Question Number 54045 by ajfour last updated on 28/Jan/19 Commented by ajfour last updated on 28/Jan/19 $${ABCD}\:{is}\:{a}\:{square},\:{and}\:{its}\:{diagonal} \\ $$$${as}\:{a}\:{side}\:{of}\:{the}\:{rectangle}.\:{Find}\:{R}/{r}. \\ $$ Answered by ajfour last…
Question Number 54022 by ajfour last updated on 28/Jan/19 Commented by ajfour last updated on 28/Jan/19 $${Given}\:{a}\:{and}\:{b},\:{find}\:\theta\:{and}\:{R}\:{in}\:{terms} \\ $$$${of}\:{a}\:{and}\:{b}.\:\:\:\: \\ $$ Commented by tanmay.chaudhury50@gmail.com last…
Question Number 185077 by emmanuelson123 last updated on 16/Jan/23 Answered by mahdipoor last updated on 16/Jan/23 $${get}\:{side}\:{of}\:{rectangle}\:{are}\:{A}\:,\:{B}\: \\ $$$$\Delta\mathrm{1}\equiv\Delta\mathrm{2}\equiv\Delta\mathrm{3}\:\:\:\:\Rightarrow \\ $$$${side}\:{of}\:\Delta\mathrm{1}:{A}\:,\:{B}\:,\:\sqrt{{A}^{\mathrm{2}} +{B}^{\mathrm{2}} } \\ $$$${and}\:{r}_{\mathrm{1}}…
Question Number 185073 by emmanuelson123 last updated on 16/Jan/23 Answered by Frix last updated on 16/Jan/23 $$\mathrm{16}°\mathrm{52}'\mathrm{30}''\:=\:\frac{\mathrm{3}\pi}{\mathrm{32}} \\ $$$$\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{8}}\:=\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$$\mathrm{sin}\:\frac{{x}}{\mathrm{2}}\:=\sqrt{\frac{\mathrm{1}−\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{x}}}{\mathrm{2}}} \\ $$$$\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{16}}\:=\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}}{\mathrm{2}} \\…
Question Number 185052 by emmanuelson123 last updated on 16/Jan/23 Commented by Frix last updated on 16/Jan/23 $$\forall{x}\in\left[\sqrt{\mathrm{2}},\:+\infty\right):\:−\mathrm{1}\leqslant\frac{\mathrm{1}}{\mathrm{1}−\lfloor{x}^{\mathrm{2}} \rfloor}<\mathrm{0}\:\Rightarrow \\ $$$$\forall{x}\in\left[\sqrt{\mathrm{2}},\:+\infty\right):\:\lfloor\frac{\mathrm{1}}{\mathrm{1}−\lfloor{x}^{\mathrm{2}} \rfloor}\rfloor=−\mathrm{1}\:\Rightarrow \\ $$$${f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{1};\:\sqrt{\mathrm{2}}\leqslant{x}<+\infty \\…
Question Number 185053 by emmanuelson123 last updated on 16/Jan/23 Answered by a.lgnaoui last updated on 18/Jan/23 $${look}\:\:{at}\:\:{the}\:{answer} \\ $$$$\mathrm{18}.\mathrm{01}.\mathrm{2023}\:\:\: \\ $$$${x}=\mathrm{31},\mathrm{445}°\:\:\: \\ $$ Terms of…
Question Number 185042 by HeferH last updated on 16/Jan/23 $${ABCD}\:{is}\:{a}\:{quadrilateral}\:{inscribed}\:{in}\:{a}\:{circle} \\ $$$$\:{if}\:\:\:\:\overset{\frown} {{AB}}\:+\:\overset{\frown} {{CD}}\:=\:\mathrm{307}°\:{and}\:{AC}\centerdot{BD}\:=\:\mathrm{6}\sqrt{\mathrm{5}} \\ $$$$\:{find}\:{the}\:{area}\:{of}\:{ABCD}.\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 119503 by liberty last updated on 25/Oct/20 $${solve}\:\lfloor\:\sqrt{{x}}\:\rfloor\:=\:\lfloor\:\sqrt[{\mathrm{3}\:}]{{x}}\:\rfloor\: \\ $$ Answered by 1549442205PVT last updated on 25/Oct/20 $$\mathrm{Graph}\:\mathrm{two}\:\mathrm{functions}\:\mathrm{y}=\left[\:^{\mathrm{3}} \sqrt{\mathrm{x}}\right]\mathrm{and} \\ $$$$\mathrm{y}=\left[\:\sqrt{\mathrm{x}}\right]\mathrm{on}\:\mathrm{the}\:\mathrm{same}\:\:\mathrm{coordinate}\:\mathrm{system} \\ $$$$\:\mathrm{Oxy}\:\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the}…
Question Number 53962 by ajfour last updated on 27/Jan/19 Commented by ajfour last updated on 27/Jan/19 $${Find}\:{maximum}\:{perimeter}\:{of}\: \\ $$$${triangle}\:{APQ}.\left({have}\:{i}\:{posted}\:{this}\right. \\ $$$$\left.{question},\:{before},\:{i}'{m}\:{not}\:{sure}!?\right) \\ $$ Commented by…
Question Number 53910 by ajfour last updated on 27/Jan/19 Commented by ajfour last updated on 27/Jan/19 $${Find}\:{maximum}\:{area}\:{of}\:{quadrilateral} \\ $$$${ABCD}. \\ $$ Commented by ajfour last…