Question Number 188359 by 073 last updated on 28/Feb/23 Answered by Rasheed.Sindhi last updated on 28/Feb/23 $$\mathrm{1}!+\mathrm{2}!+\mathrm{3}!+…+\mathrm{48}!\equiv{x}\left({mod}\:\mathrm{6}\right) \\ $$$${For}\:{k}\geqslant\mathrm{3},\:\mathrm{6}\mid{k}!\: \\ $$$$\Rightarrow\mathrm{1}!+\mathrm{2}!\equiv{x}\left({mod}\:\mathrm{6}\right) \\ $$$$\:\:\:\:\:\mathrm{1}+\mathrm{2}\equiv{x}\left({mod}\:\mathrm{6}\right. \\ $$$${x}=\mathrm{3}…
Question Number 122799 by aurpeyz last updated on 19/Nov/20 Answered by liberty last updated on 19/Nov/20 $${letting}\:{u}\:=\:\mathrm{1}+\mathrm{3}{x}^{−\mathrm{1}} \:\Rightarrow{du}=−\mathrm{3}{x}^{−\mathrm{2}} {dx} \\ $$$${L}\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{3}}\int\:{u}^{\frac{\mathrm{2}}{\mathrm{3}}} \:{du}\:\:=\:−\frac{\mathrm{1}}{\mathrm{3}}×\frac{\mathrm{3}}{\mathrm{5}}{u}^{\frac{\mathrm{5}}{\mathrm{3}}} +\:{c} \\ $$$${thus}\:\underset{\mathrm{1}}…
Question Number 122798 by aurpeyz last updated on 19/Nov/20 Commented by mohammad17 last updated on 19/Nov/20 $${I}=\int_{\mathrm{1}} ^{\:\mathrm{3}} \frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(\mathrm{1}+\frac{\mathrm{3}}{{x}}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} {dx}=−\frac{\mathrm{1}}{\mathrm{5}}\:\left(\mathrm{2}\right)^{\frac{\mathrm{5}}{\mathrm{3}}\:} \:+\frac{\mathrm{1}}{\mathrm{5}}\left(\mathrm{4}\right)^{\frac{\mathrm{5}}{\mathrm{3}}} \\ $$ Answered…
Question Number 188325 by otchereabdullai last updated on 28/Feb/23 $$\:{A}\:{polytank}\:{contains}\:\mathrm{3600}{cm}^{\mathrm{3}\:} \:{of}\: \\ $$$${water}.\:{A}\:{boy}\:{threw}\:{a}\:{stone}\:{to}\:{the}\: \\ $$$${tank}\:{and}\:{some}\:{of}\:{the}\:{water}\:{leaked}\: \\ $$$${away}.\:{When}\:{the}\:{owner}\:{realized}\:{only} \\ $$$$\mathrm{319}{cm}^{\mathrm{3}} \:{of}\:{water}\:{was}\:{left}.\:{If}\:\mathrm{1}{cm}^{\mathrm{2}\:} \\ $$$${of}\:{water}\:{cost}\:\mathrm{20}{peswas},\:{calculate}\: \\ $$$${the}\:{lost}. \\…
Question Number 122794 by rs4089 last updated on 19/Nov/20 Commented by Olaf last updated on 19/Nov/20 $$\mathrm{looks}\:\mathrm{like}\:\mathrm{Rammanujan} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 188324 by otchereabdullai last updated on 28/Feb/23 $$\:\:{A}\:{contractor}\:{wants}\:{to}\:{plaster}\:{wall} \\ $$$$\:{which}\:{is}\:\mathrm{2650}{m}.\:{He}\:{sends}\:\mathrm{50}\:{masons} \\ $$$$\:{to}\:{do}\:{the}\:{work}\:{equally},\:{what}\:{size}\:{of} \\ $$$${the}\:{work}\:{does}\:{each}\:{mason}\:{plaster}.\: \\ $$$${If}\:{a}\:{square}\:{cost}\:{Ghc}\mathrm{12}\:,\:{how}\:{much}\: \\ $$$${does}\:{each}\:{mason}\:{earn}. \\ $$ Commented by mr…
Question Number 122791 by mohammad17 last updated on 19/Nov/20 Answered by mathmax by abdo last updated on 19/Nov/20 $$\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{cos}\left(\mathrm{mx}\right)}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx}\:\Rightarrow\:\mathrm{2I}\:=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{cos}\left(\mathrm{mx}\right)}{\left(\mathrm{x}^{\mathrm{2}}…
Question Number 122789 by Khalmohmmad last updated on 19/Nov/20 Answered by MJS_new last updated on 19/Nov/20 $${A}^{−\mathrm{1}} {A}=\begin{pmatrix}{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}\end{pmatrix} \\ $$ Terms of Service Privacy Policy…
Question Number 122783 by mathocean1 last updated on 19/Nov/20 $${solve}\:: \\ $$$${x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$ Commented by Dwaipayan Shikari last updated on 19/Nov/20 $${x}={y}−\mathrm{1}…
Question Number 122773 by solstis last updated on 19/Nov/20 Answered by MJS_new last updated on 19/Nov/20 $$\frac{\mathrm{451}}{\mathrm{225}}−\frac{\mathrm{2}}{\mathrm{5}}=\frac{\mathrm{451}}{\mathrm{5}×\mathrm{45}}−\frac{\mathrm{2}}{\mathrm{5}}=\frac{\mathrm{451}}{\mathrm{5}×\mathrm{45}}−\frac{\mathrm{2}×\mathrm{45}}{\mathrm{5}×\mathrm{45}}= \\ $$$$=\frac{\mathrm{451}−\mathrm{2}×\mathrm{45}}{\mathrm{5}×\mathrm{45}}=\frac{\mathrm{361}}{\mathrm{225}} \\ $$$$\mathrm{no}\:\mathrm{other}\:\mathrm{form}\:\mathrm{to}\:\mathrm{resolve}\:\mathrm{it}.\:\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{add} \\ $$$$\mathrm{or}\:\mathrm{subtract}\:\mathrm{fractions}\:\mathrm{with}\:\mathrm{the}\:\mathrm{same}\:\mathrm{denominator} \\ $$…