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  • ...$c$", D--C, N); label("$d$",D--A,W); label("$u$",D--B,2*dir(170)); label("$v$",A--C,S); ...c, d</math>, be the sides, <math>s</math> the semiperimeter, and <math>u, v</math>, the diagonals. Then the area, <math>K</math>, is given by <cmath>K
    10 KB (1,669 words) - 17:33, 12 January 2024
  • Denote <math>D' – A' = 2\vec V.</math> ...\triangle A'C'E'</math> into <math>\triangle D'F'B'</math> is <math>2\vec {V.}</math>
    5 KB (875 words) - 00:26, 23 May 2023
  • ...<math>n=2k+1</math>, just take <math>(u,v)= (k+1,k)</math> which <math>u^2-v^2=n</math>. Thus the only answer is <math>\boxed{f\equiv 1}</math> and we a https://www.youtube.com/watch?v=Q1NUBUYvOJc
    8 KB (1,492 words) - 00:23, 24 February 2024
  • ...th>t</math> is square free and <math>gcd(u,v,w) = 1</math>, find <math>t+u+v+w</math> if <math>x=\cos 20^o</math> and <math>y=\sin 20^o</math>.
    8 KB (1,385 words) - 12:55, 23 June 2021
  • We could assume a variable <math>v</math> which equals to both <math>\log_{20x} (22x)</math> and <math>\log_{2 So that <math>(20x)^v=22x \textcircled{1}</math>
    5 KB (778 words) - 18:14, 30 January 2024
  • ...s, plus the outer region, <math>E</math> is the number of edges, and <math>V</math> is the number of vertices. Temporarily disregarding the intersection <math>V=V_{i}+210=222</math>
    12 KB (2,025 words) - 14:56, 25 January 2024
  • ...\qquad \text{(iii) }x+y\ge a\\ \\ \qquad \text{(iv) }x+a\ge y\qquad \text{(v) }y+a\ge x</math>
    665 bytes (119 words) - 18:31, 28 August 2023
  • ...t,st),E); label("M",(cu,su),N);label("P",(cu,st),S); label("C",(cos(v),sin(v)),W); //Credit to Zimbalono for the diagram </asy>
    864 bytes (169 words) - 16:47, 23 June 2021
  • ...th>C(1) = 0</math> and our recursive rules for <math>C(n)</math> and <math>V(n)</math> as follows: n & V(n) & C(n) \\ \hline
    8 KB (1,309 words) - 00:31, 6 January 2023
  • .../math> bisects the angle between <math>\mathbf{u}</math> and <math>\mathbf{v}</math>.
    16 KB (2,526 words) - 00:53, 6 May 2023
  • ...h>. Hence, <math>B = \left( - 1 , u \right)</math>, <math>D = \left( 0 , - v \right)</math>. The slope of <math>AD</math> is <math>m_{AD} = \frac{v}{3}</math>.
    5 KB (756 words) - 03:40, 23 January 2023
  • v v
    2 KB (265 words) - 22:25, 15 April 2024
  • Let <math>G</math> be a graph with the set of vertices <math>V=\{v_1,v_2,v_3,\ldots, v_n\}</math>. We define its adjacency matrix <math>\
    1 KB (264 words) - 11:26, 3 March 2022
  • ...> the formula for variance if <math>X</math> is a [[population]] is <cmath>V = \frac{1}{n}\sum_{i=1}^n (x_i - \overline{x})^2.</cmath> Additionally, <math>V = \sigma^2</math> where <math>\sigma</math> is the population [[standard de
    749 bytes (122 words) - 19:16, 3 March 2022
  • ...>, (2) <math>\hat{x}\equiv x\pmod{p^{n-k}}</math> and (3) <math>k=v(p'(x))=v(p'(\hat{x}))</math>. <center><cmath>v=u+p^{n-2k}zs\in u+p\mathbb{Z}_p\subset \mathbb{Z}_p^{\times}</cmath></cente
    13 KB (2,298 words) - 23:34, 28 May 2023
  • pair o = (0, 0), u = (1, 0), v = 0.8*dir(40), w = dir(70); draw(o--u--(u+v));
    7 KB (1,154 words) - 12:54, 20 February 2024
  • ...t{R}(\text{T}))</math> where <math>\text{R}(\text{T})=\{\text{T}(x)~|~x\in V\}</math>. In other words, rank is the dimension of the linear transformati ...aim that if <math>\beta=\{v_1,v_2,\ldots, v_n\}</math> is a basis of <math>V</math>, then
    5 KB (893 words) - 22:41, 28 May 2022
  • ...math> In accordance with Claim, <math>\angle BVD = \angle HVE \implies B', V,</math> and <math>B</math> are collinear.
