{2}ap<\frac{1}{2}r\cdot 2\pi r=T</math>. However it cannot be both <math>P>T</math> and <9 KB (1,581 words) - 18:59, 9 May 2024
- MC(90,"\mbox{semiminor axis}",7,D((0,0)--(0,3),green+linewidth(1)),E); MC("\mbox{semimajor axis}",7,D((0,0)--(5,0),red+linewidth(1)),S);5 KB (892 words) - 21:52, 1 May 2021
- ...ath>d(n)=(\alpha_{1} + 1)\cdot(\alpha_{2} + 1)\cdot\dots\cdot(\alpha_{m} + 1)</math>. It is often useful to know that this expression grows slower than * <math>{\sum_{n=1}^N d(n)=\left\lfloor\frac N1\right\rfloor+\left\lfloor\frac N2\right\rfloor1 KB (274 words) - 19:50, 29 August 2023
- ...number can be rewritten as <math>2746_{10}=2\cdot10^3+7\cdot10^2+4\cdot10^1+6\cdot10^0.</math> ...six <math>10^0</math>'s, the second digit tells us there are four <math>10^1</math>'s, the third digit tells us there are seven <math>10^2</math>'s, and4 KB (547 words) - 17:23, 30 December 2020
- ...ong computer programmers. It has just two digits: <math>0</math> and <math>1</math>. [[Hexadecimal]] is base 16. The digits in hexadecimal are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, and F. One of its common uses is f2 KB (351 words) - 10:39, 1 October 2015
- ...thcal{P}</math> is defined as <math>\|\mathcal{P}\|=\sup\{x_i-x_{i-1}\}_{i=1}^n</math> ...he set of ordered pairs <math>\mathcal{\dot{P}}=\{([x_{i-1},x_i],t_i)\}_{i=1}^n</math>.1 KB (178 words) - 20:34, 6 March 2022
- Consider the function <math>f:[0,1]\rightarrow\mathbb{R}</math> <math>f\left( \frac{1}{n}\right) =n\forall n\in\mathbb{N}</math>2 KB (401 words) - 09:46, 31 January 2018
- {{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #1]] and [[2001 AMC 10 Problems|2001 AMC 10 #3]]}}788 bytes (120 words) - 10:32, 8 November 2021
- {{AMC12 box|year=2001|num-b=1|num-a=3}}1,007 bytes (165 words) - 00:28, 30 December 2023
- A = (-1, 1); B = (1, 1);1 KB (169 words) - 01:12, 13 June 2022
- ...ngaroo is an international mathematical competition for students in grades 1 through 12. The competition consists of a single round that is taken on th Any student that is currently enrolled in grades 1 through 12 (or homeschooled equivalent) is eligible to participate. They m6 KB (936 words) - 15:38, 22 February 2024
- Contest #1 - October 10, 20191 KB (153 words) - 13:11, 14 May 2019
- <math>12 = 2^2\cdot 3^1</math> <math>15 = 3^1\cdot 5^1</math>2 KB (383 words) - 10:49, 4 September 2022
- ...llet}&&x^{2n+1}+y^{2n+1}&=(x+y)(x^{2n}-x^{2n-1}y+x^{2n-2}y^2-\ldots-xy^{2n-1}+y^{2n})\\ \text{\textbullet}&&x^{n}-y^{n}&=(x-y)(x^{n-1}+x^{n-2}y+\cdots +xy^{n-2}+y^{n-1})2 KB (327 words) - 02:06, 28 April 2024
- ...evel I</u>: 0.5 - 1<br><u>Level II</u>: 0.5 - 1.5<br><u>Level III</u>: 1 - 1.5}} * Level I: For grades 1 and 2.1 KB (197 words) - 10:59, 14 April 2024
- == Individual Round - Part 1 == * 1 point is awarded for each question left blank4 KB (644 words) - 12:56, 29 March 2017
- BD = \frac{BA \cdot DC }{AP} \; (1) ...both sides of the inequality by <math>BD</math> and using equations <math>(1) </math> and <math>(2) </math> gives us3 KB (602 words) - 09:01, 7 June 2023
- A = (1, 2); draw(B--M, StickIntervalMarker(1));1 KB (185 words) - 20:24, 6 March 2024
- ...e most surprising places, such as in the sum <math>\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}</math>. Some common [[fraction]]al approximations fo ...rmula for pi is <math>4\left( \sum_{i = 0}^\infty (-1)^i \left(\frac{1}{2n+1}\right)\right) </math>. This can be computed to the desired degree of accur8 KB (1,469 words) - 21:11, 16 September 2022
