Question Number 183773 by HeferH last updated on 30/Dec/22 $$\:{In}\:{a}\:{square}\:\left({ABCD}\right)\:{there}\:{is}\:{a}\:{quarter}\:{of} \\ $$$$\:{a}\:{circle}\:{ADC}\:\left({AD}\:=\:{DC}\right),\:{put}\:{a}\:{point}\:{N} \\ $$$$\:{in}\:{the}\:{arc}\:{AC}\:{such}\:{that}\:{AN}\:=\:\mathrm{1}\:{and}\:{NC}\:=\:\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\:{find}\:{BN}.\: \\ $$$$\: \\ $$ Answered by mr W last…
Question Number 183767 by Tawa11 last updated on 29/Dec/22 Answered by MikeH last updated on 30/Dec/22 $${P}_{\mathrm{0}} \:=\:\mathrm{1}.\mathrm{3}\:×\:\mathrm{10}^{\mathrm{5}} \:\mathrm{N}\:\mathrm{m}^{−\mathrm{2}} \\ $$$$\mathrm{P}_{\mathrm{gauge}} \:=\:\rho\mathrm{g}{h}\:+\:{P}_{\mathrm{0}} \\ $$$$\Rightarrow\:{P}_{\mathrm{gauge}} \:=\left[\:\left(\mathrm{1000}\right)\left(\mathrm{9}.\mathrm{8}\right)\left(\mathrm{10}\right)\:+\:\mathrm{1}.\mathrm{3}×\mathrm{10}^{\mathrm{5}}…
Question Number 183766 by Tawa11 last updated on 29/Dec/22 Answered by MikeH last updated on 31/Dec/22 $${P}\:=\:{s}\rho\mathrm{g}{h}\:=\left(\mathrm{1}.\mathrm{03}\right)\:\left(\mathrm{1000}\right)\left(\mathrm{10}\right)\left(\mathrm{200}\right)\:=\:\mathrm{2}.\mathrm{06}×\mathrm{10}^{\mathrm{6}} \:\mathrm{N}\:\mathrm{m}^{−\mathrm{2}} \\ $$$${P}\:=\:\frac{{F}}{{A}}\:\Rightarrow\:{F}\:=\:{PA}\:=\:\left(\mathrm{2}.\mathrm{06}×\mathrm{10}^{\mathrm{6}} \right)\left(\mathrm{0}.\mathrm{5}\right)\:=\:\mathrm{1}.\mathrm{03}×\mathrm{10}^{\mathrm{6}} \mathrm{N} \\ $$ Commented…
Question Number 183747 by HeferH last updated on 29/Dec/22 $$\:{Find}\:{the}\:{perimeter}\:{of}\:{a}\:{regular}\:{heptagon}\: \\ $$$$\:{ABCDEFG}\:{if}\:\frac{\mathrm{1}}{{AE}}\:+\:\frac{\mathrm{1}}{{AC}}\:=\:\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\: \\ $$ Answered by mr W last updated on 29/Dec/22 $$\theta=\frac{\mathrm{5}\pi}{\mathrm{7}}…
Question Number 52663 by ajfour last updated on 11/Jan/19 Commented by ajfour last updated on 11/Jan/19 $${Find}\:{radius}\:{R}\:{in}\:{terms}\:{of}\:{a}\:{and}\:{b}. \\ $$$${Also}\:{find}\:{the}\:{central}\:{area}\:{in}\:{yellow} \\ $$$${in}\:{terms}\:{of}\:{a}\:{and}\:{b}.\:\:\left({source}:\:{ajfour}\right) \\ $$ Answered by…
Question Number 183728 by mr W last updated on 29/Dec/22 Commented by mr W last updated on 29/Dec/22 $${find}\:{the}\:{smallest}\:{and}\:{the}\:{largest} \\ $$$${equilateral}\:{triangle}\:{which}\:{has}\:{two} \\ $$$${of}\:{its}\:{vertices}\:{on}\:{the}\:{parabola}\:{y}=\mathrm{2}{x}^{\mathrm{2}} \\ $$$${and}\:{the}\:{third}\:{vertex}\:{on}\:{the}\:{circle}…
Question Number 183734 by AROUNAMoussa last updated on 29/Dec/22 Answered by HeferH last updated on 29/Dec/22 Commented by HeferH last updated on 29/Dec/22 $$\bigtriangleup{ABD}\:\sim\:\bigtriangleup{EBC}\:\Rightarrow \\…
Question Number 118138 by oustmuchiya@gmail.com last updated on 15/Oct/20 Answered by 1549442205PVT last updated on 15/Oct/20 $$\mathrm{The}\:\mathrm{are}\:\mathrm{of}\:\mathrm{the}\:\mathrm{fruit}\:\:\mathrm{garden}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\mathrm{S}_{\mathrm{g}} =\pi\mathrm{r}^{\mathrm{2}} =\pi.\mathrm{6}^{\mathrm{2}} =\mathrm{36}\pi \\ $$$$\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bounded}\:\mathrm{path}\:\mathrm{equal}\:\mathrm{to} \\…
Question Number 183663 by AROUNAMoussa last updated on 28/Dec/22 Commented by mathocean1 last updated on 28/Dec/22 $${Perimetre}=\mathrm{2}\pi+\mathrm{12} \\ $$ Answered by a.lgnaoui last updated on…
Question Number 118104 by Lordose last updated on 15/Oct/20 $$\left.\mathrm{1}.\right) \\ $$$$\mathrm{A}\:\mathrm{right}\:\mathrm{circular}\:\mathrm{cone}\:\mathrm{is}\:\mathrm{circumscribed} \\ $$$$\mathrm{about}\:\mathrm{a}\:\mathrm{sphere}\:\mathrm{of}\:\mathrm{radius}\left(\boldsymbol{\mathrm{r}}\right).\:\:\mathrm{If}\:\boldsymbol{\mathrm{d}}\:\mathrm{is}\:\mathrm{the}\: \\ $$$$\mathrm{distance}\:\mathrm{from}\:\mathrm{the}\:\mathrm{center}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sphere} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{vertex}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cone},\:\mathrm{show}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cone},\boldsymbol{\mathrm{V}}=\frac{\boldsymbol{\pi\mathrm{r}}^{\mathrm{2}} \left(\boldsymbol{\mathrm{r}}+\boldsymbol{\mathrm{d}}\right)^{\mathrm{2}} }{\mathrm{3}\left(\boldsymbol{\mathrm{d}}−\boldsymbol{\mathrm{r}}\right)}. \\ $$$$\left.\mathrm{2}.\right) \\…