    10 KB (1,751 words) - 15:34, 25 November 2022
  • ...mathbf{n}\,dS,</cmath> where <math>R</math> is a region whose volume <math>V(R)</math> shrinks to <math>0</math> about the point <math>p</math>, <math>\ ...mathbf{n}\,dS,</cmath> where <math>R</math> is a region whose volume <math>V(R)</math> shrinks to <math>0</math> about the point <math>p</math>, <math>\
    7 KB (1,189 words) - 06:00, 21 December 2022
  • ...the three roots of the polynomial. The lengthened prism's volume is <cmath>V = (a+2)(b+2)(c+2) = abc+2ac+2ab+2bc+4a+4b+4c+8 = abc + 2(ab+ac+bc) + 4(a+b+ We can substitute these into the expression, obtaining <cmath>V = \frac{6}{10} + 2\left(\frac{29}{10}\right) + 4\left(\frac{39}{10}\right)
    7 KB (1,111 words) - 21:00, 21 February 2024
  • ...e R.H.S. is a not a perfect square. Thus, the number of positive <math>(u, v)</math> is equal to the number of positive divisors of <math>2 r^2</math>. ...n if <math>(a, b)</math> with <math>a = 2 r - u</math> and <math>b = 2 r - v</math>.
    7 KB (1,271 words) - 12:09, 4 January 2024
  • We denote by <math>\left( u, v \right)</math> the coordinates of circle <math>C_0</math>, and <math>r</mat u^2 + v^2 = \left( 8 - r \right)^2 . \hspace{1cm} (1)
    7 KB (1,202 words) - 01:15, 10 June 2023
  • .../math>, define the binary operation <math>\otimes</math> by<cmath>u\otimes v=ac+bdi.</cmath>Suppose <math>z</math> is a complex number such that <math>z
    13 KB (2,107 words) - 22:19, 20 April 2024
  • pair P,Q,R,S,T,U,V,W; P=(0,30); Q=(30,30); R=(40,40); S=(10,40); T=(10,10); U=(40,10); V=(30,0); W=(0,0);
    20 KB (2,846 words) - 12:06, 20 April 2024
  • <math>V=\pi{r^2}h</math> <math>V=\pi{r^2}(14-2r)</math>
    4 KB (697 words) - 17:07, 24 March 2023
  • path v=(5,-0.5)--(5,0)--(5,1)--(6,1)--(6,2)--(6,3)--(6,4)--(7,4); draw(v,p);
    11 KB (1,834 words) - 22:01, 4 January 2024
  • pair o = (0, 0), u = (1, 0), v = 0.8*dir(40), w = dir(70); draw(o--u--(u+v));
    13 KB (2,130 words) - 01:52, 31 January 2024
  • <cmath> V = \frac{1}{3} \cdot x \cdot (S_1 + \sqrt{S_1S_2} + S_2) = \frac{1}{3} \cdot Let the volume without water be <math>V,</math> volume of the pyramid <math>SCGJ</math> be <math>U.</math>
    10 KB (1,574 words) - 16:42, 8 March 2024
  • ...en so are <math>f(V)</math>, <math>V+W</math>, <math>V-W</math>, and <math>V \cdot W</math>.
