Question Number 202287 by sonukgindia last updated on 24/Dec/23 Commented by a.lgnaoui last updated on 26/Dec/23 Answered by aleks041103 last updated on 24/Dec/23 $${I}\:{guess}\:{not}\:{enough}\:{information}. \\…
Question Number 202340 by hardmath last updated on 24/Dec/23 Answered by MATHEMATICSAM last updated on 25/Dec/23 $$\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}\:+\:\sqrt{\mathrm{2}}}\:=\:\sqrt{\mathrm{3}}\:−\:\sqrt{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}}\:+\:\sqrt{\mathrm{3}}}\:=\:\sqrt{\mathrm{4}}\:−\:\sqrt{\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}\:+\:\sqrt{\mathrm{4}}}\:=\:\sqrt{\mathrm{5}}\:−\:\sqrt{\mathrm{4}} \\ $$$$. \\ $$$$.…
Question Number 202328 by MATHEMATICSAM last updated on 24/Dec/23 $$\mathrm{If}\:{n}\:\geqslant\:\mathrm{2}\:\mathrm{and}\:\mathrm{U}_{{n}} \:=\:\left(\mathrm{3}\:+\:\sqrt{\mathrm{5}}\right)^{{n}} \:+\:\left(\mathrm{3}\:−\:\sqrt{\mathrm{5}}\right)^{{n}} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{U}_{{n}\:+\:\mathrm{1}} \:=\:\mathrm{6U}_{{n}} \:−\:\mathrm{4U}_{{n}\:−\:\mathrm{1}} \:. \\ $$ Commented by aleks041103 last updated on…
Question Number 202329 by sonukgindia last updated on 24/Dec/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 202324 by hardmath last updated on 24/Dec/23 $$\mathrm{Find}:\:\:\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\:+\:\mathrm{1}\:+\:\frac{\mathrm{3}}{\mathrm{2}}\:+\:…\:+\:\mathrm{16}}{\frac{\mathrm{1}}{\mathrm{4}}\:+\:\frac{\mathrm{2}}{\mathrm{4}}\:+\:\frac{\mathrm{3}}{\mathrm{4}}\:+\:…\:+\:\mathrm{8}} \\ $$ Answered by MATHEMATICSAM last updated on 24/Dec/23 $$\frac{\frac{\mathrm{1}}{\mathrm{2}}\:+\:\mathrm{1}\:+\:\frac{\mathrm{3}}{\mathrm{2}}\:+\:….\:+\:\mathrm{16}}{\frac{\mathrm{1}}{\mathrm{4}}\:+\:\frac{\mathrm{2}}{\mathrm{4}}\:+\:\frac{\mathrm{3}}{\mathrm{4}}\:+\:….\:+\:\mathrm{8}} \\ $$$$=\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{2}}{\mathrm{2}}\:+\:\frac{\mathrm{3}}{\mathrm{2}}\:+\:\frac{\mathrm{4}}{\mathrm{2}}\:+\:….\:+\:\frac{\mathrm{32}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{4}}\:+\:\frac{\mathrm{2}}{\mathrm{4}}\:+\:\frac{\mathrm{3}}{\mathrm{4}}\:+\:\frac{\mathrm{4}}{\mathrm{4}}\:+\:….\:+\:\frac{\mathrm{32}}{\mathrm{4}}} \\ $$$$=\:\frac{\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}\:+\:\mathrm{4}\:+\:….\:+\:\mathrm{32}\right]}{\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}\:+\:\mathrm{4}\:+\:….\:+\:\mathrm{32}\right]} \\…
Question Number 202325 by Ikbal last updated on 24/Dec/23 Answered by AST last updated on 24/Dec/23 $${Let}\:{x}={re}^{{i}\theta} ={r}\left({cos}\theta+{isin}\left(\theta\right)\right); \\ $$$$\frac{\mathrm{1}}{{x}}=\frac{{e}^{−{i}\theta} }{{r}}=\frac{\mathrm{1}}{{r}}\left({cos}\left(\theta\right)−{isin}\left(\theta\right)\right) \\ $$$$\Rightarrow{x}^{{n}} ={r}^{{n}} \left[{cos}\left({n}\theta\right)+{isin}\left({n}\theta\right)\right]…
Question Number 202326 by sonukgindia last updated on 24/Dec/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 202319 by sonukgindia last updated on 24/Dec/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 202315 by MATHEMATICSAM last updated on 24/Dec/23 $$\mathrm{If}\:{x}\::\:{y}\::\:{z}\:=\:{a}\::\:{b}\::\:{c}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$$\left(\frac{{a}\:+\:{b}\:+\:{c}}{{x}\:+\:{y}\:+\:{z}}\right)^{\mathrm{3}} \:=\:\frac{{abc}}{{xyz}}\:. \\ $$ Answered by AST last updated on 24/Dec/23 $${x}={ka};{y}={kb};{z}={kc} \\ $$$$\left(\frac{{a}+{b}+{c}}{{x}+{y}+{z}}\right)^{\mathrm{3}}…
Question Number 202308 by 2024 last updated on 24/Dec/23 Commented by mr W last updated on 24/Dec/23 $$\frac{\mathrm{4}^{{x}} }{\mathrm{9}^{{x}} }=\mathrm{9} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}{x}} =\mathrm{3}^{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{{x}}…