Question Number 54647 by gunawan last updated on 08/Feb/19 $$\mathrm{show}\:\mathrm{that} \\ $$$${a}.\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\Sigma}}\:{r}^{\mathrm{3}} ._{{n}} {C}_{{r}} ={n}^{\mathrm{2}} \left({n}+\mathrm{3}\right).\mathrm{2}^{{n}−\mathrm{3}} \\ $$$${b}.\:_{{n}} {C}_{\mathrm{0}} ._{{n}} {C}_{\mathrm{1}} +_{{n}} {C}_{\mathrm{1}}…
Question Number 120160 by 77731 last updated on 29/Oct/20 Commented by JDamian last updated on 29/Oct/20 https://m.youtube.com/watch?v=9kfmk7jEuvQ Answered by mindispower last updated on 29/Oct/20 $$\frac{\mathrm{1}}{\mathrm{2}^{{m}}…
Question Number 54607 by ajfour last updated on 07/Feb/19 $${If}\:\:\frac{{u}^{\mathrm{5}} +{v}^{\mathrm{5}} }{\left({u}+{v}\right)^{\mathrm{5}} }\:=\:−\frac{\mathrm{1}}{\mathrm{5}}\:, \\ $$$${find}\:\:\:\frac{{u}^{\mathrm{3}} +{v}^{\mathrm{3}} }{\left({u}+{v}\right)^{\mathrm{3}} }\:\:=\:? \\ $$ Commented by MJS last updated…
Question Number 120118 by sarahvalencia last updated on 30/Oct/20 Commented by Dwaipayan Shikari last updated on 29/Oct/20 $${Net}\:{resistance}\:\mathrm{10}{k}\Omega \\ $$$${I}=\frac{\mathrm{50}}{\mathrm{10}.\mathrm{10}^{\mathrm{3}} }=\mathrm{5}×\mathrm{10}^{−\mathrm{3}} {A} \\ $$ Commented…
Question Number 54502 by math1967 last updated on 05/Feb/19 $${In}\:\bigtriangleup{ABC}\:\mathrm{cos}\:{A}+\mathrm{cos}\:{B}+\mathrm{cos}\:{C}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${prove}\:{that}\:{trianle}\:{is}\:{equilateral} \\ $$ Commented by tanmay.chaudhury50@gmail.com last updated on 05/Feb/19 Commented by tanmay.chaudhury50@gmail.com last…
Question Number 185553 by Noorzai last updated on 23/Jan/23 Answered by Frix last updated on 23/Jan/23 $$\mathrm{3}{a}−\mathrm{5}{b}=\mathrm{0}\:\Rightarrow\:{b}=\frac{\mathrm{3}{a}}{\mathrm{5}} \\ $$$$\mathrm{2}{b}−\mathrm{3}{c}=\mathrm{0}\:\Rightarrow\:{c}=\frac{\mathrm{2}{b}}{\mathrm{3}}=\frac{\mathrm{2}{a}}{\mathrm{5}} \\ $$$$\frac{\mathrm{2}{a}−\frac{\mathrm{12}{a}}{\mathrm{5}}+\mathrm{2}{a}}{\frac{\mathrm{3}{a}}{\mathrm{5}}−\frac{\mathrm{6}{a}}{\mathrm{5}}}=−\frac{\mathrm{8}}{\mathrm{3}} \\ $$ Terms of…
Question Number 54457 by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 185517 by Noorzai last updated on 23/Jan/23 Answered by Ar Brandon last updated on 23/Jan/23 $$\Omega=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}{x}}{{x}^{\mathrm{6}} −\mathrm{1}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}{x}}{{x}^{\mathrm{6}} −\mathrm{1}}{dx}+\int_{\mathrm{1}} ^{\infty}…
Question Number 185518 by Mastermind last updated on 23/Jan/23 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{convergence}\:\mathrm{of} \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{ne}^{−\mathrm{n}^{\mathrm{2}} } \\ $$$$ \\ $$$$\mathrm{M}.\mathrm{m} \\ $$ Commented by aleks041103 last…
Question Number 185470 by Mastermind last updated on 22/Jan/23 $$\mathrm{Determine}\:\mathrm{whether}\:\mathrm{the}\:\mathrm{series} \\ $$$$\mathrm{U}_{\mathrm{n}} =\frac{\mathrm{1}+\mathrm{2n}^{\mathrm{2}} }{\mathrm{1}+\mathrm{n}^{\mathrm{2}} }\:\mathrm{is}\:\mathrm{convergent}\:\mathrm{or}\:\mathrm{not} \\ $$$$ \\ $$$$\mathrm{M}.\mathrm{m} \\ $$ Commented by mr W…