- ...ently in mathematical writing, often to represent the constant <math>\frac{1+\sqrt{5}}{2}</math>. (The Greek letter [[Tau]] (<math>\tau</math>) was also ...as well as the positive solution of the [[quadratic equation]] <math>x^2-x-1=0</math>.2 KB (302 words) - 14:04, 1 January 2024
- ...term is the sum of the two preceding it. The first few terms are <math>1, 1, 2, 3, 5, 8, 13, 21, 34, 55,...</math>. ...[[recursion|recursively]] as <math>F_1 = F_2 = 1</math> and <math>F_n=F_{n-1}+F_{n-2}</math> for <math>n \geq 3</math>. This is the simplest nontrivial6 KB (957 words) - 23:49, 7 March 2024
- ...sequence <math>(5,1)</math> majorizes <math>(4,2)</math> (as <math>5>4, 5+1=4+2</math>), Muirhead's inequality states that for any positive <math>x,y</ <cmath>x^5y^1+y^5x^1\geq x^4y^2+y^4x^2</cmath>8 KB (1,346 words) - 12:53, 8 October 2023
- ...y of <math>\{3, 4\}</math> is 2, the cardinality of <math>\{1, \{2, 3\}, \{1, 2, 3\}\}</math> is 3, and the cardinality of the [[empty set]] is 0.2 KB (263 words) - 00:54, 17 November 2019
- ...thbb R</math>, <math>f(x)=x^2</math> is not an injection (<math>f(-1)=f(1)=1</math>), the function <math>g:[0,\infty)\to\mathbb R</math>, <math>g(x)=x^21 KB (228 words) - 01:01, 17 November 2019
- ...ve.) Interpret the distance that the object travels between times <math>t=1</math> and <math>t=2</math> geometrically, as an area under a curve. ...is the amount that <math>F</math> changes on the interval <math>[x_i, x_{i+1}]</math>, then <math>\Delta F_i \approx F'(x_i)\Delta x</math>.11 KB (2,082 words) - 15:23, 2 January 2022
- ...-- it will have "degenerated" from an <math>n</math>-gon to an <math>(n - 1)</math>-gon. (In the case of triangles, this will result in either a line2 KB (372 words) - 19:04, 30 May 2015
- ...h>A</math> and perimeter <math>P</math> then <math>\frac{4\pi A}{P^2} \leq 1</math>. This means that given a perimeter <math>P</math> for a plane figure789 bytes (115 words) - 17:08, 29 December 2021
- ...s://seasonsbeachcottage.com/mental-health-education-scholarship-2023-round-1 Mental health Education Annual Scholarship 2023] Applicants must be current ...g Scholarship] Zampi, Inc. A small business marketing platform is offering 1,000 dollar scholarship. To enter, simple create a small business marketing7 KB (1,039 words) - 18:45, 18 January 2024
- * [[Best Buy @15 Scholarship]] of <dollar/>1,000 for students in grades 9-12. [https://www.at15.com/contests_scholarship4 KB (538 words) - 00:48, 28 January 2024
- A pyramid has a square base with sides of length <math>1</math> and has lateral faces that are equilateral triangles. A cube is plac == Solution 1 ==4 KB (691 words) - 18:38, 19 September 2021
- == Solution 1==1 KB (249 words) - 13:05, 24 January 2024
- <math>[ABCD]=\frac{1}{4} \cdot \sqrt{4p^2q^2-(b^2+d^2-a^2-c^2)^2}</math>. ...vec{c} - \vec{d}</math>. The area of any such quadrilateral is <math>\frac{1}{2} |\vec{p} \times \vec{q}|</math>.3 KB (566 words) - 03:51, 12 February 2021
- pen p=linewidth(1); MA("\theta",(5,-1),(2,3),(4,6),0.3,9,yellow);7 KB (1,265 words) - 13:22, 14 July 2021
- ...bers. ex. :<cmath>((((((3^5)^6)^7)^8)^9)^{10})^{11}=\underbrace{1177\ldots 1}_{\text{793549 digits}}</cmath> would be a pain to have to calculate any ti <cmath>\log_4(5)\approx 1.160964047443681173935159715\ldots</cmath>4 KB (680 words) - 12:54, 16 October 2023
