Question Number 75296 by ajfour last updated on 09/Dec/19 Commented by ajfour last updated on 09/Dec/19 $${If}\:{perimeter}\:{of}\:\bigtriangleup{PQR}\:{is}\:{p},\:{find} \\ $$$${maximum}\:{area}\:{of}\:\bigtriangleup{PQR}\:{in} \\ $$$${terms}\:{of}\:{a},{b},{c},{p}.\:\:\:\left({p}<{a}+{b}+{c}\right)\: \\ $$ Commented by…
Question Number 75291 by 21042004 last updated on 09/Dec/19 Answered by mind is power last updated on 09/Dec/19 $$\mathrm{let}\:\mathrm{A}\left(\mathrm{a},\mathrm{0}\right),\mathrm{B}=\left(\mathrm{b},\mathrm{0}\right) \\ $$$$\mathrm{Equation}\:\mathrm{of}\:\mathrm{circl}\:\mathrm{center}\:\mathrm{in}\:\mathrm{A} \\ $$$$\mathrm{condition}\:\mathrm{3}\:\:\:\:\mathrm{AB}<\mathrm{2R} \\ $$$$\mathrm{if}\:\mathrm{not}\:\mathrm{C}_{\mathrm{A}}…
Question Number 9746 by Yozzias last updated on 30/Dec/16 $$\mathrm{Find}\:\left(\mathrm{x},\mathrm{y}\right)\in\mathbb{F}_{\mathrm{7}} ^{\mathrm{2}} \:\mathrm{such}\:\mathrm{that}\:\mathrm{x}^{\mathrm{2}} −\mathrm{3y}^{\mathrm{2}} \equiv\:\mathrm{0}\:\mathrm{mod}\:\mathrm{7}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 9717 by konen last updated on 28/Dec/16 $$\frac{\mathrm{0}.\mathrm{b}\overset{−} {\mathrm{a}}+\mathrm{0}.\mathrm{a}\overset{−} {\mathrm{b}}}{\frac{\mathrm{1}}{\mathrm{ab}−\mathrm{ba}}}=\mathrm{3}\:\Rightarrow\:\mathrm{a}^{\mathrm{2}\:} −\:\mathrm{b}^{\mathrm{2}\:} =? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 75222 by chess1 last updated on 08/Dec/19 Commented by chess1 last updated on 09/Dec/19 $$\mathrm{Sir}\:\boldsymbol{\mathrm{W}}\:\:\:\mathrm{solution}\:\mathrm{please} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 9662 by Joel575 last updated on 23/Dec/16 $$\mathrm{I}\:\mathrm{have}\:\mathrm{2}\:\mathrm{buckets}.\:\mathrm{Each}\:\mathrm{bucket}\:\mathrm{contains}\:\mathrm{green}\:\mathrm{and}\:\mathrm{blue}\:\mathrm{balls} \\ $$$$\mathrm{The}\:\mathrm{first}\:\mathrm{bucket}\:\mathrm{contains}\:\mathrm{3}\:\mathrm{green}\:\mathrm{balls}\:\mathrm{and}\:\mathrm{7}\:\mathrm{blue}\:\mathrm{balls}. \\ $$$$\mathrm{Second}\:\mathrm{bucket}\:\mathrm{contains}\:\mathrm{7}\:\mathrm{green}\:\mathrm{balls}\:\mathrm{and}\:\mathrm{8}\:\mathrm{blue}\:\mathrm{balls}. \\ $$$$\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{take}\:\mathrm{those}\:\mathrm{balls}\:\mathrm{with}\:\mathrm{coin}\:\mathrm{toss}. \\ $$$$\mathrm{If}\:\mathrm{head},\:\mathrm{I}\:\mathrm{will}\:\mathrm{take}\:\mathrm{1}\:\mathrm{ball}\:\mathrm{from}\:\mathrm{each}\:\mathrm{bucket}. \\ $$$$\mathrm{But}\:\mathrm{if}\:\mathrm{tail},\:\mathrm{I}\:\mathrm{will}\:\mathrm{take}\:\mathrm{2}\:\mathrm{balls}\:\mathrm{from}\:\mathrm{each}\:\mathrm{bucket}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{propability}\:\mathrm{if}\:\mathrm{all}\:\mathrm{the}\:\mathrm{balls}\:\mathrm{that}\:\mathrm{have}\:\mathrm{been}\:\mathrm{taken} \\ $$$$\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{color}? \\…
Question Number 9661 by geovane10math last updated on 23/Dec/16 $$\left.{a}\right)\:\mathrm{2}^{{i}} \:=\: \\ $$$$ \\ $$$$\left.{b}\right)\:\left({a}_{\mathrm{1}} \:+\:{b}_{\mathrm{1}} {i}\right)^{{a}_{\mathrm{2}} \:+\:{b}_{\mathrm{2}} {i}} \:=\: \\ $$$${Powers}\:{of}\:{complex}\:{numbers}\:??? \\ $$ Commented…
Question Number 9627 by ridwan balatif last updated on 21/Dec/16 Commented by ridwan balatif last updated on 21/Dec/16 Commented by sou1618 last updated on 22/Dec/16…
Question Number 75156 by chess1 last updated on 08/Dec/19 Answered by $@ty@m123 last updated on 08/Dec/19 $$\mathrm{2}{x}+\mathrm{3}{x}+\left(\mathrm{360}−\mathrm{230}\right)=\mathrm{180} \\ $$$$\Rightarrow\mathrm{5}{x}+\mathrm{130}=\mathrm{180} \\ $$$$\Rightarrow{x}=\mathrm{10}\:…..\left(\left(\mathrm{1}\right)\right. \\ $$$$\mathrm{58}+\mathrm{2}\left(\mathrm{20}+{y}\right)=\mathrm{180} \\ $$$$\mathrm{40}+\mathrm{2}{y}=\mathrm{122}…
Question Number 9603 by tawakalitu last updated on 20/Dec/16 $$\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} \\…