Question Number 151858 by mathdanisur last updated on 23/Aug/21 Answered by Olaf_Thorendsen last updated on 23/Aug/21 $${f}\left({x}\right)\:=\:\mathrm{sin}{x}+\frac{\mathrm{sin3}{x}}{\mathrm{3}}+\frac{\mathrm{sin5}{x}}{\mathrm{5}}+\mathrm{sin}\frac{\mathrm{7}{x}}{\mathrm{7}} \\ $$$${f}'\left({x}\right)\:=\:\mathrm{cos}{x}+\mathrm{cos3}{x}+\mathrm{cos5}{x}+\mathrm{cos7}{x} \\ $$$${f}'\left({x}\right)\:=\:\mathrm{Re}\left({e}^{{ix}} +{e}^{\mathrm{3}{ix}} +{e}^{\mathrm{5}{ix}} +{e}^{\mathrm{7}{ix}} \right)…
Question Number 151841 by mathdanisur last updated on 23/Aug/21 $$\mathrm{The}\:\mathrm{volue}\:\mathrm{of}\:\mathrm{the}\:\mathrm{limit}:\: \\ $$$$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{2}^{−\boldsymbol{\mathrm{n}}^{\mathrm{2}} } }{\underset{\boldsymbol{\mathrm{k}}=\boldsymbol{\mathrm{n}}+\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{2}^{−\boldsymbol{\mathrm{k}}^{\mathrm{2}} } }\:\:\:;\:\:\:\left(\mathrm{a}\right)\mathrm{0}\:\:\left(\mathrm{b}\right)\mathrm{some}\:\mathrm{c}\in\left(\mathrm{0};\mathrm{1}\right)\:\:\left(\mathrm{c}\right)\mathrm{1} \\ $$ Terms of Service Privacy…
Question Number 86298 by 21042009 last updated on 28/Mar/20 $${let}\:{x}^{{x}^{{x}^{\iddots} } } =\mathrm{2} \\ $$$${x}^{\mathrm{2}} =\mathrm{2} \\ $$$${x}=\pm\sqrt{\mathrm{2}} \\ $$$${then}\:{let}\:{x}^{{x}^{{x}^{\iddots} } } =\mathrm{4} \\ $$$${x}^{\mathrm{4}}…
Question Number 151826 by mathdanisur last updated on 23/Aug/21 $$\mathrm{if}\:\:\mathrm{x};\mathrm{y};\mathrm{z}>\mathrm{0}\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2}\right)\left(\mathrm{y}^{\mathrm{2}} +\mathrm{2}\right)\left(\mathrm{z}^{\mathrm{2}} +\mathrm{2}\right)\:\geqslant\:\mathrm{9}\left(\mathrm{xy}+\mathrm{yz}+\mathrm{zx}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 20739 by Tinkutara last updated on 02/Sep/17 $$\mathrm{If}\:\mid{z}_{\mathrm{1}} \mid\:=\:\mathrm{2},\:\mid{z}_{\mathrm{2}} \mid\:=\:\mathrm{3},\:\mid{z}_{\mathrm{3}} \mid\:=\:\mathrm{4}\:\mathrm{and} \\ $$$$\mid\mathrm{2}{z}_{\mathrm{1}} \:+\:\mathrm{3}{z}_{\mathrm{2}} \:+\:\mathrm{4}{z}_{\mathrm{3}} \mid\:=\:\mathrm{4},\:\mathrm{then}\:\mathrm{the}\:\mathrm{expression} \\ $$$$\mid\mathrm{8}{z}_{\mathrm{2}} {z}_{\mathrm{3}} \:+\:\mathrm{27}{z}_{\mathrm{3}} {z}_{\mathrm{1}} \:+\:\mathrm{64}{z}_{\mathrm{1}} {z}_{\mathrm{2}}…
Question Number 151806 by liberty last updated on 23/Aug/21 Commented by ghimisi last updated on 23/Aug/21 $${x}={y}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$ Answered by ghimisi last…
Question Number 20726 by Tinkutara last updated on 01/Sep/17 $$\mathrm{Five}\:\mathrm{distinct}\:\mathrm{2}-\mathrm{digit}\:\mathrm{numbers}\:\mathrm{are}\:\mathrm{in}\:\mathrm{a} \\ $$$$\mathrm{geometric}\:\mathrm{progression}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{middle} \\ $$$$\mathrm{term}. \\ $$ Answered by dioph last updated on 02/Sep/17 $$\mathrm{Let}\:\mathrm{the}\:\mathrm{GP}\:\mathrm{be}\:\left\{{a},{qa},{q}^{\mathrm{2}} {a},{q}^{\mathrm{3}}…
Question Number 20723 by rajpurohithakshay999@gmail.com last updated on 01/Sep/17 $$\int\sqrt{{x}}\:.{sinx}.{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 151789 by mathdanisur last updated on 23/Aug/21 Answered by ArielVyny last updated on 23/Aug/21 $${we}\:{have}\:\mathrm{0}\leqslant\lambda\leqslant\mathrm{1}\rightarrow\mathrm{0}\leqslant\lambda+{a}+{b}\leqslant\mathrm{1}+{a}+{b} \\ $$$${and}\:{we}\:{admit}\:{that}\:{a}+{b}\geqslant\lambda+\mathrm{1};{b}+{c}\geqslant\lambda+\mathrm{1} \\ $$$${c}+{a}\geqslant\lambda+\mathrm{1}\:\:\left({note}\:{that}\:{abc}=\mathrm{1}\right) \\ $$$$\mathrm{0}\leqslant\frac{\mathrm{1}}{\mathrm{1}+{a}+{b}}\leqslant\frac{\mathrm{1}}{\lambda+{a}+{b}}\:\leqslant\frac{\mathrm{1}}{\lambda+\mathrm{2}} \\ $$$$\mathrm{0}\leqslant\frac{\mathrm{1}}{\mathrm{1}+{b}+{c}}\leqslant\frac{\mathrm{1}}{\lambda+{b}+{c}}\leqslant\frac{\mathrm{1}}{\lambda+\mathrm{2}}…
Question Number 151791 by mathdanisur last updated on 23/Aug/21 Answered by ghimisi last updated on 23/Aug/21 $${x}_{{i}} ^{\mathrm{4}} +\mathrm{1}\geqslant\mathrm{2}{x}_{{i}} ^{\mathrm{2}} \Rightarrow{x}_{{i}} ^{\mathrm{4}} −{x}_{{i}} ^{\mathrm{2}} +\mathrm{1}\geqslant{x}_{{i}}…