- ==Proof 1== label("A",A,(1,0));6 KB (1,003 words) - 09:11, 7 June 2023
- <cmath>a^r(a-b)(a-c)+b^r(b-a)(b-c)+c^r(c-a)(c-b) \geq 0</cmath> The <math>r=1</math> case yields the well-known inequality:2 KB (398 words) - 16:57, 29 December 2021
- * <math>\sin^2x + \cos^2x = 1</math> * <math>1 + \cot^2x = \csc^2x</math>8 KB (1,397 words) - 21:55, 20 January 2024
- ...>\sqrt{2}</math> as the length of a diagonal of a square with side lengths 1 in the sixth century <math>B.C</math>. The Pythagoreans lived by the doctri <math>3\frac{10}{71}</math> <math>to</math> <math>3\frac{1}{7}</math>.3 KB (368 words) - 19:26, 6 June 2015
- ...numbers''' arise when we try to solve [[equation]]s such as <math> x^2 = -1 </math>. ...this addition, we are not only able to find the solutions of <math> x^2 = -1 </math> but we can now find ''all'' solutions to ''every'' polynomial. (Se5 KB (860 words) - 15:36, 10 December 2023
- ...ecause every integer <math>a</math> can be represented as <math>a=\frac{a}{1}</math>1 KB (207 words) - 15:51, 25 August 2022
- ...the [[complex number|complex]] [[root]]s of the [[polynomial]] <math> x^n=1 </math>. ...ik}</math>. The magnitude of the RHS is 1, making <math>r^n=1\Rightarrow r=1</math> (magnitude is always expressed as a positive real number). This lea3 KB (558 words) - 21:36, 11 December 2011
- markangle(n=1,radius=20,D,A,F,green); markangle(n=1,radius=22,F,A,B,green);3 KB (575 words) - 15:27, 19 March 2023
- * [[2005_Austrian_Mathematical_Olympiad_Final_Round-Part 1/Problem 5]]2 KB (280 words) - 15:30, 22 February 2024
- === Method 1 === real r = 1;4 KB (658 words) - 16:19, 28 April 2024
- ...<math>\mathbb{N}</math> corresponds to the sequence <math>X = (x_n) = (0, 1, 4, 9, 16, \ldots)</math>. A classic example of convergence would be to show that <math>1/n\to 0</math> as <math>n\to \infty</math>.2 KB (413 words) - 21:18, 13 November 2022
- ...ratio <math>-1/2</math>; however, <math>1, 3, 9, -27</math> and <math>-3, 1, 5, 9, \ldots</math> are not geometric sequences, as the ratio between cons ...ogression if and only if <math>a_2 / a_1 = a_3 / a_2 = \cdots = a_n / a_{n-1}</math>. A similar definition holds for infinite geometric sequences. It ap4 KB (644 words) - 12:55, 7 March 2022
- ...h>1, 2, 3, 4</math> is an arithmetic sequence with common difference <math>1</math> and <math>99, 91, 83, 75</math> is an arithmetic sequence with commo ...ogression if and only if <math>a_2 - a_1 = a_3 - a_2 = \cdots = a_n - a_{n-1}</math>. A similar definition holds for infinite arithmetic sequences. It a4 KB (736 words) - 02:00, 7 March 2024
- ...n unique factorization in [[ring]]s of the form <math>\mathbb{Z}[\sqrt[n]{-1}]</math>. Unfortunately, this is not often the case. In fact, it has now3 KB (453 words) - 11:13, 9 June 2023
- * A homothety with factor <math>-1</math> is a <math>180^\circ</math> rotation about the center.3 KB (532 words) - 01:11, 11 January 2021
- ...a piece of length <math>k_i</math> from the end of leg <math>L_i \; (i = 1,2,3,4)</math> and still have a stable table? == Solution 1 ==7 KB (1,276 words) - 20:51, 6 January 2024
- ...0 AMC 12 Problems|2000 AMC 12 #1]] and [[2000 AMC 10 Problems|2000 AMC 10 #1]]}} == Solution 1 (Verifying the Statement)==2 KB (276 words) - 05:25, 9 December 2023
- ...there will always be an infinite number of solutions when <math>\gcd(a,b)=1</math>. If <math>\gcd(a,b)\nmid c</math>, then there are no solutions to t .../math> is an [[odd]] number, then <math>m, \frac {m^2 -1}{2}, \frac {m^2 + 1}{2}</math> is a Pythagorean triple.9 KB (1,434 words) - 13:10, 20 February 2024
