Question Number 5704 by sanusihammed last updated on 24/May/16 $${Show}\:{that}\:… \\ $$$$ \\ $$$${Limit}\:\:\:\:\:\:\left[\frac{\mathrm{3}^{{x}} \:−\:\mathrm{3}^{−{x}} }{\mathrm{3}^{{x}\:} \:+\:\mathrm{3}^{−{x}} }\right]\:=\:−\:\mathrm{1} \\ $$$${x}\:\rightarrow\:−\infty \\ $$ Answered by FilupSmith…
Question Number 5695 by FilupSmith last updated on 24/May/16 Commented by FilupSmith last updated on 24/May/16 $$\mathrm{What}\:\mathrm{kinds}\:\mathrm{of}\:\mathrm{methods}\:\mathrm{can}\:\mathrm{be}\:\mathrm{used} \\ $$$$\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:{ABC}? \\ $$$$\left(\mathrm{Both}\:\mathrm{circles}\:\mathrm{are}\:\mathrm{identical}\right) \\ $$ Commented by…
Question Number 5672 by sanusihammed last updated on 23/May/16 $${Differentiate}\:\:\:\:\frac{{lnx}}{{e}^{{x}} }\:\:\:\:{fom}\:{the}\:{first}\:{principle}. \\ $$$$ \\ $$$${Please}\:{help}\:{me}. \\ $$ Answered by Yozzii last updated on 23/May/16 $${Let}\:{y}=\frac{{lnx}}{{e}^{{x}}…
Question Number 71184 by ajfour last updated on 12/Oct/19 Commented by ajfour last updated on 12/Oct/19 $${Each}\:{semicircle}\:{have}\:{radius} \\ $$$${unity}.\:{Find}\:{maximum}\:{value}\:{of}\:{x}. \\ $$ Commented by ajfour last…
Question Number 5630 by Rasheed Soomro last updated on 23/May/16 Commented by Rasheed Soomro last updated on 23/May/16 Commented by Yozzii last updated on 23/May/16…
Question Number 5626 by Rasheed Soomro last updated on 23/May/16 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\boldsymbol{\mathrm{length}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{chord}}\:\mathrm{in}\:\mathrm{a}\:\boldsymbol{\mathrm{circle}}\:\boldsymbol{\mathrm{of}} \\ $$$$\boldsymbol{\mathrm{radius}}\:\boldsymbol{\mathrm{r}}\:\mathrm{which}\:\:\mathrm{divides}\:\mathrm{the}\:\boldsymbol{\mathrm{circumference}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{circle}}\:\mathrm{in}\:\boldsymbol{\mathrm{m}}\::\:\boldsymbol{\mathrm{n}}\:? \\ $$ Answered by mrW last updated on 17/Nov/16 $$\theta=\mathrm{2}\pi\centerdot\frac{{m}}{{m}+{n}} \\ $$$${L}=\mathrm{2}\centerdot{r}\centerdot\mathrm{sin}\:\left(\theta/\mathrm{2}\right)=\mathrm{2}\centerdot{r}\centerdot\mathrm{sin}\:\left(\frac{{m}\pi}{{m}+{n}}\right)…
Question Number 5624 by Rasheed Soomro last updated on 23/May/16 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\boldsymbol{\mathrm{length}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{chord}}\:\mathrm{in}\:\mathrm{a}\:\boldsymbol{\mathrm{circle}}\:\boldsymbol{\mathrm{of}}\: \\ $$$$\boldsymbol{\mathrm{radius}}\:\boldsymbol{\mathrm{r}}\:\mathrm{which}\:\mathrm{divides}\:\mathrm{the}\:\boldsymbol{\mathrm{region}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{circle}}\:\mathrm{in}\:\boldsymbol{\mathrm{m}}\::\:\boldsymbol{\mathrm{n}}\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 5620 by sanusihammed last updated on 22/May/16 $${Find}\:{the}\:{resolved}\:{part}\:{of}\:{the}\:{vector}\:{a}\:=\:\mathrm{6}{i}\:−\:\mathrm{3}{j}\:+\:\mathrm{9}{k}\: \\ $$$${in}\:{the}\:{diection}\:{of}\:{b}\:=\:\mathrm{2}{i}\:+\:\mathrm{2}{j}\:−\:{k} \\ $$$$ \\ $$$${please}\:{help}. \\ $$$$ \\ $$$${i}\:{got}\:{the}\:{answer}\:{to}\:{be}\:\left(−\mathrm{1}\right) \\ $$ Commented by prakash…
Question Number 5556 by Rasheed Soomro last updated on 20/May/16 $$\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{is}\:\mathrm{diretly}\: \\ $$$$\mathrm{proportional}\:\mathrm{to}\:\mathrm{the}\:\mathrm{square}\:\mathrm{of} \\ $$$$\mathrm{its}\:\mathrm{diameter}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{constant}\:\mathrm{of}\:\mathrm{proportionality}? \\ $$ Commented by Yozzii last updated on…
Question Number 5561 by Rasheed Soomro last updated on 20/May/16 $$\mathrm{A}\:\mathrm{chord}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{in}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{that} \\ $$$$\mathrm{divides}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{in}\:\mathrm{1}:\mathrm{2}\:\left(\mathrm{ratio}\right). \\ $$$$\mathrm{In}\:\mathrm{what}\:\mathrm{ratio}\:\mathrm{the}\:\mathrm{chord}\:\mathrm{divides} \\ $$$$\mathrm{the}\:\mathrm{diameter}\:\mathrm{perpendicular}\:\mathrm{to} \\ $$$$\mathrm{the}\:\mathrm{chord}? \\ $$ Commented by Rasheed Soomro…