    5 KB (772 words) - 08:50, 23 February 2023
  • Plugging in <math>A</math> and <math>B</math> gives <math>u=v=0</math>. Plugging in <math>P</math> gives Since it passes through <math>A</math>, for some <math>v, w</math>, the equation of circle <math>(ABP)</math> is,
    7 KB (1,280 words) - 01:22, 6 February 2024
  • Plugging in <math>A</math> and <math>B</math> gives <math>u=v=0</math>. Plugging in <math>P</math> gives
    4 KB (684 words) - 01:55, 6 February 2024
  • <cmath>u = 5y - 1, v = 2x - 3y + 1 \implies u^2 - 5 v^2 = - 4 \implies</cmath>
    15 KB (2,351 words) - 08:28, 20 April 2023
  • https://youtube.com/watch?v=LAuyU2OuVzE ...and <math>w=\frac{2b^3cx_0z_0}{c^2y_0^2-b^2z_0^2}</math>, so <cmath>\frac{v}{w}=\frac{b^2z_0}{c^2y_0},</cmath> which is the required condition for <mat
    14 KB (2,600 words) - 23:37, 10 March 2024
  • ...n c&oacute; biết tại sao SXMN lại c&oacute; sức hấp dẫn như vậy hay kh&ocirc;ng, h&atilde;y c&ugrave;ng l&yacute; giải qua b&agrave;i <p><strong><span style="color: #000000;">XSMN ph&aacute;t triển với mục đ&iacute;ch &Iacute;ch nước lợi nh&agrave;</span></strong><
    11 KB (2,230 words) - 23:24, 9 May 2023
  • 231. Gmaas once demanded Epic Games to give him 5,000,000 V-bucks for his 569823rd birthday. EDIT: This is why Gmaas no longer has an E
    88 KB (14,927 words) - 01:36, 16 April 2024
  • ...math>AB</math> and <math>AC</math> meet the line <math>AU</math> and <math>V</math> and <math>W</math>, respectively. The lines <math>BV</math> and <ma
    736 bytes (124 words) - 23:58, 28 March 2024
  • ...th>f(u) = f(t^{2}+v) = f^{-1}(v)+f(t)^{2}> f^{-1}(v) = f(f(f^{-1}(v))) = f(v)</math>. Hence f is strictly increasing. It is now clear that since <math>f
    1 KB (282 words) - 14:53, 12 March 2024
  • ...sed as <math>(2u-3w, v+4w)</math> with <math>0\le u\le1</math>, <math>0\le v\le1,</math> and <math>0\le w\le1</math>?
    13 KB (1,959 words) - 10:29, 4 April 2024
  • <math>V = \sqrt{(a^2 + b^2 - c^2)(b^2 + c^2 - a^2)(a^2 + c^2 - b^2)/72}</math> ...umes <cmath>[ABCI] + [ABDI] + [ACDI] + [BCDI] = V,</cmath> We have <cmath>V = \frac{4AR}{3}.</cmath> Plugging in and simplifying, we get <math>R = \fra
    13 KB (2,042 words) - 09:34, 3 April 2024
  • label("$v'$", (0, 2.18), NE); ...eed to find <math>v</math> with the largest imaginary part such that <math>v^2</math> lies on <math>\omega</math>.
