Question Number 134899 by 0731619177 last updated on 08/Mar/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 134890 by Khalmohmmad last updated on 08/Mar/21 Answered by bemath last updated on 08/Mar/21 $$\mathrm{let}\:{a}\:=\:\mathrm{16}.\mathrm{2}\overset{−} {\mathrm{7}}\:;\:\mathrm{10}{a}\:=\:\mathrm{162}.\overset{−} {\mathrm{7}} \\ $$$$\Rightarrow\:{a}\:=\:\frac{\mathrm{146}.\mathrm{5}}{\mathrm{9}} \\ $$$$\mathrm{let}\:{b}\:=\:\mathrm{15}.\mathrm{1}\overset{−} {\mathrm{7}}\:;\:\mathrm{10}{b}\:=\:\mathrm{151}.\overset{−} {\mathrm{7}}…
Question Number 134846 by mohammad17 last updated on 07/Mar/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 134839 by SOMEDAVONG last updated on 07/Mar/21 $$\mathrm{Solve}\:,\:\mathrm{S}_{\mathrm{n}} =\:\mathrm{1}+\mathrm{2}×\mathrm{2}+\mathrm{3}×\mathrm{2}^{\mathrm{2}} +\mathrm{4}×\mathrm{2}^{\mathrm{3}} +…+\mathrm{n}×\mathrm{2}^{\mathrm{n}−\mathrm{1}} . \\ $$ Commented by mr W last updated on 07/Mar/21 $${see}\:{Q}\mathrm{134200}…
Question Number 134831 by abdurehime last updated on 07/Mar/21 $${proof}\:{that}\:{if}\:\mathrm{R}\:\mathrm{is}\:\mathrm{a}\:\mathrm{commetative}\:\mathrm{ring}\:\mathrm{with}\:\mathrm{unity}\:\mathrm{then}\:\mathrm{an}\:\mathrm{ideal}\: \\ $$$$\mathrm{M}\:\mathrm{and}\:\mathrm{R}\:\mathrm{is}\:\mathrm{maximal}\:\mathrm{iff}\:\frac{\mathrm{R}}{\mathrm{M}}\:\mathrm{is}\:\mathrm{a}\:\mathrm{field}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 134812 by mohammad17 last updated on 07/Mar/21 $$\int\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{cosh}\left({t}\right)\right.}\:\:{dt} \\ $$ Answered by mathmax by abdo last updated on 07/Mar/21 $$\mathrm{I}=\int\sqrt{\frac{\mathrm{1}+\mathrm{ch}\left(\mathrm{t}\right)}{\mathrm{2}}}\mathrm{dt}\:=\int\sqrt{\mathrm{ch}^{\mathrm{2}} \left(\frac{\mathrm{t}}{\mathrm{2}}\right)}\mathrm{dt}\:=\int\:\mathrm{ch}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)\mathrm{dt} \\ $$$$=\mathrm{2sh}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)\:+\mathrm{C}…
Question Number 134822 by abdurehime last updated on 07/Mar/21 Answered by greg_ed last updated on 07/Mar/21 $$\mathrm{p}=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{4}} −\sqrt{{x}}}{\:\sqrt{{x}}−\mathrm{1}} \\ $$$$\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{L}}'\boldsymbol{\mathrm{Hopital}}'\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{rule}} \\ $$$$\mathrm{p}=\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\mathrm{4}{x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}}{\frac{\mathrm{1}}{\mathrm{2}\:\sqrt{{x}}}}\:…
Question Number 134814 by Ndala last updated on 07/Mar/21 Commented by Ndala last updated on 07/Mar/21 $$\mathrm{I}\:\mathrm{need}\:\mathrm{your}\:\mathrm{help}\:\mathrm{my}\:\mathrm{friends}! \\ $$ Answered by Ar Brandon last updated…
Question Number 69268 by A8;15: last updated on 22/Sep/19 Answered by mr W last updated on 22/Sep/19 $${x}^{{x}^{\mathrm{20}} } =\mathrm{2}^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \\ $$$${x}^{{x}} =\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{20}\sqrt{\mathrm{2}}}} ={a} \\…
Question Number 134780 by CutieJanab last updated on 07/Mar/21 Terms of Service Privacy Policy Contact: info@tinkutara.com