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Quadratic formula
<cmath>x+\frac{b}{2a}=\pm\sqrt{\frac{b^{2}-4ac}{4a^{2}}}=\frac{\pm\sqrt{b^{2}-4ac}}{2a}</cmath> <cmath>x=-\frac{b}{2a}+\frac{\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}</cmath>

2 KB (290 words) - 16:55, 19 February 2024
Sieve of Eratosthenes
...umber | composite]], then it has a prime [[divisor]] not exceeding <math>\sqrt n</math>.

1 KB (212 words) - 21:16, 7 December 2007
Imaginary unit
...ex number itself. It has a [[magnitude]] of 1, and can be written as <math>1 \text{cis } \left(\frac{\pi}{2}\right)</math>. Any [[complex number]] can b #<math>i^1=\sqrt{-1}</math>

2 KB (321 words) - 15:57, 5 September 2008
Absolute value
* (Alternative definition) <math>|x| = \sqrt{x^2}</math> ...ex number]]s <math>z</math>, the absolute value is defined as <math>|z| = \sqrt{x^2+y^2}</math>, where <math>x</math> and <math>y</math> are the real and i

2 KB (368 words) - 10:37, 5 January 2009
Rational approximation of famous numbers
...<+\infty</math>, the [[inequality]] <math>\left|x-\frac pq\right|\ge \frac 1{q^M}</math> holds for all sufficiently large denominators <math>q</math>. Suppose that there exist <math>0<\beta<\gamma<1</math>, <math>1<Q<+\infty</math>

8 KB (1,431 words) - 13:48, 26 January 2008
LaTeX:Symbols
...setminus</math>||\setminus||<math>\wr</math>||\wr||<math>\sqrt{x}</math>||\sqrt{x} ...rc}||<math>\triangledown</math>||\triangledown||<math>\sqrt[n]{x}</math>||\sqrt[n]{x}

16 KB (2,324 words) - 16:50, 19 February 2024
2006 AMC 12B Problems/Problem 21
\mathrm{(A)}\ \frac {16\sqrt {2006}}{\pi} \mathrm{(C)}\ 8\sqrt {1003}

2 KB (339 words) - 13:15, 12 July 2015
3D Geometry
...he 30-60-90 triangle, or the Pythagorean Theorem, to find that <math> x = \sqrt{3} </math> units.

2 KB (263 words) - 12:29, 30 December 2023
Srinivasa Ramanujan
<math>\frac{1}{\pi} = \frac{2\sqrt{2}}{9801} \sum^\infty_{k=0} \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}</mat

496 bytes (62 words) - 23:16, 24 February 2007
2004 AIME II Problems/Problem 11
...xact opposite side of the cone whose distance from the vertex is <math>375\sqrt{2}.</math> Find the least distance that the fly could have crawled. ...<math>R=\sqrt{r^2+h^2}=\sqrt{600^2+(200\sqrt{7})^2}=\sqrt{360000+280000}=\sqrt{640000}=800</math>. Setting <math>\theta R=C\implies 800\theta=1200\pi\impl

2 KB (268 words) - 22:20, 23 March 2023
1986 AIME Problems/Problem 1
...he sum of the solutions to the equation <math>\sqrt[4]{x} = \frac{12}{7 - \sqrt[4]{x}}</math>? Let <math>y = \sqrt[4]{x}</math>. Then we have

688 bytes (104 words) - 13:34, 22 July 2020
1987 AIME Problems/Problem 9
C = (21*sqrt(3),0);

1 KB (200 words) - 18:44, 5 February 2024
1989 AIME Problems/Problem 12
...ontsize(10)); label("$7$",A--C,2*dir(210),fontsize(10)); label("$18$",A--D,1.5*dir(30),fontsize(10)); label("$36$",(3,0),up,fontsize(10)); ...h>\triangle CDM</math>. The formula for the length of a median is <math>m=\sqrt{\frac{2a^2+2b^2-c^2}{4}}</math>, where <math>a</math>, <math>b</math>, and

2 KB (376 words) - 13:49, 1 August 2022
2003 AIME I Problems/Problem 8
<cmath>d=\frac{-3a + \sqrt{9a^2+120a}}{2}.</cmath> <cmath>a=\frac{-120 + \sqrt{14400+36x^2}}{18}</cmath>

5 KB (921 words) - 23:21, 22 January 2023
2003 AIME II Problems/Problem 10
...qrt{b}+\sqrt{b+60}=\sqrt{c}</math>. Square both sides to get <math>2b+60+2\sqrt{b^2+60b}=c</math>. Thus, <math>b^2+60b</math> must be a square, so we have

1 KB (218 words) - 14:14, 25 June 2021
2005 AMC 10B Problems/Problem 4
...al numbers <math>a</math> and <math>b</math>, define <math>a \diamond b = \sqrt{a^2 + b^2}</math>. What is the value of ...\textbf{(B) } \frac{17}{2} \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 13\sqrt{2} \qquad \textbf{(E) } 26</math>

833 bytes (110 words) - 13:58, 24 July 2022
Incircle
...formula above can be simplified with Heron's Formula, yielding <math>r = \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}.</math> *The [[area]] of the [[triangle]] by [[Heron's Formula]] is <math>A=\sqrt{s(s-a)(s-b)(s-c)}</math>.

2 KB (384 words) - 18:38, 9 March 2023
Hypotenuse
...a triangle with legs of length <math>a</math> and <math>b</math> is <math>\sqrt{a^2 + b^2}</math>.

810 bytes (133 words) - 19:02, 15 October 2018
1969 Canadian MO Problems/Problem 9
...>1,</math> the length of the shortest side is less than or equal to <math>\sqrt{2}</math>. ...ossible. Thus, at least one of the sides must have length less than <math>\sqrt 2</math>, so certainly the shortest side must.

795 bytes (138 words) - 22:42, 17 November 2007
Parallel
<math>\sqrt{\frac{(y_2-y_1)^2}{m^2+1}}</math>

2 KB (326 words) - 12:11, 21 May 2009
Furman University Wylie Mathematics Tournament/1996 Senior Exam/Solutions
== Problem 1 == Multiplying the denominator by <math> 1-2i </math> gives us that the expression is <math> 2+3i, </math> the coeffic

9 KB (1,364 words) - 15:59, 21 July 2006
University of South Carolina High School Math Contest/1993 Exam/Problem 27
...mathrm{(C) \ }36 \qquad \mathrm{(D) \ }12\sqrt{2} \qquad \mathrm{(E) \ }12\sqrt{3} </math></center>

1 KB (149 words) - 16:27, 18 August 2006
University of South Carolina High School Math Contest/1993 Exam/Problem 26
...Then the first nonzero digit in the decimal expansion of <math>\sqrt{n^2 + 1} - n</math> is <center><math> \mathrm{(A) \ }1 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }3 \qquad \mathrm{(D) \ }4 \qq

2 KB (258 words) - 16:34, 18 August 2006
University of South Carolina High School Math Contest/1993 Exam/Problem 16
...r><math> \mathrm{(A) \ }1 \qquad \mathrm{(B) \ }4/3 \qquad \mathrm{(C) \ }\sqrt{2} \qquad \mathrm{(D) \ }3/2 \qquad \mathrm{(E) \ }2 </math></center> ...P</math> is [[parallel]] to <math>AB</math> and thus <math>\frac{CQ}{QN} = 1</math> and <math>CQ = QN = 4</math>. Then <math>CN = CQ + QN = 8</math>.

1 KB (221 words) - 11:45, 17 August 2006
Distance formula
...math>(a_1,a_2,...,a_n)</math> and <math>(b_1,b_2,...,b_n)</math> is <math>\sqrt{(a_1-b_1)^2+(a_2-b_2)^2+\cdots+(a_n-b_n)^2}</math>. <cmath>\dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}</cmath>

2 KB (340 words) - 18:34, 8 September 2018
2006 Romanian NMO Problems/Grade 8/Problem 1
...e and a point <math>Q</math> on the circle, <math>PQ=\sqrt{PO_1^2+O_1Q^2}=\sqrt{PO_1^2+r_1^2}</math>, which does not depend on the point <math>Q</math> cho

3 KB (509 words) - 23:22, 15 August 2012
1969 Canadian MO Problems/Problem 3
...have lengths <math>a</math> and <math>b</math>. Prove that <math>a+b\le c\sqrt{2}</math>. When does the equality hold?

790 bytes (129 words) - 22:39, 17 November 2007
Equilateral triangle
...f an equilateral triangle can be found in terms of a side: <math>\frac{s^2\sqrt{3}}{4}</math>.

1 KB (186 words) - 19:57, 15 September 2022
Notation
...easier to express using notation: <math>\displaystyle x+x^2=\frac{x}{2}+x\sqrt{x}</math>. The latter is easier to understand, and we can immediately jump

692 bytes (109 words) - 10:28, 4 August 2006
Galois theory
...ath>. Then the map <math>f:K\to K</math> given by <math>f(a+b\sqrt{2})=a-b\sqrt{2}</math> is a field automorphism; that is, <math>f(\alpha\beta)=f(\alpha)f ...he case. For example, if <math>K=\mathbb{Q}</math> and <math>L=\mathbb{Q}(\sqrt[3]{2})</math>, then <math>Gal(L/K)</math> is the trivial group, so every el

2 KB (385 words) - 14:09, 5 May 2008
1969 Canadian MO Problems
== Problem 1 == [[1969 Canadian MO Problems/Problem 1 | Solution]]

3 KB (536 words) - 12:46, 8 October 2007
Mock USAMO by probability1.01 dropped problems
== Problem 1 == Let <math>n>1</math> be a fixed positive integer, and let <math>a_1,a_2,\ldots,a_n</math>

3 KB (572 words) - 02:46, 16 May 2009
Finite Fourier Analysis
...th>f(0) = f(1)</math> and <math>\frac{d^m f}{dx^m}(0) = \frac{d^m f}{dx^m}(1)</math> for all <math>m \in \mathbb{Z}^+</math> . <math>S</math> is a vect ...rac1N</math>, <math>x=\frac2N</math>, <math>\ldots</math>, <math>x=\frac{N-1}N </math>. For each <math>n \in \mathbb{Z}</math>, let <math>f_n</math> be

4 KB (724 words) - 19:15, 9 September 2006
Real part
...can be written in the form <math>z = a + bi</math> where <math>i = \sqrt{-1}</math> is the [[imaginary unit]] and <math>a</math> and <math>b</math> ar ...athrm{Re}(4(\cos \frac \pi6 + i \sin \frac\pi 6)) = 4 \cos \frac \pi 6 = 2\sqrt 3</math>

2 KB (281 words) - 15:56, 5 September 2008
Imaginary part
...can be written in the form <math>z = a + bi</math> where <math>i = \sqrt{-1}</math> is the [[imaginary unit]] and <math>a</math> and <math>b</math> ar * <math>\mathrm{Im}((1 + i)\cdot(2 + i)) = \mathrm{Im}(1 + 3i) = 3</math>. Note in particular that <math>\mathrm Im</math> is ''not

2 KB (269 words) - 15:56, 5 September 2008
Euclidean metric
...{y} = (y_1, y_2, \ldots, y_n)</math> is given by <math>d(\mathbf{x, y}) = \sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2 + \ldots + (x_n - y_n)^2}</math>. It is stra

813 bytes (132 words) - 17:49, 28 March 2009
Mock AIME 5 2005-2006 Problems
== Problem 1 == ...5</math> and <math>f(101) = 0</math>). Evaluate the remainder when <math>f(1)+f(2)+\cdots+f(99)</math> is divided by <math>1000</math>.

