Question Number 141574 by sarkor last updated on 20/May/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 76037 by Crabby89p13 last updated on 22/Dec/19 $${Prove}\:{That} \\ $$$$\mathrm{sin}\:\mathrm{3}°\mathrm{sin}\:\mathrm{39}°\mathrm{sin}\:\mathrm{75}°=\mathrm{sin}\:\mathrm{9}°\mathrm{sin}\:\mathrm{24}°\mathrm{sin}\:\mathrm{30}° \\ $$ Answered by MJS last updated on 23/Dec/19 $$\mathrm{sin}\:\mathrm{3}\:=\mathrm{sin}\:\left(\mathrm{75}−\mathrm{2}×\mathrm{36}\right)\:= \\ $$$$=\mathrm{2cos}^{\mathrm{2}} \:\mathrm{36}\:\mathrm{sin}\:\mathrm{75}\:−\mathrm{2sin}\:\mathrm{36}\:\mathrm{cos}\:\mathrm{36}\:\mathrm{cos}\:\mathrm{75}\:−\mathrm{sin}\:\mathrm{75}\:=…
Question Number 141569 by SOMEDAVONG last updated on 20/May/21 $$\underset{\mathrm{n}\rightarrow+\propto} {\mathrm{lim}}\left(\frac{\mathrm{110}^{\mathrm{2}} }{\left(\mathrm{11}−\mathrm{10}\right)\left(\mathrm{11}^{\mathrm{2}} −\mathrm{10}^{\mathrm{2}} \right)}\:+\:\frac{\mathrm{110}^{\mathrm{2}} }{\left(\mathrm{11}^{\mathrm{2}} −\mathrm{10}^{\mathrm{2}} \right)\left(\mathrm{11}^{\mathrm{3}} −\mathrm{10}^{\mathrm{3}} \right)}\:+\:….+\:\frac{\mathrm{110}^{\mathrm{2}} }{\left(\mathrm{11}^{\mathrm{n}} −\mathrm{10}^{\mathrm{n}} \right)\left(\mathrm{11}^{\mathrm{n}+\mathrm{1}} −\mathrm{10}^{\mathrm{n}+\mathrm{1}} \right)}\right) \\…
Question Number 76034 by hmamarques1994@gmail.com last updated on 22/Dec/19 $$\: \\ $$$$\:\mathrm{53}^{\boldsymbol{\mathrm{log}}_{\boldsymbol{\mathrm{x}}} \left(\mathrm{7}\right)} \:=\:\sqrt{\boldsymbol{\mathrm{x}}} \\ $$$$\: \\ $$$$\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$$$\: \\ $$ Answered by MJS…
Question Number 141568 by sarkor last updated on 20/May/21 Commented by Dwaipayan Shikari last updated on 20/May/21 $${Area}\:{between}\:{this}\:{curves} \\ $$$$\sqrt{{x}}={x}^{\mathrm{2}} \Rightarrow{x}=\mathrm{1}\:\left({Meeting}\:{point}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{{x}}−{x}^{\mathrm{2}}…
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Question Number 141570 by ayb last updated on 21/May/21 Answered by MathematicalUser2357 last updated on 29/Dec/23 $$\mathrm{And}\:\mathrm{that}'\mathrm{s}\:\mathrm{your}\:\mathrm{answer}. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 10495 by ajfour last updated on 14/Feb/17 $$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \sqrt{{R}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{2}{Rr}\mathrm{cos}\:\theta}\:{d}\theta \\ $$ Answered by robocop last updated on 14/Feb/17 $${todo}\:{en}\:{funcion}\:{de}\:\theta \\…
Question Number 10493 by ABD last updated on 14/Feb/17 $$\frac{\mathrm{1}}{\mathrm{2}!}+\frac{\mathrm{2}}{\mathrm{3}!}+\frac{\mathrm{3}}{\mathrm{4}!}+\frac{\mathrm{4}}{\mathrm{5}!}+…+\frac{\mathrm{17}}{\mathrm{18}!}=? \\ $$ Answered by mrW1 last updated on 14/Feb/17 $${since}\:\frac{{n}}{\left({n}+\mathrm{1}\right)!}=\frac{\mathrm{1}}{{n}!}−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)!} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}!}+\frac{\mathrm{2}}{\mathrm{3}!}+\frac{\mathrm{3}}{\mathrm{4}!}+\frac{\mathrm{4}}{\mathrm{5}!}+…+\frac{\mathrm{17}}{\mathrm{18}!} \\ $$$$=\left(\frac{\mathrm{1}}{\mathrm{1}!}−\frac{\mathrm{1}}{\mathrm{2}!}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}!}−\frac{\mathrm{1}}{\mathrm{3}!}\right)+\left(\frac{\mathrm{1}}{\mathrm{3}!}−\frac{\mathrm{1}}{\mathrm{4}!}\right)+\centerdot\centerdot\centerdot+\left(\frac{\mathrm{1}}{\mathrm{17}!}−\frac{\mathrm{1}}{\mathrm{18}!}\right) \\…
Question Number 10492 by ABD last updated on 14/Feb/17 $$\mathrm{3}^{{logx}} −\mathrm{2}^{{logx}−\mathrm{1}} =\mathrm{2}^{{logx}+\mathrm{1}} −\mathrm{2}×\mathrm{3}^{{logx}−\mathrm{1}} \Rightarrow{x}=? \\ $$ Answered by mrW1 last updated on 14/Feb/17 $$\mathrm{3}^{{logx}} −\frac{\mathrm{2}^{\mathrm{log}\:{x}}…