Question Number 124071 by zarminaawan last updated on 30/Nov/20 $${Given}\:{the}\:{system}\:{of}\:{linear}\:{equation} \\ $$$${x}_{\mathrm{1}} +\mathrm{2}{x}_{\mathrm{2}} −\mathrm{3}{x}_{\mathrm{3}} =\mathrm{4} \\ $$$$\mathrm{3}{x}_{\mathrm{1}} −{x}_{\mathrm{2}} +\mathrm{5}{x}_{\mathrm{3}} =\mathrm{2} \\ $$$$\mathrm{4}{x}_{\mathrm{1}} +{x}_{\mathrm{2}} +\left({a}^{\mathrm{2}} −\mathrm{14}\right){x}_{\mathrm{3}}…
Question Number 124069 by zarminaawan last updated on 30/Nov/20 $${let}\:{us}\:{S}=\left\{{v}_{\mathrm{1}} ,{v}_{\mathrm{2}} ,{v}_{\mathrm{3},} {v}_{\mathrm{4}} \right\}\:{be}\:{the}\:{vector}\:{of}\:{R}^{\mathrm{3}} ?{where} \\ $$$${v}_{\mathrm{1}} =\begin{bmatrix}{\mathrm{1}}\\{−\mathrm{2}}\\{\mathrm{1}}\end{bmatrix},\:{v}_{\mathrm{2}} =\begin{bmatrix}{\mathrm{2}}\\{−\mathrm{3}}\\{−\mathrm{5}}\end{bmatrix},\:{v}_{\mathrm{3}} =\begin{bmatrix}{\mathrm{3}}\\{−\mathrm{1}}\\{\mathrm{1}}\end{bmatrix},\:{v}_{\mathrm{4}} =\begin{bmatrix}{−\mathrm{0}}\\{\mathrm{4}}\\{\mathrm{3}}\end{bmatrix} \\ $$$${Determine}\:{whether}\:{the}\:{set}\:{S}\:{spans}\:{R}^{\mathrm{3}} \\ $$…
Question Number 58508 by azizullah last updated on 24/Apr/19 Answered by tanmay last updated on 24/Apr/19 $${L}.{P}={x} \\ $$$${S}.{P}={x}−\frac{{x}×\mathrm{10}}{\mathrm{100}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{9}{x}}{\mathrm{10}} \\ $$$${C}.{P}={S}.{P}−{Profit} \\ $$$${C}.{P}=\frac{\mathrm{9}{x}}{\mathrm{10}}−\mathrm{150}…
Question Number 58500 by Kunal12588 last updated on 24/Apr/19 $${change}\:{in}\:{simplest}\:{form}\:: \\ $$$$\mathrm{tan}^{−\mathrm{1}} \frac{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$ Answered by MJS last updated on…
Question Number 58501 by naka3546 last updated on 24/Apr/19 $${a}\:\:{and}\:\:{b}\:\:{are}\:\:{roots}\:\:{of}\:\:{this}\:\:{equation}\:\:: \\ $$$$\:\:\:\:\:\:\:\:{x}^{\mathrm{2018}} \:−\:\mathrm{2}{x}\:+\:\mathrm{1}\:\:=\:\:\mathrm{0} \\ $$$${Calculate}\:\:{the}\:\:{value}\:\:\:{of} \\ $$$$\:\:\mathrm{2}\:+\:\left({a}+{b}\right)\:+\:\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\:+\:\left({a}^{\mathrm{3}} \:+\:{b}^{\mathrm{3}} \right)\:+\:\ldots\:+\:\left({a}^{\mathrm{2017}} \:+\:{b}^{\mathrm{2017}} \right) \\ $$…
Question Number 189559 by 073 last updated on 18/Mar/23 $$\:\left(−\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =? \\ $$$$\sqrt[{\mathrm{3}}]{−\mathrm{1}}=? \\ $$ Answered by Frix last updated on 18/Mar/23 $$−\mathrm{1}=\mathrm{e}^{\mathrm{i}\pi} \\ $$$$\Rightarrow…
Question Number 124009 by joki last updated on 30/Nov/20 Answered by mindispower last updated on 30/Nov/20 $${f}\left({z}\right)=\frac{{z}^{\mathrm{4}} }{{e}^{{z}} −\mathrm{1}},{is}\:{holomorphic}\:{withe}\:{no}\:{pols}\:{in}\:\Gamma \\ $$$${e}^{{z}} −\mathrm{1}=\mathrm{0}\Rightarrow{z}=\:\mathrm{2}{ik}\pi,{k}\in\mathbb{Z} \\ $$$$\Rightarrow\int_{\Gamma} {f}\left({z}\right){dz}=\mathrm{0}…
Question Number 189536 by 073 last updated on 18/Mar/23 Commented by 073 last updated on 18/Mar/23 $$\mathrm{solution}\:\mathrm{please} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 58467 by azizullah last updated on 23/Apr/19 Commented by azizullah last updated on 23/Apr/19 $$\:\:\:\:\:\:\:\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{easy}}\:\boldsymbol{\mathrm{method}}. \\ $$ Commented by tanmay last updated on…
Question Number 124000 by joki last updated on 30/Nov/20 Answered by liberty last updated on 30/Nov/20 $$\:\int\:\frac{{dx}}{\left({x}−\mathrm{2}\right)\sqrt{\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }}\:\:;\left[\:{x}−\mathrm{2}=\mathrm{2tan}\:{r}\:\right] \\ $$$$\mu\left({x}\right)=\int\:\frac{\mathrm{2sec}\:^{\mathrm{2}} {r}\:{dr}}{\mathrm{2tan}\:{r}\:\sqrt{\mathrm{4sec}\:^{\mathrm{2}} {r}}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\:{dr}}{\mathrm{sin}\:{r}} \\ $$$$\mu\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\int\:\mathrm{cosec}\:{r}\:{dr}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\ell{n}\:\mid\:\mathrm{cosec}\:{r}\:−\:\mathrm{cot}\:{r}\:\mid\:+\:{c}…