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- 51 bytes (8 words) - 18:14, 29 October 2024
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- Let <math>H </math> be the perpendicular to side <math>AB </math> from <math>{} C </math>. ...reas of similar triangles are proportional to the squares of corresponding side lengths,5 KB (885 words) - 22:03, 5 October 2024
- A rectangular box has integer side lengths in the ratio <math>1: 3: 4</math>. Which of the following could be Let the smallest side length be <math>x</math>. Then the volume is <math>x \cdot 3x \cdot 4x =12x1 KB (216 words) - 23:23, 5 September 2024
- ...led. The first intersects the side of length 5; By letting the bottom left side of the triangle be defined as (0,0), its slope is <math>m = \frac{1-4}{8-0} ...shaded part from [0,1) can also be found using an integral. Assume the top side of the rectangle is defined as y = 5. We can find the area between the line8 KB (1,016 words) - 23:17, 30 December 2023
- ...anual, published by AoPS in 2005, and more generally manages the technical side of AoPS publishing endeavours. Patrick also teaches several AoPS courses,3 KB (379 words) - 10:18, 27 September 2024
- ...is much like solving polynomial equations -- bringing all the terms to one side and finding the roots. A correct solution would be to move everything to the left side of the inequality, and form a common denominator. Then, it will be simple12 KB (1,806 words) - 05:07, 19 June 2024
- where <math>a,b</math> and <math>c</math> are side lengths of a triangle, lies in the interval <math>(p,q]</math>, where <math3 KB (583 words) - 20:20, 2 August 2024
- ...]], <math>(x-3)^2=x^2-6x+9</math>. To do this, add <math>7</math> to each side of the equation to get2 KB (422 words) - 15:20, 5 March 2023
- ...formula]] for finding the [[area]] of a [[triangle]] given only the three side lengths. For any triangle with side lengths <math>{a}, {b}, {c}</math>, the area <math>{A}</math> can be found4 KB (675 words) - 22:45, 6 October 2024
- ...to subtract or add the <math>x, y,</math> and <math>xy</math> terms to one side so you can isolate the constant and make the equation factorable. It can be4 KB (679 words) - 20:05, 16 October 2024
- Move <math>\frac{c}{a}</math> to the other side: Finally, move the <math>\frac{b}{2a}</math> to the other side:2 KB (290 words) - 15:55, 19 February 2024
- ...Euler made the leap of claiming that the [[polynomial]] on the right hand side can be factored as<br> ...e 0 at the same places. Dividing both sides by x and simplifying the right side, we get<br>2 KB (314 words) - 05:45, 1 May 2014
- ...f{a} \cdot \mathbf{b})^2 = a^2 b^2 (\cos\theta) ^2</math> .The right hand side of the inequality is equal to <math> \left( ||\mathbf{a}|| * ||\mathbf{b}|| ...and <math>\mathbf{a} = \lambda \mathbf{b}</math>. Lastly, multiplying each side by <math>||\mathbf{a}||||\mathbf{b}||</math>, we have <cmath>|\langle \math13 KB (2,048 words) - 14:28, 22 February 2024
- A [[quadrilateral]] has one [[vertex]] on each side of a [[square]] of side-length 1. Show that the lengths <math>a</math>, <math>b</math>, <math>c</ma914 bytes (148 words) - 08:08, 20 March 2008
- We imbed a [[hypercube]] of side length <math>a</math> in <math>\mathbb{R}^p</math> (the <math>p</math>-th d16 KB (2,660 words) - 22:42, 28 August 2024
- ...es of a non[[degenerate]] triangle is greater than the length of the third side. This inequality is particularly useful and shows up frequently on Intermed7 KB (1,300 words) - 00:11, 28 October 2024
- ...math>a</math>, <math>b</math>, <math>c</math>, <math>d</math> are the four side lengths and <math>s = \frac{a+b+c+d}{2}</math>. ...gives a formula for the area of a non-cyclic quadrilateral given only the side lengths; applying [[Ptolemy's Theorem]] to Bretschneider's.3 KB (543 words) - 18:35, 29 October 2024
- '''Ptolemy's theorem''' gives a relationship between the side lengths and the diagonals of a [[cyclic quadrilateral]]; it is the [[equali Given a [[cyclic quadrilateral]] <math>ABCD</math> with side lengths <math>{a},{b},{c},{d}</math> and [[diagonal]]s <math>{e},{f}</math>7 KB (1,198 words) - 23:28, 19 September 2024
- ...a_1)(b_2-b_1) \geq 0</math>. Expanding and taking some terms to the other side of the inequality, we get <math>a_1b_1+a_2b_2 \geq a_1b_2+a_2b_1</math>, as5 KB (804 words) - 12:54, 26 January 2023
- ...[[inscribed angle theorem]], their circumcenters are the midpoints of the side lengths of <math>\triangle ABC</math>, which we know to be on the nine-poin8 KB (1,408 words) - 08:39, 10 July 2024
- ...re split into six categories; three by their [[angle]]s and three by their side lengths. ...on right triangles, and the famous [[Pythagorean Theorem]] deals with the side lengths of the right triangle. Note that it is impossible to have an equila4 KB (631 words) - 20:16, 8 October 2024