Question Number 190079 by uchihayahia last updated on 26/Mar/23 $$ \\ $$$$ \\ $$$$\:{F}\left({t}\right)=\left(\mathrm{4}{t}^{\mathrm{3}} ,\mathrm{2}{cos}\left(\mathrm{2}{t}\right),\mathrm{3}{e}^{\mathrm{3}{t}} \right) \\ $$$$\:{find}\:{F}\:'\left({t}\right) \\ $$$$\:{F}\:'\left({t}\right)=\left(\mathrm{12}{t}^{\mathrm{2}} ,-\mathrm{4}{sin}\left(\mathrm{2}{t}\right),\mathrm{9}{e}^{\mathrm{3}{t}} \right) \\ $$$$\:{is}\:{my}\:{answer}\:{correct}? \\…
Question Number 58971 by hovea cw last updated on 02/May/19 Commented by hovea cw last updated on 02/May/19 $$\mathrm{hd}\:\mathrm{plz} \\ $$ Answered by tanmay last…
Question Number 124485 by snipers237 last updated on 03/Dec/20 $${Let}\:{S}^{{n}} =\left\{\left({x}_{\mathrm{0}} ,….,{x}_{{n}} \right)\in\mathbb{R}^{{n}+\mathrm{1}} \:/\:\:\:\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}{x}_{{i}} ^{\mathrm{2}} \:\leqslant\mathrm{1}\right\} \\ $$$${Find}\:\:{Vol}\left({S}^{{n}} \right)\:{for}\:{all}\:{n}\geqslant\mathrm{3} \\ $$ Terms of…
Question Number 124483 by snipers237 last updated on 03/Dec/20 $$\:{Develop}\:{the}\:{function}\:{f}\left({x}\right)={e}^{{x}} {sinx} \\ $$$${Then}\:{Deduce}\:{that}\: \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} {\sum}}\left(−\mathrm{1}\right)^{{k}} {C}_{{n}} ^{\mathrm{2}{k}+\mathrm{1}} \:=\:\mathrm{2}^{\frac{{n}}{\mathrm{2}}} {sin}\left(\frac{{n}\pi}{\mathrm{4}}\right) \\ $$ Answered by…
Question Number 124415 by snipers237 last updated on 03/Dec/20 $${Let}\:{a}>\mathrm{0},\:\:\:{A}=\left\{{f}\in{C}^{\mathrm{2}} \left(\left[\mathrm{0},{a}\right],\mathbb{R}\right)\:,\:{f}\left(\mathrm{0}\right)={f}'\left(\mathrm{0}\right)=\mathrm{0}\right\} \\ $$$${N}_{\mathrm{1}} \left({f}\right)=\:{sup}\left\{\mid{f}\left({x}\right)\mid+\mid{f}''\left({y}\right)\mid\:\:\:,{x},{y}\in\left[\mathrm{0},{a}\right]\right\} \\ $$$${N}_{\mathrm{2}} \left({f}\right)={sup}\left\{\mid{f}\left({x}\right)+{f}''\left({x}\right)\mid\:\:\:,{x}\in\left[\mathrm{0},{a}\right]\right\} \\ $$$${Prove}\:{that}\:{N}_{\mathrm{1}} {and}\:{N}_{\mathrm{2}} \:{are}\:{equivalents}\:{norms} \\ $$ Commented by…
Question Number 58720 by Forkum Michael Choungong last updated on 28/Apr/19 $${find}\:\frac{{dy}}{{dx}}\:{given}\:{that}\:\:{y}\:=\:{cos}\left({x}°\right) \\ $$ Answered by tanmay last updated on 28/Apr/19 $$\mathrm{180}^{{o}} =\pi\:{radian} \\ $$$${x}^{{o}}…
Question Number 124244 by bramlexs22 last updated on 02/Dec/20 $${Find}\:\frac{{dy}}{{dx}}\:{of}\:{function}\:{y}=\mathrm{5}^{\sqrt{{x}}} \\ $$$${by}\:{first}\:{principle}. \\ $$ Answered by liberty last updated on 02/Dec/20 Terms of Service Privacy…
Question Number 124169 by bramlexs22 last updated on 01/Dec/20 $${A}\:{long}\:{strip}\:{of}\:{sheet}\:{metal}\:\mathrm{12}\:{inches} \\ $$$${wide}\:{is}\:{to}\:{be}\:{made}\:{into}\:{a}\:{small}\: \\ $$$${trough}\:{by}\:{turning}\:{up}\:{two}\: \\ $$$${sides}\:{at}\:{right}\:{angles}\:{to}\:{the}\:{base}\: \\ $$$${If}\:{trough}\:{is}\:{to}\:{have}\:{maximum} \\ $$$${capasit}\bar {{y}},\:{how}\:{many}\:{inches}\:{should}\:{be} \\ $$$${turned}\:{up}\:{on}\:{each}\:{side}?\: \\ $$$$\left({a}\right)\:\mathrm{6}\:{in}\:\:\:\:\left({b}\right)\:\mathrm{4}\:{in}\:{on}\:{one}\:{side},\:\mathrm{5}\:{in}\:{on}\:{the}\:{other}…
Question Number 124133 by bramlexs22 last updated on 01/Dec/20 $${Given}\:{equation}\:{of}\:{tangent}\:{line} \\ $$$${of}\:{the}\:{curve}\:{y}\:=\:\frac{{b}}{{x}^{\mathrm{2}} }\:{at}\:{point}\:\left({x},{y}\right) \\ $$$${is}\:{bx}−\mathrm{4}{y}=−\mathrm{21}.\:{The}\:{value}\:{of}\:{b}\:=? \\ $$ Answered by mr W last updated on 01/Dec/20…
Question Number 124064 by mnjuly1970 last updated on 30/Nov/20 $$\:\:\:\:\:\:\:\:\:…\:{nice}\:\:{calculus}… \\ $$$$\:\:\:\:\:{prove}\:\:\:{that}::: \\ $$$$\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \left\{\frac{\mathrm{1}+\left(\mathrm{1}−{x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{{x}}\:+\frac{\mathrm{2}}{{ln}\left(\mathrm{1}−{x}\right)}\right\}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\overset{???} {=}\mathrm{2}\left(\gamma−\mathrm{1}+{log}\left(\mathrm{2}\right)\right) \\ $$ Answered by mindispower…