Question Number 99994 by Dwaipayan Shikari last updated on 24/Jun/20 $$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{16}}+\frac{\mathrm{1}}{\mathrm{81}}+\frac{\mathrm{1}}{\mathrm{256}}+…..\infty \\ $$ Answered by smridha last updated on 24/Jun/20 $$\boldsymbol{\zeta}\left(\mathrm{4}\right)=\frac{\boldsymbol{\pi}^{\mathrm{4}} }{\mathrm{90}} \\ $$ Terms…
Question Number 99997 by ghiniboss last updated on 24/Jun/20 Commented by Dwaipayan Shikari last updated on 24/Jun/20 $$\mathrm{3}^{{x}} =\mathrm{4}^{{x}} \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{x}} =\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{0}} \\ $$$${x}=\mathrm{0} \\…
Question Number 99995 by Dwaipayan Shikari last updated on 24/Jun/20 $$\sqrt{\mathrm{1}\sqrt{\mathrm{3}\sqrt{\mathrm{5}\sqrt{\mathrm{7}\sqrt{\mathrm{9}}}}}}…\infty \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 34445 by Rio Mike last updated on 06/May/18 $$\left.\:{A}\mathrm{1}\right)\:\mathrm{Events}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{are}\:\mathrm{such}\:\mathrm{that} \\ $$$$\:\:{P}\left({A}\right)\:=\:\frac{\mathrm{9}}{\mathrm{30}},\:{P}\left(\boldsymbol{{B}}\right)=\:\frac{\mathrm{2}}{\mathrm{5}}\:{and}\: \\ $$$${P}\left({A}\:\cup\:{B}\right)\:=\:\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\left.{find}\:{a}\right)\:{P}\left({A}\:\cap\:{B}\right) \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:{b}\right)\:{P}\left({A}\mid{B}\right) \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:{c}\right)\:{P}\left(\bar {{B}}\right) \\ $$$$\left.\mathrm{2}\right)\:\mathrm{A}\:\mathrm{bag}\:\mathrm{contains}\:\mathrm{3}\:\mathrm{black}\:\mathrm{balls}\:\mathrm{and} \\…
Question Number 99975 by Dwaipayan Shikari last updated on 24/Jun/20 $$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}….\infty} } =? \\ $$ Answered by bachamohamed last updated on 24/Jun/20 $$\boldsymbol{{solution}}:\: \\ $$$$\boldsymbol{{on}}\:\boldsymbol{{a}}\:\:\boldsymbol{{a}}_{\boldsymbol{{n}}}…
Question Number 99962 by Rio Michael last updated on 24/Jun/20 $$\mathrm{A}\:\mathrm{particle}\:{Q}\:\mathrm{moves}\:\mathrm{in}\:\mathrm{a}\:\mathrm{plane}\:\mathrm{and}\:\mathrm{its}\:\mathrm{polar}\:\mathrm{coordinate}\:\left({r},\theta\right) \\ $$$$\mathrm{are}\:\mathrm{described}\:\mathrm{by}\:{r}\:=\:{at}^{\mathrm{2}} \:\mathrm{and}\:\theta\:=\:\frac{\mathrm{1}}{\mathrm{3}}{t}^{\mathrm{4}} \:\mathrm{find}\:\mathrm{its} \\ $$$$\mathrm{speed}\:\mathrm{at}\:{t}\:=\:\mathrm{2s} \\ $$ Answered by smridha last updated on…
Question Number 34408 by Rio Mike last updated on 05/May/18 $${what}\:{is}\:{the}\:{value}\:{of}\: \\ $$$$\:\sqrt{\mathrm{5}\sqrt{\mathrm{5}\sqrt{\mathrm{5}\sqrt{\mathrm{5}\sqrt{\mathrm{5}\sqrt{\mathrm{5}}}}}}} \\ $$$${ignoring}\:\mathrm{64}. \\ $$ Commented by candre last updated on 06/May/18 $${x}=\sqrt{\mathrm{5}\sqrt{\mathrm{5}\sqrt{\mathrm{5}\sqrt{\mathrm{5}\sqrt{\mathrm{5}\sqrt{\mathrm{5}}}}}}}…
Question Number 34376 by oshochandra.79@gmail.com last updated on 05/May/18 $$\boldsymbol{{H}}{ow}\:{to}\:{do}\:{A}\mathrm{4}\:{size}\:{page}\:{layout}\:{in}\:{equation}\:{editor}\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 99892 by Dwaipayan Shikari last updated on 23/Jun/20 $${solve}\:{the}\:{equation} \\ $$$${xa}^{\frac{\mathrm{1}}{{x}}} +\frac{\mathrm{1}}{{x}}{a}^{{x}} =\mathrm{2}{a} \\ $$$${Where}\:{a}\left\{−\mathrm{1},\mathrm{0},\mathrm{1}\right\} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 99889 by Dwaipayan Shikari last updated on 23/Jun/20 $$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{7}}+…….\infty\left\{\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\right\} \\ $$ Answered by smridha last updated on 23/Jun/20 $$\boldsymbol{{this}}\:\boldsymbol{{is}}\:\boldsymbol{{called}}\:\boldsymbol{{H}}{a}\boldsymbol{{rmonic}}\:\boldsymbol{{series}} \\ $$$$\boldsymbol{{and}}\:\boldsymbol{{H}}{a}\boldsymbol{{rmonic}}\:\boldsymbol{{series}}\:\boldsymbol{{always}} \\ $$$$\boldsymbol{{diverge}}\:\boldsymbol{{so}}\:\boldsymbol{{it}}\:\boldsymbol{{has}}\:\boldsymbol{{no}}\:\boldsymbol{{finite}}\:\boldsymbol{{value}}…