Question Number 2499 by Syaka last updated on 21/Nov/15 $${ABCD}.{EFGH}\:{is}\:{cube},\:{X}\:{is}\:{midpoint}\:{EF} \\ $$$${if}\:{AB}\:=\:\mathrm{6}\:{cm},\:{how}\:{distance}\:{AX}\:{to}\:{BD}?? \\ $$ Commented by prakash jain last updated on 21/Nov/15 $$\mathrm{What}\:\mathrm{is}\:\mathrm{AM}?\:\mathrm{Is}\:\mathrm{X}=\mathrm{M}? \\ $$…
Question Number 2466 by prakash jain last updated on 20/Nov/15 $$\sqrt{−\frac{\mathrm{1}}{\mathrm{5}}\:}−\frac{\mathrm{1}}{\:\sqrt{−\mathrm{5}}}=? \\ $$ Answered by Yozzi last updated on 20/Nov/15 $$\sqrt{\frac{\mathrm{1}}{−\mathrm{5}}}−\frac{\mathrm{1}}{\:\sqrt{−\mathrm{5}}}={i}\sqrt{\frac{\mathrm{1}}{\mathrm{5}}}−\frac{\mathrm{1}}{{i}\sqrt{\mathrm{5}}} \\ $$$$={i}\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}+\frac{{i}}{\:\sqrt{\mathrm{5}}}=\mathrm{2}\frac{{i}}{\:\sqrt{\mathrm{5}}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{−\mathrm{5}}}=\frac{\sqrt{\mathrm{1}}}{\:\sqrt{−\mathrm{5}}}=\sqrt{\frac{\mathrm{1}}{−\mathrm{5}}}\Rightarrow\sqrt{\frac{\mathrm{1}}{−\mathrm{5}}}−\frac{\mathrm{1}}{\:\sqrt{−\mathrm{5}}}\overset{?}…
Question Number 2458 by alib last updated on 20/Nov/15 $${The}\:{medians}\:{of}\:{a}\:{triangle} \\ $$$${are}\:{m}_{\mathrm{1}} ,\:{m}_{\mathrm{2}} ,\:{m}_{\mathrm{3}} . \\ $$$${Find}\:{the}\:{length}\:{of}\:{each}\:{sides}\: \\ $$$${the}\:{triangle}. \\ $$ Answered by prakash jain…
Question Number 67977 by behi83417@gmail.com last updated on 02/Sep/19 Commented by behi83417@gmail.com last updated on 02/Sep/19 $$\boldsymbol{\mathrm{AD}}=\boldsymbol{\mathrm{DC}},\angle\boldsymbol{\mathrm{AEB}}=\mathrm{90}^{\bullet} \:\:. \\ $$$$\boldsymbol{\mathrm{find}}:\:\:\:\boldsymbol{\mathrm{ED}}\:\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{terms}}\:\boldsymbol{\mathrm{of}}:\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}},\boldsymbol{\mathrm{c}}. \\ $$ Answered by $@ty@m123…
Question Number 67973 by behi83417@gmail.com last updated on 02/Sep/19 Commented by behi83417@gmail.com last updated on 02/Sep/19 $$\mathrm{A}\overset{\bigtriangleup} {\mathrm{B}C},\mathrm{is}\:\mathrm{given}. \\ $$$$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}: \\ $$$$\mathrm{1}.\:\:\:\:\:\frac{\boldsymbol{\mathrm{BX}}}{\boldsymbol{\mathrm{XC}}}\:×\:\:\frac{\boldsymbol{\mathrm{CY}}}{\boldsymbol{\mathrm{YA}}}\:×\:\frac{\boldsymbol{\mathrm{AZ}}}{\boldsymbol{\mathrm{ZB}}}=\mathrm{1}. \\ $$$$\mathrm{2}.\:\:\:\:\:\frac{\boldsymbol{\mathrm{PX}}}{\boldsymbol{\mathrm{XA}}}\:+\:\:\frac{\boldsymbol{\mathrm{PY}}}{\boldsymbol{\mathrm{YB}}}\:+\:\frac{\boldsymbol{\mathrm{PZ}}}{\boldsymbol{\mathrm{ZC}}}=\mathrm{1}. \\…
