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  • {{iTest box|year=2006|num-b=10|num-a=12|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Proble
    2 KB (232 words) - 20:49, 21 January 2020
  • {{iTest box|year=2006|num-b=15|num-a=17|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Proble
    2 KB (277 words) - 13:03, 29 November 2018
  • {{iTest box|year=2006|num-b=12|num-a=14|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Proble
    2 KB (216 words) - 23:35, 3 November 2023
  • {{iTest box|year=2006|num-b=16|num-a=18|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Proble
    2 KB (290 words) - 18:27, 30 November 2018
  • {{iTest box|year=2006|num-b=13|num-a=15|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Proble
    2 KB (238 words) - 20:27, 30 November 2018
  • <math>(k-n)(k+n) = 3^n</math> Let <math>k-n = 3^a, k+n = 3^{n-a}</math>
    2 KB (330 words) - 20:24, 21 February 2024
  • Similarly, since <math>PQNF</math> is also a cyclic-quadrilateral, reasoning as above, <math>\angle QNP = \angle PFQ = \angle CF
    4 KB (729 words) - 22:53, 13 December 2018
  • D = A+1/4*(B-A); E = A+1/4*(C-A);
    7 KB (1,053 words) - 14:58, 14 January 2024
  • ...her odometer showed <math>15951</math> (miles). This number is a palindrome-it reads the same forward and backward. Then <math>2</math> hours later, the A three-quarter sector of a circle of radius <math>4</math> inches together with its
    16 KB (2,496 words) - 20:09, 19 May 2024
  • 817 bytes (173 words) - 22:34, 11 April 2019
  • ...led diagram. Failure to meet this requirement will result in an automatic 1-point deduction. ...ath>j</math> to bowl <math>j-1</math>, provided that the difference <math>i-j</math> is even. We permit multiple fruits in the same bowl at the same tim
    4 KB (652 words) - 14:21, 10 March 2024
  • ...ath>j</math> to bowl <math>j-1</math>, provided that the difference <math>i-j</math> is even. We permit multiple fruits in the same bowl at the same tim ...s one apple and one pear are both moved from odd-numbered positions to even-numbered positions. If <math>i</math> and <math>j</math> are even, then <mat
    12 KB (2,156 words) - 17:44, 4 September 2022
  • ...math>, and for <math>a=-b</math> let <math>f(x)=-x</math> and <math>g(x)=-x-a</math>, so <math>|a|=|b|</math> does work and are the only solutions, as ...a contradition. Thus, <math>f(j-b)=k-b</math>. Continuing, <math>g(k-2a)=j-a</math>, and so on: <math>f(j+nb)=k+na</math> and <math>g(k+na)=j+(n+1)b</m
    4 KB (816 words) - 18:53, 7 April 2021
  • ...led diagram. Failure to meet this requirement will result in an automatic 1-point deduction. ...at <cmath>\frac{P(x)}{yz}+\frac{P(y)}{zx}+\frac{P(z)}{xy}=P(x-y)+P(y-z)+P(z-x)</cmath>holds for all nonzero real numbers <math>x,y,z</math> satisfying <
    3 KB (495 words) - 13:47, 22 November 2023
  • ...total ways for all <math>S_{n-1, j}</math>, so doing the same thing <math>n-1</math> more times yields a final answer of <math>(2n)!\cdot 2^{\left(n^2\r ...ke <math>S_{1,1}</math> and add 2 new elements, getting you <math>\binom{2n-2}{2}</math> ways to generate <math>S_{2,2}</math>. And you can keep going d
    7 KB (1,288 words) - 19:17, 26 April 2023
  • Let's first work out the slope-intercept form of all three lines: ...plies <math>y=-x+10</math>. Thus the lines are <math>y=\frac{x}{2} +1, y=2x-2,</math> and <math>y=-x+10</math>.
    7 KB (1,079 words) - 22:24, 10 November 2023
  • ...feet wide and <math>17</math> feet long is tiled with <math>170</math> one-foot square tiles. A bug walks from one corner to the opposite corner in a s https://youtu.be/Z-sUMqZH0j4
    3 KB (529 words) - 02:24, 8 September 2023
  • ...e of the points <math>(a-1,b)</math>, <math>(a,b-1)</math>, or <math>(a-1,b-1)</math>, each with probability <math>\tfrac{1}{3}</math>, independently of <cmath> (x-z_1)^3(x-z_2)^3 \cdots (x-z_{673})^3 </cmath>
    8 KB (1,331 words) - 06:57, 4 January 2021
  • Let <math>ABC</math> be an acute-angled triangle with <math>AB\neq AC</math>. The circle with diameter <math>
    3 KB (488 words) - 04:36, 21 February 2021
  • x_{n+1} - x_n = \frac{x_n^2 + 5x_n + 4 - x_n(x_n+6)}{x_n+6} = \frac{4-x_n}{x_n+6} < 0. x_{n+1} - 4 = (x_n-4)\cdot\frac{x_n + 5}{x_n+6}.
    8 KB (1,290 words) - 22:06, 3 October 2023

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