Question Number 142347 by mathmax by abdo last updated on 30/May/21 $$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{2}} \mathrm{logx}}{\mathrm{x}^{\mathrm{6}} \:+\mathrm{1}}\mathrm{dx} \\ $$ Commented by mathmax by abdo last updated…
Question Number 76808 by aliesam last updated on 30/Dec/19 $$\int\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}}{dx} \\ $$ Commented by redmiiuser last updated on 28/Mar/20 $${sir}\:{please}\:{check}\:{my}\: \\ $$$${answer} \\ $$…
Question Number 142346 by mathmax by abdo last updated on 30/May/21 $$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{log}\left(\mathrm{1}+\mathrm{x}^{\mathrm{3}} \right)}{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 76809 by Rio Michael last updated on 30/Dec/19 Commented by Rio Michael last updated on 30/Dec/19 $$\mathrm{I}_{\mathrm{q}} \:\mathrm{is}\:\mathrm{the}\:\mathrm{moment}\:\mathrm{of}\:\mathrm{inertia}\:\mathrm{of}\:\mathrm{a}\:\mathrm{uniform}\:\mathrm{rod} \\ $$$$\mathrm{AB}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{m}\:\mathrm{about}\:\mathrm{its}\:\mathrm{centre}\:\mathrm{and}\:\mathrm{I}_{\mathrm{p}} \:\mathrm{is}\:\mathrm{the}\: \\ $$$$\mathrm{moment}\:\mathrm{of}\:\mathrm{inertia}\:\mathrm{about}\:\mathrm{the}\:\mathrm{axis}\:\mathrm{shown}\:\mathrm{above}.…
Question Number 11272 by chux last updated on 18/Mar/17 Answered by mrW1 last updated on 18/Mar/17 $$\left(\mathrm{1}\right)\Rightarrow{y}+{z}=\mathrm{1}−{x} \\ $$$$\left(\mathrm{2}\right)\Rightarrow{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{35}−{x}^{\mathrm{2}} \\ $$$$\left({y}+{z}\right)^{\mathrm{2}} =\left(\mathrm{1}−{x}\right)^{\mathrm{2}} \\…
Question Number 142341 by mohammad17 last updated on 30/May/21 Commented by mohammad17 last updated on 30/May/21 $${how}\:{can}\:{solve}\:{this} \\ $$ Answered by Dwaipayan Shikari last updated…
Question Number 76807 by aliesam last updated on 30/Dec/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 11268 by tawa last updated on 18/Mar/17 $$\mathrm{Solve}\:\mathrm{x},\:\mathrm{y}\:\mathrm{and}\:\mathrm{z}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{p},\:\mathrm{q}\:\mathrm{and}\:\mathrm{r} \\ $$$$\mathrm{yz}\:=\:\mathrm{py}\:+\:\mathrm{qz}\:\:\:\:\:……..\:\left(\mathrm{i}\right) \\ $$$$\mathrm{zx}\:=\:\mathrm{qz}\:+\:\mathrm{rx}\:\:\:\:\:\:…….\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{xy}\:=\:\mathrm{rx}\:+\:\mathrm{py}\:\:\:\:\:…….\:\left(\mathrm{iii}\right) \\ $$ Answered by mrW1 last updated on 18/Mar/17…
Question Number 76802 by kavinila last updated on 30/Dec/19 $$\left.\mathrm{3}\right)\alpha\mathrm{particle}\:\mathrm{of}\:\mathrm{energy}\:\mathrm{5MeV}\:\mathrm{pass} \\ $$$$\mathrm{through}\:\mathrm{an}\:\mathrm{ionisation}\:\mathrm{chamber}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{rate}\:\mathrm{of}\:\mathrm{10}\:\mathrm{pwe}\:\mathrm{second}\:.\:\mathrm{Assum}\:\mathrm{that}\:\mathrm{all} \\ $$$$\mathrm{the}\:\mathrm{energy}\:\mathrm{is}\:\mathrm{used}\:\mathrm{in}\:\mathrm{producing}\:\mathrm{ion}\: \\ $$$$\mathrm{pairs},\mathrm{calculate}\:\mathrm{the}\:\mathrm{current}\:\mathrm{produced}. \\ $$$$\left(\mathrm{35MeV}\:\mathrm{is}\:\mathrm{required}\:\mathrm{for}\:\mathrm{producing}\:\:\mathrm{an}\:\right. \\ $$$$\left.\mathrm{ion}\:\mathrm{pair}\:\mathrm{and}\:\mathrm{e}=\mathrm{1}.\mathrm{6}×\mathrm{10}^{-\mathrm{19}} \:\mathrm{C}\right) \\ $$$$\boldsymbol{\mathrm{solution}}:…
Question Number 11266 by ketto last updated on 18/Mar/17 Commented by tawa last updated on 19/Mar/17 $$\mathrm{probability}\:\mathrm{that}\:\mathrm{juma}\:\mathrm{pass}\:=\:\mathrm{50\%}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{probability}\:\mathrm{that}\:\mathrm{juma}\:\mathrm{fail}\:=\:\mathrm{1}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{probability}\:\mathrm{that}\:\mathrm{gidi}\:\mathrm{fail}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\: \\ $$$$\mathrm{probability}\:\mathrm{that}\:\mathrm{gidi}\:\mathrm{pass}\:=\:\mathrm{1}\:−\:\frac{\mathrm{1}}{\mathrm{4}}\:=\:\frac{\mathrm{3}}{\mathrm{4}}…