8 KB (1,370 words) - 21:52, 27 February 2007
Hexagon
[[Area]]: <math>\frac{3s^2\sqrt{3}}{2}</math> Where <math>s</math> is the side length of the hexagon. [[Apothem]], or [[inradius]]: <math>\dfrac{s\sqrt{3}}{2}</math>

688 bytes (94 words) - 21:00, 14 December 2018
Coefficient
...</math>, and <math>c</math> in the [[Quadratic Formula]], <math>\frac{b\pm\sqrt{b^2-4ac}}{2a}</math> are all coefficients of the polynomial <math>ax^2+bx+c

1 KB (227 words) - 22:38, 6 October 2020
Rectangular prism
...e interior [[diagonal]]s can be determined by using the formula <math>d = \sqrt{l^2 + w^2 + h^2}</math>. Proof: To get a base diagonal, we use the [[pythagorean theorem]]: <math> \sqrt{l^2+w^2}</math>. We call that v. Then we use the pythagorean theorem again

883 bytes (132 words) - 09:04, 12 September 2007
2007 BMO Problems
== Problem 1 == [[2007 BMO Problems/Problem 1 | Solution]]

2 KB (297 words) - 23:44, 4 May 2007
Specimen Cyprus Seniors Provincial/2nd grade/Problem 3
...lambda^2} = \sqrt{(\kappa\lambda)^2(\nu^2 + 4\nu + 4 - 4)} = \kappa\lambda\sqrt{\nu^2+4\nu}</math> <math>(\nu+1)^2 < \nu^2+4\nu < (\nu+2)^2</math>.

1 KB (184 words) - 22:54, 22 May 2009
Specimen Cyprus Seniors Provincial/2nd grade/Problem 4
If <math>\rho_{1}, \rho_{2}</math> are the roots of equation <math>x^2-x+1=0</math> then: a) Prove that <math>\rho_{1}^3=\rho_{2}^3 = -1</math> and

2 KB (217 words) - 23:28, 22 May 2009
2005 Alabama ARML TST Problems/Problem 3
...ath>r = 10</math> and the side length of the triangle is equal to <math>20\sqrt 3</math>.

1 KB (221 words) - 19:38, 6 February 2010
Mock AIME 6 Pre 2005/Problems
<math>1</math>. Find the remainder when <math>\displaystyle 11^{2005}</math> is div <math>5</math>. Let <math>\displaystyle f(x)=\sqrt{x+2005\sqrt{x+2005\sqrt{\cdots}}}</math> where <math>\displaystyle f(x)>0</math>. Find the remainde

7 KB (1,110 words) - 05:15, 31 December 2006
Mock AIME 4 2006-2007 Problems
==Problem 1== ...first digit of 1. He writes <math>1, 10, 11, 12, \ldots</math> but by the 1,000th digit he (finally) realizes that the list would contain an infinite n

6 KB (923 words) - 14:17, 16 January 2007
Mock AIME 4 2006-2007 Problems/Problem 11
...e area of <math>\triangle ACF</math> is 3. If <math>\frac{CE}{EA}=\frac{p+\sqrt{q}}{r}</math>, where <math>p</math>, <math>q</math>, and <math>r</math> are ...}-3}{6}</math>, so <math>\frac{a}{b} = \frac{6}{\sqrt{129}-3} = \frac{3 + \sqrt{129}}{20}</math>.

2 KB (325 words) - 19:33, 9 February 2017
2005 IMO Shortlist Problems/A1
...ystyle p(x)q(x) </math> is a polynomial having all coefficients <math> \pm 1 </math>. ...the constant term of <math> \displaystyle p(x) </math> must be <math> \pm 1 </math>.

2 KB (363 words) - 22:57, 31 January 2007
Mock AIME 3 Pre 2005 Problems/Problem 1
...>, <math>3+r</math>, and <math>7+r</math>. From Heron's Formula, <math>84=\sqrt{(10+r)(r)(7)(3)}</math>, or <math>84*84=r(10+r)*21</math>, or <math>84*4=r(

795 bytes (129 words) - 10:22, 4 April 2012
Mock AIME 4 2005-2006/Problems
== Problem 1 == 1. A 5-digit number is leet if and only if the sum of the first 2 digits, the

7 KB (1,176 words) - 04:44, 26 February 2007
Inverse of a function
...ule <math>\displaystyle f(x) = x^3 + 6</math>. The function <math>g(x) = \sqrt[3]{x-6}</math> has the property that <math>f(g(x)) = x</math>. In this cas ...on <math>f</math> is denoted by <math>f^{-1}</math>. Note that the <math>-1</math> does ''not'' indicate an [[exponent]].

1 KB (269 words) - 13:50, 5 March 2007
2007 AIME II Problems/Problem 11
...rts. The distance <math>x</math> can be expressed in the form <math>a\pi+b\sqrt{c},</math> where <math>a,</math> <math>b,</math> and <math>c</math> are [[i ...48\sqrt{3}</math>. Thus, the total horizontal distance covered is <math>96\sqrt{3}</math>.

2 KB (407 words) - 16:31, 29 February 2020
1976 IMO Problems/Problem 1
...n any such case it is clear from the Pythagorean theorem that <math>AC = 8\sqrt 2</math>. Therefore the other diagonal has only one possible length: <math>8\sqrt 2</math>.

1 KB (255 words) - 21:05, 26 September 2009
2007 Cyprus MO/Lyceum/Problem 11
...2006}+2006 \sqrt{2007}}</math> and <math>Y=\frac{1}{\sqrt{2006}}-\frac{1}{\sqrt{2007}}</math>, which of the following is correct? ...= \frac{\sqrt{b}}{b} - \frac{\sqrt{a}}{a} = \frac{1}{\sqrt{b}} - \frac{1}{\sqrt{a}} = Y</math>

808 bytes (139 words) - 16:56, 6 May 2007
2007 Cyprus MO/Lyceum/Problem 7
...B) \ } \frac{d^2}{2}\qquad \mathrm{(C) \ } 2d^2\qquad \mathrm{(D) \ } d^2 \sqrt{2}\qquad \mathrm{(E) \ } \mathrm{None\;of\;these}</math> ...the sides of the rectangle are <math>\frac{d}{2}</math> and <math>\frac{d\sqrt{3}}{2}</math>.

736 bytes (112 words) - 20:43, 6 May 2007
2007 Cyprus MO/Lyceum/Problem 6
...qrt{3}\qquad \mathrm{(D) \ } \sqrt{3} - 1\qquad \mathrm{(E) \ } \sqrt{2} - 1</math> ...triangle]]s. <math>AF = BG = \sqrt{3}</math>, and so <math>DF = CG = 2 - \sqrt{3} \Longrightarrow \mathrm{C}</math>.

1,013 bytes (152 words) - 10:39, 8 May 2007
2007 Cyprus MO/Lyceum/Problem 14
...c{1}{\sqrt{2}}\qquad \mathrm{(D) \ } \sqrt2-1\qquad \mathrm{(E) \ } \frac{\sqrt{2}}4</math> ...of <math>S_1:S_2</math> is <math>x\sqrt{2}-x:x</math>, or <math>\sqrt{2}-1:1\Longrightarrow\mathrm{ D}</math>.

868 bytes (143 words) - 23:02, 8 May 2007
2007 Cyprus MO/Lyceum/Problem 15
...\mathrm{(B) \ } 8\qquad \mathrm{(C) \ } 1\qquad \mathrm{(D) \ } \frac{6+2\sqrt{3}}{3}\qquad \mathrm{(E) \ } \mathrm{None\;of\;these}</math> pair a=(0,1), c=(1,0), bb=(a+c)/2, b=bb+dir(225)/sqrt(6);

2 KB (388 words) - 14:26, 29 January 2009
2006 Cyprus MO/Lyceum/Problems
== Problem 1 == ...ntity of milk with <math>4\%</math> fat adds a quantity of milk with <math>1\%</math> fat and produces <math>1200</math>kg of milk with <math>2\%</math>

11 KB (1,672 words) - 10:56, 27 April 2008
2007 Cyprus MO/Lyceum/Problem 30
A coin with a shape of a regular hexagon of side 1 is tangent to a square of side 6, as shown in the figure. ...) \ } 24\qquad \mathrm{(C) \ } \frac{28\pi}{3}\qquad \mathrm{(D) \ } 6 \pi\sqrt{2}\qquad \mathrm{(E) \ } \mathrm{None\;of\;these}</math>

1 KB (168 words) - 00:22, 10 May 2007
2004 IMO Shortlist Problems/N1
=== Solution 1 === ...each divisor <math> d > \sqrt{n} </math>, a divisor <math> \frac{n}{d} < \sqrt{n} </math>. Hence we obtain the inequality

7 KB (1,166 words) - 22:14, 10 May 2007
2005 IMO Shortlist Problems/G1
=== Solution 1 === ...h> are congruent. This means that <math> NI = MP = \frac{1}{2} DI = \frac{1}{2} KI </math>, so <math> \displaystyle PN </math> is the perpendicular bis

7 KB (1,088 words) - 16:57, 30 May 2007
2007 Cyprus MO/Lyceum/Problem 22
...rt{2}}\qquad \mathrm{(C) \ } \frac{5a}{2}\qquad \mathrm{(D) \ } \frac{3a}{\sqrt{3}}\qquad \mathrm{(E) \ } 2a</math> ...5-45-90</math> triangle, the length of <math>AM</math> is <math>\frac{3a}{\sqrt{2}}</math>, and the answer is <math>\boxed{\mathrm{B}}</math>.

