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Question Number 59975 by vajpaithegrate@gmail.com last updated on 16/May/19 Answered by tanmay last updated on 16/May/19 $${T}.{S}.{A}\:{of}\:{cone}=\pi{rl}+\pi{r}^{\mathrm{2}} \\ $$$$\left.{T}\left..{S}.{A}=\pi{r}\left({r}+\sqrt{{h}^{\mathrm{2}} +{r}^{\mathrm{2}} }\:\right)\:\:\:\:\:\right]\right]{when}\:\left[{l}^{\mathrm{2}} ={r}^{\mathrm{2}} +{h}^{\mathrm{2}} \right] \\…
Question Number 59974 by vajpaithegrate@gmail.com last updated on 16/May/19 Answered by tanmay last updated on 16/May/19 $${lateral}\:{surface}\:{area}=\pi{rl} \\ $$$${given}\:{cos}\mathrm{45}^{{o}} =\frac{{h}}{{l}} \\ $$$${l}=\frac{{h}}{{cos}\mathrm{45}^{{o}} }={h}\sqrt{\mathrm{2}} \\ $$$${tan}\mathrm{45}^{{o}}…
Question Number 59936 by Sardor2211 last updated on 16/May/19 Answered by tanmay last updated on 16/May/19 $${x}\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\:+\frac{{dy}}{{dx}}.{y}.\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:=\mathrm{0} \\ $$$${x}\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }\:{dx}+{y}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:{dy}=\mathrm{0} \\ $$$$\frac{{xdx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}}…
Question Number 190927 by alcohol last updated on 14/Apr/23 Answered by MikeH last updated on 15/Apr/23 $${I}\:=\:\int{t}^{\mathrm{7}} \mathrm{sin}\left(\mathrm{2}{t}^{\mathrm{4}} \right){dt} \\ $$$$\mathrm{let}\:{u}\:=\mathrm{2}\:{t}^{\mathrm{4}} \:\Rightarrow\:{du}\:=\:\mathrm{8}{t}^{\mathrm{3}} {dt}\:\:\Rightarrow\:{dt}\:=\:\frac{{du}}{\mathrm{8}{t}^{\mathrm{3}} } \\…
Question Number 59838 by necx1 last updated on 15/May/19 $${What}\:{is}\:{the}\:{nth}\:{derivative}\:{of}\:{sinx}\:{in} \\ $$$${terms}\:{of}\:{the}\:{sine}\:{function}? \\ $$ Commented by maxmathsup by imad last updated on 15/May/19 $${if}\:{f}\left({x}\right)={sinx}\:\:\:,\:\:{f}^{\left({n}\right)} \left({x}\right)\:={sin}\left({x}+\frac{{n}\pi}{\mathrm{2}}\right)\:\:\:{with}\:{n}\geqslant\mathrm{1}\:\:{and}\:\:{f}^{\left(\mathrm{0}\right)}…
Question Number 125368 by bemath last updated on 10/Dec/20 $$\:{How}\:{to}\:{derive}\:{the}\:{formula}\: \\ $$$${for}\:{finding}\:{the}\:{volume}\:{of}\: \\ $$$${spherical}\:{curl}\:? \\ $$ Commented by Ar Brandon last updated on 10/Dec/20 https://www.therightgate.com/deriving-curl-in-cylindrical-and-spherical/…
Question Number 125322 by john_santu last updated on 10/Dec/20 Commented by john_santu last updated on 10/Dec/20 $${question}\:\left({c}\right) \\ $$ Commented by john_santu last updated on…
Question Number 125297 by mnjuly1970 last updated on 09/Dec/20 $$\:::::{prove}\:\:{that}\:: \\ $$$$\:\:\:\:{i}:\:\:\:\Omega=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{x}^{{s}−\mathrm{1}} }{\mathrm{1}+{x}}{dx}=\psi\left({s}\right)\:−\:\psi\left(\frac{{s}}{\mathrm{2}}\right)−{ln}\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:{ii}:\:\psi\left(\mathrm{2}{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\psi\left({x}\right)+\frac{\mathrm{1}}{\mathrm{2}}\psi\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)+{ln}\left(\mathrm{2}\right) \\ $$ Answered by Bird last updated on…
Question Number 59753 by Aditya789 last updated on 14/May/19 $$\mathrm{x}=\left(\mathrm{a}+\mathrm{b}\right)\mathrm{cosx}−\mathrm{bcos}\left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{b}}\right)\mathrm{x} \\ $$$$\mathrm{y}=\left(\mathrm{a}+\mathrm{b}\right)\mathrm{sinx}−\mathrm{bsin}\left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{b}}\right)\mathrm{x} \\ $$$$\mathrm{find}\:\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{tan}\left(\frac{\mathrm{a}}{\mathrm{2b}}+\mathrm{1}\right)\mathrm{x} \\ $$ Answered by tanmay last updated on 14/May/19 $${x}+{y}=\left({a}+{b}\right)\left({sinx}+{cosx}\right)−{b}\left[{sin}\left(\frac{{a}+{b}}{{b}}\right){x}+{cos}\left(\frac{{a}+{b}}{{b}}\right){x}\right] \\…