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Author: Tinku Tara

Let-P-t-denote-a-given-cubic-polynomial-Find-the-constants-a-1-u-1-a-2-and-u-2-such-that-1-1-P-t-dt-a-1-P-u-1-a-2-P-u-2-

Question Number 1778 by 112358 last updated on 23/Sep/15 $${Let}\:{P}\left({t}\right)\:{denote}\:{a}\:{given}\:{cubic} \\ $$$${polynomial}.\:{Find}\:{the}\:{constants} \\ $$$${a}_{\mathrm{1}} ,{u}_{\mathrm{1}} ,{a}_{\mathrm{2}} \:{and}\:{u}_{\mathrm{2}} \:{such}\:{that} \\ $$$$\int_{−\mathrm{1}} ^{\:\mathrm{1}} {P}\left({t}\right){dt}={a}_{\mathrm{1}} {P}\left({u}_{\mathrm{1}} \right)+{a}_{\mathrm{2}} {P}\left({u}_{\mathrm{2}}…

The-n-positive-numbers-x-1-x-2-x-n-where-n-3-satisfy-x-1-1-1-x-2-x-2-1-1-x-3-x-n-1-1-1-x-n-and-x-n-1-1-x-1-Show-that-x-1-x-2-x-3-x-n-

Question Number 1777 by 112358 last updated on 22/Sep/15 $${The}\:{n}\:{positive}\:{numbers}\:{x}_{\mathrm{1}} ,{x}_{\mathrm{2}} ,…{x}_{{n}} \\ $$$${where}\:{n}\geqslant\mathrm{3},\:{satisfy}\: \\ $$$${x}_{\mathrm{1}} =\mathrm{1}+\frac{\mathrm{1}}{{x}_{\mathrm{2}} },{x}_{\mathrm{2}} =\mathrm{1}+\frac{\mathrm{1}}{{x}_{\mathrm{3}} },\:…\:,\:{x}_{{n}−\mathrm{1}} =\mathrm{1}+\frac{\mathrm{1}}{{x}_{{n}} } \\ $$$${and}\:{x}_{{n}} =\mathrm{1}+\frac{\mathrm{1}}{{x}_{\mathrm{1}}…

P-k-x-N-n-N-x-gt-0-n-x-1-k-C-k-3-P-k-0-1-C-P-N-proof-or-give-a-counter-example-that-x-y-z-C-3-x-2-yz-x-y-x-z-

Question Number 1776 by 123456 last updated on 20/Sep/15 $$\boldsymbol{\mathrm{P}}_{{k}} =\left\{{x}\in\mathbb{N},{n}\in\mathbb{N}:{x}>\mathrm{0},\underset{{n}\mid{x}} {\sum}\mathrm{1}={k}\right\} \\ $$$$\boldsymbol{\mathrm{C}}=\underset{{k}\geqslant\mathrm{3}} {\cup}\boldsymbol{\mathrm{P}}_{{k}} \\ $$$$\left\{\mathrm{0},\mathrm{1}\right\}\cup\boldsymbol{\mathrm{C}}\cup\mathbb{P}=\mathbb{N} \\ $$$$\mathrm{proof}\:\mathrm{or}\:\mathrm{give}\:\mathrm{a}\:\mathrm{counter}\:\mathrm{example}\:\mathrm{that} \\ $$$$\left({x},{y},{z}\right)\in\boldsymbol{\mathrm{C}}^{\mathrm{3}} ,{x}^{\mathrm{2}} \mid{yz}\Rightarrow{x}\mid{y}\vee{x}\mid{z} \\ $$…

7-couples-are-there-in-a-society-The-selection-of-pairs-are-to-be-done-for-a-mixed-double-badminton-tournament-In-how-many-ways-we-can-choose-these-pairs-

Question Number 1775 by lakshaysethi039 last updated on 26/Sep/15 $$\mathrm{7}\:{couples}\:{are}\:{there}\:{in}\:{a}\:{society}.\:{The}\:{selection}\:{of} \\ $$$$\:{pairs}\:{are}\:{to}\:{be}\:{done}\:{for}\:{a}\:{mixed}\:{double}\:{badminton} \\ $$$${tournament}.\:{In}\:{how}\:{many}\:{ways}\:{we}\:{can}\:{choose} \\ $$$$\:{these}\:{pairs}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

2x-y-2z-4-x-10y-3z-10-

Question Number 1771 by hareem ali last updated on 19/Sep/15 $$\mathrm{2}{x}−{y}+\mathrm{2}{z}=\mathrm{4} \\ $$$${x}+\mathrm{10}{y}−\mathrm{3}{z}=\mathrm{10} \\ $$ Answered by 123456 last updated on 19/Sep/15 $$\begin{bmatrix}{\mathrm{2}}&{−\mathrm{1}}&{\mathrm{2}}\\{\mathrm{1}}&{\mathrm{10}}&{−\mathrm{3}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\end{bmatrix}\begin{bmatrix}{{x}}\\{{y}}\\{{z}}\end{bmatrix}=\begin{bmatrix}{\mathrm{4}}\\{\mathrm{10}}\\{\mathrm{0}}\end{bmatrix} \\ $$$$\begin{bmatrix}{\mathrm{2}}&{−\mathrm{1}}&{\mathrm{2}}&{\mathrm{4}}\\{\mathrm{1}}&{\mathrm{10}}&{−\mathrm{3}}&{\mathrm{10}}\end{bmatrix}…

Question-67307

Question Number 67307 by aliesam last updated on 25/Aug/19 Commented by mathmax by abdo last updated on 25/Aug/19 $${S}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{cos}\left({n}\theta\right)}{\mathrm{2}^{{n}} }\:\Rightarrow\:{S}\:={Re}\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{e}^{{in}\theta} }{\mathrm{2}^{{n}}…