979 bytes (166 words) - 02:33, 19 January 2024
2006 Cyprus MO/Lyceum/Problem 2
The value of the expression <math>K=\left[\left(1+\sqrt{3}\right)*2\right]*\sqrt{2}</math> is ...B)}\ 0\qquad\mathrm{(C)}\ \sqrt{3}\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 1</math>

552 bytes (78 words) - 10:56, 27 April 2008
2006 Cyprus MO/Lyceum/Problem 3
The [[domain]] of the [[function]] <math>f(x)=\sqrt{4+2x}</math> is

477 bytes (74 words) - 10:50, 27 April 2008
2006 Cyprus MO/Lyceum/Problem 4
\alpha x^2+9x+\frac{81}{4\alpha}=\left(\sqrt{\alpha}x+\frac{9}{2\sqrt{\alpha}}\right)^2 ...th>x</math> has the unique root of <math>-\frac{\frac{9}{2\sqrt{\alpha}}}{\sqrt{\alpha}} = \frac{-9}{2\alpha}</math>, so it touches the x-axis, <math>\math

1 KB (216 words) - 10:46, 27 April 2008
2006 Cyprus MO/Lyceum/Problem 6
The value of the expression <math>K=\sqrt{19+8\sqrt{3}}-\sqrt{7+4\sqrt{3}}</math> is <math>\mathrm{(A)}\ 4\qquad\mathrm{(B)}\ 4\sqrt{3}\qquad\mathrm{(C)}\ 12+4\sqrt{3}\qquad\mathrm{(D)}\ -2\qquad\mathrm{(E)}\ 2</math>

887 bytes (124 words) - 10:43, 27 April 2008
2006 Cyprus MO/Lyceum/Problem 7
...h>\Delta E\perp A\Gamma</math>, <math>EZ\perp B\Gamma</math>. If <math>EZ=\sqrt{3}</math>, then the length of the side of the triangle <math>AB\Gamma</math <math>\mathrm{(A)}\ \frac{3\sqrt{3}}{2}\qquad\mathrm{(B)}\ 8\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 3\qquad

1 KB (163 words) - 10:42, 27 April 2008
2006 Cyprus MO/Lyceum/Problem 9
If <math>x=\sqrt[3]{4}</math> and <math>y=\sqrt[3]{6}-\sqrt[3]{3}</math>, then which of the following is correct? ...[3]{18}+\sqrt[3]{9})&:(\sqrt[3]{6}-\sqrt[3]{3})(\sqrt[3]{36}+\sqrt[3]{18}+\sqrt[3]{9})\\

1 KB (182 words) - 10:40, 27 April 2008
Surface area
For cones, the surface area is <math>\pi \cdot r \cdot (r+\sqrt{h^2+r^2})</math> ...the surface area is <math>lw+l \cdot \sqrt{(\frac{w}{2})^2+h^2}+w^2 \cdot \sqrt{(\frac{l}{2})^2+h^2}</math>.

1 KB (209 words) - 01:34, 25 January 2016
Root-mean-square
...n of [[real number]]s <math>x_1,\dots , x_n</math> is defined to be <math>\sqrt{\frac{x^2_1+x^2_2+\dots+x^2_n}{n}}</math>. This is the second [[power mean

543 bytes (86 words) - 13:41, 8 May 2013
Octagon
[[Area]]: <math>2s^2(1 + \sqrt{2})</math> [[Apothem]]: <math>\frac{s(1+ \sqrt2 )}{2}</math>

489 bytes (60 words) - 20:40, 14 October 2007
2005 AMC 12A Problems/Problem 22
...iameter of the sphere is the space diagonal of the prism, which is <cmath>\sqrt{l^2 + w^2 +h^2}.</cmath> <cmath>\sqrt{l^2 + w^2 +h^2} = \sqrt{400} = 20.</cmath> The radius is half of the diameter, so

2 KB (334 words) - 10:20, 16 September 2022
2005 AMC 12A Problems/Problem 3
...ath>. The area of the rectangle is <math>(w)(2w) = 2w^2 = 2\left(\frac{x}{\sqrt{5}}\right)^2 = \frac{2}{5}x^2 \ \mathrm{(B)}</math>.

739 bytes (113 words) - 21:15, 3 July 2013
2006 Cyprus MO/Lyceum/Problem 14
...Delta</math> and to the square <math>ZB\Gamma E</math>, so that <math>AE=2\sqrt{5}\ \text{m}</math> and the shaded area of the triangle <math>\Delta BE</ma ...)}\ 16\ \text{m}^2\qquad\mathrm{(D)}\ 32\ \text{m}^2\qquad\mathrm{(E)}\ 10\sqrt{5}\ \text{m}^2</math>

1 KB (229 words) - 09:11, 12 August 2008
2006 Cyprus MO/Lyceum/Problem 15
...7} + \frac{1}{\sqrt7+\sqrt{10}}+ \frac{1}{\sqrt{10}+\sqrt{13}} + \frac{1}{\sqrt{13}+4}</math> equals ...{(B)}\ \frac{3}{2}\qquad\mathrm{(C)}\ \frac{2}{5}\qquad\mathrm{(D)}\ \frac{1}{2}\qquad\mathrm{(E)}\ \frac{2}{3}</math>

985 bytes (124 words) - 10:29, 27 April 2008
2006 Cyprus MO/Lyceum/Problem 22
<math>\mathrm{(A)}\ \frac{5}{9}\qquad\mathrm{(B)}\ \frac{1}{3}\qquad\mathrm{(C)}\ \frac{9}{5}\qquad\mathrm{(D)}\ \frac{3}{5}\qquad\mat ...{\sqrt{9x^4 + 9y^4 + 82x^2y^2}}{18}</math>. Thus the ratio is <math>\frac{\sqrt{9x^4 + 9y^4 + 82x^2y^2}}{18xy}</math>. There is no way we can simplify this

2 KB (285 words) - 13:19, 26 April 2008
Magnitude
...agnitude <math>|z|</math> of a complex number <math>z</math> equals <math>\sqrt{\mathrm {Re}(z)^2 + \mathrm{Im}(z)^2}</math>. Both types of magnitude are b <cmath>\begin{align*} |z\omega| &= \sqrt{(ac - bd)^2 + (ad + bc)^2} \\

2 KB (246 words) - 18:27, 2 March 2023
Torricelli's Law
<cmath>v_2=\sqrt{2gh+v_1^2}</cmath> ...2 v_1^2)=\frac 12 mv_2^2</math>; solving for <math>v_2</math>, <cmath>v_2=\sqrt{2gh+v_1^2}</cmath>

1 KB (184 words) - 13:58, 22 December 2007
1998 IMO Shortlist Problems/A2
...math>r_1, r_2, \dotsc, r_n</math> be real numbers greater than or equal to 1. Prove that ...} + \dotsb + \frac{1}{r_n + 1} \ge \frac{n}{\sqrt[n]{r_1 r_2 \dotsm r_n} + 1} . </cmath>

2 KB (272 words) - 22:03, 30 December 2007
1998 BMO Problems/Problem 2
If <math>n \ge 2</math> is an integer and <math>0 < a_1 < a_2 \dotsb < a_{2n+1}</math> are real numbers, prove the inequality: ...n}} + \sqrt[n]{a_{2n+1}} < \sqrt[n]{a_1 -a_2 +a_3 - \dotsb -a_{2n} + a_{2n+1}} . </cmath>

2 KB (346 words) - 00:08, 31 December 2007
2007 AMC 10A Problems/Problem 17
...of <math>m</math> is <math>3^2 \cdot 5 = 45</math>, which makes <math>n = \sqrt[3]{3^3 \cdot 5^3} = 15</math>. The minimum possible value for the sum of <m

1 KB (201 words) - 08:04, 11 February 2023
1995 AHSME Problems/Problem 23
<cmath>15^2 >11^2 + k^2 \Longrightarrow k < \sqrt{104}</cmath> <cmath>k^2 >11^2 + 15^2 \Longrightarrow k > \sqrt{346}</cmath>

1 KB (183 words) - 14:05, 5 July 2013
1970 AMC 12 Problems/Problem 2
<math>\mathrm{(A)}\ \frac{2}{\pi}\qquad \mathrm{(B)}\ \frac{\pi}{\sqrt{2}}\qquad \mathrm{(C)}\ \frac{4}{5}\qquad \mathrm{(D)}\ \frac{5}{6}\qquad \

613 bytes (104 words) - 15:49, 9 January 2008
1951 AHSME Problems/Problem 16
...nly one root (applying the quadratic formula, we get <math>x = \frac{-b + \sqrt{0}}{2a} = -b/2a</math>). Thus it follows that <math>f(x)</math> touches the

1 KB (166 words) - 12:20, 5 July 2013
Successor set
:(i) <math>1\in S</math> :(ii) <math>\forall n\in S</math>; <math>n+1\in S</math>

546 bytes (92 words) - 11:36, 26 January 2008
Bounded
...lane]], <math>\mathbb{R}^2</math> with distance <math>d((x, y), (w, z)) = \sqrt{(x - w)^2 + (y - z)^2}</math>), is bounded if for some <math>x \in X</math>

847 bytes (159 words) - 17:37, 15 February 2008
Momentum
...math>\vec{p} = \gamma m\vec{v}</math>, where <math>\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}</math> is the [[Lorentz factor]] dependent on the magnitu

2 KB (271 words) - 09:02, 11 March 2008
2005 Alabama ARML TST Problems/Problem 9
...=6+3\sqrt{2}</math>. We multiply and divide by 2 to get <math>\boxed{18+18\sqrt{2}}</math>.

776 bytes (136 words) - 00:19, 4 March 2008
Mock AIME 1 Pre 2005 Problems/Problem 5
<cmath>\sqrt[3]{x} + \sqrt[3]{20 - x} = 2</cmath> The smaller of the two values can be expressed as <math>p - \sqrt{q}</math>, where <math>p</math> and <math>q</math> are integers. Compute <m

979 bytes (159 words) - 17:46, 21 March 2008
2008 AIME I Problems/Problem 5
...tations. The value of <math>h/r</math> can be written in the form <math>m\sqrt {n}</math>, where <math>m</math> and <math>n</math> are positive integers a ...h>\sqrt {r^{2} + h^{2}}</math>. Thus, the length of the path is <math>2\pi\sqrt {r^{2} + h^{2}}</math>.

