Create the page "A- 1" on this wiki! See also the search results found.

• 2003 AMC 12B Problems
  == Problem 1 == ...xt {(B) } -\frac{2}{3} \qquad \text {(C) } \frac{2}{3} \qquad \text {(D) } 1 \qquad \text {(E) } \frac{14}{3}
  13 KB (1,987 words) - 18:53, 10 December 2022
• 2004 AMC 12B Problems
  == Problem 1 == [[2004 AMC 12B Problems/Problem 1|Solution]]
  13 KB (2,049 words) - 13:03, 19 February 2020
• 2005 AMC 12B Problems
  == Problem 1 == .../math> dollars. They sell all the candy bars at the price of two for <math>1</math> dollar. What was their profit, in dollars?
  12 KB (1,781 words) - 12:38, 14 July 2022
• 2006 AMC 12B Problems/Problem 1
  What is <math>( - 1)^1 + ( - 1)^2 + \cdots + ( - 1)^{2006}</math>? ...} - 2006 \qquad \text {(B) } - 1 \qquad \text {(C) } 0 \qquad \text {(D) } 1 \qquad \text {(E) } 2006
  468 bytes (70 words) - 10:40, 15 September 2017
• 2006 AMC 12B Problems/Problem 2
  {{AMC12 box|year=2006|ab=B|num-b=1|num-a=3}}
  473 bytes (71 words) - 10:44, 4 July 2013
• 2006 AMC 12B Problems/Problem 3
  ===Solution 1===
  910 bytes (136 words) - 13:39, 13 February 2016
• 2006 AMC 12B Problems/Problem 4
  ...e items are <math>7.99</math>, <math>4.99</math>, <math>2.99</math>, <math>1.99</math>, and <math>0.99</math>. Mary will pay with a twenty-dollar bill. The total price of the items is <math>(8-.01)+(5-.01)+(3-.01)+(2-.01)+(1-.01)=19-.05=18.95</math>
  1 KB (152 words) - 16:11, 8 December 2013
• 2006 AMC 12B Problems/Problem 5
  ...b is also walking east, but at a speed of 5 miles per hour. If Bob is now 1 mile west of John, how many minutes will it take for Bob to catch up to Joh ...iles per hour. Since Bob is one mile behind John, it will take <math>\frac{1}{2} \Rightarrow \text{(A)}</math> of an hour to catch up to John.
  654 bytes (115 words) - 21:47, 1 August 2020
• 2006 AMC 12B Problems/Problem 7
  == Solution 1 == Now there are <math>2 \times 1</math> ways to seat the adults.
  1 KB (213 words) - 15:33, 9 April 2024
• 2006 AMC 12B Problems/Problem 8
  ...a</math> and <math>y = \frac 14x + b</math> intersect at the point <math>(1,2)</math>. What is <math>a + b</math>?<!-- don't remove the following tag, \text {(A) } 0 \qquad \text {(B) } \frac 34 \qquad \text {(C) } 1 \qquad \text {(D) } 2 \qquad \text {(E) } \frac 94
  1 KB (235 words) - 00:46, 6 January 2022
• 2006 AMC 12B Problems/Problem 9
  ==Solution 1 (Alcumus Edition)== ...then <math>a</math> and <math>b</math> must each be chosen from the digits 1, 2, and 3. Therefore there are <math>\binom{3}{2}=3</math> choices for <mat
  3 KB (409 words) - 17:10, 30 April 2024
• 2006 AMC 12B Problems/Problem 11
  \text {(A) } \frac 67 \qquad \text {(B) } \frac {13}{14} \qquad \text {(C) } 1 \qquad \text {(D) } \frac {14}{13} \qquad \text {(E) } \frac 76 ...and drank <math>\frac{1}{7}</math> of it. Therefore, she drank <math>\frac{1}{7}</math> of her cream, leaving her <math>2*\frac{6}{7}</math>.
  927 bytes (137 words) - 10:45, 4 July 2013
• 2006 AMC 12B Problems/Problem 12
  == Solution 1==
  1 KB (239 words) - 22:20, 17 October 2020
• 2006 AMC 12B Problems/Problem 13
  pair A=origin, B=(2,0), C=(3, sqrt(3)), D=(1, sqrt(3)), E=(1, 1/sqrt(3)), F=(2, 2/sqrt(3)); == Solution 1==
  3 KB (447 words) - 03:49, 16 January 2021
• 2006 AMC 12B Problems/Problem 14
  ..., <math>J</math> and <math>N</math> are all positive integers with <math>N>1</math>. What is the cost of the jam Elmo uses to make the sandwiches? \mathrm{(A)}\ 1.05
  1 KB (227 words) - 17:21, 8 December 2013
• 2006 AMC 12B Problems/Problem 15
  O[1] = (6,0); draw(Circle(O[1],2));
  3 KB (458 words) - 16:40, 6 October 2019
• 2006 AMC 12B Problems/Problem 16
  ...tices <math>A</math> and <math>C</math> at <math>(0,0)</math> and <math>(7,1)</math>, respectively. What is its area? ...ore, the side length of the hexagon is <math>\frac{5\sqrt{2}}{2}\cdot\frac{1}{\sqrt{3}}\cdot2 = \frac{5\sqrt{6}}{3}</math>.
  1 KB (203 words) - 16:36, 18 September 2023
• 2006 AMC 12B Problems/Problem 17
  ...ath>, <math>5</math> and <math>6</math> on each die are in the ratio <math>1:2:3:4:5:6</math>. What is the probability of rolling a total of <math>7</ma ...>1</math> on the first and <math>6</math> on the second die is <math>\frac 1{21}\cdot\frac 6{21}</math>. Similarly we can express the probabilities for
  1 KB (188 words) - 22:10, 9 June 2016
• 2006 AMC 12B Problems/Problem 18
  == Solution 1 == If the object only makes <math>1</math> move, it is obvious that there are only 4 possible points that the o
  2 KB (354 words) - 16:57, 28 December 2020
• 2006 AMC 12B Problems/Problem 19
  If <math>b=1</math>, the number is not divisible by <math>2</math> (unless it's <math>18 ...e is even, because you can only discard one number from the integers <math>1</math> through <math>9</math>, inclusive (<math>8</math> children, oldest i
  4 KB (696 words) - 09:47, 10 August 2015

View (previous 20 | next 20) (20 | 50 | 100 | 250 | 500)