Question Number 184490 by Ml last updated on 07/Jan/23 Commented by Ml last updated on 07/Jan/23 $$\mathrm{solution}\:\mathrm{please}? \\ $$ Answered by Frix last updated on…
Question Number 184481 by yaslm last updated on 07/Jan/23 Answered by mahdipoor last updated on 07/Jan/23 $$\mathrm{i}={e}^{\mathrm{0}} \left({cos}\frac{\pi}{\mathrm{2}}+\mathrm{i}{sin}\frac{\pi}{\mathrm{2}}\right)={e}^{\frac{\pi}{\mathrm{2}}\mathrm{i}} \Rightarrow \\ $$$${Z}_{\mathrm{1}} =\left(\mathrm{i}^{\frac{\mathrm{100}}{\mathrm{i}}} \right)^{\mathrm{0}.\mathrm{5}} =\left({e}^{\left(\frac{\pi}{\mathrm{2}}\mathrm{i}\right)\left(\frac{\mathrm{100}}{\mathrm{i}}\right)} \right)^{\mathrm{0}.\mathrm{5}}…
Question Number 118944 by mathdave last updated on 20/Oct/20 Answered by Dwaipayan Shikari last updated on 21/Oct/20 $${For}\:{stationary}\:{boats}\: \\ $$$$\Sigma{P}_{{boat},{man}} =\mathrm{0}\:\:\:\left({As}\:{the}\:{boat}\:{and}\:{the}\:{man}\:{system}\:{was}\:{stationary}\right) \\ $$$${MV}+{M}'\Psi=\mathrm{0}\:\:\:\:\left({Using}\:{conservation}\:{of}\:{Momentum}\right) \\ $$$$\Psi=−\frac{{MV}}{{M}'}\left(\Psi={velocity}\:{boatV}={velocity}\:{man}\:{M},{M}'\left({mass}\:{of}\:{man}\:{and}\:{boat}\right)\right.…
Question Number 184470 by SANOGO last updated on 07/Jan/23 $$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{lnx}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{dx} \\ $$ Answered by mr W last updated on 07/Jan/23 $$=−\int_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 53391 by Kunal12588 last updated on 21/Jan/19 $${if}\:{A}+{B}+{C}=\mathrm{2}{S},\:{prove}\:{that}\: \\ $$$$\mathrm{4}\:\mathrm{sin}\:{S}\:\mathrm{sin}\left({S}−{A}\right)\mathrm{sin}\left({S}−{B}\right)\mathrm{sin}\left({S}−{C}\right) \\ $$$$=\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:{A}−\mathrm{cos}^{\mathrm{2}} \:{B}−\mathrm{cos}^{\mathrm{2}} \:{C}+\mathrm{2}\:\mathrm{cos}\:{A}\:\mathrm{cos}\:{B}\:\mathrm{cos}\:{C}\: \\ $$ Commented by Kunal12588 last updated on…
Question Number 184431 by SulaymonNorboyev last updated on 06/Jan/23 $${If}\:\:\:\begin{cases}{\mathrm{a}+\frac{\mathrm{1}}{\mathrm{b}}=\frac{\mathrm{7}}{\mathrm{3}}}\\{\mathrm{b}+\frac{\mathrm{1}}{\mathrm{c}}=\mathrm{4}}\\{\mathrm{c}+\frac{\mathrm{1}}{\mathrm{a}}=\mathrm{1}}\end{cases}\:\:\:\mathrm{find}\:\:\mathrm{2022}\centerdot\mathrm{abc} \\ $$ Answered by mr W last updated on 06/Jan/23 $${c}=\mathrm{1}−\frac{\mathrm{1}}{{a}}=\frac{{a}−\mathrm{1}}{{a}} \\ $$$${b}=\mathrm{4}−\frac{{a}}{{a}−\mathrm{1}}=\frac{\mathrm{3}{a}−\mathrm{4}}{{a}−\mathrm{1}} \\ $$$$\Rightarrow{abc}={a}×\frac{\mathrm{3}{a}−\mathrm{4}}{{a}−\mathrm{1}}×\frac{{a}−\mathrm{1}}{{a}}=\mathrm{3}{a}−\mathrm{4}…
Question Number 184424 by Fridunatjan08 last updated on 06/Jan/23 $${Help}\:{please}: \\ $$$$\begin{cases}{{x}+{x}^{−\mathrm{1}} −\sqrt{\mathrm{2}}=\mathrm{0}}\\{{x}^{\mathrm{2017}} +{x}^{\mathrm{2017}} +\sqrt{\mathrm{2}}=?}\end{cases} \\ $$ Commented by Rasheed.Sindhi last updated on 06/Jan/23 $${x}^{\mathrm{2017}}…
Question Number 118876 by bemath last updated on 20/Oct/20 $${Find}\:{the}\:{value}\:{of}\:\lfloor\:\left(\frac{\mathrm{3}+\sqrt{\mathrm{17}}}{\mathrm{2}}\right)^{\mathrm{6}} \:\rfloor\:. \\ $$ Answered by bobhans last updated on 20/Oct/20 $$\mathrm{2040} \\ $$ Answered by…
Question Number 118872 by ZiYangLee last updated on 20/Oct/20 Commented by ZiYangLee last updated on 20/Oct/20 $${P}\:\mathrm{is}\:\mathrm{a}\:\mathrm{point}\:\mathrm{in}\:\mathrm{square}\:{ABCD}.\: \\ $$$$\mathrm{Given}\:\mathrm{that}\:{AP}=\mathrm{7},\:{PB}=\mathrm{6},\:{CP}=\mathrm{11}\:\mathrm{and} \\ $$$$\angle{APB}\:\mathrm{is}\:\theta.\:\mathrm{Find}\:\mathrm{tan}\:\theta. \\ $$ Answered by…
Question Number 118875 by Cristina last updated on 20/Oct/20 $$\mathrm{CAS} \\ $$$$\:^{\mathrm{1}} \mathrm{H}=\mathrm{1}.\mathrm{008}\:\mathrm{u}\:\:\:\:\:\mathrm{99}.\mathrm{985\%} \\ $$$$\:^{\mathrm{2}} \mathrm{H}=\mathrm{2}.\mathrm{014}\:\mathrm{u}\:\:\:\:\:\mathrm{0}.\mathrm{013\%} \\ $$$$\mathrm{1}.\mathrm{008}\bullet\frac{\mathrm{99}.\mathrm{985}}{\mathrm{100}}+\mathrm{2}.\mathrm{014}\bullet\frac{\mathrm{0}.\mathrm{013}}{\mathrm{100}}=\mathrm{1}.\mathrm{0078u}+\mathrm{0}.\mathrm{0003}=\mathrm{1}.\mathrm{0081u} \\ $$$$ \\ $$ Terms of Service…