Question Number 125444 by ajfour last updated on 11/Dec/20 Commented by ajfour last updated on 11/Dec/20 $${Find}\:{radius}\:{of}\:{the}\:{quarter}\:{circle}. \\ $$ Answered by ajfour last updated on…
Question Number 125337 by ajfour last updated on 10/Dec/20 Commented by ajfour last updated on 10/Dec/20 $${If}\:{all}\:{coloured}\:{areas}\:{are}\:\mathrm{10}\:{units} \\ $$$${each},\:{find}\:{coordinates}\:{of}\:{G}. \\ $$$$\left({FGE}\:\:\:{and}\:{CGD}\:{are}\:{straight}\right. \\ $$$$\left.{segments}\right) \\ $$…
Question Number 190845 by Rupesh123 last updated on 13/Apr/23 Answered by som(math1967) last updated on 13/Apr/23 Commented by som(math1967) last updated on 13/Apr/23 $${ar}\bigtriangleup{ABC}=\frac{\mathrm{1}}{\mathrm{2}}{ADEB}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{10}=\mathrm{5}{squ} \\…
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Question Number 190790 by Rupesh123 last updated on 11/Apr/23 Commented by Frix last updated on 11/Apr/23 $$\frac{{a}}{{b}}=\pi−\mathrm{1}+\sqrt{\pi^{\mathrm{2}} −\mathrm{2}\pi} \\ $$ Terms of Service Privacy Policy…
Question Number 125215 by ajfour last updated on 09/Dec/20 Commented by ajfour last updated on 09/Dec/20 $${A}\:{cylindrical}\:{hole}\:{of}\:{radius}\:{half} \\ $$$${that}\:{of}\:{the}\:{sphere}\:{is}\:{drilled}\:{out}\:, \\ $$$${touching}\:{the}\:{equatorial}\:{plane}\:{of} \\ $$$${sphere}.\:{Find}\:{fraction}\:{of}\:{volume} \\ $$$${drilled}\:{out}.…
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Question Number 125108 by peter frank last updated on 08/Dec/20 Commented by mr W last updated on 08/Dec/20 $${where}\:{did}\:{you}\:{get}\:{these}\:{questions}? \\ $$$${they}\:{are}\:{wrong},\:{i}.{e}.\:{there}\:{is}\:{no} \\ $$$${solution}. \\ $$$${the}\:{last}\:{digit}\:{of}\:\mathrm{2}^{{any}\:{integer}}…
Question Number 125084 by ajfour last updated on 08/Dec/20 Commented by ajfour last updated on 08/Dec/20 $${Find}\:{x},\:{hence}\:{radius}\:{of}\:{circle}. \\ $$ Commented by ajfour last updated on…
Question Number 125021 by Ggjj last updated on 07/Dec/20 Answered by mathmax by abdo last updated on 07/Dec/20 $$\mathrm{s}\Rightarrow\begin{cases}{\mid\mathrm{x}−\mathrm{1}\mid+\mid\mathrm{y}−\mathrm{5}\mid=\mathrm{1}}\\{\mid\mathrm{x}−\mathrm{1}\mid−\left(\mathrm{y}−\mathrm{5}\right)=\mathrm{0}\:\:\:\:\mathrm{due}\:\mathrm{to}\:\mid\mathrm{x}−\mathrm{1}\mid\geqslant\mathrm{0}\:\mathrm{we}\:\mathrm{get}\:\mathrm{y}\geqslant\mathrm{5}\:\Rightarrow}\end{cases} \\ $$$$\mathrm{s}\Rightarrow\begin{cases}{\mid\mathrm{x}−\mathrm{1}\mid+\mathrm{y}−\mathrm{5}=\mathrm{1}}\\{\mid\mathrm{x}−\mathrm{1}\mid−\left(\mathrm{y}−\mathrm{5}\right)=\mathrm{0}\:\Rightarrow\mathrm{2}\mid\mathrm{x}−\mathrm{1}\mid=\mathrm{1}\:\Rightarrow\mid\mathrm{x}−\mathrm{1}\mid=\frac{\mathrm{1}}{\mathrm{2}}}\end{cases} \\ $$$$\Rightarrow\mathrm{x}−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{or}\:\mathrm{x}−\mathrm{1}=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{x}=\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{or}\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{x}=\frac{\mathrm{3}}{\mathrm{2}}\Rightarrow\mathrm{y}=\mathrm{5}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{11}}{\mathrm{2}}\:\Rightarrow\mathrm{x}+\mathrm{y}=\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{11}}{\mathrm{2}}=\mathrm{7}…