- ...ices to the [[circumcenter]]. This creates a triangle that is <math>\frac{1}{n},</math> of the total area (consider the regular [[octagon]] below as an ..._b^{-1})(H-h_c^{-1})}</math>, where <math>H=\frac{(h_a^{-1}+h_b^{-1}+h_c^{-1})}{2}</math> and the triangle has altitudes <math>h_a</math>, <math>h_b</ma6 KB (1,181 words) - 22:37, 22 January 2023
- ...the denominator such as <math>\frac{3}{3}</math> is always equal to <math>1</math>. ...d denominator by the same quantity (since we're essentially multiplying by 1).3 KB (432 words) - 19:34, 11 June 2020
- ...'.) Often the inverse of a function <math>f</math> is denoted by <math>f^{-1}</math>. ...le of such a function is <math>f(x) = 1/x</math> because <math>f(f(x)) = f(1/x) = x</math>. Cyclic functions can significantly help in solving function2 KB (361 words) - 14:40, 24 August 2021
- A=(0,1); C=(1,0);1 KB (238 words) - 22:51, 20 February 2022
- ...loor+\left\lfloor a+\frac{1}{n}\right\rfloor+\ldots+\left\lfloor a+\frac{n-1}{n}\right\rfloor</cmath> ...te the largest integer not exceeding <math>x</math>. For example, <math>[2.1]=2</math>, <math>[4]=4</math> and <math>[5.7]=5</math>. How many positive i3 KB (508 words) - 21:05, 26 February 2024
- ...sum of the two numbers appearing above it. For example, <math>{5 \choose 1}+{5 \choose 2} = 5 + 10 = 15 = {6 \choose 2}</math>. This property allows ...math>{6 \choose 0}+{5 \choose 1}+{4 \choose 2}+{3 \choose 3} = 1 + 5 + 6 + 1 = 13 = F(7)</math>. A "shallow diagonal" is plotted in the diagram.5 KB (838 words) - 17:20, 3 January 2023
- <center><math> P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0</math></center> <cmath>a_nP_1 + a_{n-1} = 0</cmath>4 KB (690 words) - 13:11, 20 February 2024
- ...words, we want to divide by 2 four times. Therefore, <math> 2^{-4}=\frac 1{2^4}.</math> * <math>b^0 = 1</math> if <math>b \neq 0</math>5 KB (803 words) - 16:25, 10 August 2020
- ...en denominator <math>q\ge 1</math> with an error not exceeding <math>\frac 1{2q}</math>. ...interval <math>\left[qx-\frac12,qx+\frac12\right]</math> has length <math>1</math> and, therefore, contains at least one integer. Choosing <math>p</mat7 KB (1,290 words) - 12:18, 30 May 2019
- ...d as the reciprocal of the sine of <math>A</math>. <cmath>\csc (A) = \frac{1}{\sin (x)} = \frac{\textrm{hypotenuse}}{\textrm{opposite}} = \frac{c}{a}.</ ...as the reciprocal of the cosine of <math>A</math>. <cmath>\sec (A) = \frac{1}{\cos (x)} = \frac{\textrm{hypotenuse}}{\textrm{adjacent}} = \frac{c}{b}.</8 KB (1,217 words) - 20:15, 7 September 2023
- ...>e</math> such that <math>\ln e=1</math>, where <math>\ln x=\int_1^x \frac{1}{t} \, dt</math>. It has been shown to be both [[irrational]] and [[transce ...tion such that <math>\frac{d}{dx} \exp(x)=\exp(x)</math> and <math>\exp(0)=1</math> is equal to <math>e^x</math>.4 KB (764 words) - 21:09, 13 March 2022
- :<math>f\left( x \right) = a\left( {{1 \over 2}} \right)^{{x \over h}} :If <math>b > 1</math>, then the function will show growth.2 KB (312 words) - 15:57, 6 March 2022
- ...three instructional sessions: 8:30 AM - 10:00 AM, 10:15 AM - 11:45 AM, and 1:15 PM - 2:45 PM. Classes usually consist of a lecture followed by a problem ...ered in place of the afternoon lecture, and is graded with comments within 1-2 days.6 KB (936 words) - 10:37, 27 November 2023