1 KB (230 words) - 20:18, 4 July 2013
2006 IMO Shortlist Problems/A5
...\sqrt{c}+\sqrt{a} - \sqrt{b}} + \frac{\sqrt{a+b-c}}{\sqrt{a} + \sqrt{b} - \sqrt{c}} \le 3. </cmath> <cmath> \sum_{\rm cyc} \left( \frac{y}{x} - 1 \right) \left( \frac{z}{x} -1 \right) \ge 0 . </cmath>

3 KB (529 words) - 08:03, 29 March 2008
1966 AHSME Problems/Problem 3
<cmath>\sqrt{\eta\zeta}=10\Rightarrow \eta\zeta=100</cmath>.

942 bytes (143 words) - 01:27, 26 June 2016
Mock AIME 1 2007-2008 Problems/Problem 5
Let <math>S = (1+i)^{17} - (1-i)^{17}</math>, where <math>i=\sqrt{-1}</math>. Find <math>|S|</math>. ...m, <math>1+i = \sqrt{2}\,\text{cis}\,\frac{\pi}{4}</math> and <math>1-i = \sqrt{2}\,\text{cis}\,-\frac{\pi}{4}</math>, where <math>\text{cis}\,\theta = \co

1,008 bytes (141 words) - 21:13, 2 April 2008
2007 Alabama ARML TST Problems
==Problem 1== Compute: <math>\sqrt{2000\cdot 2007\cdot 2008\cdot 2015+784}.</math>

5 KB (841 words) - 11:33, 28 January 2009
2008 Mock ARML 1 Problems
= Set 1 = == Problem 1 ==

3 KB (409 words) - 16:41, 29 May 2008
2007 Alabama ARML TST Problems/Problem 1
Compute: <math>\sqrt{2000\cdot 2007\cdot 2008\cdot 2015+784}.</math> <center><math>\sqrt{2000(2000+7)(2000+8)(2000+15)+784}</math>

454 bytes (40 words) - 09:26, 18 June 2008
2006 Alabama ARML TST Problems/Problem 7
...s <math>x^2 = 10^2 + 10^2 - 2\cdot10\cdot10\cdot \cos{150^\circ} = 200+100\sqrt{3}</math>.

1 KB (185 words) - 01:57, 17 March 2021
KGS math club/solution 3 1
...of a little sphere is sqrt(12+12+...12) = sqrt(n) The radius of the big sphere is sqrt(n)-1

221 bytes (41 words) - 13:33, 1 July 2008
1966 AHSME Problems/Problem 8
<math>\text{(A)} \ 27 \qquad \text{(B)} \ 21 \qquad \text{(C)} \ \sqrt {389} \qquad \text{(D)} \ 15 \qquad \text{(E)} \ \text{undetermined}</math> ...\sqrt {10^2 - (\frac{16}{2})^2} = \sqrt {100 - 8^2} = \sqrt {100 - 64} = \sqrt {36} = 6</math>

2 KB (256 words) - 01:45, 26 June 2016
1985 IMO Problems/Problem 4
...square. Let the elements be <math>x,y</math> and let <math>x=\sqrt{ab},y=\sqrt{cd}</math> where <math>a,b,c,d\in M</math>. Then, we have <math>xy=z^2</ma

1 KB (261 words) - 23:56, 29 January 2021
2008 AMC 12B Problems/Problem 18
...}\ 196\sqrt {6} \qquad \textbf{(C)}\ 392\sqrt {2} \qquad \textbf{(D)}\ 392\sqrt {3} \qquad \textbf{(E)}\ 784</math> Therefore, <math>h=\sqrt{15^2-9^2}=12</math>, and the volume of the pyramid is <math>\frac{bh}{3}=\f

1 KB (214 words) - 12:01, 2 February 2015
2002 AMC 10A Problems/Problem 23
...ath>, we find <math>AE=\sqrt{x^2+12x+100}</math>. From symmetry, <math>DE=\sqrt{x^2+12x+100}</math> as well. Now, we use the fact that the perimeter of <ma We have <math>2\sqrt{x^2+12x+100}+2x+12=2(32)</math> so <math>\sqrt{x^2+12x+100}=26-x</math>. Squaring both sides, we have <math>x^2+12x+100=67

2 KB (378 words) - 21:38, 19 July 2023
2002 AMC 10B Problems/Problem 14
Taking the root, we get <math>N=\sqrt{25^{64}\cdot 64^{25}}=5^{64}\cdot 8^{25}</math>.

1 KB (148 words) - 08:08, 27 November 2020
1972 USAMO Problems/Problem 4
<center><math>\left|\frac{aR^2+bR+c}{dR^2+eR+f}-\sqrt[3]{2}\right|<|R-\sqrt[3]{2}|</math></center> .../math>, <math>\frac{aR^2+bR+c}{dR^2+eR+f}</math> must also approach <math>\sqrt[3]{2}</math> for the given inequality to hold. Therefore

1 KB (245 words) - 15:09, 2 June 2018
1962 IMO Problems/Problem 2
<math>\sqrt{\sqrt{3-x}-\sqrt{x+1}}>\dfrac{1}{2}</math> Obviously we need <math>\sqrt{3-x} \geq \sqrt{x+1}</math> for the outer square root to be defined, <math>x\leq 3</math> for t

3 KB (445 words) - 14:43, 2 February 2009
1998 AHSME Problems/Problem 23
...th>, which is a circle centered at <math>(2,6)</math> with radius <math>r=\sqrt{40+k}</math>. ...>\sqrt{40+k} \geq 2</math> we get that <math>k\geq -36</math>. From <math>\sqrt{40+k}\leq 12</math> we get <math>k\leq 104</math>.

1 KB (193 words) - 09:12, 2 December 2018
1999 AHSME Problems/Problem 16
Using [[Pythagorean theorem]] we can compute <math>a=\sqrt{ (d_1/2)^2 + (d_2/2)^2 }=13</math>. pair A=(-3,0), B=(0,-1.25), C=(3,0), D=(0,1.25);

2 KB (308 words) - 14:35, 5 July 2013
2009 Zhautykov International Olympiad Problems
==Day 1== ===Problem 1===

2 KB (402 words) - 13:52, 3 February 2009
2009 AMC 12A Problems/Problem 21
*<math>x^4 - 2009 = 0</math>: The real roots are <math>\pm \sqrt [4]{2009}</math>, and there are two complex roots. *<math>x^4 - 9002 = 0</math>: The real roots are <math>\pm \sqrt [4]{9002}</math>, and there are two complex roots.

2 KB (322 words) - 10:25, 29 July 2020
2009 AMC 12A Problems/Problem 13
...^\circ</math>, hence <math>BD_2 = BC_2 / 2</math> and <math>D_2C_2 = BC_2 \sqrt 3 / 2</math>. label("$20/\sqrt 2$",midpoint(B--D1),S);

5 KB (739 words) - 10:24, 9 February 2015
2009 AMC 12B Problems/Problem 13
...BC</math>. Then <math>BD = \sqrt {13^2 - 12^2} = 5</math> and <math>DC = \sqrt {15^2 - 12^2} = 9</math>. Thus <math>BC = BD + BC = 5 + 9 = 14</math> or

883 bytes (137 words) - 18:42, 23 February 2017
1966 AHSME Problems/Problem 4
...\quad \text{(C) } \sqrt{3} \quad \text{(D) } 2\sqrt{2} \quad \text{(E) } 2\sqrt{3}</math> ...onal to be <math>2\sqrt{2}x</math>. Half of that is the radius, or <math>x\sqrt{2}</math>. Using the same equation as before, you get the area of the large

1 KB (191 words) - 22:09, 14 January 2018
2006 Alabama ARML TST Problems/Problem 14
Thus, <math>|x+yi|^2 = \sqrt[3]{2^3 \cdot 37^3} = 2\cdot37 = 74</math>. <math>-1 < \tan{(3\theta)} = -\dfrac{182}{610} < 0</math>.

2 KB (318 words) - 22:55, 29 March 2009
2007 Alabama ARML TST Problems/Problem 3
...2 cents is <math>\frac{144}{a+2}</math> cents a dozen. Their difference is 1, so we have <math>\frac{144}{a+2}+1=\frac{144}{a}</math>

1,020 bytes (167 words) - 13:14, 6 May 2009
2010 AIME II Problems/Problem 5
...{10}x)(\log_{10} yz) + (\log_{10}y) (\log_{10}z) = 468</math>. Find <math>\sqrt {(\log_{10}x)^2 + (\log_{10}y)^2 + (\log_{10}z)^2}</math>. ...>\ a+b+c</math> and <math>\ 2ab+2ac+2bc</math>, and we want to find <math>\sqrt {a^2 + b^2 + c^2}</math>.

2 KB (392 words) - 23:49, 5 February 2022
1961 AHSME Problems/Problem 7
...ified, the third term in the expansion of <math>(\frac{a}{\sqrt{x}}-\frac{\sqrt{x}}{a^2})^6</math> is: <cmath>\binom{6}{2}(\frac{a}{\sqrt{x}})^{4}(\frac{-\sqrt{x}}{a^2})^2</cmath>

723 bytes (112 words) - 20:09, 17 May 2018
1961 AHSME Problems/Problem 10
<math>\textbf{(A)}\ \sqrt{18} \qquad \textbf{(B)}\ \sqrt{28} \qquad

1 KB (176 words) - 23:27, 26 May 2018
Euclidean space
...he [[distance formula|distance]] [[metric]], <math>d(\bold{x},\bold{y}) = \sqrt{(x_1-y_1)^2 + \cdots + (x_n - y_n)^2}</math>. Similarly, the Euclidean spa

1,018 bytes (164 words) - 12:46, 13 March 2010
2010 AMC 12B Problems/Problem 22
...{389}{2}} \qquad \textbf{(D)}\ \sqrt{\dfrac{425}{2}} \qquad \textbf{(E)}\ \sqrt{\dfrac{533}{2}}</math> ...+ 49 + 36 + 144 = 425</math>, so our answer is <math>\boxed{\textbf{(D)} \sqrt{\frac{425}{2}}}</math>.