Question Number 67969 by behi83417@gmail.com last updated on 02/Sep/19 $$\boldsymbol{\mathrm{Two}}\:\boldsymbol{\mathrm{triangles}}\:\bigtriangleup_{\mathrm{1}} \:\boldsymbol{\mathrm{and}}\:\bigtriangleup_{\mathrm{2}} \:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{given}},\boldsymbol{\mathrm{such}}\: \\ $$$$\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{length}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{sides}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{triangle}}\:\mathrm{1},\boldsymbol{\mathrm{are}}\: \\ $$$$\boldsymbol{\mathrm{equail}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{length}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{medians}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{triangle}}\:\mathrm{2}. \\ $$$$\mathrm{1}.\boldsymbol{\mathrm{find}}\:\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{ratio}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{areas}}\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{triangles}}. \\ $$$$\mathrm{2}.\boldsymbol{\mathrm{given}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{small}}\:\boldsymbol{\mathrm{side}}\:\boldsymbol{\mathrm{of}}\:\bigtriangleup_{\mathrm{1}} ,\:\boldsymbol{\mathrm{be}}\:\boldsymbol{\mathrm{equail}}\:\boldsymbol{\mathrm{to}}:\sqrt{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{one}}\:\boldsymbol{\mathrm{angle}}\:\boldsymbol{\mathrm{be}}:\mathrm{90}^{\bullet} . \\…
Question Number 67948 by A8;15: last updated on 02/Sep/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 67860 by TawaTawa last updated on 01/Sep/19 Commented by TawaTawa last updated on 01/Sep/19 $$\mathrm{Find}\:\mathrm{angle}\:\:\mathrm{a} \\ $$ Commented by mr W last updated…
Question Number 2329 by 123456 last updated on 16/Nov/15 $${a}_{{n}} =\left(−\mathrm{1}\right)^{{n}} \left(\lfloor\frac{{n}}{\mathrm{9}}\rfloor+\mathrm{1}\right) \\ $$$$\underset{{n}\geqslant\mathrm{0}} {\sum}{a}_{{n}} =? \\ $$$${b}_{{n}} =\frac{\left({a}_{{n}} −\mathrm{1}\right)\left(\mathrm{9}−{a}_{{n}} \right)}{{n}+\mathrm{1}} \\ $$$$\underset{{n}\geqslant\mathrm{0}} {\sum}{b}_{{n}} =?…
Question Number 67852 by mr W last updated on 01/Sep/19 Commented by Prithwish sen last updated on 01/Sep/19 $$ \\ $$$$\boldsymbol{\mathrm{A}}=\left(\mathrm{0},\boldsymbol{\mathrm{a}}\right),\:\boldsymbol{\mathrm{B}}\:=\:\left(\frac{−\mathrm{4}}{\boldsymbol{\mathrm{a}}},\mathrm{0}\right),\boldsymbol{\mathrm{C}}=\left(\frac{\mathrm{20}}{\boldsymbol{\mathrm{a}}},−\mathrm{9}\boldsymbol{\mathrm{a}}\right),\boldsymbol{\mathrm{D}}=\left(\frac{\mathrm{2}}{\boldsymbol{\mathrm{a}}},\mathrm{0}\right) \\ $$$$\boldsymbol{\mathrm{E}}=\left(\mathrm{0},\frac{−\mathrm{3}\boldsymbol{\mathrm{a}}}{\mathrm{2}}\right)\: \\ $$$$\mathrm{Equation}\:\mathrm{of}\:\mathrm{line}\:\mathrm{passing}\:\mathrm{through}\:\mathrm{A}\:\mathrm{and}\:\mathrm{D}…