Question Number 9603 by tawakalitu last updated on 20/Dec/16 $$\mathrm{A}\:\mathrm{regular}\:\mathrm{hexagon}\:\mathrm{has}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{lenght}\:\mathrm{8}\:\mathrm{cm}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{perpendicular}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{two} \\ $$$$\mathrm{opposite}\:\mathrm{faces}. \\ $$ Answered by mrW last updated on 20/Dec/16 $$\mathrm{2}×\frac{\mathrm{8}}{\mathrm{2}}×\sqrt{\mathrm{3}}=\mathrm{8}\sqrt{\mathrm{3}}\:\mathrm{cm} \\…
Question Number 75104 by chess1 last updated on 07/Dec/19 Answered by mr W last updated on 07/Dec/19 Commented by mr W last updated on 08/Dec/19…
Question Number 75105 by chess1 last updated on 07/Dec/19 Commented by MJS last updated on 07/Dec/19 $$\mathrm{arctan}\:\frac{\mathrm{40}−\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{21}} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 75058 by behi83417@gmail.com last updated on 07/Dec/19 $$\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{triangle}}:\:\:\boldsymbol{\mathrm{ABC}}: \\ $$$$\boldsymbol{\mathrm{a}}=\sqrt{\mathrm{2}\:},\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{c}}=\frac{\sqrt{\mathrm{2}}+\mathrm{1}}{\mathrm{2}},\overset{} {\boldsymbol{\mathrm{B}}}−\overset{} {\boldsymbol{\mathrm{C}}}=\frac{\boldsymbol{\pi}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{find}}:\:\:\boldsymbol{\mathrm{h}}_{\boldsymbol{\mathrm{a}}} ,\:\:\boldsymbol{\mathrm{S}}_{\boldsymbol{\mathrm{ABC}}\:\:} ,\boldsymbol{\mathrm{d}}_{\boldsymbol{\mathrm{a}}\:\:\:} ,\:\boldsymbol{\mathrm{R}}\:\:\:\:,\overset{} {\boldsymbol{\mathrm{A}}}. \\ $$ Commented by mr…
Question Number 9523 by Joel575 last updated on 12/Dec/16 $$\mathrm{3}{a}\:=\:\left({b}\:+\:{c}\:+\:{d}\right)^{\mathrm{2014}} \\ $$$$\mathrm{3}{b}\:=\:\left({a}\:+\:{c}\:+\:{d}\right)^{\mathrm{2014}} \\ $$$$\mathrm{3}{c}\:=\:\left({a}\:+\:{b}\:+\:{d}\right)^{\mathrm{2014}} \\ $$$$\mathrm{3}{d}\:=\:\left({a}\:+\:{b}\:+\:{c}\right)^{\mathrm{2014}} \\ $$$$\mathrm{Find}\:\mathrm{all}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{of}\:\left({a},\:{b},\:{c},\:{d}\right)\:\mathrm{if}\:{a},\:{b},\:{c},\:{d}\:\in\:\mathbb{R} \\ $$ Answered by mrW last updated…
Question Number 140545 by I want to learn more last updated on 09/May/21 Commented by mr W last updated on 09/May/21 $${AL}={AE}=\mathrm{3} \\ $$ Answered…
Question Number 9453 by Raja Naik last updated on 09/Dec/16 $$\mathrm{find}\:\mathrm{dc}'\mathrm{s}\:\mathrm{dr}'\mathrm{s}\:\mathrm{of}\:\mathrm{a}\:\mathrm{mormal}\:\mathrm{to}\:\mathrm{the}\: \\ $$$$\mathrm{plane}\:\mathrm{2x}+\frac{\mathrm{5}}{\mathrm{2}}\mathrm{y}+\frac{\mathrm{7}}{\mathrm{8}}\mathrm{z}=\mathrm{23} \\ $$ Answered by mrW last updated on 10/Dec/16 $$\mathrm{d}.\mathrm{r}.'\mathrm{s}\:\mathrm{of}\:\mathrm{the}\:\mathrm{normal}\:\mathrm{are}\:\left(\mathrm{2},\frac{\mathrm{5}}{\mathrm{2}},\frac{\mathrm{7}}{\mathrm{8}}\right) \\ $$$$\sqrt{\mathrm{2}^{\mathrm{2}}…
Question Number 9438 by tawakalitu last updated on 08/Dec/16 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x},\:\mathrm{y}\:\mathrm{and}\:\mathrm{z} \\ $$$$\mathrm{2xy}\:=\:\mathrm{x}\:+\:\mathrm{y}\:\:\:\:\:\:……\:\left(\mathrm{i}\right) \\ $$$$\mathrm{6xz}\:=\:\mathrm{6z}\:−\:\mathrm{2x}\:\:\:\:…….\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{3yz}\:=\:\mathrm{3y}\:+\:\mathrm{4z}\:\:\:\:………\:\left(\mathrm{iii}\right) \\ $$$$ \\ $$$$\mathrm{That}\:\mathrm{is}\:\mathrm{the}\:\mathrm{correct}\:\mathrm{question}\:\mathrm{sir}. \\ $$ Answered by mrW…
Question Number 9436 by geovane10math last updated on 08/Dec/16 $$\mathrm{If}\:\mathrm{they}\:\mathrm{said}\:\mathrm{that}\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:{k}\:\mathrm{diverges},\:\mathrm{why} \\ $$$$\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}\:+\:\mathrm{4}\:+\:…\:=\:−\:\frac{\mathrm{1}}{\mathrm{12}}\:? \\ $$ Answered by mrW last updated on 08/Dec/16 $$\:\:\:\mathrm{T}=\mathrm{1}−\mathrm{2}+\mathrm{3}−\mathrm{4}+\mathrm{5}−\mathrm{6}+… \\…
Question Number 9348 by An2812 last updated on 02/Dec/16 $$\mathrm{pH}\:=\:\mathrm{pK}_{\mathrm{a}} \:+\:\mathrm{lg}\frac{\mathrm{C}_{\mathrm{b}} }{\mathrm{C}_{\mathrm{a}} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com