Question Number 227446 by mr W last updated on 31/Jan/26 Commented by mr W last updated on 31/Jan/26 $${find}\:{the}\:{blue}\:{area}. \\ $$ Commented by Spillover last…
Question Number 227357 by mr W last updated on 18/Jan/26 Answered by A5T last updated on 18/Jan/26 $$\mathrm{Let}\:\parallel\:\mathrm{be}\:\mathrm{x}\:,\:\mathrm{tangent}\:\mathrm{be}\:\mathrm{t}\:\mathrm{and}\:\mathrm{radius}\:\mathrm{be}\:\mathrm{r}. \\ $$$$\mathrm{x}\left(\mathrm{2x}\right)=\mathrm{t}^{\mathrm{2}} \Rightarrow\mathrm{t}=\mathrm{x}\sqrt{\mathrm{2}} \\ $$$$\sqrt{\left(\mathrm{2r}\right)^{\mathrm{2}} −\mathrm{r}^{\mathrm{2}} }=\mathrm{t}−\mathrm{r}\Rightarrow\mathrm{t}=\mathrm{r}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)…
Question Number 227200 by Tawa11 last updated on 06/Jan/26 Commented by Tawa11 last updated on 07/Jan/26 $$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$ Commented by fantastic2 last…
Question Number 227155 by zahraa last updated on 03/Jan/26 Answered by Ghisom_ last updated on 03/Jan/26 $${x}=\mathrm{2}{y}'+\mathrm{ln}\:{y}' \\ $$$$\mathrm{e}^{{x}} =\mathrm{e}^{\mathrm{2}{y}'} {y}' \\ $$$$\mathrm{2e}^{{x}} =\mathrm{e}^{\mathrm{2}{y}'} \mathrm{2}{y}'…
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Question Number 227032 by mr W last updated on 27/Dec/25 Commented by mr W last updated on 27/Dec/25 $${draw}\:{a}\:{straight}\:{line}\:{only}\:{with}\:{ruler} \\ $$$${and}\:{compass},\:{which}\:{divides}\:{the}\: \\ $$$${yellow}\:{figure}\:{into}\:{two}\:{parts}\:{with}\: \\ $$$${equal}\:{area}.…
Question Number 226910 by Estevao last updated on 18/Dec/25 Answered by Estevao last updated on 18/Dec/25 $$>{Ashed}\:{Area} \\ $$ Answered by fantastic2 last updated on…
Question Number 226908 by mr W last updated on 18/Dec/25 Commented by mr W last updated on 18/Dec/25 $${find}\:{the}\:{area}\:{of}\:{the}\:{circle}. \\ $$ Answered by A5T last…
Question Number 226850 by cherokeesay last updated on 17/Dec/25 Answered by TonyCWX last updated on 17/Dec/25 $${y}\:=\:{x}\left({x}−\mathrm{1}\right)\:\Rightarrow\:{x}\:=\:\mathrm{0}\:\mathrm{or}\:{x}\:=\:\mathrm{1} \\ $$$$\mathrm{A}>\mathrm{0},\:\mathrm{A}=\mathrm{1} \\ $$$$ \\ $$$${x}\left({x}−\mathrm{1}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}}{x} \\ $$$${x}^{\mathrm{2}}…