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- Suppose <math>15\%</math> of <math>x</math> equals <math>20\%</math> of <math>y.</math> What ...g the given condition by <math>5</math> shows that <math>y</math> is <math>15 \cdot 5 = \boxed{\textbf{(C) }75}</math> percent of <math>x</math>.3 KB (455 words) - 16:34, 26 January 2024
- 5 KB (806 words) - 00:13, 20 October 2023
- 331 bytes (50 words) - 19:09, 23 October 2021
- 929 bytes (153 words) - 10:07, 29 January 2021
- 760 bytes (133 words) - 15:40, 29 January 2021
- 611 bytes (86 words) - 21:51, 31 January 2021
- 679 bytes (90 words) - 09:58, 2 February 2021
- == Problem 15 ==1 KB (207 words) - 21:10, 13 February 2021
- The root(s) of <math>\frac {15}{x^2 - 4} - \frac {2}{x - 2} = 1</math> is (are): Use a common denominator: <math>\frac{15 - 2(x+2)}{x^2-4} = 1</math>. After moving the <math>1</math> to the left si645 bytes (100 words) - 20:07, 12 February 2021
- ...l(\phantom{ }\underbrace{(k+2)^2-2q}_{n}\phantom{ }\bigr)&=k+2q+2. \hspace{15.5mm}(4) \\ ...th the least <math>k</math> and <math>m</math> are <math>(2,5), (5,10), (8,15),</math> and <math>(11,20).</math>14 KB (2,569 words) - 09:28, 28 March 2024
- ...8 \quad \textbf{(C)} \; 12 \quad \textbf{(D)} \; 13 \quad \textbf{(E)} \; 15 </math>900 bytes (122 words) - 11:24, 9 March 2021
- 760 bytes (111 words) - 23:16, 27 March 2021
- 1 KB (250 words) - 16:19, 1 April 2021
- 806 bytes (122 words) - 13:44, 26 April 2021
- \text{(D) }\text{ between 15 and 16}\qquad ...s between 4 and 5. <math>y=11</math>. So <math>x+y</math> is between <math>15</math> and <math>16</math>. Select <math>\boxed{D}</math>.1 KB (227 words) - 22:41, 16 August 2022
- 9 bytes (1 word) - 13:08, 29 September 2021
- ...<math>D</math>, as shown. Suppose that <math>AB = 2</math>, <math>O_1O_2 = 15</math>, <math>CD = 16</math>, and <math>ABO_1CDO_2</math> is a convex hexag point O1=(0,0),O2=(15,0),B=9*dir(30);14 KB (2,217 words) - 00:28, 29 June 2023
- 984 bytes (129 words) - 20:54, 7 September 2021
- 2 KB (342 words) - 00:02, 12 July 2021
- == Problem 15 == \textbf{(B) }15\qquad1 KB (198 words) - 14:46, 20 July 2021
Page text matches
- ...sides of the triangle have lengths <math>AB=13, BC=14,</math> and <math>CA=15,</math> and the [[tangent]] of angle <math>PAB</math> is <math>m/n,</math> pair A=(0,0),B=(13,0),C=IP(circle(A,15),circle(B,14));7 KB (1,184 words) - 13:25, 22 December 2022
- &= 0.5(\cos 0 - \cos 10 + \cos 5 - \cos 15 + \cos 10 \ldots + \cos 165 - \cos 175+ \cos 170 - \cos 180)4 KB (614 words) - 04:38, 8 December 2023
- Round 7: <math>b</math> to <math>b</math>, <math>15</math> to right, <math>16</math> left in deck, <math>n = -2 + 8k</math>, be ...ans our sieving process will return to normal after Round 7, with <math>31-15=16</math> cards remaining. Since <math>16</math> is a perfect power of <mat15 KB (2,673 words) - 19:16, 6 January 2024
- ...20*dir(75),p); MP("x",P,17*dir(245),p); MP("2x",Q,15*dir(70),p); MP("2x",A,15*dir(-90),p); MP("2y",P,2*left,p); MP("3x",P,10*dir(-95),p); MP("x+y",C,5*di {{AIME box|year=2000|n=I|num-b=13|num-a=15|t=721509}}8 KB (1,275 words) - 03:04, 27 February 2022