- ...h> f^{(n)}(z_0)=\frac{n!}{2\pi i} \oint_\Gamma \frac{f(z)\; dz}{(z-z_0)^{n+1}}. </cmath>2 KB (271 words) - 22:06, 12 April 2022
- ...cdots+x_a^{n_4}}{a}}</cmath> where <math>n_1>1,~~0<n_2<1,~~-1<n_3<0,~~n_4<-1</math>, and <math>n</math> is the root mean power. ...s greater than or equal to 1. This creates the indeterminate form of <math>1^{\infty}</math>. Then, we can say that the limit as x goes to 0, the result5 KB (912 words) - 20:06, 14 March 2023
- <math>\sum_{i=1}^{\infty}\frac{1}{i}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots</math> <math>\sum_{i=1}^{\infty}\frac{(-1)^{i+1}}{i}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots</math> , though, approaches <math>\ln 2</math>.2 KB (334 words) - 20:52, 13 March 2022
- He considered 1 to be a prime number, a [[mathematical convention|convention]] subsequently ...m</math> and <math>n-m</math> simultaneously being prime to be <math>\frac{1}{\ln m \cdot \ln (n-m)}</math>. This heuristic is non-rigorous for a numbe7 KB (1,201 words) - 16:59, 19 February 2024
- ...1}{5}+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+\frac{1}{13}+\frac{1}{17}+\frac{1}{19}+\cdots</math></center> ...y if one of variables is 0 will the factorization be trivial (contain only 1 and itself).2 KB (308 words) - 02:27, 1 May 2024
- === 2 = 1 === <div style="text-align:center"><math> 2 = 1 </math> (dividing by <math>a^2-ab</math>)</div>2 KB (429 words) - 08:27, 5 June 2013
- ...[[root |zero]]s of the [[Riemann zeta function]] have [[real part]] <math>1/2</math>. From the [[functional equation]] for the zeta function, it is eas ...atement of the Riemann hypothesis is that <math>\pi(x)=\mathrm{Li}(x)+O(x^{1/2}\ln(x))</math>.2 KB (425 words) - 12:01, 20 October 2016
- ...\mathrm{ and }\ a\ \mathrm{\ is\ a\ quadratic\ residue\ modulo\ }\ p, \\ -1 & \mathrm{if }\ p\nmid a\ \mathrm{ and }\ a\ \mathrm{\ is\ a\ quadratic\ no ...left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{\frac{p-1}{2}\frac{q-1}{2}}</math>. This is known as the [[Quadratic Reciprocity Theorem]].5 KB (778 words) - 13:10, 29 November 2017
- ''Proof.'' We have already proven the theorem for a <math>1</math>-sphere (i.e., a circle), so it only remains to prove the theorem for ==== Problem 1 ====5 KB (827 words) - 17:30, 21 February 2024
- ...loss of generality, suppose that <math>P</math> is [[monic]]. Then <math>1/P(z)</math> is an [[entire]] function; we wish to show that it is bounded. ...absolute values of the coefficients of <math>P</math>, so that <math>R \ge 1</math>. Then for <math>\lvert z \rvert \ge R</math>,5 KB (832 words) - 14:22, 11 January 2024
- ...>, the sequence <math>\{n,f(n),f(f(n)),f(f(f(n))),\ldots\}</math> contains 1. This conjecture is still open. Some people have described it as the easies Observing that if <math>n=2m+1</math> then <math>3n+1=6m+4</math>, as well as: <cmath>{6m+4\over 2}=3m+2</cmath> we can then obse1 KB (231 words) - 19:45, 24 February 2020
- ...ynomial]] with [[rational]] [[coefficient]]s. Examples include <math>\frac{1}{3}</math>, <math>\sqrt{2}+\sqrt{3}</math>, <math>i</math>, and <math>\frac1,006 bytes (151 words) - 21:56, 22 April 2022
- D=intersectionpoint(A+0.1*expi((angle(B-A)+angle(C-A))/2)--A+20*expi((angle(B-A)+angle(C-A))/2), circ dot(A^^B^^C^^D^^I);label("A",A,(0,1));label("B",B,(-1,0));label("C",C,(1,0));2 KB (381 words) - 19:38, 24 November 2011
