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- ...e trying to get <math>a^2+b^2-2f^2+2de</math> on the LHS, because then the RHS would be <math>c^2</math>.6 KB (1,003 words) - 23:02, 19 May 2024
- ...>, to get <math>r^ne^{ni\theta} = e^{2\pi ik}</math>. The magnitude of the RHS is 1, making <math>r^n=1\Rightarrow r=1</math> (magnitude is always express ...ta} </math>, we see that <math> r^n = 1</math>, since the magnitude of the RHS of <math> x^n=1 </math> is 1, and for two complex numbers to be equal, both3 KB (558 words) - 20:36, 11 December 2011
- We can see that the LHS is <math>cis(n(90^{\circ}-t))</math>, and the RHS is <math>cis(90^{\circ}-nt)</math> So, <math>n(90-t) \equiv 90-nt \mod 360<6 KB (1,154 words) - 02:30, 11 January 2024
- ...ath>, we see that <math>a_{k}a_{k+1} = 37\cdot 72 - 3k</math>. Setting the RHS of this equation equal to <math>3</math>, we find that <math>m</math> must3 KB (499 words) - 17:52, 21 November 2022
- The leading coefficient of the RHS is <math>-1</math> because the leading coefficient of the LHS is <math>-1</6 KB (1,051 words) - 18:35, 1 August 2024
- ...ike <math>1,-1,-1,-1,1</math> we find the LHS to be <math>5</math> and the RHS to be <math>1.</math> What happened? There were more negatives than positiv Straight off, we notice that the RHS must be greater than or equal to 19, because an absolute value only gives n2 KB (394 words) - 09:21, 27 January 2024
- ...this equation matches the LHS equation that I said was important. So, the RHS of both equations are equal, and thus <math>145k-2 = k+34</math> We move al5 KB (766 words) - 23:46, 8 November 2024
- ...ath>du</math> and <math>dv</math> as <math>v</math> in the integral on the RHS, <math>u</math> should be chosen such that it has an "easy" (or "easier") [1 KB (235 words) - 16:01, 11 March 2022
- Subtracting the LHS from the RHS,1 KB (214 words) - 14:06, 18 October 2015
- Since <math> 1+2+...+n = \frac{n(n+1)}{2} </math> we may substitute the [[RHS]] in the above [[fraction]]. So the problem asks us for how many [[positive1 KB (220 words) - 11:54, 14 December 2021
- ...tions we will have a total of 4 of each variable on the [[LHS]]. On the [[RHS]] we will have <math>4+8+12+16+20 = 60</math>. Thus <center><math> 4(a+b+c ...d equation leaves <math>d</math> on the LHS and <math>15-8=7</math> on the RHS. And thus we continue on in this way to find that <math>(a,b,c,d,e)=(-5,-15 KB (786 words) - 15:11, 7 December 2024
- ...ine <math>\theta</math> such that <math> x = \tan{\theta}</math>. Then the RHS becomes2 KB (312 words) - 09:38, 4 April 2012
- Adding LHS of <math>(1)</math> with RHS of <math>(2)</math> (and vice-versa), we get We know that the RHS is <math>2007</math> by previous work. Therefore, <math>b_{n+1}b_{n-1}-b_n^13 KB (2,214 words) - 16:39, 28 November 2024
- We must prove that the RHS of this equation is less than or equal to <math>a^2 + b^2 + c^2</math>. and the RHS becomes <math>4\sqrt{3}\sqrt{(x+y+z)xyz}</math> If we use Heron's formula.4 KB (849 words) - 06:29, 30 July 2024
- Obviously, <math>\deg Q = 2</math> and the RHS has roots <math>1, 2, 3.</math> Thus, the only requirement is that <math>Q<5 KB (848 words) - 22:42, 29 September 2024
- As a [[function]] of <math>h_x</math>, the [[RHS]] of this equation is strictly decreasing, so it takes each value in its [[6 KB (994 words) - 12:40, 3 December 2024
- After cancelling the <math> a^{\frac{14}{3}}</math> term, we apply AM-GM to RHS and obtain7 KB (1,194 words) - 05:11, 22 October 2024
- ...nd the sum of the imaginary parts of complex conjugates is zero. Hence the RHS is zero.3 KB (517 words) - 13:13, 5 September 2021
- ...\equiv 1\pmod{3}</math>: Then the LHS is <math>1\pmod{3}</math>, while the RHS isn't. ...\equiv 2\pmod{3}</math>: Then the LHS is <math>1\pmod{3}</math>, while the RHS isn't.1 KB (250 words) - 23:38, 27 October 2015
- ...gets divided out (leaving a <math>5</math> for multiplication), and on the RHS, the <math>5</math> gets divided out (leaving a <math>3</math> for multipli4 KB (562 words) - 17:49, 8 November 2020