Question Number 132358 by Engr_Jidda last updated on 13/Feb/21 $${find}\:{the}\:{equation}\:{whose}\:{roots}\:{are} \\ $$$$\alpha\:{and}\:\beta.\:{find}\:\alpha\:{and}\:\beta\:{if}\:\:\alpha−\beta=\mathrm{2}\:{and}\:\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} =\mathrm{3} \\ $$ Commented by Dwaipayan Shikari last updated on 13/Feb/21 $${x}^{\mathrm{2}}…
Question Number 66816 by mathmax by abdo last updated on 20/Aug/19 $${let}\:{x}>\mathrm{0}\:{and}\:{f}\left({x}\right)=\int_{\mathrm{1}} ^{\mathrm{2}} \left({t}+\mathrm{1}\right)\sqrt{{t}^{\mathrm{2}} −\mathrm{2}{xt}−\mathrm{1}}{dt} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{determine}\:{also}\:{g}\left({x}\right)=\int_{\mathrm{1}} ^{\mathrm{2}} \frac{{t}^{\mathrm{2}} \:+{t}}{\:\sqrt{{t}^{\mathrm{2}} −\mathrm{2}{xt}−\mathrm{1}}}{dt} \\ $$$$\left.\mathrm{3}\right){find}\:{the}\:{value}\:{of}\:{integrals}\:\:\int_{\mathrm{1}}…
Question Number 1280 by Rasheed Soomro last updated on 19/Jul/15 $$\:{f}\:\left(\:\frac{\mathrm{1}}{{f}\left({x}\right)}\right)={f}\left({x}\right)\: \\ $$$${f}\left({x}\right)=? \\ $$ Commented by 123456 last updated on 19/Jul/15 $${f}\left({x}\right)=\frac{\mathrm{1}}{{x}} \\ $$…
Question Number 66814 by ~ À ® @ 237 ~ last updated on 20/Aug/19 $${Let}\:{consider}\:{an}\:{integer}\:{serie}\:\left\{{a}_{{n}} {x}^{{n}} \right\}\:{given}\:{by}\:\:{a}_{{n}} \:=\:{H}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}}\: \\ $$$$\left.\mathrm{1}\right)\:{Find}\:{out}\:{the}\:{largest}\:{domain}\:{D}\:{of}\:{convergence}\:{of}\:{that}\:{integer}\:{serie} \\ $$$$\left.\mathrm{2}\right)\:\forall\:{x}\in{D}\:\:,\:{explicit}\:{the}\:{sum}\:{S}\left({x}\right)\:{of}\:{the}\:\left\{{a}_{{n}}…
Question Number 66815 by Rio Michael last updated on 20/Aug/19 $${solve}\:{the}\:{congruence}\:{equation}\: \\ $$$$\:\:\mathrm{6}{x}\:\equiv\:\mathrm{4}\:\left({mod}\:\mathrm{5}\right)\:\:{i}\:{need}\:{help}\:{please}\:{with}\:{some}\:{explanations} \\ $$ Commented by mathmax by abdo last updated on 20/Aug/19 $${if}\:{we}\:{consider}\:{congruence}\:{modulo}\:\mathrm{5}\:\:{we}\:{have}\:…
Question Number 1278 by Rasheed Soomro last updated on 19/Jul/15 $${Determine}\:{f}\left({x}\right)\:{when}\:{f}\left(\:{f}\left({x}\right)\right)={x}\:\:. \\ $$ Commented by prakash jain last updated on 19/Jul/15 $${f}\left({x}\right)=\frac{{ax}+{b}}{{cx}+{d}} \\ $$$${f}\left({f}\left({x}\right)\right)=\frac{{a}\frac{{ax}+{b}}{{cx}+{d}}+{b}}{{c}\frac{{ax}+{b}}{{cx}+{d}}+{d}}={x} \\…
Question Number 132347 by Study last updated on 13/Feb/21 $${sin}\left(\mathrm{2}{x}+\frac{\pi}{\mathrm{4}}\right)−{sin}\left(\mathrm{2}{x}+\frac{\pi}{\mathrm{3}}\right)=\mathrm{0}\:\:\:{x}=? \\ $$ Answered by EDWIN88 last updated on 13/Feb/21 $$\mathrm{sin}\:\mathrm{X}−\mathrm{sin}\:\mathrm{Y}\:=\:\mathrm{2cos}\:\left(\frac{\mathrm{X}+\mathrm{Y}}{\mathrm{2}}\right)\mathrm{sin}\:\:\left(\frac{\mathrm{X}−\mathrm{Y}}{\mathrm{2}}\right) \\ $$$$\mathrm{where}\:\begin{cases}{\mathrm{X}=\mathrm{2x}+\frac{\pi}{\mathrm{4}}}\\{\mathrm{Y}=\mathrm{2x}+\frac{\pi}{\mathrm{3}}}\end{cases} \\ $$$$\mathrm{sin}\:\left(\mathrm{2x}+\frac{\pi}{\mathrm{4}}\right)−\mathrm{sin}\:\left(\mathrm{2x}+\frac{\pi}{\mathrm{3}}\right)=\mathrm{0} \\…
Question Number 132346 by Study last updated on 13/Feb/21 $$\left({y}^{\mathrm{4}} −\mathrm{3}{x}^{\mathrm{2}} \right){dy}+{xydx}=\mathrm{0}\:\:\:\:{y}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 1268 by 314159 last updated on 18/Jul/15 $$\boldsymbol{\mathrm{Prove}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{AM}}\:>\:\boldsymbol{\mathrm{HM}}. \\ $$ Answered by prakash jain last updated on 18/Jul/15 $$\mathrm{AM}=\frac{{a}+{b}}{\mathrm{2}},\:\mathrm{HM}=\frac{\mathrm{2}{ab}}{{a}+{b}} \\ $$$$\mathrm{AM}−\mathrm{HM}=\frac{{a}+{b}}{\mathrm{2}}−\frac{\mathrm{2}{ab}}{{a}+{b}}=\:\frac{\left({a}−{b}\right)^{\mathrm{2}} }{\mathrm{2}\left({a}+{b}\right)} \\…
Question Number 132337 by I want to learn more last updated on 13/Feb/21 Answered by physicstutes last updated on 14/Feb/21 $$\mathrm{First}\:\mathrm{we}\:\mathrm{determine}\:\mathrm{the}\:\mathrm{limiting}\:\mathrm{reagent}: \\ $$$$\mathrm{mass}\:\mathrm{of}\:\mathrm{BaCl}_{\mathrm{2}} \:=\:\mathrm{40}.\mathrm{8}\:\mathrm{g} \\…