- ...+ \ldots + a_n = \left(\sum_{j=1}^n a_j\right) - a_i</math> for <math>i = 1, 2 \ldots, n</math>. Expressing the inequality in this form leads to <math2 KB (268 words) - 03:02, 3 January 2021
- ...for all <math> 1 \le k \le n </math>, <math> \sum_{i=1}^{k}a_i \ge \sum_{i=1}^{k}b_i </math>, with equality when <math>k = n </math>. If <math>\{a_i\} ...h>\sum_{i=k}^n a_i \le \sum_{i=k}^n b_i</math>, with equality when <math>k=1 </math>. An interesting consequence of this is that the finite sequence <m2 KB (288 words) - 22:48, 5 July 2023
- ...function <math>f: \mathbb N \to\mathbb N</math> defined by <math>f(x) = x+1</math> is not surjective because there exists a [[natural number]] which is794 bytes (131 words) - 22:39, 13 May 2020
- ...|region=International|type=Proof|difficulty=5.5 - 10|breakdown=<u>Problem 1/4</u>: 6.5<br><u>Problem 2/5</u>: 7.5-8<br><u>Problem 3/6</u>: 9.5<br><u>Pr * Gold - the top 1/12 of individual scores.3 KB (490 words) - 03:32, 23 July 2023
- <cmath> \lim_{x\to \infty} \frac{\pi(x) \log x}{x} = 1 . </cmath> with <math>A(x)</math> tending to some constant number around 1.08366.10 KB (1,729 words) - 19:52, 21 October 2023
- ...a_a, \lambda_b, \dotsc, \lambda_z</math> are nonnegative reals with sum of 1, then ...f{a}</math> and <math>\mathbf{b}</math>, and <math>\lambda_a = \lambda_b = 1/2</math>, this is the elementary form of the [[Cauchy-Schwarz Inequality]].4 KB (774 words) - 12:12, 29 October 2016
- ...ot empty and cannot be put into bijection with any set of the form <math>\{1, 2, \ldots, n\}</math> for a [[positive integer]] <math>n</math>. <math>\sum_{i = 3}^{\infty}{(2i - 1)}</math>1 KB (186 words) - 23:19, 16 August 2013
- {{asy image|<asy>draw((0,1)--(1,5)--(3,5)--(4,1)--cycle);</asy>|right|An isosceles trapezoid.}}577 bytes (81 words) - 10:33, 18 April 2019
- ! scope="row" | '''Mock AMC #1''' | [http://www.artofproblemsolving.com/Forum/viewtopic.php?t=9572 1-5]51 KB (6,175 words) - 20:58, 6 December 2023
- ...Y2QwOTc3NWZiYjY0LnBkZg==&rn=TWlsZG9yZiBNb2NrIEFJTUUucGRm Mildorf Mock AIME 1] ** [[Mock AIME 1 2006-2007]]8 KB (906 words) - 17:30, 26 April 2024
- ...n <math>AB</math> and <math>BA</math> are matrices of sizes <math>1 \times 1</math> and <math>2 \times 2</math>, respectively.2 KB (301 words) - 17:46, 16 March 2012
- ...ath>. We often drop the brackets and commas, so the permutation <math>\{2,1,3\}</math> would just be represented by <math>213</math>. ...ent objects we can choose from. For the second element, there are <math>n-1</math> objects we can choose, <math>n-2</math> for the third, and so on. I3 KB (422 words) - 11:01, 25 December 2020
- ...ith <math>g(a,0)=p(a)</math>, <math>g(a,1)=q(a)</math>, and <math>g(0,b)=g(1,b)=x</math>. We call <math>g</math> a [[homotopy]]. Now define <math>\pi_1( ...>g\cdot h(a)=\begin{cases} g(2a) & 0\le a\le 1/2, \\ h(2a-1) & 1/2\le a\le 1.\end{cases}</math> One can check that this is indeed [[well-defined]].3 KB (479 words) - 15:35, 1 December 2015
- <math>\frac{2^{2n}}{(2n+1)}\le{\binom{2n}{n}}\le <math>\left(\prod_{n<p\le{2n}}p\right)\ge \frac{4^{\frac n3}}{(2n+1)(2n)^{\sqrt {2n}}}</math>2 KB (309 words) - 21:43, 11 January 2010
- <cmath>\zeta (s)=\sum_{n=1}^{\infty}\frac{1}{n^s}= 1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\cdots</cmath>9 KB (1,547 words) - 03:04, 13 January 2021
- ...ww.artofproblemsolving.com/Forum/viewtopic.php?p=423926#p423926 Mock USAMO 1 2006] ...http://www.artofproblemsolving.com/Forum/viewtopic.php?t=200572 Mock USAMO 1 2008]2 KB (205 words) - 19:56, 4 March 2020
