Question Number 193073 by Tawa11 last updated on 03/Jun/23 Answered by Subhi last updated on 03/Jun/23 $$ \\ $$$${the}\:{system}\:{is}\:{in}\:{equalibrium} \\ $$$$\Sigma{f}\:=\:\mathrm{0}\:{at}\:{x},{y}\:{axis} \\ $$$${p}.{cos}\left(\mathrm{30}\right)+\mathrm{2}.{sin}\left(\mathrm{45}\right)−{Q}.{sin}\left(\mathrm{60}\right)=\mathrm{0} \\ $$$$\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}.{p}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{Q}\:=\:−\sqrt{\mathrm{2}}…
Question Number 127492 by Dwaipayan Shikari last updated on 30/Dec/20 $$\frac{{i}!}{\pi!}\:\left({Exact}\:{form}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 192999 by Mastermind last updated on 01/Jun/23 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{following}\:\mathrm{functions} \\ $$$$\mathrm{are}\:\mathrm{continous}\:\mathrm{on}\:\mathrm{a}\:\mathrm{close}\:\mathrm{interval} \\ $$$$\left[\mathrm{0},\:\mathrm{1}\right]. \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\left\{_{\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}=\mathrm{1}} ^{\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{2}}{\mathrm{x}−\mathrm{1}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}\neq\mathrm{1}} \right. \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$…
Question Number 61923 by naka3546 last updated on 12/Jun/19 $${Find}\:\:{all}\:\:{solutions}\:\:{of} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{3}} \:−\:\mathrm{12}{x}\:+\:\mathrm{8}\:=\:\:\mathrm{0} \\ $$ Answered by MJS last updated on 12/Jun/19 $${x}^{\mathrm{3}} +{px}+{q}=\mathrm{0} \\…
Question Number 61915 by aseerimad last updated on 11/Jun/19 $$\mathrm{1}+{iw}+\left({iw}\right)^{\mathrm{2}} +\left({iw}\right)^{\mathrm{3}} +………\left({iw}\right)^{\mathrm{989}} =? \\ $$$$ \\ $$$${ans}=\:\:\:\:\frac{\mathrm{2}}{\mathrm{1}−{iw}}\:\:\:\:\:\:{answer}\:{is}\:{correct}. \\ $$$${pls}\:{help}\:..\:{how}\:{to}\:{do}\:{this}? \\ $$$${TIA} \\ $$ Answered by…
Question Number 61843 by psyche last updated on 10/Jun/19 $$\boldsymbol{{let}}\:\boldsymbol{{V}}\:\:\:\boldsymbol{{be}}\:\boldsymbol{{a}}\:\boldsymbol{{vector}}\:\boldsymbol{{space}}\:\boldsymbol{{and}}\:\boldsymbol{{let}}\:\boldsymbol{{H}}\:\boldsymbol{{and}}\:\boldsymbol{{K}}\:\boldsymbol{{be}}\: \\ $$$$\boldsymbol{{subspace}}\:\boldsymbol{{of}}\:\boldsymbol{{V}}.\:\boldsymbol{{show}}\:\boldsymbol{{that}}\:, \\ $$$${H}+{K}=\left\{\boldsymbol{{x}}:\boldsymbol{{x}}=\boldsymbol{{h}}+\boldsymbol{{k}},\:\boldsymbol{{where}}\:\boldsymbol{{h}}\in{H}\:\boldsymbol{{and}}\:\:\boldsymbol{{k}}\in{K}\right\}\:\boldsymbol{{is}}\:\:\boldsymbol{{a}}\:\boldsymbol{{subspace}}\:\boldsymbol{{of}}\:\boldsymbol{{V}}.\: \\ $$ Commented by arcana last updated on 10/Jun/19 $$\mathrm{if}\:\mathrm{V}\:\mathrm{is}\:\mathrm{a}\:\mathrm{vector}\:\mathrm{space}\:\mathrm{over}\:\mathrm{a}\:\mathrm{field}\:\mathrm{K}. \\…
Question Number 61840 by psyche last updated on 10/Jun/19 $$\boldsymbol{{consider}}\:\boldsymbol{{the}}\:\boldsymbol{{triple}}\:\boldsymbol{{of}}\:\boldsymbol{{real}}\:\boldsymbol{{numbers}}\:\left(\boldsymbol{{x}},{y},{z}\right) \\ $$$${defined}\:{by}\:{the}\:{addittion}\:\left(\boldsymbol{{x}},{y},{z}\right)+\left({x}',{y}',{z}'\right)=\left({x}+{x}',{y}+{y}',{z}+{z}'\right) \\ $$$$\boldsymbol{{and}}\:\boldsymbol{{scalar}}\:\boldsymbol{{multiplication}}\:\boldsymbol{{by}}\:\:\:\boldsymbol{\alpha}\left({x},{y},{z}\right)=\left(\mathrm{0},\mathrm{0},\mathrm{0}\right).\: \\ $$$$\boldsymbol{{S}}{how}\:{that}\:{all}\:{axioms}\:{for}\:{a}\:{vector}\:{space}\:{are}\:{satisfied}\:{except}\:{axiom}\:\mathrm{8}. \\ $$ Answered by arcana last updated on 10/Jun/19…
Question Number 61767 by Learner last updated on 08/Jun/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 127295 by Dwaipayan Shikari last updated on 28/Dec/20 $$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{tanh}^{−\mathrm{1}} {x}}{\:\sqrt[{\mathrm{5}}]{{x}}}{dx}\:\:\:\:\:\:\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 61755 by arcana last updated on 08/Jun/19 $$\mathrm{if}\:{f}\left({z}\right)=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{k}} {z}^{{k}} ,{a}_{{k}} ,{z}\in\mathbb{C}.\mathrm{Prove} \\ $$$$ \\ $$$${a}_{{k}} =\frac{\mathrm{1}}{\mathrm{2}\pi{i}}\underset{\mid{z}\mid={r}\:} {\int}\frac{{f}\left({z}\right)}{{z}^{{k}+\mathrm{1}} }{dz} \\ $$$$ \\…