Question Number 136697 by bramlexs22 last updated on 25/Mar/21 Commented by bramlexs22 last updated on 25/Mar/21 $${what}\:{the}\:{missing}\:{number} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 5627 by Rasheed Soomro last updated on 23/May/16 $$\mathrm{Determine}\:\mathrm{interval}\:\mathrm{in}\:\mathrm{which} \\ $$$$\mathrm{2}^{\mathrm{x}} \geqslant\mathrm{x}^{\mathrm{2}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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 136693 by liberty last updated on 25/Mar/21 Commented by Olaf last updated on 25/Mar/21 $${f}\left({x}\right)\:=\:{mx}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{2}} +{kx}−\mathrm{18} \\ $$$$=\:\left(\mathrm{6}−{x}−{x}^{\mathrm{2}} \right)\left(−{mx}−\mathrm{3}\right) \\ $$$$=\left({x}^{\mathrm{2}} +{x}−\mathrm{6}\right)\left({mx}+\mathrm{3}\right)…
Question Number 5620 by sanusihammed last updated on 22/May/16 $${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 5616 by sanusihammed last updated on 22/May/16 $${Find}\: \\ $$$$ \\ $$$$\left[\mathrm{1}\:+\:\mathrm{3}{n}^{−\mathrm{1}} \right]^{{n}} \\ $$$$ \\ $$$${Limit}\:{as}\:{n}\:\rightarrow\infty \\ $$$$ \\ $$$$ \\ $$$${Please}\:{help}.…
Question Number 5615 by Rasheed Soomro last updated on 22/May/16 $$\mathrm{The}\:\mathrm{tangent}\:\mathrm{at}\:\mathrm{the}\:\mathrm{point}\:\mathrm{P}\left({a},{b}\right)\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{curve}\:\:\mathrm{y}=\frac{{ab}}{{x}}\:\:\mathrm{meets}\:\mathrm{the}\:\mathrm{x}-\mathrm{axis}\:\mathrm{and}\:\mathrm{y}-\mathrm{axis} \\ $$$$\mathrm{at}\:\mathrm{Q}\:\mathrm{and}\:\mathrm{R}\:\mathrm{respectively}.\:\mathrm{Show}\:\mathrm{that} \\ $$$$\mathrm{PQ}=\mathrm{RP}\:. \\ $$ Commented by prakash jain last updated…
Question Number 5614 by Rasheed Soomro last updated on 22/May/16 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{cordinates}\:\mathrm{of}\:\mathrm{the}\:\mathrm{two}\:\mathrm{points} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{y}=\mathrm{4}−\mathrm{x}^{\mathrm{2}} \:\:\mathrm{whose}\:\mathrm{tangents} \\ $$$$\mathrm{pass}\:\mathrm{through}\:\mathrm{the}\:\mathrm{point}\:\left(−\mathrm{1},\:\mathrm{7}\right)\:. \\ $$ Commented by Rasheed Soomro last updated on…
Question Number 71149 by naka3546 last updated on 12/Oct/19 Commented by MJS last updated on 17/Oct/19 $$\mathrm{if}\:\mathrm{the}\:\mathrm{hexagon}\:\mathrm{is}\:\mathrm{regular}\:\mathrm{it}'\mathrm{s}\:\mathrm{symmetric} \\ $$$${BE}\:\mathrm{is}\:\mathrm{the}\:\mathrm{symmetry}\:\mathrm{axis} \\ $$$${M}\in\left({BE}\right) \\ $$$${A}={C}'\:\wedge\:{F}={D}' \\ $$$$\Rightarrow\:\left({AF}\right)=\left({CD}\right)'\:\mathrm{and}\:\mathrm{similar}\:\mathrm{for}\:\mathrm{all}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{lines}…