2 KB (262 words) - 16:43, 15 February 2021
2011 AMC 10A Problems/Problem 11
...\,\frac78 \qquad\text{(D)}\,\frac{5\sqrt{2}}{8} \qquad\text{(E)}\,\frac{\sqrt{14}}{4} </math> ...s^2}=\sqrt{50s^2}</math>, and hence the ratio in the areas is <math>\frac{\sqrt{50s^2}^2}{(8s)^2}=\frac{50}{64} = \boxed{\frac{25}{32} \ \mathbf{(B)}}</mat

991 bytes (155 words) - 03:42, 23 January 2023
2012 IMO Problems/Problem 1
...dot \frac{\sqrt{\frac{(s-a)(s-b)}{ab}}}{\sqrt{\frac{(s-b)(s-c)}{bc}}} = c \sqrt{\frac{(s-a)(s-c)}{ac}} = c \sin \frac{B}{2}.</cmath>

7 KB (1,189 words) - 01:22, 19 November 2023
2001 AMC 10 Problems/Problem 24
draw(circle((0.2,4.92), 1.3)); draw(circle((1.04,1.58), 2.14));

5 KB (772 words) - 17:58, 3 March 2024
LaTeX Cheat Sheet
\sqrt{x} represents <math>\sqrt{x}</math> \sqrt[n]{x} represents <math>\sqrt[n]{x}</math>

2 KB (318 words) - 13:47, 4 May 2024
1998 AHSME Problems/Problem 6
...o be as small as possible, then the two numbers must be as close to <math>\sqrt{1998}</math> as possible.

1 KB (157 words) - 14:28, 5 July 2013
1998 AHSME Problems/Problem 15
<math> \mathrm{(A) \ }\sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }\sqrt{6} \qquad \mathrm{(D) \ }3 \qquad \mathrm{(E) \ }6 </math> <math>A_{\triangle} = \frac{s_t^2\sqrt{3}}{4}</math>

764 bytes (123 words) - 14:29, 5 July 2013
1984 AHSME Problems/Problem 25
...rm{(B) \ }\sqrt{12} \qquad \mathrm{(C) \ } \sqrt{13}\qquad \mathrm{(D) \ }\sqrt{14} \qquad \mathrm{(E) \ } \text{Not uniquely determined} </math> The length of the interior diagonal is <math> \sqrt{a^2+b^2+c^2} </math>, so if we can find <math> a^2+b^2+c^2 </math>, we can

1 KB (243 words) - 12:52, 5 July 2013
Radical
'''Radicals''' are basically the opposite of exponents. <math>\sqrt{4}</math> can be denoted as <math>x^2=4</math>.

185 bytes (23 words) - 18:50, 4 July 2011
Gauss Grade 7 Year 2011 Problem 2
The value of <math> \sqrt {9 + 16} </math> is <math> \sqrt {9 + 16} = \sqrt {25} = \sqrt {5^2} = \boxed {\textbf {(E) 5}} </math>

293 bytes (29 words) - 15:59, 5 July 2011
1996 AHSME Problems/Problem 23
...ce) is given by <math>\sqrt{x^2 + y^2 + z^2}</math>. Thus, we have <math>\sqrt{x^2 + y^2 + z^2} = 21</math>, so <math>x^2 + y^2 + z^2 = 21^2</math>.

1 KB (220 words) - 16:24, 28 June 2021
1992 AJHSME Problems/Problem 16
draw((0,0)--(7,-3*sqrt(51)/10)); label("10",(7/2,-3*sqrt(51)/20),NE);

2 KB (266 words) - 00:09, 5 July 2013
1993 AJHSME Problems/Problem 17
fill((20,0)--(20+5*sqrt(2),5*sqrt(2))--(20+5*sqrt(2),5+5*sqrt(2))--(20,5)--cycle,lightgray); draw((0,5)--(5*sqrt(2),5+5*sqrt(2))--(20+5*sqrt(2),5+5*sqrt(2))--(20,5));

1 KB (187 words) - 09:35, 1 December 2020
1993 AJHSME Problems/Problem 25
A checkerboard consists of one-inch squares. A square card, <math>1.5</math> inches on a side, is placed on the board so that it covers part or ...gorean Theorem]], the diagonal of the square <math>\sqrt{(1.5)^2+(1.5)^2}=\sqrt{4.5}>2</math>. Because this is longer than <math>2</math>( length of the si

1 KB (171 words) - 23:49, 12 March 2023
Mock AIME I 2015 Problems/Problem 3
...e, we have <cmath>BC=\sqrt{25^2+29^2-2\cdot25\cdot29\cdot\dfrac{17}{145}}=\sqrt{1296}=\boxed{036}.</cmath>

2 KB (278 words) - 23:31, 10 June 2015
Mock Geometry AIME 2011 Problems/Problem 7
...D=75,</math> <math>CP-BP</math> can be expressed in the form <math>\frac{a\sqrt{b}} {c},</math> where <math>a,b,c</math> are positive integers and <math>a, ...}{4})(\frac{25}{4})(\frac{5}{13})</math>. Solving yields <math>BP=\frac{15\sqrt{26}}{13}</math>.

2 KB (311 words) - 21:38, 3 January 2012
2010 AMC 8 Problems/Problem 16
<math> \textbf{(A)}\ \frac{\sqrt{\pi}}{2}\qquad\textbf{(B)}\ \sqrt{\pi}\qquad\textbf{(C)}\ \pi\qquad\textbf{(D)}\ 2\pi\qquad\textbf{(E)}\ \pi^ ...th>, we have <math>\frac{s}{r}=\sqrt{\pi}\Rightarrow \boxed{\textbf{(B)}\ \sqrt{\pi}}</math>

868 bytes (150 words) - 22:40, 21 April 2024
2012 AIME II Problems/Problem 8
...we find the two possible values of <math>zw</math> to be <math>7i-15 \pm \sqrt{(15-7i)^2 + 240}</math> = <math>6+2i,</math> <math>12i - 36.</math> The sma ...en, we substitute that into <math>a-bi</math> which is the value of <math>\sqrt{416-210i}</math> and continue from there.

2 KB (299 words) - 21:54, 14 January 2024
Mock AIME II 2012 Problems/Problem 8
...h>D</math> to meet the circle at <math>E</math>. The length <math>EC^2=m+k\sqrt{n}</math>, where <math>m</math>,<math>n</math>, and <math>k</math> are posi ...ath>, <math>EC^2=9^2+9^2-2(\cos 135)\cdot 9\cdot 9=2(81)+81\sqrt{2}=162+81\sqrt{2}</math>.

1 KB (184 words) - 03:11, 5 April 2012
1951 AHSME Problems/Problem 49
...drawn from the vertices of the acute angles are <math>5</math> and <math>\sqrt{40}</math>. The value of the hypotenuse is: ...\textbf{(B)}\ 2\sqrt{40}\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 2\sqrt{13}\qquad\textbf{(E)}\ \text{none of these} </math>

1 KB (198 words) - 12:44, 22 February 2015
1951 AHSME Problems/Problem 22
...t {233}}{2} \qquad\textbf{(B)}\ 20, - 5 \qquad\textbf{(C)}\ \frac {15 \pm \sqrt {305}}{2}</math> ...dratic formula to solve for <math>a</math>, and we get <math>a=\frac{15\pm\sqrt{625}}{2}\implies a=\boxed{\textbf{(B)}\ 20, - 5}</math>

703 bytes (104 words) - 23:01, 13 March 2015
1980 AHSME Problems/Problem 9
...ft and walks 3 miles in the new direction. If he finishes a a point <math>\sqrt{3}</math> from his starting point, then <math>x</math> is <math>\text{(A)} \ \sqrt 3 \qquad \text{(B)} \ 2\sqrt{5} \qquad \text{(C)} \ \frac 32 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ \

978 bytes (160 words) - 16:00, 20 March 2018
1953 AHSME Problems/Problem 45
<math>\textbf{(A)}\ \frac{a+b}{2} > \sqrt{ab} \qquad \textbf{(B)}\ \frac{a+b}{2} < \sqrt{ab} \qquad

689 bytes (111 words) - 23:02, 14 February 2020
1951 AHSME Problems/Problem 32
...}AB\sqrt{2}\qquad\textbf{(C)}\ \geq AB\sqrt{2}\qquad\textbf{(D)}\ \leq AB\sqrt{2} </math> <math> Therefore, the only possible case is <math>\boxed{\textbf{(D)}\ \leq AB\sqrt{2}}</math>

2 KB (263 words) - 12:26, 5 July 2013
2013 AMC 12A Problems/Problem 9
pair A=(10,sqrt(28^2-100)),B=origin,C=(20,0),D=(A.x*r,A.y*r);

2 KB (352 words) - 16:25, 12 August 2023
2013 AMC 12A Problems/Problem 12
...4,5,x</math>. The sum of the possible values of x equals <math>a+\sqrt{b}+\sqrt{c}</math> where <math>a, b</math>, and <math>c</math> are positive integers <math> x = \frac {4 + \sqrt{16 + 36}}{2} </math>, therefore <math> x = 2 + \sqrt{13} </math>.

2 KB (338 words) - 19:06, 25 December 2020
1991 AHSME Problems/Problem 6
If <math>x\geq 0</math>, then <math>\sqrt{x\sqrt{x\sqrt{x}}}=</math> <math>\textbf{(A) } x\sqrt{x}\qquad

886 bytes (130 words) - 18:13, 5 September 2021
1991 AHSME Problems/Problem 8
...ed, it spreads out on the surface of water to form a circular film <math>0.1</math>cm thick. A rectangular box measuring <math>6</math>cm by <math>3</ma ...\sqrt{216}}{\pi} \quad \text{(B)} \sqrt{\frac{216}{\pi}} \quad \text{(C)} \sqrt{\frac{2160}{\pi}} \quad \text{(D)} \frac{216}{\pi} \quad \text{(E)} \frac{2

1 KB (165 words) - 16:44, 14 March 2023
Euler's inequality
...be the distance between the circumcenter and the incenter. Then <cmath>d=\sqrt{R(R-2r)}</cmath> From this formula, Euler's Inequality follows as <cmath>d^

597 bytes (106 words) - 00:44, 25 June 2015
1966 AHSME Problems/Problem 22
...a^2+b^2}=ab</math>, (III) <math>\sqrt{a^2+b^2}=a+b</math>, (IV) <math>\sqrt{a^2+b^2}=a - b</math>, where we allow <math>a</math> and <math>b</math> to

717 bytes (118 words) - 22:10, 11 March 2024
1967 AHSME Problems/Problem 3
...{s^2}{6}\qquad\text{(C)}\ \frac{s^2\sqrt{2}}{6}\qquad\text{(D)}\ \frac{s^2\sqrt{3}}{6}\qquad\text{(E)}\ \frac{s^2}{3} </math> ...c{\sqrt{3}}{6}s</math>. This means the square has side length <math>\frac{\sqrt{6}}{6}s</math> which has area of <math>\frac{s^2}{6}=\fbox{B}</math>.

