Question Number 141910 by dhgt last updated on 24/May/21 $${Q}\mathrm{137026} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 10836 by okhema last updated on 27/Feb/17 $${solve}\:{cos}\mathrm{2}\theta−\mathrm{3}{cos}\theta=\mathrm{1} \\ $$$${for}\:{o}\leqslant\theta\leqslant\mathrm{2}\pi \\ $$ Answered by ridwan balatif last updated on 27/Feb/17 $$\mathrm{cos2}\theta−\mathrm{3cos}\theta=\mathrm{1} \\ $$$$\left(\mathrm{2cos}^{\mathrm{2}}…
Question Number 141905 by dhgt last updated on 24/May/21 $${Q}\mathrm{137503} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 141907 by dhgt last updated on 24/May/21 $${Q}\mathrm{140000} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 76368 by Rio Michael last updated on 26/Dec/19 $${prove}\:{that} \\ $$$$\mathrm{1}.\:\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\:{r}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{n}\left({n}+\mathrm{1}\right) \\ $$$$\mathrm{2}.\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\:{r}^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{6}}{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}\:+\:\mathrm{1}\right) \\ $$$$\mathrm{3}.\:\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r}^{\mathrm{3}} =\:\frac{\mathrm{1}}{\mathrm{4}}{n}^{\mathrm{2}}…
Question Number 141901 by gsk2684 last updated on 24/May/21 Commented by gsk2684 last updated on 24/May/21 $${does}\:{this}\:{series}\:{converges}\:{or}\:{diverges} \\ $$$${by}\:{using}\:{comparison}\:{test}\:{or}\:{limit} \\ $$$${comparison}\:{test}?\:{explanation}\:{please} \\ $$ Terms of…
Question Number 141900 by iloveisrael last updated on 24/May/21 $$\:\mathcal{L}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{e}^{{x}} \:\mathrm{cos}\:{x}−{x}−\mathrm{1}}{{x}^{\mathrm{3}} }\:=? \\ $$ Answered by Dwaipayan Shikari last updated on 24/May/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{e}^{{x}}…
Question Number 76367 by Rio Michael last updated on 26/Dec/19 $${prove}\:{that}\:\:\underset{{r}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$ Commented by mathmax by abdo last updated on…
Question Number 10830 by Saham last updated on 26/Feb/17 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\frac{\mathrm{3}^{\mathrm{x}} \:−\:\mathrm{3}^{−\mathrm{x}} }{\mathrm{3}^{\mathrm{x}} \:+\:\mathrm{3}^{−\mathrm{x}} } \\ $$ Answered by bahmanfeshki last updated on 26/Feb/17 $$\underset{{x}\rightarrow\infty}…
Question Number 10829 by Saham last updated on 26/Feb/17 $$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\int_{\:\mathrm{1}} ^{\:\mathrm{x}} \:\:\:\frac{\mathrm{e}^{\mathrm{t}^{\mathrm{2}} } \:\left(\mathrm{dt}\right)}{\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{1}}\: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{1}\:\left(\mathrm{b}\right)\:\mathrm{0}\:\left(\mathrm{c}\right)\:\mathrm{e}/\mathrm{2}\:\left(\mathrm{d}\right)\:\mathrm{e} \\ $$ Answered by bahmanfeshki last updated…