Question Number 213330 by MrGaster last updated on 02/Nov/24 Commented by MathematicalUser2357 last updated on 06/Nov/24 $${I}\:{need}\:{to}\:{send}\:{the}\:{equation}\:{to}\:{my}\:{chinese}\:{lover} \\ $$ Answered by MrGaster last updated on…
Question Number 213322 by Spillover last updated on 02/Nov/24 Commented by Spillover last updated on 03/Nov/24 Answered by Spillover last updated on 03/Nov/24 Answered by…
Question Number 213323 by Spillover last updated on 02/Nov/24 Commented by Frix last updated on 03/Nov/24 $$\mathrm{No}\:\mathrm{nice}\:\mathrm{numbers} \\ $$$$\mathrm{Let}\:{r}=\mathrm{1} \\ $$$$\Rightarrow \\ $$$$\rho_{{q}.{c}.} \approx\mathrm{4}.\mathrm{87164313433}\:\left[\:_{\mathrm{polynome}\:\mathrm{of}\:\mathrm{8}^{\mathrm{th}} \:\mathrm{degree}\:}…
Question Number 213281 by ajfour last updated on 02/Nov/24 Commented by ajfour last updated on 02/Nov/24 $$\:\:\:\:{Find}\:{r}\:{in}\:{terms}\:{of}\:{a},\:{b}. \\ $$ Answered by mr W last updated…
Question Number 213201 by ajfour last updated on 01/Nov/24 Commented by ajfour last updated on 01/Nov/24 $${Find}\:{R} \\ $$ Answered by A5T last updated on…
Question Number 213250 by ajfour last updated on 01/Nov/24 Answered by a.lgnaoui last updated on 02/Nov/24 $$\mathrm{Calcul}\:\mathrm{de}\:\mathrm{l}\:\mathrm{aire} \\ $$$$ \\ $$$$\mathrm{BC}=\mathrm{a}\sqrt{\mathrm{2}}\:\:\:\:\:\:\:\mathrm{DM}=\mathrm{a}\sqrt{\mathrm{2}}\:+\frac{\mathrm{a}}{\mathrm{2}}. \\ $$$$\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{ABDMC}}\right)=\mathrm{S}\left(\boldsymbol{\mathrm{ABC}}\right)+\mathrm{2}\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{BDE}}\right)+\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{BCEF}}\right) \\ $$$$…
Question Number 213173 by ajfour last updated on 31/Oct/24 Commented by ajfour last updated on 31/Oct/24 $${Find}\:{r} \\ $$ Answered by ajfour last updated on…
Question Number 213021 by Spillover last updated on 28/Oct/24 Commented by Ghisom last updated on 29/Oct/24 $${r}_{\mathrm{1}} =\mathrm{1}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\approx.\mathrm{351153302} \\ $$$${r}_{\mathrm{2}} =\frac{\left(\mathrm{1}−\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\right)\sqrt{\mathrm{2}}}{\mathrm{2}}\approx.\mathrm{165910681} \\ $$$$\frac{{r}_{\mathrm{1}} }{{r}_{\mathrm{2}} }=\mathrm{2}−\sqrt{\mathrm{2}}+\mathrm{2}\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\approx\mathrm{2}.\mathrm{11652017}…
Question Number 213020 by Spillover last updated on 28/Oct/24 Commented by Ghisom last updated on 29/Oct/24 $$\mathrm{we}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{only}\:\mathrm{if}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{special}\:\mathrm{one} \\ $$$$\mathrm{rectangular}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{fit}\:\mathrm{but}\:\mathrm{isosceles}\:\mathrm{fits} \\ $$$${a}={b}=\mathrm{8}\wedge{c}=\frac{\mathrm{48}}{\mathrm{5}} \\ $$$$\mathrm{I}\:\mathrm{used}\:\mathrm{Heron}'\mathrm{s}\:\mathrm{Formula}\:\mathrm{and}\:\mathrm{its}\:\mathrm{further}…
Question Number 213022 by Spillover last updated on 28/Oct/24 Answered by TonyCWX08 last updated on 31/Oct/24 $${FC}\:=\:{a}+{b} \\ $$$${height}\:{of}\:{trapezium}_{{AFBC}} \: \\ $$$$=\:\sqrt{{b}^{\mathrm{2}} −\left(\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} } \\…