- <math>1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, \ldots </math> This is the way in which we count in '''modulo 12'''. When we add <math>1</math> to <math>11</math>, we arrive back at <math>0</math>. The same is t15 KB (2,396 words) - 20:24, 21 February 2024
- ...th> so that <math>m^2=n</math>. The first few perfect squares are <math>0, 1, 4, 9, 16, 25, 36</math>. ...th> square numbers (starting with <math>1</math>) is <math>\frac{n(n+1)(2n+1)}{6}</math>954 bytes (155 words) - 01:14, 29 November 2023
- ...Equivalently, <math>f </math> is convex if for every <math> \lambda \in [0,1] </math> and every <math> x,y \in I</math>, <center><math>\lambda f(x) + (1-\lambda)f(y) \ge f\left( \lambda x + (1-\lambda) y \right) </math>.</center>2 KB (417 words) - 00:10, 20 February 2016
- Suppose that the set <math>A=\{x\in\mathbb{R}:0<x< 1\}</math> is countable. Let <math>\{\omega_1, \omega_2, \omega_3, ...\}</mat ...ath>. However, <math>\omega</math> is clearly a real number between 0 and 1, a [[contradiction]]. Thus our assumption that <math>A</math> is countable2 KB (403 words) - 20:53, 13 October 2019
- ...<math>\Re(z)>0</math>, we define <cmath>\Gamma(z)=\int_0^\infty e^{-t}t^{z-1}\; dt</cmath> ...f by one. Since <math>\Gamma(1)=1</math>, we therefore have <math>\Gamma(n+1)=n!</math> for nonnegative integers <math>n</math>. But with the integral,840 bytes (137 words) - 22:26, 22 June 2009
- * Subtraction: <math>a-c\equiv b-d\pmod {n}</math>. <math>\{ \ldots, -5, -2, 1, 4, 7, \ldots \}</math>14 KB (2,317 words) - 19:01, 29 October 2021
- A = (0, 1); B = (1, 0);3 KB (499 words) - 23:41, 11 June 2022
- ...thbb{C} \subset \mathbb{C}\cup\{\textrm{Groucho, Harpo, Chico}\} \supset \{1, 2, i, \textrm{Groucho}\}</math>1 KB (217 words) - 09:32, 13 August 2011
- * [[2006 AMC 10A Problems/Problem 1]]2 KB (180 words) - 18:06, 6 October 2014
- ...the value of the function <math>f(x)</math> at <math>x = 0</math> is <math>1</math>, the limit <math>\lim_{x\rightarrow 0} f(x)</math> is, in fact, zero ...ons <math>d_A(a,b)</math> and <math>d_B(a,b)</math> are both simply <math>|a-b|</math>. We then obtain the following definition commonly found in calculu7 KB (1,325 words) - 13:51, 1 June 2015
- == Problem 1 == [[2006 AIME I Problems/Problem 1|Solution]]7 KB (1,173 words) - 03:31, 4 January 2023
- ...</math> and <math> |x_k|=|x_{k-1}+3| </math> for all integers <math> k\ge 1, </math> find the minimum possible value of <math> |x_1+x_2+\cdots+x_{2006} === Solution 1 ===6 KB (910 words) - 19:31, 24 October 2023
- ...s <math>4</math> feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let <math> h </math> be the height in feet of t == Solution 1 ==6 KB (980 words) - 21:45, 31 March 2020
- ...or each positive integer <math> n, </math> let <math> S_n=\sum_{k=1}^{2^{n-1}}g(2k). </math> Find the greatest integer <math> n </math> less than 1000 s == Solution 1 ==10 KB (1,702 words) - 00:45, 16 November 2023
- ...e with [[edge]]-[[length]] <math> k </math> for each [[integer]] <math> k, 1 \le k \le 8. </math> A tower is to be built using all 8 cubes according to ...th>, so there are <math>2\cdot 3^6 = 1458</math> towers using blocks <math>1, 2, \ldots, 8</math>, so the answer is <math>\boxed{458}</math>.3 KB (436 words) - 05:40, 4 November 2022
- ...math> c </math> are positive integers whose [[greatest common divisor]] is 1. Find <math> a^2+b^2+c^2. </math> for(i=0; i<3; i=i+1) {4 KB (731 words) - 17:59, 4 January 2022
