Question Number 223655 by Rojarani last updated on 01/Aug/25 Commented by MathematicalUser2357 last updated on 01/Aug/25 $${is}\:{this}\:{like}: \\ $$ Commented by MathematicalUser2357 last updated on…
Question Number 223666 by MASANJAJJ last updated on 01/Aug/25 Answered by mr W last updated on 01/Aug/25 Commented by mr W last updated on 02/Aug/25…
Question Number 223619 by Tawa11 last updated on 01/Aug/25 Answered by Frix last updated on 01/Aug/25 $$\mid{C}_{\mathrm{1}} {C}_{\mathrm{2}} \mid=\sqrt{\mathrm{2}} \\ $$$$\Rightarrow\:{C}_{\mathrm{1}} ,\:{C}_{\mathrm{2}} \:\mathrm{form}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{each}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{intersection}\:\mathrm{points}\:\mathrm{with}…
Question Number 223615 by fantastic last updated on 31/Jul/25 $$\mathrm{40}^{{x}−\mathrm{1}} =\mathrm{2}^{\mathrm{2}{x}+\mathrm{1}} \\ $$ Answered by mr W last updated on 31/Jul/25 $$\frac{\mathrm{40}^{{x}} }{\mathrm{40}}=\mathrm{2}×\mathrm{2}^{\mathrm{2}{x}} =\mathrm{2}×\mathrm{4}^{{x}} \\…
Question Number 223595 by ajfour last updated on 31/Jul/25 Commented by ajfour last updated on 31/Jul/25 $$\:\:\:\:\:\:\:\:\:\:{Given}\:\theta.\:{Find}\:\phi. \\ $$ Answered by mr W last updated…
Question Number 223591 by Tawa11 last updated on 31/Jul/25 Commented by ajfour last updated on 31/Jul/25 $${Very}\:{nice}\:{Question}! \\ $$ Commented by Tawa11 last updated on…
Question Number 223580 by Nicholas666 last updated on 30/Jul/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}\:} \:\frac{{e}^{−\boldsymbol{{r}}^{\mathrm{2}} } \boldsymbol{\mathrm{sin}}\left(\mathrm{1}/\boldsymbol{{r}}^{\mathrm{2}} \right)\boldsymbol{\mathrm{ln}}\left(\boldsymbol{{r}}+\mathrm{1}\right)}{\boldsymbol{{r}}^{\mathrm{2}} }\:\boldsymbol{\mathrm{d}{r}} \\ $$$$ \\ $$ Answered by MathematicalUser2357…
Question Number 223585 by hardmath last updated on 30/Jul/25 Commented by hardmath last updated on 30/Jul/25 $$\mathrm{Radius}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:=\:\mathrm{R} \\ $$$$\mathrm{AB}\:\bot\:\mathrm{CD} \\ $$$$\mathrm{AC}\:=\:\mathrm{a} \\ $$$$\mathrm{BD}\:=\:\mathrm{b} \\ $$$$\mathrm{Prove}\:\mathrm{that}:\:\mathrm{R}\:=\:\frac{\sqrt{\mathrm{a}^{\mathrm{2}}…
Question Number 223571 by hardmath last updated on 30/Jul/25 $$\mathrm{S}_{\mathrm{1}} \:=\:\mathrm{1}\centerdot\mathrm{1}!\:+\:\mathrm{2}\centerdot\mathrm{2}!\:+\:\mathrm{3}\centerdot\mathrm{3}!\:+…+\:\mathrm{16}\centerdot\mathrm{16}! \\ $$$$\mathrm{S}_{\mathrm{2}} \:=\:\mathrm{1}\centerdot\mathrm{1}!\:+\:\mathrm{2}\centerdot\mathrm{2}!\:+\:\mathrm{3}\centerdot\mathrm{3}!\:+…+\:\mathrm{14}\centerdot\mathrm{14}! \\ $$$$\mathrm{Find}:\:\:\:\frac{\mathrm{S}_{\mathrm{1}} }{\mathrm{S}_{\mathrm{2}} }\:=\:? \\ $$ Answered by parthasc last updated…
Question Number 223560 by fantastic last updated on 29/Jul/25 Commented by fantastic last updated on 29/Jul/25 $${area}=? \\ $$ Commented by fantastic last updated on…