    6 KB (1,055 words) - 21:58, 28 February 2024
  • https://www.youtube.com/watch?v=dxYw1wYHid4&t=12s ...t, to find the amount of vertices we can use Euler's characteristic, <math>V - E + F = 2</math>, and therefore the amount of vertices is <math>12</math>
    12 KB (2,101 words) - 10:08, 31 March 2024
  • ...nvex polyhedra, <math>V-E+F=2</math>. In our case, <math>V-24+12=2\implies V=14.</math> We know that <math>A+B</math> is the total number of vertices as Note that Euler's formula is <math>V+F-E=2</math>. We know <math>F=12</math> from the question. We also know <ma
    7 KB (1,175 words) - 16:45, 28 January 2024
  • pair K, L, M, N, O, P, Q, R, S, T, U, V; pair K, L, M, N, O, P, Q, R, S, T, U, V;
    19 KB (2,967 words) - 16:56, 24 February 2024
  • ...sed as <math>(2u-3w, v+4w)</math> with <math>0\le u\le1</math>, <math>0\le v\le1,</math> and <math>0\le w\le1</math>? ...h <math>x</math> and <math>y</math>. We first fix <math>u</math> and <math>v</math> to <math>0</math>, and graph <math>(-3w,4w)</math> from <math>0\le w
    6 KB (983 words) - 13:53, 4 April 2024
  • V
    814 bytes (128 words) - 10:36, 11 November 2023
  • Denote <math>m = k u</math> and <math>n = k v</math>. Thus, <math>{\rm gcd} \left( u, v \right) = 1</math>.
    4 KB (616 words) - 14:14, 29 March 2024
  • pair P,Q,R,S,T,U,V,W; P=(0,30); Q=(30,30); R=(40,40); S=(10,40); T=(10,10); U=(40,10); V=(30,0); W=(0,0);
    3 KB (426 words) - 01:50, 18 April 2024
  • For complex number <math>u = a+bi</math> and <math>v = c+di</math> (where <math>i=\sqrt{-1}</math>), define the binary operation <math>u \otimes v = ac + bdi</math>
    1 KB (205 words) - 02:52, 20 November 2023
  • Given the problem \( (f(x) + f(y))(f(u) + f(v)) = f(xu - yv) + f(xv - yu) \), we aim to find a function that satisfies it We start by considering the case when \( x = y = u = v = 0 \).
    2 KB (369 words) - 16:12, 26 March 2024
  • * [[V Olimpiada Iberoamericana de Matemáticas (OIM) ]] Valladolid, España , 19 * [[X Olimpiada Iberoamericana de Matemáticas (OIM) ]] Región V, Chile , 1995
    14 KB (1,323 words) - 12:18, 13 December 2023
  • ...at if <math>x+yt+zt^2 \ne 0</math>, then there exist <math>u</math>, <math>v</math>, and <math>w</math> as rational numbers such that:
    511 bytes (84 words) - 13:28, 13 December 2023
  • <math>V=\sum_{i}^{}2^ik_i=13^3</math> where <math>k_i</math> is the quantity of cub
    2 KB (387 words) - 10:29, 23 December 2023
  • ...lar polygon with <math>n</math> sides, <math>n \ge 4</math>, and let <math>V</math> be a subset of <math>r</math> vertices of the polygon. Show that if vertices are in <math>V</math>.
    396 bytes (66 words) - 04:02, 14 December 2023
  • ...h> by 2000, and <math>y'</math> is the reminder of the division of <math>y+v'</math> by 2001. There are two ships on the board: the Martian one and the
    1 KB (256 words) - 04:26, 14 December 2023
  • ...h> is greater than the ''dragonian distance'' from <math>C</math> to <math>V</math>.
    1,015 bytes (177 words) - 16:49, 14 December 2023
  • 458 bytes (44 words) - 17:51, 14 December 2023
  • X [[Olimpiada Iberoamericana de Matemáticas]] was held in Región V, Chile , 1995
    455 bytes (45 words) - 17:53, 14 December 2023
  • ...here <math>d(v)</math> indicates the distance from <math>A</math> to <math>v</math>. Consider the outgoing edges <math>(A,X)</math> and <math>(A,M)</mat https://www.youtube.com/watch?v=RRTxlduaDs8
    5 KB (887 words) - 14:09, 3 April 2024
  • DVI is an exam in mathematics at the Moscow State University named after M.V. Lomonosov. The first four problems have a standard level. Denote <math>u = \frac {x}{4}, v = \frac {t}{4}.</math>
    24 KB (4,039 words) - 10:06, 26 April 2024
  • https://archifol.io/v/tjqbn89d https://archifol.io/v/brzmkkgh
    273 bytes (44 words) - 21:35, 1 April 2024

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