Question Number 73221 by aliesam last updated on 08/Nov/19 Answered by MJS last updated on 08/Nov/19 $${A}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\:{B}=\begin{pmatrix}{{a}}\\{\mathrm{0}}\end{pmatrix}\:\:{C}=\begin{pmatrix}{{a}}\\{{a}}\end{pmatrix}\:\:{D}=\begin{pmatrix}{\mathrm{0}}\\{{a}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{E}=\begin{pmatrix}{{b}}\\{\mathrm{0}}\end{pmatrix}\:\:{F}=\begin{pmatrix}{{b}}\\{{b}}\end{pmatrix}\:\:{G}=\begin{pmatrix}{\mathrm{0}}\\{{b}}\end{pmatrix} \\ $$$${T}=\begin{pmatrix}{\frac{{a}+{b}}{\mathrm{2}}}\\{\mathrm{0}}\end{pmatrix}\:\:{I}=\begin{pmatrix}{\frac{{a}+{b}}{\mathrm{2}}}\\{\frac{{a}−{b}}{\mathrm{2}}}\end{pmatrix} \\ $$$${M}=\begin{pmatrix}{{b}}\\{\frac{{a}−{b}}{\mathrm{2}}}\end{pmatrix}\:\:{N}=\begin{pmatrix}{\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left({a}+{b}\right)+{b}}\\{\frac{\mathrm{3}{a}−{b}}{\mathrm{4}}}\end{pmatrix}\:\:{L}=\begin{pmatrix}{{b}}\\{{a}}\end{pmatrix} \\ $$$$\mathrm{radius}\:\mathrm{of}\:\mathrm{incircle}\:\mathrm{of}\:{ETIM}=\frac{{r}}{\mathrm{2}}=\frac{{a}−{b}}{\mathrm{4}}…
Question Number 138388 by ajfour last updated on 13/Apr/21 $${Find}\:{maximum}\:{volume}\:{of}\:{a} \\ $$$${cylinder}\:{within}\:{a}\:{unit}\:{cube}\: \\ $$$${whose}\:{axis}\:{is}\:{along}\:{a}\:{diagonal} \\ $$$${of}\:{the}\:{cube}. \\ $$ Answered by mr W last updated on…
Question Number 72668 by aliesam last updated on 31/Oct/19 Answered by mind is power last updated on 01/Nov/19 $$\theta=\mathrm{DAB}? \\ $$$$ \\ $$ Terms of…
Question Number 72671 by aliesam last updated on 31/Oct/19 Commented by kaivan.ahmadi last updated on 31/Oct/19 $$\left.{a}\right) \\ $$$$\mid{x}_{{n}+\mathrm{1}} −{x}_{{n}} \mid\leqslant{c}\mid{x}_{{n}} −{x}_{{n}−\mathrm{1}} \mid\leqslant{c}^{\mathrm{2}} \mid{x}_{{n}−\mathrm{1}} −{x}_{{n}−\mathrm{2}}…
Question Number 72666 by aliesam last updated on 31/Oct/19 Answered by mind is power last updated on 31/Oct/19 $$\mathrm{let}\:\alpha=\angle\mathrm{CDA}=\mathrm{BAE}\:\:,\mathrm{OCA}=\mathrm{2}\alpha \\ $$$$\mathrm{E}=\left(\mathrm{AD}\right)\cap\mathrm{BC} \\ $$$$\frac{\mathrm{3}}{\mathrm{5}}=\frac{\mathrm{BE}}{\mathrm{EC}},\mathrm{BE}+\mathrm{EC}=\mathrm{4} \\ $$$$\Rightarrow\mathrm{BE}=\frac{\mathrm{3}}{\mathrm{2}}…
Question Number 72603 by aliesam last updated on 30/Oct/19 Answered by mr W last updated on 30/Oct/19 Commented by aliesam last updated on 30/Oct/19 $${thank}\:{you}\:{sir}…
Question Number 72434 by aliesam last updated on 28/Oct/19 Answered by mind is power last updated on 28/Oct/19 $$\frac{\mathrm{c}}{\mathrm{sin}\left(\mathrm{x}\right)}=\frac{\mathrm{d}}{\mathrm{sin}\left(\mathrm{y}\right)}=\frac{\mathrm{BD}}{\mathrm{sin}\left(\mathrm{40}\right)} \\ $$$$\mathrm{BD}^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{2abcos}\left(\mathrm{40}\right) \\…
Question Number 6893 by Tawakalitu. last updated on 01/Aug/16 Commented by Rasheed Soomro last updated on 02/Aug/16 $$\bullet{Surface}\:{area}\:{of}\:{sphere}=\mathrm{4}\pi{r}^{\mathrm{2}} \\ $$$$\:{Radius}\:{of}\:{the}\:\:{sphere}\:=\frac{\mathrm{12}}{\mathrm{2}}=\mathrm{6}\:{cm} \\ $$$${Surface}\:{area}\:{of}\:{sphere}=\mathrm{4}\pi\left(\mathrm{6}\right)^{\mathrm{2}} =\mathrm{144}\pi……….\left({i}\right) \\ $$$$…
Question Number 6880 by Tawakalitu. last updated on 01/Aug/16 $${A}\:{cone}\:{of}\:{height}\:\mathrm{10}\:{cm}\:{and}\:{base}\:{radius}\:\mathrm{3}\:{cm}\:{is}\:{held}\:{vertex}\: \\ $$$${down}\:{and}\:{is}\:{filled}\:{with}\:{sand}\:.\:{A}\:{hole}\:{in}\:{the}\:{bottom}\:{of}\:{the}\:{cone} \\ $$$${allows}\:{the}\:{sand}\:{to}\:{escape}\:{at}\:{a}\:{steady}\:{rate}\:{of}\:\mathrm{0}.\mathrm{5}\:{cm}^{\mathrm{3}} /{s}.\:{find}\:{the} \\ $$$${rate}\:{that}\:{the}\:{depth}\:{is}\:{decreasing}\:{when}\:{the}\:{depth}\:{is}\:\mathrm{2}\:{cm}. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 72414 by aliesam last updated on 28/Oct/19 Commented by aliesam last updated on 28/Oct/19 $${a},{b}\:{and}\:{d}\:{are}\:{centres}\:{of}\:{three}\:{circles}\: \\ $$$${find}\: \\ $$$$\frac{{AC}}{{AB}} \\ $$ Answered by…