685 bytes (108 words) - 01:34, 16 August 2023
1967 AHSME Problems/Problem 5
...ue of} \; r\qquad\text{(B)}\ \frac{\sqrt{2}}{r}\qquad\text{(C)}\ \frac{2}{\sqrt{r}}\qquad\text{(D)}\ \frac{2}{r}\qquad\text{(E)}\ \frac{r}{2} </math>

833 bytes (137 words) - 19:33, 28 September 2023
1967 AHSME Problems/Problem 11
\textbf{(B)}\ \sqrt{50}\qquad \textbf{(D)}\ \sqrt{200}\qquad

849 bytes (126 words) - 20:00, 28 September 2023
1962 AHSME Problems/Problem 6
...teral triangle have equal perimeters. The area of the triangle is <math>9 \sqrt{3}</math> square inches. Expressed in inches the diagonal of the square is: ...bf{(B)}\ 2\sqrt{5}\qquad\textbf{(C)}\ 4\sqrt{2}\qquad\textbf{(D)}\ \frac{9\sqrt{2}}{2}\qquad\textbf{(E)}\ \text{none of these} </math>

1 KB (174 words) - 15:55, 18 April 2018
1962 AHSME Problems/Problem 13
...math>. Find <math>S</math> when <math>R = \sqrt {48}</math> and <math>T = \sqrt {75}</math>. When <math>R=\sqrt{48}</math> and <math>T=\sqrt{75}</math> we have

969 bytes (143 words) - 22:19, 10 April 2023
1962 AHSME Problems/Problem 16
...}\ \frac{135}{2}\qquad\textbf{(D)}\ 9\sqrt{10}\qquad\textbf{(E)}\ \frac{27\sqrt{10}}{4} </math> <cmath>a=\frac{3\sqrt{10}}2</cmath>

1 KB (170 words) - 22:16, 3 October 2014
1962 AHSME Problems/Problem 18
...ac{3r^2\sqrt{3}}{4}\qquad\textbf{(D)}\ r^2\sqrt{3}\qquad\textbf{(E)}\ 3r^2\sqrt{3} </math>

902 bytes (147 words) - 22:17, 3 October 2014
1962 AHSME Problems/Problem 25
...\ 4\qquad\textbf{(B)}\ 4\sqrt{2}\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 5\sqrt{2}\qquad\textbf{(E)}\ 6 </math> ...i of the same circle, <math>-x+8=OE=OA=AD=\sqrt{x^2+16}</math>. So, <math>\sqrt{x^2+16}=-x+8</math>. Squaring both sides, we obtain <math>x^2+16=x^2-16x+64

1 KB (216 words) - 21:22, 25 November 2017
1951 AHSME Problems/Problem 39
<cmath>t_1=\frac{\sqrt{d}}4</cmath> So <math>\frac{d}{1120}+\frac{\sqrt{d}}4-7.7=0</math>. If we let <math>u=\sqrt{d}</math>, this becomes a quadratic.

2 KB (325 words) - 13:59, 19 April 2014
1988 AHSME Problems/Problem 1
<math> \sqrt{8}+\sqrt{18}= </math> ...\sqrt{3})\qquad\text{(C)}\ 7\qquad\text{(D)}\ 5\sqrt{2}\qquad\text{(E)}\ 2\sqrt{13} </cmath>

585 bytes (72 words) - 01:42, 23 October 2014
1952 AHSME Problems/Problem 6
...\textbf{(C) \ }+9 \qquad \textbf{(D) \ }2\sqrt{85} \qquad \textbf{(E) \ }\sqrt{85} </math> ...=-9 </math>. Thus, <math> r_1-r_2=\sqrt{49+4\cdot 9}=\boxed{\textbf{(E)}\ \sqrt{85}} </math>.

649 bytes (95 words) - 21:55, 1 January 2014
Square root property
<math>x=\pm\sqrt{18}</math> so the [[root|roots]] are <math>\sqrt{18}</math> and <math>-\sqrt{18}</math>.

626 bytes (98 words) - 23:55, 15 May 2014
2014 AMC 12B Problems/Problem 14
...extbf{(C)}\ 16\sqrt{3}\qquad\textbf{(D)}\ 20\sqrt{2}\qquad\textbf{(E)}\ 40\sqrt{2} </math> <cmath>4 \sqrt{x^2+y^2+z^2}</cmath>

1 KB (167 words) - 00:38, 4 August 2023
1964 AHSME Problems/Problem 33
...quad\textbf{(B) }3\sqrt{2}\qquad\textbf{(C) }3\sqrt{3}\qquad\textbf{(D) }4\sqrt{2}\qquad \textbf{(E) }2</math> P = (2.5, 1.5);

2 KB (337 words) - 00:46, 15 June 2022
1967 AHSME Problems/Problem 21
<math>\textbf{(A)}\ \frac{3\sqrt{6}}{4}\qquad \textbf{(B)}\ \frac{3\sqrt{5}}{4}\qquad

858 bytes (128 words) - 01:40, 16 August 2023
1993 AHSME Problems/Problem 29
...e <math>\sqrt{a^2 + b^2},</math> <math>\sqrt{b^2 + c^2},</math> and <math>\sqrt{a^2 + c^2}.</math> If we square each of these to obtain <math>a^2 + b^2,</m

2 KB (322 words) - 15:07, 6 July 2022
1993 AHSME Problems/Problem 13
<math>\text{(A) } \sqrt{58}\quad \text{(B) } \frac{7\sqrt{5}}{2}\quad

976 bytes (146 words) - 21:42, 7 October 2018
1992 AHSME Problems/Problem 30
Thus, <math>DM=\sqrt{19\cdot 23}</math>. Furthermore, <math>x^2=AM^2-DM^2=46^2-19\cdot 23=1679,

1 KB (213 words) - 16:00, 1 January 2015
1991 AHSME Problems/Problem 22
draw(circle((0,3sqrt(2)),sqrt(2)),black+linewidth(.75)); MP("B",(-8/3,16*sqrt(2)/3),W);MP("B'",(8/3,16*sqrt(2)/3),E);

2 KB (295 words) - 19:09, 11 October 2016
1990 AHSME Problems/Problem 5
<math>\text{(A) } \sqrt{\sqrt[3]{5\cdot 6}}\quad \text{(B) } \sqrt{6\sqrt[3]{5}}\quad

598 bytes (81 words) - 04:05, 4 February 2016
1990 AHSME Problems/Problem 28
...the inradius and <math>s</math> is the semiperimeter, <math>r=\frac{3}{2}\sqrt{1001}</math>.

2 KB (335 words) - 11:51, 5 October 2019
1990 AHSME Problems/Problem 23
<math>\text{(A) } 12\sqrt{2}\quad \text{(B) } 13\sqrt{3}\quad

903 bytes (147 words) - 13:41, 4 February 2016
1969 AHSME Problems/Problem 25
<math>\text{(A) } 2\sqrt{6}\quad \text{(C) } 8\sqrt{2}\quad

786 bytes (120 words) - 17:59, 21 April 2020
1970 AHSME Problems/Problem 35
\text{(B) } \frac{(p-q)^2}{2\sqrt{ab}}\quad \text{(E) } \sqrt{(a-b)(p-q)}</math>

2 KB (296 words) - 04:42, 2 June 2021
1952 AHSME Problems/Problem 45
<math>\text{(A) } \frac{2ab}{a+b}>\sqrt{ab}>\frac{a+b}{2}\qquad \text{(B) } \sqrt{ab}>\frac{2ab}{a+b}>\frac{a+b}{2} \\

624 bytes (104 words) - 17:36, 9 July 2015
1952 AHSME Problems/Problem 46
...n, by pythagorean theorem, the diagonal of the rectangle has length <math>\sqrt{x^{2}+y^{2}}</math>. ...he problem, the area of the new rectangle is <math>(\sqrt{x^{2}+y^{2}}+y)(\sqrt{x^{2}+y^{2}}-y)</math>

1 KB (230 words) - 18:59, 20 April 2020
1952 AHSME Problems/Problem 35
...nominator, the expression <math>\frac {\sqrt {2}}{\sqrt {2} + \sqrt {3} - \sqrt {5}}</math> is equivalent to: <math>\textbf{(A)}\ \frac {3 + \sqrt {6} + \sqrt {15}}{6} \qquad

954 bytes (119 words) - 08:54, 1 May 2016
1952 AHSME Problems/Problem 39
<math>\textbf{(A)}\ \frac {\sqrt {8d^2 - p^2}}{2} \qquad \textbf{(B)}\ \frac {\sqrt {8d^2 + p^2}}{2} \qquad

2 KB (267 words) - 22:36, 15 April 2020
1980 AHSME Problems/Problem 19
\text{(B)} \ 4\sqrt{7} \qquad \text{(C)} \ \frac{5\sqrt{65}}{3} \qquad

2 KB (350 words) - 16:41, 12 September 2021
1980 AHSME Problems/Problem 23
\text{(C)} \ \frac{3\sqrt{5}}{5} \qquad \text{(D)}\ \frac{2\sqrt{5}}{3}\qquad

2 KB (273 words) - 22:06, 11 November 2015
1958 AHSME Problems/Problem 16
\textbf{(C)}\ 200\sqrt{2}\qquad \textbf{(D)}\ 200\sqrt{3}\qquad

1 KB (159 words) - 01:38, 12 March 2017
1958 AHSME Problems/Problem 42
...ath>BX = \sqrt{144 - h^2}</math>. <math>CE = CX - EX = \sqrt{144 - h^2} - \sqrt{64 - h^2}</math>. Using Power of a Point on <math>E</math>, <math>(BE)(EC) <cmath>(\sqrt{144 - h^2} - \sqrt{64 - h^2})(\sqrt{144 - h^2} + \sqrt{64 - h^to hard problem forthis easy content to hard sweet2}) = 8(ED)</cmath

1 KB (204 words) - 23:46, 19 April 2024
2007 iTest Problems/Problem 27
...qrt{2}</math>. Thus, the volume of the cube is <math>(2 \sqrt{2})^3 = 16 \sqrt{2}</math>, so <math>n = \boxed{16}</math>.

