Question Number 117739 by bemath last updated on 13/Oct/20 $$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle} \\ $$$$\mathrm{with}\:\mathrm{radius}\:\mathrm{4}\sqrt{\mathrm{2}}\:\mathrm{that}\:\mathrm{can}\:\mathrm{be}\: \\ $$$$\mathrm{inscribed}\:\mathrm{in}\:\mathrm{the}\:\mathrm{parabola}\: \\ $$$$\mathrm{y}=\mathrm{x}^{\mathrm{2}} −\mathrm{16x}+\mathrm{128}? \\ $$ Answered by bobhans last updated on…
Question Number 52034 by somil last updated on 02/Jan/19 Commented by afachri last updated on 02/Jan/19 $$\mathrm{Number}\:\mathrm{of}\:\mathrm{discs}\:\mathrm{can}\:\mathrm{be}\:\mathrm{calculated}\:\mathrm{by} \\ $$$$\mathrm{dividing}\:\mathrm{area}\:\mathrm{of}\:\mathrm{sheet}\:\mathrm{and}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{disc}.\: \\ $$ Answered by tanmay.chaudhury50@gmail.com last…
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Question Number 51774 by ajfour last updated on 30/Dec/18 Commented by ajfour last updated on 30/Dec/18 $${Find}\:{maximum}\:{area}\:{A}\:{in}\:{light}\:{blue}. \\ $$$$\left({source}\::\:{ajfour}\right) \\ $$ Commented by tanmay.chaudhury50@gmail.com last…
Question Number 182437 by akolade last updated on 09/Dec/22 Answered by qaz last updated on 09/Dec/22 $$\begin{pmatrix}{{n}+\mathrm{2}}\\{\:\:\:\:{r}}\end{pmatrix}=\left[{x}^{{r}} \right]\left(\mathrm{1}+{x}\right)^{{n}+\mathrm{2}} =\left[{x}^{{r}} \right]\left(\left(\mathrm{1}+{x}\right)^{{n}} +\mathrm{2}{x}\left(\mathrm{1}+{x}\right)^{{n}} +{x}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)^{{n}} \right) \\…
Question Number 182109 by amin96 last updated on 04/Dec/22 $$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\mathrm{3}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{x}}\sqrt{\mathrm{3}}−\mathrm{8}\:\:\:\:\:\boldsymbol{{g}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{x}}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\boldsymbol{{gof}}^{−\mathrm{1}} \left(\mathrm{18}\right)=? \\ $$ Answered by FelipeLz last updated on 04/Dec/22 $${f}\left({x}\right)\:=\:\mathrm{3}{x}^{\mathrm{2}}…
Question Number 50936 by Rio Michael last updated on 22/Dec/18 Commented by Rio Michael last updated on 22/Dec/18 $${the}\:{figure}\:{above}\:{is}\:{a}\:{solid}\:{with}\:{trapezium}\:{PQRS}\:{as}\:{its}\:{uniform} \\ $$$${cross}−{section}. \\ $$$$\left.{a}\right)\:{Find}\:{the}\:{area}\:{of}\:{the}\:{trapezuim}\:{PQRS} \\ $$$$\left.{b}\right)\:{calculate}\:{the}\:{volume}\:{of}\:{the}\:{solid}\:{figure}.…
Question Number 50818 by ajfour last updated on 20/Dec/18 $${If}\:{a}\:{right}\:{angled}\:{triangle}\:{has} \\ $$$${same}\:{area}\:{and}\:{double}\:{perimeter} \\ $$$${as}\:{that}\:{of}\:{a}\:{circle}\:{of}\:{unit}\:{radius}, \\ $$$${find}\:{the}\:{mutually}\:{perpendicular} \\ $$$${sides}\:{of}\:{the}\:{triangle}. \\ $$ Answered by mr W last…
Question Number 50640 by ajfour last updated on 18/Dec/18 Commented by mr W last updated on 18/Dec/18 $${when}\:{the}\:{angle}\:{between}\:{k}\:{and}\:{h}\:{is} \\ $$$${fixed},\:{then}\:{the}\:{shape}\:{and}\:{therefore} \\ $$$${also}\:{the}\:{area}\:{of}\:{the}\:{quadrilateral}\:{is} \\ $$$${fixed}.\:{the}\:{area}\:{is}\:{a}\:{constant}. \\…
Question Number 50630 by peter frank last updated on 18/Dec/18 Answered by ajfour last updated on 18/Dec/18 $$\alpha+\beta\:=\:\mathrm{0}\:\:,\:\gamma+\delta\:=\:−{b}\:\:\:\:…\left({i}\right) \\ $$$$\alpha\beta+\left(\alpha+\beta\right)\gamma+\gamma\delta+\left(\alpha+\beta\right)\delta\:=\:{c} \\ $$$$\Rightarrow\:\:\alpha\beta+\gamma\delta\:=\:{c}\:\:\:\:\:\:\:\:\:\:\:….\left({ii}\right) \\ $$$$\alpha\beta.\gamma\delta\:=\:{e}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\left({iii}\right) \\…