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Category: Geometry

I-have-2-buckets-Each-bucket-contains-green-and-blue-balls-The-first-bucket-contains-3-green-balls-and-7-blue-balls-Second-bucket-contains-7-green-balls-and-8-blue-balls-I-want-to-take-those-balls-

Question Number 9662 by Joel575 last updated on 23/Dec/16 $$\mathrm{I}\:\mathrm{have}\:\mathrm{2}\:\mathrm{buckets}.\:\mathrm{Each}\:\mathrm{bucket}\:\mathrm{contains}\:\mathrm{green}\:\mathrm{and}\:\mathrm{blue}\:\mathrm{balls} \\ $$$$\mathrm{The}\:\mathrm{first}\:\mathrm{bucket}\:\mathrm{contains}\:\mathrm{3}\:\mathrm{green}\:\mathrm{balls}\:\mathrm{and}\:\mathrm{7}\:\mathrm{blue}\:\mathrm{balls}. \\ $$$$\mathrm{Second}\:\mathrm{bucket}\:\mathrm{contains}\:\mathrm{7}\:\mathrm{green}\:\mathrm{balls}\:\mathrm{and}\:\mathrm{8}\:\mathrm{blue}\:\mathrm{balls}. \\ $$$$\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{take}\:\mathrm{those}\:\mathrm{balls}\:\mathrm{with}\:\mathrm{coin}\:\mathrm{toss}. \\ $$$$\mathrm{If}\:\mathrm{head},\:\mathrm{I}\:\mathrm{will}\:\mathrm{take}\:\mathrm{1}\:\mathrm{ball}\:\mathrm{from}\:\mathrm{each}\:\mathrm{bucket}. \\ $$$$\mathrm{But}\:\mathrm{if}\:\mathrm{tail},\:\mathrm{I}\:\mathrm{will}\:\mathrm{take}\:\mathrm{2}\:\mathrm{balls}\:\mathrm{from}\:\mathrm{each}\:\mathrm{bucket}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{propability}\:\mathrm{if}\:\mathrm{all}\:\mathrm{the}\:\mathrm{balls}\:\mathrm{that}\:\mathrm{have}\:\mathrm{been}\:\mathrm{taken} \\ $$$$\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{color}? \\…

Question-75156

Question Number 75156 by chess1 last updated on 08/Dec/19 Answered by $@ty@m123 last updated on 08/Dec/19 $$\mathrm{2}{x}+\mathrm{3}{x}+\left(\mathrm{360}−\mathrm{230}\right)=\mathrm{180} \\ $$$$\Rightarrow\mathrm{5}{x}+\mathrm{130}=\mathrm{180} \\ $$$$\Rightarrow{x}=\mathrm{10}\:…..\left(\left(\mathrm{1}\right)\right. \\ $$$$\mathrm{58}+\mathrm{2}\left(\mathrm{20}+{y}\right)=\mathrm{180} \\ $$$$\mathrm{40}+\mathrm{2}{y}=\mathrm{122}…

A-regular-hexagon-has-sides-of-lenght-8-cm-Find-the-perpendicular-distance-between-two-opposite-faces-

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} \\…

in-triangle-ABC-a-2-b-c-2-1-2-B-C-2-find-h-a-S-ABC-d-a-R-A-

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…

3a-b-c-d-2014-3b-a-c-d-2014-3c-a-b-d-2014-3d-a-b-c-2014-Find-all-the-solution-of-a-b-c-d-if-a-b-c-d-R-

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…