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  • <math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math> 10^x\cdot100^{2x} & =10^x\cdot(10^2)^{2x} \
    1 KB (190 words) - 09:58, 16 June 2023
  • 190 bytes (26 words) - 05:13, 16 February 2024
  • given (a) <math>A=\sqrt{2}</math>, (b) <math>A=1</math>, (c) <math>A=2</math>, where only non-negative real numbers are admitted for square roots? The square roots imply that <math>x\ge \frac{1}{2}</math>.
    3 KB (464 words) - 00:29, 5 November 2024
  • {{duplicate|[[2001 AMC 12 Problems|2001 AMC 12 #2]] and [[2001 AMC 10 Problems|2001 AMC 10 #6]]}} <math>\text{(A)}\ 2\qquad \text{(B)}\ 3\qquad \text{(C)}\ 6\qquad \text{(D)}\ 8\qquad \text{(E)
    1,007 bytes (165 words) - 19:31, 27 October 2024
  • ...]] <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 elements The smallest <math>S</math> is <math>1+2+ \ldots +90 = 91 \cdot 45 = 4095</math>. The largest <math>S</math> is <mat
    1 KB (189 words) - 19:05, 4 July 2013
  • 473 bytes (71 words) - 09:44, 4 July 2013
  • #REDIRECT [[2006 AMC 12A Problems/Problem 2]]
    45 bytes (5 words) - 13:45, 14 January 2016
  • ...C 12B Problems|2005 AMC 12B #2]] and [[2005 AMC 10B Problems|2005 AMC 10B #2]]}} \textbf{(A) }\ 2 \qquad
    1 KB (145 words) - 12:56, 14 December 2021
  • 289 bytes (45 words) - 12:14, 16 July 2017
  • 1 KB (164 words) - 13:58, 14 April 2020
  • *Person 2: <math>\frac{6 \cdot 4 \cdot 2}{6 \cdot 5 \cdot 4} = \frac 25</math> Our answer is thus <math>\frac{9}{28} \cdot \frac{2}{5} = \frac{9}{70}</math>, and <math>m + n = \boxed{79}</math>.
    4 KB (628 words) - 10:28, 14 April 2024
  • #REDIRECT [[2006 AMC 12A Problems/Problem 2]]
    45 bytes (5 words) - 09:57, 20 February 2016
  • #REDIRECT [[2006 AMC 12A Problems/Problem 2]]
    45 bytes (5 words) - 10:00, 20 February 2016
  • ...th>(334,6)</math>, <math>(501,4)</math>, <math>(668,3)</math>, <math>(1002,2)</math> and <math>(2004,1)</math>, and each of these gives a possible value ...that there are <math>(2 + 1)(1 + 1)(1 + 1) = 12</math> divisors of <math>2^2\cdot 3^1\cdot 167^1</math>.
    2 KB (303 words) - 00:31, 5 December 2022
  • ...Therefore, the largest element in <math>A</math> is <math>2 + \frac{m-1}{2}</math>. ...ath>m < 3</math>, which is clearly not possible, thus <math>2 + \frac{m-1}{2} < m</math>.
    8 KB (1,437 words) - 20:53, 19 May 2023
  • <cmath>2 \cdot \frac{14}{323} + \frac{90}{323} = \frac{118}{323}</cmath> ...e two red candies will have the form <math>\frac{{8\choose 2}}{{18 \choose 2}}</math>, and the probability that Terry chooses two different candies will
    2 KB (330 words) - 12:42, 1 January 2015
  • ...nct from <math>J</math>. The value of <math>B - J</math> is at least <math>2</math> with a probability that can be expressed in the form <math>\frac{m}{ ...e J</math>, so the probability that <math>B-J < 0</math> is <math>\frac{1}{2}</math> by symmetry.
    5 KB (830 words) - 21:15, 28 December 2023
  • === Solution 2 ===
    1 KB (184 words) - 21:19, 20 October 2024
  • == Solution 2== ...uiv 8 \cdot (-5)^k \equiv 5 \pmod{15}</math> for all integers <math>k \geq 2</math>. Since we are restricted to only the digits <math>8,0</math>, becaus
    2 KB (260 words) - 00:24, 18 July 2024
  • ...\frac{1}{a^3(b+c)} + \frac{1}{b^3(c+a)} + \frac{1}{c^3(a+b)} \geq \frac{3}{2}. </cmath> &= \frac{x^2}{y+z} + \frac{y^2}{z+x} + \frac{z^2}{x+y} .
    7 KB (1,164 words) - 00:01, 26 July 2024

