Question Number 123179 by Dwaipayan Shikari last updated on 23/Nov/20 $$\frac{\mathrm{1}^{\mathrm{7}} }{{e}^{\mathrm{2}\pi} −\mathrm{1}}+\frac{\mathrm{2}^{\mathrm{7}} }{{e}^{\mathrm{4}\pi} −\mathrm{1}}+\frac{\mathrm{3}^{\mathrm{7}} }{{e}^{\mathrm{6}\pi} −\mathrm{1}}+…. \\ $$ Commented by Lordose last updated on…
Question Number 123167 by Dwaipayan Shikari last updated on 23/Nov/20 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\sqrt{{n}}}{{n}^{\mathrm{2}} +{n}+\mathrm{1}} \\ $$ Commented by mathmax by abdo last updated on 23/Nov/20…
Question Number 57619 by tanmay.chaudhury50@gmail.com last updated on 08/Apr/19 Answered by math1967 last updated on 09/Apr/19 $$\begin{vmatrix}{{q}+{r}}&{{r}+{p}}&{{p}+{q}}\\{{y}+{z}}&{{z}+{x}}&{{x}+{y}}\end{vmatrix} \\ $$$$\begin{vmatrix}{{b}+{c}−{c}−{a}−{a}−{b}}&{{c}+{a}}&{{a}+{b}}\\{{q}+{r}−{r}−{p}−{p}−{q}}&{{r}+{p}}&{{p}+{q}}\\{{y}+{z}−{z}−{x}−{x}−{y}}&{{z}+{x}}&{{x}+{y}}\end{vmatrix}{C}_{\mathrm{1}} −{C}_{\mathrm{2}} −{C}_{\mathrm{3}} \\ $$$$−\mathrm{2}\begin{vmatrix}{{a}}&{{c}+{a}}&{{a}+{b}}\\{{p}}&{{r}+{p}}&{{p}+{q}}\\{{x}}&{{z}+{x}}&{{x}+{y}}\end{vmatrix} \\ $$$$=−\mathrm{2}\begin{vmatrix}{{a}}&{{c}+{a}−{a}}&{{a}+{b}−{a}}\\{{p}}&{{r}+{p}−{p}}&{{p}+{q}−{p}}\\{{x}}&{{z}+{x}−{x}}&{{x}+{y}−{x}}\end{vmatrix}{C}_{\mathrm{2}}…
Question Number 57617 by tanmay.chaudhury50@gmail.com last updated on 08/Apr/19 Commented by tanmay.chaudhury50@gmail.com last updated on 08/Apr/19 $${source}\:{Hall}\:{andknight}\:\:{algebra} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on…
Question Number 57618 by tanmay.chaudhury50@gmail.com last updated on 08/Apr/19 Answered by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19 $$\mid{b}+{c}\:\:\:\:{a}−{b}\:\:\:\:{a}\mid \\ $$$$\mid{c}+{a}\:\:\:\:\:{b}−{c}\:\:\:\:{b}\mid \\ $$$$\mid{a}+{b}\:\:\:\:\:{c}−{a}\:\:\:\:{c}\mid \\ $$$$\mid{b}\:\:\:\:\:\:\:{a}\:\:\:\:\:\:{a}\mid+\mid{b}\:\:\:−{b}\:\:\:\:{a}\mid+\mid{c}\:\:\:\:{a}\:\:\:\:{a}\mid+\mid{c}\:\:\:\:−{b}\:\:{a}\:\mid \\ $$$$\mid{c}\:\:\:\:\:\:\:{b}\:\:\:\:\:\:{b}\:\mid\:\:\:\:\:\mid{c}\:\:\:\:\:−{c}\:\:{b}\mid\:\:\:\:\mid{a}\:\:\:{b}\:\:\:\:\:{b}\mid\:\:\:\:\mid{a}\:\:−{c}\:\:\:\:\:{b}\mid…
Question Number 57601 by ketto2 last updated on 08/Apr/19 Commented by Tawa1 last updated on 08/Apr/19 $$\:\:\:\mathrm{a}\::\:\mathrm{b}\:=\:\mathrm{4}\::\:\mathrm{9} \\ $$$$\therefore\:\:\:\:\:\frac{\mathrm{a}}{\mathrm{b}}\:\:=\:\:\frac{\mathrm{4}}{\mathrm{9}} \\ $$$$\therefore\:\:\:\:\:\:\mathrm{9a}\:\:=\:\:\mathrm{4b} \\ $$$$\therefore\:\:\:\:\:\mathrm{a}\:=\:\frac{\mathrm{4b}}{\mathrm{9}}\:\:\:\:\:\:\:\:\:\:>>>>>\:\:\mathrm{note} \\ $$$$\mathrm{Second}\:\mathrm{one}…
Question Number 57599 by ketto2 last updated on 08/Apr/19 Answered by MJS last updated on 08/Apr/19 $$\frac{{a}}{{b}}×\frac{{b}}{{c}}=\frac{{a}}{{c}} \\ $$$$\frac{\mathrm{4}}{\mathrm{9}}×\frac{\mathrm{3}}{\mathrm{7}}=\frac{\mathrm{4}}{\mathrm{21}} \\ $$ Answered by Tawa1 last…
Question Number 123135 by Dwaipayan Shikari last updated on 23/Nov/20 $$\int_{\mathrm{0}} ^{\infty} \frac{{ta}−{bt}^{\mathrm{3}} }{\left(\mathrm{1}+{ct}^{\mathrm{2}} \right)\left({e}^{\mathrm{2}\pi{t}} −\mathrm{1}\right)}{dt} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 188657 by Ajetunmobi last updated on 04/Mar/23 Answered by Ajetunmobi last updated on 04/Mar/23 $${please}\:{solve}\:{with}\:{detail}\:{solution} \\ $$ Answered by witcher3 last updated on…
Question Number 57491 by Joel578 last updated on 05/Apr/19 $$\mathrm{If}\:{R}\:\mathrm{is}\:\mathrm{a}\:\mathrm{region}\:\mathrm{enclosed}\:\mathrm{by}\:{y}\:=\:{f}\left({x}\right),\:{y}\:=\:{g}\left({x}\right),\:{x}\:=\:{a},\:{x}\:=\:{b}, \\ $$$$\mathrm{is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{have}\:{f}\left({x}\right)\:\mathrm{and}\:{g}\left({x}\right)\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{center}\:\mathrm{of}\:\mathrm{gravity}\:\left(\bar {{x}},\:\bar {{y}}\right)\:\mathrm{is}\:\mathrm{not}\:\mathrm{inside}\:{R}\:? \\ $$ Answered by ajfour last updated on 05/Apr/19…