Question Number 208381 by hardmath last updated on 14/Jun/24 $$\mathrm{g}\left(\mathrm{x}\right)\:=\:\mathrm{lnx}^{\mathrm{2}} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\sqrt[{\mathrm{3}}]{\mathrm{x}\:+\:\mathrm{25}} \\ $$$$\mathrm{Find}:\:\:\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\boldsymbol{\mathrm{e}}} {\mathrm{lim}}\:\left(\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}\right)\right)\:=\:?\right. \\ $$ Answered by A5T last updated on 14/Jun/24 $${f}\left({g}\left({x}\right)\right)=\sqrt[{\mathrm{3}}]{{ln}\left({x}^{\mathrm{2}}…
Question Number 208377 by hardmath last updated on 14/Jun/24 $$\mathrm{sin}\:\mathrm{x}\:−\:\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\:>\:\mathrm{0} \\ $$$$\mathrm{x}\:=\:? \\ $$ Commented by hardmath last updated on 14/Jun/24 $$ \\ $$Solve the…
Question Number 208370 by lmcp1203 last updated on 14/Jun/24 $${if}\:\:\:\left({fof}\right)\left({x}\right)={f}\left({x}\right)+{x}\:\:{and}\:{f}\left(\mathrm{1}\right)=\mathrm{1}\:\:\: \\ $$$${find}\:\:{fofofofofofofofofof}\left(\mathrm{1}\right) \\ $$ Commented by A5T last updated on 14/Jun/24 $${f}\left({f}\left({x}\right)\right)={f}\left({x}\right)+{x}\Rightarrow\:{f}\left({f}\left(\mathrm{1}\right)\right)={f}\left(\mathrm{1}\right)+\mathrm{1} \\ $$$$\Rightarrow{f}\left(\mathrm{1}\right)={f}\left(\mathrm{1}\right)+\mathrm{1}\Rightarrow\mathrm{1}\overset{?} {=}\mathrm{0}…
Question Number 208362 by hardmath last updated on 13/Jun/24 $$\mathrm{P}\left(\mathrm{x}\right)\:\:\mathrm{is}\:\mathrm{polynomial} \\ $$$$\mathrm{P}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{x}^{\mathrm{4}} \:+\:\mathrm{2ax}^{\mathrm{3}} \:−\:\mathrm{bx}\:−\:\mathrm{5}}{\left(\mathrm{x}\:+\:\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\mathrm{Find}:\:\:\:\boldsymbol{\mathrm{b}}\:=\:? \\ $$ Answered by mr W last updated…
Question Number 208342 by hardmath last updated on 13/Jun/24 $$\mathrm{a},\mathrm{b},\mathrm{c}\in\mathbb{N} \\ $$$$\mathrm{x}\:=\:\mathrm{4}\left(\mathrm{2a}+\mathrm{5}\right)\:=\:\mathrm{6}\left(\mathrm{b}+\mathrm{9}\right)\:=\:\mathrm{9}\left(\mathrm{c}−\mathrm{1}\right) \\ $$$$\mathrm{find}:\:\:\:\boldsymbol{\mathrm{min}}\left(\mathrm{x}+\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\:=\:? \\ $$ Answered by A5T last updated on 13/Jun/24 $${c}=\frac{\mathrm{6}\left({b}+\mathrm{9}\right)}{\mathrm{9}}+\mathrm{1}=\frac{\mathrm{2}{b}}{\mathrm{3}}+\mathrm{7}\Rightarrow{b}=\mathrm{3}{k} \\…
Question Number 208332 by essaad last updated on 12/Jun/24 Answered by Frix last updated on 12/Jun/24 $$\underset{{k}=\mathrm{0}} {\overset{{n}} {\prod}}\:\left({k}+{a}\right)\:={a}\left({a}+\mathrm{1}\right)\left({a}+\mathrm{2}\right)…\left({a}+{n}\right)=\frac{\left({a}+{n}\right)!}{\left({a}−\mathrm{1}\right)!} \\ $$$$\Rightarrow \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\prod}}\:\frac{{k}^{\mathrm{2}}…
Question Number 208282 by hardmath last updated on 10/Jun/24 $$\mathrm{If}\:\:\:\mathrm{cos}\alpha−\mathrm{cos}\beta\:=\:\frac{\mathrm{1}}{\mathrm{5}}\:\mathrm{sin}\alpha\:+\:\mathrm{sin}\beta\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{Find}\:\:\:\mathrm{cos}\left(\alpha\:+\:\beta\right)\:=\:? \\ $$ Commented by mr W last updated on 12/Jun/24 $${what}\:{you}\:{wrote}\:{means} \\ $$$$\mathrm{cos}\alpha−\mathrm{cos}\beta\:=\frac{\mathrm{1}}{\mathrm{2}}…
Question Number 208292 by universe last updated on 10/Jun/24 $$\:\mathrm{let}\:\mathrm{T}\:\mathrm{be}\:\mathrm{a}\:{n}×{n}\:\mathrm{matrix}\:\mathrm{with}\:\mathrm{integral}\: \\ $$$$\:\mathrm{entries}\:\mathrm{and}\:\:\mathrm{Q}\:=\:\mathrm{T}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{I}\:\:\:\mathrm{where}\:\mathrm{I}\:\mathrm{denote} \\ $$$$\:\:\mathrm{the}\:\mathrm{n}×\mathrm{n}\:\mathrm{identity}\:\mathrm{matrix}\:\mathrm{then}\:\mathrm{prove} \\ $$$$\:\:\mathrm{that}\:\mathrm{matrix}\:\mathrm{Q}\:\mathrm{is}\:\mathrm{invertible} \\ $$ Answered by Berbere last updated on 10/Jun/24…
Question Number 208259 by hardmath last updated on 09/Jun/24 Answered by cherokeesay last updated on 09/Jun/24 Answered by mr W last updated on 09/Jun/24 Commented…
Question Number 208241 by Frix last updated on 08/Jun/24 $$\mathrm{Solve}\:\mathrm{for}\:{p},\:{q},\:{r} \\ $$$${p}+{q}+{r}=\alpha \\ $$$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} =\beta \\ $$$${pq}={r} \\ $$ Answered by mr W…