- == Solution 1 == ...s <math>1 + \frac{1001-11}{22}=1 + \frac{990}{22}=46</math>. (We must add 1 because both endpoints are being included.) So the answer is <math>\boxed{04 KB (651 words) - 18:27, 22 May 2021
- ...1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1,0), J=(2.1,-3.2), K=(2.1,-1.6); draw(F--G--(2.1,0));5 KB (730 words) - 15:05, 15 January 2024
- for(int i=0; i<4; i=i+1) { fill((2*i,0)--(2*i+1,0)--(2*i+1,6)--(2*i,6)--cycle, mediumgray);4 KB (709 words) - 01:50, 10 January 2022
- == Solution 1 == ...ple, <math>.\overline{123}+.\overline{876}=.\overline{999} \Longrightarrow 1</math>.2 KB (237 words) - 19:14, 20 November 2023
- == Solution 1 ==3 KB (439 words) - 18:24, 10 March 2015
- ...th>'s at the right end of the decimal representation of the product <math> 1!2!3!4!\cdots99!100!. </math> Find the remainder when <math> N </math> is di ...rom <math>100</math>. Thus, our final total is <math>970 + 76 + 51 + 26 + 1 = 1124</math>, and the answer is <math>\boxed{124}</math>.2 KB (278 words) - 08:33, 4 November 2022
- ...en its leftmost [[digit]] is deleted, the resulting integer is <math>\frac{1}{29}</math> of the original integer. === Solution 1 ===4 KB (622 words) - 03:53, 10 December 2022
- ...et]] <math> \mathcal{A} </math> be a 90-[[element]] [[subset]] of <math> \{1,2,3,\ldots,100\}, </math> and let <math> S </math> be the sum of the elemen ...ible values of S, so the number of possible values of S is <math>4995-4095+1=901</math>.1 KB (189 words) - 20:05, 4 July 2013
- ==Solution 1 (Linear Polynomials)== P(16) &= &512 + 16a + b &= 54, \hspace{20mm}&&(1) \\4 KB (670 words) - 13:03, 13 November 2023
- ==Problem 1== What is the value of <cmath>\dfrac{20}{2\cdot1} - \dfrac{2+0}{2/1}?</cmath>12 KB (1,784 words) - 16:49, 1 April 2021
- == Problem 1 == What is <math>( - 1)^1 + ( - 1)^2 + \cdots + ( - 1)^{2006}</math>?13 KB (2,058 words) - 12:36, 4 July 2023
- * [[2006 AMC 12A Problems/Problem 1]]1 KB (168 words) - 21:51, 6 October 2014
- * [[2004 AMC 12A Problems/Problem 1]]2 KB (186 words) - 17:35, 16 December 2019
- ** [[2005 AMC 12A Problems/Problem 1|Problem 1]]2 KB (202 words) - 21:30, 6 October 2014
- * [[2002 AMC 12A Problems/Problem 1]]1 KB (158 words) - 21:33, 6 October 2014
- * [[2003 AMC 12A Problems/Problem 1]]1 KB (162 words) - 21:52, 6 October 2014
- == Problem 1 == [[2006 AMC 12A Problems/Problem 1|Solution]]15 KB (2,223 words) - 13:43, 28 December 2020
- == Problem 1 == (\mathrm {A}) \ 1 \qquad (\mathrm {B}) \ 2 \qquad (\mathrm {C})\ 5 \qquad (\mathrm {D}) \ 1013 KB (1,971 words) - 13:03, 19 February 2020
- == Problem 1 == Alicia earns <math> 20</math> dollars per hour, of which <math>1.45\%</math> is deducted to pay local taxes. How many cents per hour of Ali13 KB (1,953 words) - 00:31, 26 January 2023
- == Problem 1 == <math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 2003\qquad \mathrm{(E) \ } 4013 KB (1,955 words) - 21:06, 19 August 2023
- == Problem 1 == [[2002 AMC 12A Problems/Problem 1|Solution]]12 KB (1,792 words) - 13:06, 19 February 2020
- == Problem 1 == [[2000 AMC 12 Problems/Problem 1|Solution]]13 KB (1,948 words) - 12:26, 1 April 2022
- == Problem 1 == [[2001 AMC 12 Problems/Problem 1|Solution]]13 KB (1,957 words) - 12:53, 24 January 2024
- == Problem 1 == [[2002 AMC 12B Problems/Problem 1|Solution]]10 KB (1,547 words) - 04:20, 9 October 2022