- Add all powers of 2: 15 Or, <math>156*1875+15*1248</math> <math>=311220</math>4 KB (667 words) - 13:58, 31 July 2020
- ...>5|a,b</math>, so there are 2 pairs of <math>a</math> and <math>b: (20,5),(15,10)</math>. ...ack in both. Thus, <math>P(\text{both white}) = \frac{1}{10}\cdot \frac{6}{15} = \frac{1}{25}</math>. <math>m + n = 26</math>.7 KB (1,011 words) - 20:09, 4 January 2024
- ...,Arrows(5)); D((0,-15)--(0,15),linewidth(0.6),Arrows(5)); D((-15,-15)--(15,15),linewidth(0.6),Arrows(5));3 KB (434 words) - 22:43, 16 May 2021
- n&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17&18&19\\\hline Our sub-cases are still the same. However, our equations become <math>2x+3y=15,16,17.</math> Computing yields <math>28+37+49=114</math> sequences.13 KB (2,298 words) - 19:46, 9 July 2020
- *[[2005 AMC 10B Problems/Problem 15]]1 KB (165 words) - 12:40, 14 August 2020
- In [[triangle]] <math>ABC</math>, <math>AB=13</math>, <math>BC=15</math> and <math>CA=17</math>. Point <math>D</math> is on <math>\overline{A pair A=(0,0),B=(13,0),C=IP(CR(A,17),CR(B,15)), D=A+p*(B-A), E=B+q*(C-B), F=C+r*(A-C);4 KB (673 words) - 20:15, 21 February 2024
- ...that the sum is equal to <math>\sum_{i = 0}^{5}{i+2\choose2}(6 - i) = 6 + 15 + 24 + 30 + 30 + 21 = 126</math>. The requested probability is <math>\frac{ #Now, for the first case, there are <math>{6\choose4} = 15</math> ways for this. We do not have to consider the order because the comb11 KB (1,729 words) - 20:50, 28 November 2023
- ...ity (again, higher ones give <math>b > 9</math>), giving us a sum of <math>15</math>. ...= 0</math> as we have been doing so far, then the sum is <math>495 + 120 + 15 = \boxed{630}</math>.4 KB (687 words) - 18:37, 27 November 2022
- {{AIME box|year=2002|n=I|num-b=13|num-a=15}}2 KB (267 words) - 19:18, 21 June 2021
- <cmath>(10^{2002} + 1)^{\frac {10}7} = 10^{2860}+\dfrac{10}{7}10^{858}+\dfrac{15}{49}10^{-1144}+\cdots</cmath>2 KB (316 words) - 19:54, 4 July 2013
- ...25}(x_2)=(3+\sqrt{5})+(3-\sqrt{5})=6</math> <math>\Rightarrow x_1x_2=225^6=15^{12}</math>. Thus, <math>\log_{30}\left(x_1y_1x_2y_2\right) = \log_{30}\left(15^{12}\cdot2^{12} \right) = \log_{30}\left(30^{12} \right) = \boxed{012}</mat1 KB (194 words) - 19:55, 23 April 2016
- ...es. So in total, we have overcounted by <math>9+6=15</math>, and <math>198-15=\fbox{183}</math>. ...ing we counted them 6 times. Therefore, our answer is <math>198-3(6-1)=198-15=\boxed{183}.</math>1 KB (220 words) - 20:50, 12 November 2022
- {{AIME box|year=2003|n=I|num-b=13|num-a=15}}3 KB (477 words) - 14:23, 4 January 2024
- ...3ad + 4d^2 = 30a + 30d</math>, which upon rearranging yields <math>2d(2d - 15) = 3a(10 - d)</math>. ...th>a = 18</math>. Alternatively, note that <math>3|2d</math> or <math>3|2d-15</math> implies that <math>3|d</math>, so only <math>9</math> may work. Henc5 KB (921 words) - 23:21, 22 January 2023
- pair O=(0,0),A=(-15,0),B=(-6,0),C=(15,0),D=(0,8); MP("6",B/2); MP("15",C/2); MP("9",(A+B)/2);3 KB (490 words) - 18:13, 13 February 2021
- {{AIME box|year=2003|n=II|num-b=13|num-a=15}}9 KB (1,461 words) - 15:09, 18 August 2023
