Question Number 196595 by sonukgindia last updated on 27/Aug/23 Answered by Rasheed.Sindhi last updated on 28/Aug/23 $${f}\left({x}\right)=\frac{{ax}^{\mathrm{2}} +{bx}+{c}}{{dx}+{e}} \\ $$$${f}\left({x}\right)+{f}\left(−{x}\right)=\frac{{ax}^{\mathrm{2}} +{bx}+{c}}{{dx}+{e}}+\frac{{ax}^{\mathrm{2}} −{bx}+{c}}{−{dx}+{e}} \\ $$$$=\frac{\left({ax}^{\mathrm{2}} +{bx}+{c}\right)\left({e}−{dx}\right)+\left({ax}^{\mathrm{2}}…
Question Number 196594 by sonukgindia last updated on 27/Aug/23 Answered by JDamian last updated on 27/Aug/23 $${really}? \\ $$ Answered by sniper237 last updated on…
Question Number 196582 by MATHEMATICSAM last updated on 27/Aug/23 $$\mathrm{If}\:{x}\:=\:\mathrm{log}_{{a}} {bc},\:{y}\:=\:\mathrm{log}_{{b}} {ca}\:\mathrm{and}\:{z}\:=\:\mathrm{log}_{{c}} {ab} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:{x}\:+\:{y}\:+\:{z}\:=\:{xyz}\:−\:\mathrm{2}. \\ $$ Answered by som(math1967) last updated on 27/Aug/23 $${x}={log}_{{a}}…
Question Number 196576 by sonukgindia last updated on 27/Aug/23 Commented by Frix last updated on 27/Aug/23 $$\mathrm{Infinite}\:\mathrm{solutions}\:\mathrm{for}\:{x},\:{y}\:\in\mathbb{R} \\ $$$$\mathrm{No}\:\mathrm{solution}\:\mathrm{for}\:{x},\:{y}\:\in\mathbb{Z} \\ $$ Answered by AST last…
Question Number 196571 by cortano12 last updated on 27/Aug/23 Answered by som(math1967) last updated on 27/Aug/23 $$\:{x}=\sqrt{\mathrm{6}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} }=\sqrt{\mathrm{121}}=\mathrm{11}{cm} \\ $$ Answered by hafeez…
Question Number 196565 by mokys last updated on 27/Aug/23 $${find}\:{the}\:{power}\:{series}\:{exponition}\:{of}\: \\ $$$${f}\left({z}\right)=\frac{\mathrm{2}{z}+\mathrm{1}}{{z}^{\mathrm{2}} −\mathrm{3}{z}+\mathrm{2}}\:\:{about}\:{z}_{{o}} \:=\:{i} \\ $$ Answered by aleks041103 last updated on 27/Aug/23 $$\frac{\mathrm{2}{z}+\mathrm{1}}{{z}^{\mathrm{2}} −\mathrm{3}{z}+\mathrm{2}}=\frac{\mathrm{2}{z}+\mathrm{1}}{\left({z}−\mathrm{2}\right)\left({z}−\mathrm{1}\right)}=\frac{{A}}{{z}−\mathrm{1}}+\frac{{B}}{{z}−\mathrm{2}}…
Question Number 196567 by pete last updated on 27/Aug/23 $$\mathrm{Show}\:\mathrm{that}\:\left(\frac{\mathrm{1}+\:\mathrm{itan}\theta}{\mathrm{1}−\:\mathrm{itan}\theta}\right)^{\mathrm{n}} =\frac{\mathrm{1}+\:\mathrm{itan}\left(\mathrm{n}\theta\right)}{\mathrm{1}−\:\mathrm{tan}\left(\mathrm{n}\theta\right)} \\ $$ Commented by mokys last updated on 27/Aug/23 $$\left(\frac{\mathrm{1}+{itan}\theta}{\mathrm{1}−{itan}\theta}\right)^{{n}} =\left(\frac{\frac{{cos}\theta+{isin}\theta}{{cos}\theta}}{\frac{{cos}\theta−{isin}\theta}{{cos}\theta}}\right)^{{n}} =\left(\frac{{cos}\left({n}\theta\right)+{isin}\left({n}\theta\right)}{{cos}\left({n}\theta\right)−{isin}\left({n}\theta\right)}\right) \\ $$$$…
Question Number 196560 by peter frank last updated on 27/Aug/23 Commented by Spillover last updated on 27/Aug/23 Check your solution https://m.facebook.com/story.php?story_fbid=pfbid0QHNhoZjesQzGqy7jHAgbvRSRUx25yXHezsGoh2qHLCapd9KB4z5PFi88czsFYxZ6l&id=100084816875916&mibextid=Nif5oz Answered by MM42 last updated on 27/Aug/23…
Question Number 196557 by BHOOPENDRA last updated on 27/Aug/23 Answered by qaz last updated on 27/Aug/23 $$\int_{\mathrm{0}} ^{{t}} {e}^{−{u}} \mathrm{sin}\:{udu}=−\Im\int_{\mathrm{0}} ^{{t}} {e}^{−\left(\mathrm{1}+{i}\right){u}} {du}=\Im\frac{\mathrm{1}}{\mathrm{1}+{i}}\left({e}^{−\left(\mathrm{1}+{i}\right){t}} −\mathrm{1}\right) \\…
Question Number 196559 by peter frank last updated on 27/Aug/23 Terms of Service Privacy Policy Contact: info@tinkutara.com