Question Number 38562 by shakirhsp@gmail.com last updated on 27/Jun/18 $$\left(\sqrt{\mathrm{2}}\:+{i}\right)\left(\mathrm{1}−\sqrt{\mathrm{2}{i}}\:\right) \\ $$ Answered by MJS last updated on 27/Jun/18 $$\left(\sqrt{\mathrm{2}}+\mathrm{i}\right)\left(\mathrm{1}−\sqrt{\mathrm{2i}}\right)=\left(\sqrt{\mathrm{2}}+\mathrm{i}\right)\left(\mathrm{1}−\left(\mathrm{1}+\mathrm{i}\right)\right)= \\ $$$$=\left(\sqrt{\mathrm{2}}+\mathrm{i}\right)\left(−\mathrm{i}\right)=\mathrm{1}−\mathrm{i}\sqrt{\mathrm{2}} \\ $$$$\left(\sqrt{\mathrm{2}}+\mathrm{i}\right)\left(\mathrm{1}−\sqrt{\mathrm{2}}\mathrm{i}\right)=\sqrt{\mathrm{2}}−\mathrm{2i}+\mathrm{i}+\sqrt{\mathrm{2}}=\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{i} \\…
Question Number 38559 by nishant last updated on 27/Jun/18 $${in}\:{a}\:{geometric}\:{series},\:{the}\:{first}\:{term} \\ $$$$={a},\:{common}\:{ratio}={r}.\:{If}\:{S}_{{n}} \:{denotes} \\ $$$${the}\:{sum}\:{of}\:{the}\:{n}\:{terms}\:{and}\:{U}_{{n}} =\underset{{n}=\mathrm{1}} {\overset{{n}} {\sum}}{S}_{{n},} \\ $$$${then}\:{rS}_{{n}} +\left(\mathrm{1}−{r}\right){U}_{{n}\:\:} {equals}\:{to} \\ $$$$\left({a}\right)\:\:\mathrm{0}\:\:\:\:\:\:\left({b}\right)\:\:{n}\:\:\:\:\:\left({c}\right)\:\:\:\:{na}\:\:\:\:\left({d}\right){nar} \\…
Question Number 169612 by Mastermind last updated on 04/May/22 $${Differentiate}\:{w}.{r}.{t}\:'{x}'\: \\ $$$${x}^{{y}} +{y}^{{x}} ={c} \\ $$$$ \\ $$$${Mastermind} \\ $$ Answered by thfchristopher last updated…
Question Number 38535 by Rio Mike last updated on 27/Jun/18 $${Given}\:{the}\:{function} \\ $$$${f}\left({x}\right)\:{where}\: \\ $$$$ \\ $$$${f}\left({x}\right)=\:\begin{cases}{\int{x}^{\mathrm{2}} \:+\:\mathrm{1}\:,{for}\:\left\{{x}:{x}\:{D}\left({f}\right)\:\mathrm{2}\right.}\\{\int{x}^{\mathrm{3}} \:−\:\mathrm{1},{for}\:{y}\:=\:{f}'\left({x}\right)}\end{cases} \\ $$$$\left.{a}\right)\:{Evaluate}\:{f}\left(\mathrm{2}\right) \\ $$$${if}\:{f}\left({a}\right)=\:\mathrm{2}\:+\:{a}^{{n}−\mathrm{1}} \\ $$$${find}\:{the}\:{value}\:{of}\:{a}…
Question Number 38534 by Rio Mike last updated on 27/Jun/18 $$\int\underset{{R}} {\int}\left(\mathrm{2}{x}\:+\:\mathrm{3}{y}\right)^{\mathrm{2}} \:{dA}=?? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 38517 by math khazana by abdo last updated on 26/Jun/18 $${simlify} \\ $$$${A}=\:\frac{\mathrm{1}}{\left(\mathrm{2}−\sqrt{\mathrm{5}}\right)^{\mathrm{4}} }\:+\:\frac{\mathrm{1}}{\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{4}} } \\ $$$${B}\:=\:\frac{\mathrm{1}}{\left(\mathrm{3}−\sqrt{\mathrm{2}}\right)^{\mathrm{6}} }\:+\frac{\mathrm{1}}{\left(\mathrm{3}+\sqrt{\mathrm{2}}\right)^{\mathrm{6}} } \\ $$ Answered by…
Question Number 38515 by Rio Mike last updated on 26/Jun/18 $$\:{Question}\:; \\ $$$${x}^{\mathrm{3}} \:+\:{x}^{\mathrm{3}} \:=\: \\ $$$$\left.{A}\right)\:{x}^{\mathrm{9}} \\ $$$$\left.{B}\right)\:{x}^{\mathrm{6}} \\ $$$$\left.{C}\right)\:{x}^{\mathrm{3}} \\ $$$$\left.{D}\right)\:\mathrm{1} \\ $$$${Give}\:{a}\:{reason}\:{for}\:{your}\:{answer}.…
Question Number 38495 by kunal1234523 last updated on 26/Jun/18 $${prove}\:{that} \\ $$$$\boldsymbol{\mathrm{tan}}\:\mathrm{3}\boldsymbol{{a}}\:\boldsymbol{\mathrm{tan}}\:\mathrm{2}\boldsymbol{{a}}\:\boldsymbol{\mathrm{tan}}\:\boldsymbol{{a}}\:=\:\:\boldsymbol{\mathrm{tan}}\:\mathrm{3}\boldsymbol{{a}}\:−\:\boldsymbol{\mathrm{tan}}\:\mathrm{2}\boldsymbol{{a}}\:−\:\boldsymbol{\mathrm{tan}}\:\boldsymbol{{a}} \\ $$ Answered by kunal1234523 last updated on 26/Jun/18 $${tan}\:\mathrm{3}{a}\:=\:{tan}\:\left({a}+\mathrm{2}{a}\right) \\ $$$$\Rightarrow{tan}\:\mathrm{3}{a}\:=\:\frac{{tan}\:{a}\:+\:{tan}\:\mathrm{2}{a}}{\mathrm{1}\:−\:{tan}\:{a}\:{tan}\:\mathrm{2}{a}} \\…
Question Number 38488 by Rio Mike last updated on 26/Jun/18 $$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}\:\mathrm{if}\: \\ $$$$\mathrm{3}^{{x}} \:=\:\mathrm{9}{x} \\ $$ Answered by $@ty@m last updated on 26/Jun/18 $$\mathrm{3} \\…
Question Number 38420 by Rio Mike last updated on 25/Jun/18 $${In}\:{the}\:{figure}\:{below},{a}\:{particle}\:{A}\:{of} \\ $$$${mass}\:\mathrm{2}{kg}\:{is}\:{lying}\:{on}\:{a}\:{rough}\:{wooden} \\ $$$${block}.{The}\:{particle}\:{A}\:{is}\:{connected}\:{by} \\ $$$${a}\:{light}\:{inextensible}\:{horizontal}\:{string} \\ $$$${passing}\:{over}\:{a}\:{smooth}\:{light}\:{fixed} \\ $$$${pulley}\:{at}\:{the}\:{edge}\:{of}\:{the}\:{block},{to}\:{a} \\ $$$${particle}\:{B}\:{of}\:{mass}\:\mathrm{3}{kg}\:{which}\:{hangs} \\ $$$${freely}.\:{The}\:{coefficent}\:{of}\:{friction}…