Question Number 128845 by Dwaipayan Shikari last updated on 10/Jan/21 $$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}.\mathrm{4}}{\left(\mathrm{5}.\mathrm{1}!\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}}.\frac{\mathrm{1}.\mathrm{4}.\mathrm{6}.\mathrm{9}}{\left(\mathrm{5}^{\mathrm{2}} .\mathrm{2}!\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}.\frac{\mathrm{1}.\mathrm{4}.\mathrm{6}.\mathrm{9}.\mathrm{11}.\mathrm{14}}{\left(\mathrm{5}^{\mathrm{3}} .\mathrm{3}!\right)^{\mathrm{2}} }+…=\frac{{b}^{\mathrm{2}} \sqrt{\frac{{b}−\sqrt{{b}}}{\mathrm{2}}}}{{a}\pi} \\ $$$${Find}\:\mathrm{5}{a}−\mathrm{8}{b} \\ $$ Answered by mindispower last…
Question Number 63296 by Rio Michael last updated on 02/Jul/19 $${A}\:{random}\:{Variable}\:{Y}\:{has}\:{probability}\:{function}\:{P},\:{defined}\:{by} \\ $$$$\:{P}\left({y}\right)\:=\:\begin{cases}{\frac{{y}^{\mathrm{2}} }{{k}}\:,\:{y}=\:\mathrm{1},\mathrm{2},\mathrm{3}}\\{\frac{\left({y}−\mathrm{7}\right)^{\mathrm{2}} }{{k}}\:,\:{y}=\:\mathrm{4},\mathrm{5},\mathrm{6}}\\{\mathrm{0}\:\:\:\:,\:{otherwise}.}\end{cases} \\ $$$${Find}\: \\ $$$$\left({i}\right)\:{The}\:{value}\:{of}\:{the}\:{constant}\:{k}. \\ $$$$\left({ii}\right)\:{the}\:{mean}\:{and}\:{varriance}\:{of}\:{Y}. \\ $$$$\left({iii}\right)\:{The}\:{variance}\:{R},\:{where}\:{R}=\:\mathrm{2}{Y}\:−\mathrm{3}. \\ $$…
Question Number 63288 by Tawa1 last updated on 02/Jul/19 Answered by Rio Michael last updated on 02/Jul/19 $$\:{if}\:{AB}\:=\:{AC}\:\: \\ $$$${then},\:\angle{B}\left(\mathrm{73}°\right)\:=\:\angle{C} \\ $$$$\Rightarrow\:\mathrm{73}°\:=\:\mathrm{34}°\:+\:{x}° \\ $$$${x}°\:=\:\mathrm{39}^{°} \:\:{please}\:{check}.…
Question Number 63267 by Tawa1 last updated on 01/Jul/19 $$\:\:\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\:\left(\frac{\mathrm{n}^{\mathrm{3}} \:+\:\mathrm{1}}{\mathrm{n}^{\mathrm{3}} \:−\:\mathrm{1}}\right)^{\mathrm{2n}\:−\:\mathrm{n}^{\mathrm{3}} } \\ $$ Commented by Tawa1 last updated on 01/Jul/19 $$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\…
Question Number 128731 by TITA last updated on 09/Jan/21 $$\mathrm{for}\:\:\mathrm{a}>\mathrm{b}>\mathrm{0}\:\:\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{b}<\frac{\mathrm{ax}^{\mathrm{x}} +\mathrm{bx}^{−\mathrm{x}} }{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} }<\mathrm{a} \\ $$ Commented by TITA last updated on 09/Jan/21…
Question Number 128725 by Dwaipayan Shikari last updated on 09/Jan/21 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{coth}\left({n}\pi\right)}{{n}^{\mathrm{4}} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 128566 by Dwaipayan Shikari last updated on 08/Jan/21 $$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{1}!}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}.\mathrm{3}}{\mathrm{2}^{\mathrm{2}} }.\frac{\mathrm{1}}{\mathrm{2}!}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}}{\mathrm{2}^{\mathrm{3}} }.\frac{\mathrm{1}}{\mathrm{3}!}\right)^{\mathrm{2}} +…=_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};\mathrm{2};\mathrm{1}\right)=\frac{\mathrm{4}}{\pi} \\ $$ Commented by Dwaipayan Shikari last…
Question Number 128561 by 777316 last updated on 08/Jan/21 Answered by Dwaipayan Shikari last updated on 08/Jan/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{cosx}+\left(\frac{\mathrm{1}−{cosx}}{{sinx}}\right)=\mathrm{1}+\frac{\mathrm{2}{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{{sinx}}=\mathrm{1} \\ $$ Terms of Service…
Question Number 128557 by AgnibhoMukhopadhyay last updated on 08/Jan/21 $$\: \: \: \: \: \: \: \: \: \\ $$$$\: \: \: \\ $$ Answered…
Question Number 128522 by BHOOPENDRA last updated on 08/Jan/21 $${f}\left({t}\right)={t}+\mathrm{1}\:\:\:\:\:\:\mathrm{0}\leqslant{t}\leqslant\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{3}\:\:\:\:\:\:\:\:\:\:{t}>\mathrm{2} \\ $$$${find}\:{laplace}\:{transformation}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com