699 bytes (105 words) - 19:15, 10 June 2018
1970 Canadian MO Problems/Problem 8
<math>b=\frac{3a \pm \sqrt{80 - a^2}}{5}</math> <math>x=\frac{b+a}{2}=\frac{8a \pm \sqrt{80 - a^2}}{10}</math>

1 KB (242 words) - 02:44, 28 November 2023
Mock AIME 5 2005-2006 Problems/Problem 12
...ying pythagoras theorem on <math>\triangle AED</math> , We get, <math>AE=\sqrt{160}</math>

2 KB (294 words) - 16:24, 24 August 2022
2013 UMO Problems/Problem 1
D(shift(sqrt(2),0)*S,black+linewidth(.75)); D(shift(2*sqrt(2),0)*S,black+linewidth(.75));

709 bytes (98 words) - 15:25, 14 October 2014
2011 UNCO Math Contest II Problems/Problem 5
A=(0,4*sqrt(10)); B=(4*sqrt(10),4*sqrt(10));

769 bytes (132 words) - 00:54, 11 March 2019
1993 UNCO Math Contest II Problems/Problem 10
draw((0,0)--(52,0)--(24,sqrt(3)*26)--cycle); draw((24,0)--(24,sqrt(3)*26));

2 KB (246 words) - 23:57, 3 June 2022
1988 AHSME Problems/Problem 21
...<math>|z|^{2}</math>? Note: if <math>z = a + bi</math>, then <math>|z| = \sqrt{a^{2} + b^{2}}</math>. ...math>b = 8</math>. Plugging this in back to our equation gives us <math>a+\sqrt{a^2+64} = 2</math>.

1 KB (193 words) - 20:58, 8 June 2016
1987 AHSME Problems/Problem 8
...th>5 + \sqrt{116}</math>, and as <math>100 < 116 < 121</math>, <math>10 < \sqrt{116} < 11</math>, so that the sum is between <math>5+10</math> and <math>5+

1 KB (139 words) - 14:02, 1 March 2018
1987 AHSME Problems/Problem 22
\textbf{(D)}\ 8\sqrt{3} \qquad \textbf{(E)}\ 6\sqrt{6} </math>

1 KB (173 words) - 18:59, 15 January 2024
1986 AHSME Problems/Problem 11
...<math>AH</math>. By Heron's Formula, <math>s=\frac{13+14+15}{2}=21, [ABC]=\sqrt{21*(21-13)(21-14)(21-15)}=84</math>. Using the area formula <math>A=0.5bh</

2 KB (290 words) - 11:41, 1 December 2022
1973 AHSME Problems/Problem 20
<math> \textbf{(A)}\ 4+\sqrt{185} \qquad \textbf{(E)}\ \sqrt{32}+\sqrt{137}</math>

798 bytes (124 words) - 14:02, 20 February 2020
2015 AMC 12B Problems/Problem 13
...(B)}\; 6 \qquad\textbf{(C)}\; \dfrac{9}{2}\sqrt{2} \qquad\textbf{(D)}\; 8-\sqrt{2} \qquad\textbf{(E)}\; 7</math>

821 bytes (120 words) - 16:06, 6 March 2015
2015 AMC 12B Problems/Problem 14
...{2} \qquad\textbf{(D)}\; 4(\pi-\sqrt{3}) \qquad\textbf{(E)}\; 2\pi-\dfrac{\sqrt{3}}{2}</math> ...n), so the answer is <math>4\pi-4\sqrt{3} = \boxed{\textbf{(D)}\; 4(\pi-\sqrt{3})}</math>.

1 KB (197 words) - 15:02, 4 March 2015
2015 AIME II Problems/Problem 4
...d <math>\log 64</math> being 6. Then the 2 legs of the trapezoid is <math>\sqrt{3^2+4^2}=5=\log 32</math>.

5 KB (766 words) - 14:17, 28 June 2023
1954 AHSME Problems/Problem 5
...ext{ sq. in.}\qquad\textbf{(D)}\ \text{600 sq. in.}\qquad\textbf{(E)}\ 300\sqrt{3}\text{ sq. in.} </math> ...\implies 100\cdot3\sqrt{3}\pi=2A\pi\implies 50\cdot3\sqrt{3}=A\implies 150\sqrt{3}\ \boxed{(\textbf{A})}</math>

657 bytes (97 words) - 20:36, 17 February 2020
1960 AHSME Problems/Problem 40
...2\sqrt{3}-9}{13}\qquad\textbf{(C)}\ 6\sqrt{3}-8\qquad\textbf{(D)}\ \frac{5\sqrt{10}}{6}\qquad\textbf{(E)}\ \frac{25}{12}\qquad</math> ...>, <math>RB</math> has length <math>\frac{3}{4}x</math>. <cmath>RC+RB=BC=x\sqrt{3}+\frac{3}{4}x=3</cmath>

1 KB (220 words) - 22:29, 28 December 2023
1960 AHSME Problems/Problem 39
<cmath>a = \frac{-b \pm \sqrt{b^2 - 4b^2}}{2}</cmath> <cmath>a = \frac{-b \pm \sqrt{-3b^2}}{2}</cmath>

1 KB (247 words) - 22:28, 28 December 2023
1983 AHSME Problems/Problem 5
...qrt 5}{3}\qquad \textbf{(C)}\ \frac{2}{\sqrt 5}\qquad \textbf{(D)}\ \frac{\sqrt{5}}{2}\qquad \textbf{(E)}\ \frac{5}{3}</math> A = (0,1.7);

1 KB (171 words) - 00:42, 20 February 2019
Proofs
...}=0 \Rightarrow x~+~\frac{b}{2a}=\pm\sqrt{\frac{b^2-4ac}{4a^2}}=\frac{\pm \sqrt{b^2-4ac}}{2a}</cmath> <cmath>\boxed{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}</cmath>

755 bytes (135 words) - 14:16, 15 May 2017
1953 AHSME Problems/Problem 8
<math>\textbf{(A)}\ -2+\sqrt{5} \qquad \textbf{(B)}\ -2-\sqrt{5} \qquad

599 bytes (91 words) - 20:40, 1 April 2017
2016 UNCO Math Contest II Problems/Problem 6
(a) <math>\frac{3}{4}</math> (b)<math>\frac{15-\sqrt{65}}{8}</math>

974 bytes (151 words) - 04:03, 13 January 2019
1979 AHSME Problems/Problem 9
The product of <math>\sqrt[3]{4}</math> and <math>\sqrt[4]{8}</math> equals <math>\textbf{(A) }\sqrt[7]{12}\qquad

663 bytes (83 words) - 13:17, 5 January 2017
1979 AHSME Problems/Problem 13
The inequality <math>y-x<\sqrt{x^2}</math> is satisfied if and only if <math>\sqrt{x^2} = \pm x</math>, so the inequality is just <math>y-x<\pm x</math>. Ther

808 bytes (134 words) - 13:05, 6 January 2017
2013 Mock AIME I Problems/Problem 12
...math> to each side gives us <math>(t-12)^{2}=46</math>. Thus <math>t=12\pm\sqrt{46}</math>. ...r of <math>\overline{AC}</math>, so we consider the case where <math>t=12-\sqrt{46}</math>.

3 KB (570 words) - 22:37, 8 June 2022
2017 USAMO Problems/Problem 3
...ega</math> be the circle centered at <math>S</math> with radius <math>R = \sqrt {SK \cdot SM}.</math>

4 KB (711 words) - 18:24, 8 May 2023
1983 AHSME Problems/Problem 28
\textbf{(D)}\ \frac{5}{3}\sqrt{10}\qquad

2 KB (365 words) - 01:11, 20 February 2019
1981 AHSME Problems/Problem 9
...rt{3}}.</math> The surface area of the cube is therefore <math>6(\frac{a}{\sqrt{3}})^2=6 \cdot \frac{a^2}{3}=\boxed{2a^2},</math> which is answer choice <m

345 bytes (60 words) - 21:14, 12 March 2018
1954 AHSME Problems/Problem 26
\textbf{(D)}\ CB\sqrt{3}\\

1 KB (184 words) - 17:31, 19 September 2022
1960 AHSME Problems/Problem 4
...} 8\qquad \textbf{(C)} 4\sqrt{3}\qquad \textbf{(D)} 4\qquad \textbf{(E)} 2\sqrt{3} </math> label("$2\sqrt{3}$",(20,15));

1 KB (154 words) - 18:54, 17 May 2018
1960 AHSME Problems/Problem 6
<math>\textbf{(A) }\frac{25\sqrt{2}}{\pi}\qquad \textbf{(B) }\frac{50\sqrt{2}}{\pi}\qquad

1 KB (178 words) - 18:57, 17 May 2018
1960 AHSME Problems/Problem 15
...frac{A^2\sqrt{3}}{4}</math>. By using 30-60-90 triangles, <math>R=\frac{A\sqrt{3}}{3}</math>. ...ath>p=3a</math>, <math>k=\frac{a^2\sqrt{3}}{4}</math>, and <math>r=\frac{a\sqrt{3}}{3}</math>.

2 KB (276 words) - 19:01, 17 May 2018
1960 AHSME Problems/Problem 21
\textbf{(B)}\ \sqrt{2}(a+b)^2\qquad \textbf{(D)}\ \sqrt{8}(a+b) \qquad

1,008 bytes (182 words) - 02:00, 1 February 2024
1960 AHSME Problems/Problem 35
...onal]], <math>t = \sqrt{mn}</math>. Since <math>m+n=10</math>, <math>t = \sqrt{m(10-m)}</math>.