Page text matches

  • ...length <math>c</math> we have the relationship <math>{a}^{2}+{b}^{2}={c}^{2}</math>. This theorem has been know since antiquity and is a classic to pr <math> \frac{[ABC]}{AB^2} = \frac{[CBH]}{CB^2} = \frac{[ACH]}{AC^2} </math>.
    5 KB (885 words) - 22:03, 5 October 2024
  • draw(O3--P3,linetype("2 2")); ..."text-align:center;"><math>(r-3)^2 + (r-1)^2 = (r+1)^2</math><br /><math>r^2 - 10r + 9 = 0</math><br /><math>r = 1, 9</math></div>
    2 KB (307 words) - 23:58, 17 November 2024
  • <math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math> 10^x\cdot100^{2x} & =10^x\cdot(10^2)^{2x} \
    1 KB (190 words) - 09:58, 16 June 2023
  • ==Solution 2== {{AMC10 box|year=2016|ab=A|num-b=2|num-a=4}}
    1 KB (176 words) - 09:58, 16 June 2023
  • ...x}{y}</math>. What is the value of <math>\text{rem} (\tfrac{3}{8}, -\tfrac{2}{5} )</math>? \text{rem}\left(\frac{3}{8},-\frac{2}{5}\right)
    2 KB (257 words) - 09:57, 16 June 2023
  • ...h>. Then the volume is <math>x \cdot 3x \cdot 4x =12x^3</math>. If <math>x=2</math>, then <math>12x^3 = 96 \implies \boxed{\textbf{(D) } 96.}</math> == Solution 2 ==
    1 KB (216 words) - 23:23, 5 September 2024
  • ...Emilio copies Star's numbers, replacing each occurrence of the digit <math>2</math> by the digit <math>1</math>. Star adds her numbers and Emilio adds h ...its digit 2, we subtract 1. Because 2 appears 10 times as a tens digit and 2 appears 3 times as a units digit, the answer is <math>10\cdot 10+1\cdot 3=\
    967 bytes (143 words) - 02:18, 27 June 2023
  • ==Solution 2==
    2 KB (268 words) - 17:19, 27 September 2023
  • ...(\frac{1}{2}\right)=30 , \Rightarrow (+40)=70 , \Rightarrow \left(\frac{1}{2}\right)=\boxed{\textbf{(C) }35}</cmath> == Solution 2 ==
    1 KB (169 words) - 13:59, 8 August 2021
  • ...of <math>2016</math> coins has <math>1</math> coin in the first row, <math>2</math> coins in the second row, <math>3</math> coins in the third row, and ...-1)+N=\frac{N(N+1)}{2}=2016.</cmath> Noticing that <math>\frac{63\cdot 64}{2}=2016,</math> we have <math>N=63,</math> so our answer is <math>\boxed{\tex
    2 KB (395 words) - 22:29, 3 December 2024
  • filldraw(rectangle((2,2),(5,3)),white); label("$1$",(0.5,2.5));
    2 KB (335 words) - 04:52, 18 December 2024
  • label("$1$",(1/2,5),dir(90)); label("$7$",(9/2,5),dir(90));
    8 KB (1,016 words) - 23:17, 30 December 2023
  • A triangle with vertices <math>A(0, 2)</math>, <math>B(-3, 2)</math>, and <math>C(-3, 0)</math> is reflected about the <math>x</math>-ax ==Video Solution 2==
    1 KB (235 words) - 13:52, 25 June 2023
  • ...t is slightly less than <math>\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}</math>, so <math>\boxed{\textbf{(A) }p<\dfrac{1}{8}}</math>. ==Solution 2 (Disregards Order)==
    2 KB (302 words) - 00:57, 18 October 2024
  • <math>\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad \textbf{(E) } 5</math> ...at 1. Ceci must have been in seat 3 to keep seat 1 open, which leaves seat 2.
    2 KB (402 words) - 13:54, 25 June 2023
  • ...problem solving contest for elementary and middle school students. [Grades 2-8]
    5 KB (512 words) - 08:27, 17 November 2024
  • ...onal Competition]]: The largest online math contest for students in grades 2-8. ...g Math Contest]]: a popular problem-solving contest for students in grades 2-8.
    7 KB (805 words) - 15:17, 14 December 2024
  • ...ate 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 gallim
    5 KB (706 words) - 22:49, 29 January 2024
  • * [[Beestar National Competition]](BNC) - Grades 2 - 7 (http://www.beestar.org)
    4 KB (410 words) - 18:21, 19 November 2024
  • ...er), 2-2.5 (State/National)<br><u>Target:</u> 1.5 (School), 2 (Chapter), 2-2.5 (State/National)}} 8 problems are given 2 at a time. Students have 6 minutes to complete each set of two problems. S
    10 KB (1,504 words) - 13:10, 1 December 2024

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