Question Number 119177 by help last updated on 22/Oct/20 Answered by Dwaipayan Shikari last updated on 22/Oct/20 $$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{−{k}} =\mathrm{1}+\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{3}^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}} }+…=\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}}=\mathrm{4} \\ $$…
Question Number 119160 by zakirullah last updated on 22/Oct/20 Commented by zakirullah last updated on 22/Oct/20 $$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{position}}\:\boldsymbol{\mathrm{vecter}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{{p}\mathrm{oint}}\:\boldsymbol{\mathrm{R}} \\ $$$$\boldsymbol{{which}}\:\boldsymbol{{divides}}\:\boldsymbol{{the}}\:\boldsymbol{{line}}\:\boldsymbol{{joining}} \\ $$$$\boldsymbol{{the}}\:\boldsymbol{{point}}\:\boldsymbol{{whose}}\:\boldsymbol{{the}}\:\boldsymbol{{position}} \\ $$$$\boldsymbol{{vecters}}\:\boldsymbol{{are}}\:\boldsymbol{{P}}\left(\boldsymbol{{i}}+\mathrm{2}\boldsymbol{{j}}−\boldsymbol{{k}}\right),\boldsymbol{{Q}}\left(−\boldsymbol{{i}}+\boldsymbol{{j}}+\boldsymbol{{k}}\right)\:\boldsymbol{{in}} \\ $$$$\boldsymbol{{the}}\:\boldsymbol{{ratio}}\:\mathrm{2}:\mathrm{1}\:\boldsymbol{{internaly}}\:\boldsymbol{{and}}\:\boldsymbol{{externaly}}…
Question Number 53621 by kaivan.ahmadi last updated on 23/Jan/19 $$\mathrm{How}\:\mathrm{many}\:\mathrm{group}\:\mathrm{homomorphism}\:\mathrm{exist} \\ $$$$\mathrm{from}\:\mathrm{A}_{\mathrm{4}\:} \:\mathrm{to}\:\mathbb{Z}_{\mathrm{2}} ×\mathbb{Z}_{\mathrm{2}} \:\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 119159 by mathocean1 last updated on 22/Oct/20 $${We}\:{are}\:{in}\:\mathbb{C}. \\ $$$${Given}\:{Z}_{\mathrm{0}\:\:} =\mathrm{1}\:;\:\:\:\:\:{Z}_{{n}+\mathrm{1}\:} =\frac{\mathrm{1}}{\mathrm{2}}{Z}_{{n}\:} +\frac{\mathrm{1}}{\mathrm{2}}{i} \\ $$$${n}\:\in\:\mathbb{N}. \\ $$$${Show}\:{that}\:\forall\:{n}\:\in\:\mathbb{N}^{\ast\:} ,\:\mid{Z}_{{n}} \mid<\mathrm{1}. \\ $$ Answered by…
Question Number 119147 by Engr_Jidda last updated on 22/Oct/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 119141 by Engr_Jidda last updated on 22/Oct/20 Answered by TANMAY PANACEA last updated on 23/Oct/20 $$\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }+\frac{{dy}}{{dt}}=−{sint} \\ $$$$\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }+\mathrm{4}\frac{{dx}}{{dt}}+\mathrm{3}{x}−{sint}=−{sint} \\…
Question Number 184669 by Shrinava last updated on 10/Jan/23 Commented by mr W last updated on 10/Jan/23 $${it}\:{is}\:{meant}: \\ $$$${before}:\:−\mathrm{2}\: \\ $$$$\left({e}.{g}.\:{your}\:{loss}\:{last}\:{year}\:{was}\:\mathrm{2}\:{mio}.\:\$\right) \\ $$$${reduced}\:{by}\:\mathrm{2}\:{times} \\…
Question Number 119133 by Cristina last updated on 22/Oct/20 $$\mathrm{Informatica} \\ $$$$\left(\mathrm{11110000}\right)_{\mathrm{2}} =\mathrm{0}\bullet\mathrm{2}^{\mathrm{0}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{1}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{2}} +\mathrm{0}\bullet\mathrm{2}^{\mathrm{3}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{4}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{5}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{6}} +\mathrm{1}\bullet\mathrm{2}^{\mathrm{7}} = \\ $$$$=\mathrm{0}\bullet\mathrm{1}+\mathrm{0}\bullet\mathrm{2}+\mathrm{0}\bullet\mathrm{4}+\mathrm{0}\bullet\mathrm{8}+\mathrm{1}\bullet\mathrm{16}+\mathrm{1}\bullet\mathrm{32}+\mathrm{1}\bullet\mathrm{64}+\mathrm{1}\bullet\mathrm{128}= \\ $$$$\mathrm{0}+\mathrm{0}+\mathrm{0}+\mathrm{0}+\mathrm{16}+\mathrm{32}+\mathrm{64}+\mathrm{128}=\left(\mathrm{240}\right)_{\mathrm{10}}…
Question Number 184668 by Shrinava last updated on 10/Jan/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 119119 by benjo_mathlover last updated on 22/Oct/20 Answered by mindispower last updated on 22/Oct/20 $$\Rightarrow{abc}+\mathrm{1}=\frac{{a}}{{a}+{b}+{c}}…\mathrm{1} \\ $$$${abc}+\mathrm{2}=\frac{{b}}{{a}+{b}+{c}}….\mathrm{2} \\ $$$${abc}+\mathrm{7}=\frac{{c}}{{a}+{b}+{c}}…\mathrm{3} \\ $$$${abc}+\mathrm{10}=\mathrm{1} \\ $$$$\Rightarrow{abc}=−\mathrm{9}…\mathrm{2}…