1 KB (218 words) - 11:03, 15 May 2018
1961 AHSME Problems/Problem 36
\textbf{(B)}\ \sqrt{17} \qquad \textbf{(D)}\ 2\sqrt{5} \qquad

2 KB (232 words) - 11:15, 2 June 2018
1963 AHSME Problems/Problem 10
\textbf{(D)}\ \frac{a\sqrt{2}}{2}\qquad

1 KB (184 words) - 08:22, 5 June 2018
1963 AHSME Problems/Problem 35
By [[Heron's Formula]], the area of the triangle is <math>\sqrt{24 \cdot 3(24-b)(b-3)}</math>. Plug in the answer choices for <math>b</mat ...sults in the area being <math>\sqrt{6 \cdot 4 \cdot 3 \cdot 14 \cdot 7} = \sqrt{2^4 \cdot 3^2 \cdot 7^2}</math>, so <math>10</math> works. The third side

1 KB (217 words) - 17:58, 7 June 2018
Positive root
...th>n</math>, the positive root (if there is a positive solution) of <math>\sqrt[x]{n}</math> is the solution greater than zero. The positive root of a poly

311 bytes (58 words) - 20:56, 21 February 2018
1953 AHSME Problems/Problem 26
<math>\textbf{(A)}\ 5\sqrt{6}\text{ inches} \qquad \textbf{(C)}\ 4\sqrt{3}\text{ inches}\qquad

1 KB (166 words) - 22:51, 24 February 2020
1973 AHSME Problems/Problem 4
...</math>. The area of the shared region is <math>\tfrac12 \cdot 12 \cdot 2\sqrt{3} = \boxed{\textbf{(D)}\ 12\sqrt3}</math>.

1 KB (157 words) - 13:57, 20 February 2020
1973 AHSME Problems/Problem 25
label("6",(-2.598,1.5),SW); label("3",(0,1.5),E);

1 KB (216 words) - 14:03, 20 February 2020
2008 iTest Problems/Problem 55
Thus, the length of <math>XY</math> is <math>\sqrt{1156} = \boxed{34}</math>.

1 KB (189 words) - 00:19, 13 July 2018
1973 AHSME Problems/Problem 15
...ity while <math>3\sec \tfrac12 \theta</math> would approach <math>\tfrac{3\sqrt{2}}{2}</math>. A super large circle would definitely not be a circumcircle

2 KB (306 words) - 14:00, 20 February 2020
2008 iTest Problems/Problem 32
...<math>\tfrac{2008}{2} = 1004.</math> The radius of the incircle is <math>\sqrt{100 \pi^2} = 10\pi</math>. That means the area of the triangle is <math>10

945 bytes (144 words) - 13:30, 28 February 2020
2008 iTest Problems/Problem 66
pair C=(0,0), B=(3.864,0), A=(0,1.035); dot((1.035,0));

2 KB (359 words) - 09:39, 8 August 2018
2003 Indonesia MO Problems/Problem 8
<cmath>ab = 3(a+b+\sqrt{a^2 + b^2})</cmath> ab-3(a+b) &= 3\sqrt{a^2 + b^2} \\

1 KB (204 words) - 01:12, 11 August 2018
2008 iTest Problems/Problem 86
...Thus, <math>a^2 + 2ab + b^2 = 4008</math>, so <math>a+b = \sqrt{4008} = 2\sqrt{1002}</math>. ...same steps to show that <math>c+d = 2\sqrt{1002}</math>. Thus, <math>S = 4\sqrt{1002} \approx 126.62</math>, so <math>\lfloor S\rfloor = \boxed{126}</math>

1 KB (173 words) - 21:36, 21 November 2018
2008 iTest Problems/Problem 89
...rt{54^2 - 21^2} = 15\sqrt{11}</math> and <math>YM = \sqrt{66^2 - 21^2} = 3\sqrt{435}</math>. ...math>OY \perp YM</math>, <math>OXMY</math> is a rectangle, so <math>OX = 3\sqrt{435}</math>.

2 KB (250 words) - 18:53, 23 November 2018
2006 iTest Problems/Problem 8
...uad\mathrm{(C)}\,6\sqrt{3}\quad\mathrm{(D)}\,4\sqrt{3}\quad\mathrm{(E)}\,4\sqrt{2}\quad\mathrm{(F)}\,12\quad\mathrm{(G)}\,6\quad\mathrm{(H)}\,\text{none of

2 KB (307 words) - 20:50, 29 November 2018
2006 iTest Problems/Problem 22
By the [[Pythagorean Theorem]], <math>BE = \sqrt{60^2 - 48^2} = 36</math>. Therefore, the area of triangle <math>BDE</math>

2 KB (262 words) - 22:30, 30 November 2018
The Quadratic Formula
The quadratic formula is <math>\frac{-b\pm\sqrt{b^2-4ac}}{2a}.</math> This is used to find the roots of a quadratic, <math>

588 bytes (107 words) - 19:26, 29 March 2019
Proof of Quadratic Theorem
<cmath>x+\frac{b}{2a}=\frac{\pm\sqrt{b^2-4ac}}{2a}</cmath> <cmath>x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}</cmath>

1 KB (264 words) - 17:45, 28 April 2019
2015 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 5
...that for any point <math>(x, y) \in S</math>, <math>\sqrt{x^2 + y^2} = 3 \sqrt{(x - b)^2 + y^2} \implies x^2 + y^2 = 9((x - b)^2 + y^2) \implies x^2 + y^2

868 bytes (152 words) - 13:29, 21 January 2024
2017 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 5
...^4}{3}})</math> or <math>(\sqrt[3]{20},-\frac{3\sqrt[3]{20}}{ 2} , \frac{5\sqrt[3]{20}}{ 2})</math>

561 bytes (78 words) - 04:14, 19 January 2019
2018 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 4
Suppose ABCD is a parallelogram with area <math>39\sqrt{95}</math> square units and <math>\angle{DAC}</math> is a right angle. If t Then one diagonal is <math>\sqrt{y^{2} - x^{2}}</math>

729 bytes (117 words) - 06:39, 9 August 2019
2018 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 5
...h> and <math>y</math> be two real numbers satisfying <math>x-4\sqrt{y} = 2\sqrt{x-y}</math>. What are all the possible values of <math>x</math>?

313 bytes (47 words) - 04:46, 20 January 2019
1988 IMO Problems/Problem 5
<cmath> \frac{E}{E_1}=\frac{AB \cdot AC}{AD^2} = \frac{BC}{AD} =\frac{BD+CD}{\sqrt{BD \cdot CD}}\geq 2 </cmath>

3 KB (578 words) - 11:38, 30 January 2021
Proof that the square root of any nonperfect square positive integer is irrational
...a nonperfect square positive integer is a perfect square. Therefore <math>\sqrt{n}</math> is irrational.

435 bytes (72 words) - 14:17, 19 June 2019
2006 SMT/General Problems/Problem 12
...can divide 8091 by 9, yielding 899. Testing all prime numbers up to <math>\sqrt{899} \approx 30</math>, we see that 899 is divisible by 29.

330 bytes (46 words) - 18:06, 14 January 2020
2019 Mock AMC 10B Problems/Problem 3
...h></cmath>C) <math>3^2+\sqrt{2}^2+6\sqrt{2}</math> which equals <math>11+6\sqrt{2}</math> which is irrational <cmath></cmath>E) <math>(3-\sqrt{2})(3+\sqrt{2})=9-2=7</math> which is rational

585 bytes (85 words) - 11:47, 5 November 2019
1953 AHSME Problems/Problem 42
\textbf{(D)}\ 40.1\text{ inches}\\ \textbf{(E)}\ 40\text{ inches} </math> ...Theorem]] on right triangle <math>ABE</math>, <math>CD=AE=\sqrt{41^2-9^2}=\sqrt{1600}=40</math>. The length of the common internal tangent is <math>\boxed{

2 KB (264 words) - 01:59, 14 February 2020
1955 AHSME Problems/Problem 31
...quad\textbf{(D)}\ \frac{2+\sqrt{2}}{2}\qquad\textbf{(E)}\ \frac{2\sqrt{3}-\sqrt{6}}{2}</math> ...of the trapezoids, so it works out to be <math>\boxed{\textbf{(D)}\frac{2+\sqrt{2}}{2}}</math> units long.

1 KB (166 words) - 19:17, 18 November 2020
2005 Indonesia MO Problems/Problem 3
...ath> be positive integers such that <math>\frac12\left(\sqrt{k+4\sqrt{m}}-\sqrt{k}\right)</math> is an integer. (a) Prove that <math> \sqrt{k}</math> is rational.

1 KB (196 words) - 12:33, 17 September 2019
2019 Mock AMC 10B Problems/Problem 15
Solving for m gives m= 12+sqrt(72)/2, m = 6+3sqrt2, a=6, b=3, c=2. a+b+c=11. (A)

283 bytes (49 words) - 00:23, 14 December 2019
1953 AHSME Problems/Problem 37
<math>\textbf{(A)}\ \frac{7\sqrt{15}}{5} \qquad \textbf{(B)}\ 4\sqrt{3} \qquad

2 KB (245 words) - 01:13, 25 January 2020
1972 AHSME Problems/Problem 25
This means that the diameter is <math>\sqrt{4225}=65</math> so our answer is <math>\boxed{C}</math>.

1 KB (237 words) - 10:14, 29 January 2021
1955 AHSME Problems/Problem 18
The discriminant of the equation <math>x^2+2x\sqrt{3}+3=0</math> is zero. Hence, its roots are: ...The distinct solution^ to the equation, as we can clearly see, is <math>-\sqrt{3}</math>, which is <math> \textbf{(A)} \text{real and equal}</math>

846 bytes (130 words) - 17:55, 2 August 2020
1955 AHSME Problems/Problem 20
The expression <math>\sqrt{25-t^2}+5</math> equals zero for: ...owing equation: <math>\sqrt{25-t^2}+5 = 0</math>, which means that <math>\sqrt{25-t^2} = -5</math> Square both sides, and we get <math>25 - t^2 = 25</math

873 bytes (137 words) - 12:38, 3 August 2020
1981 AHSME Problems/Problem 2
...t{3}\qquad\textbf{(B)}\ \sqrt{5}\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 2\sqrt{3}\qquad\textbf{(E)}\ 5</math> ...ight triangle. Thus, we do Pythagorean theorem to find that side <math>BC=\sqrt{3}</math>. Since this is the side length of the square, the area of <math>A

589 bytes (101 words) - 22:16, 7 August 2020
1955 AHSME Problems/Problem 36
...tbf{(D)}\ 3+\sqrt{5}\\ \textbf{(E)}\ \text{either }3-\sqrt{5}\text{ or }3+\sqrt{5} </math>

883 bytes (140 words) - 11:56, 11 August 2020

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