Question Number 50012 by ggny last updated on 13/Dec/18 $$\mathrm{help}\:\mathrm{me}\:\mathrm{sir}\:\mathrm{plz} \\ $$$$ \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 115544 by bemath last updated on 26/Sep/20 $${Given}\:{x}^{\mathrm{2}} +\mathrm{12}\sqrt{{x}}\:=\:\mathrm{5} \\ $$$${then}\:{x}+\mathrm{2}\sqrt{{x}}\:? \\ $$ Commented by MJS_new last updated on 26/Sep/20 $${x}^{\mathrm{2}} +\mathrm{12}\sqrt{{x}}=\mathrm{5} \\…
Question Number 181079 by Agnibhoo98 last updated on 21/Nov/22 $$\mathrm{Solve}\:\mathrm{for}\:{x}\:: \\ $$$$\left(\frac{{x}\:+\:{a}}{{x}\:+\:{b}}\right)^{\mathrm{3}} =\:\frac{{x}\:+\:\mathrm{2}{a}\:−\:{b}}{{x}\:−\:{a}\:+\:\mathrm{2}{b}} \\ $$ Commented by CElcedricjunior last updated on 21/Nov/22 $$\left(\frac{\boldsymbol{{x}}+\boldsymbol{{a}}}{\boldsymbol{{x}}+\boldsymbol{{b}}}\right)^{\mathrm{3}} =\frac{\boldsymbol{{x}}+\mathrm{2}\boldsymbol{{a}}−\boldsymbol{{b}}}{\boldsymbol{{x}}−\boldsymbol{{a}}+\mathrm{2}\boldsymbol{{b}}} \\…
Question Number 181066 by Shrinava last updated on 21/Nov/22 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{expression}: \\ $$$$\mathrm{2arcctg}\left(\mathrm{tg3}\right)\:−\:\mathrm{3arctg}\left(\mathrm{ctg2}\right) \\ $$$$\left.\mathrm{a}\left.\right)\left.\mathrm{2}\left.\pi\left.−\mathrm{3}\:\:\mathrm{b}\right)\mathrm{12}\:\:\mathrm{c}\right)−\mathrm{6}\:\:\mathrm{d}\right)\pi−\mathrm{4}/\mathrm{2}\:\:\mathrm{e}\right)\mathrm{3}\pi/\mathrm{2} \\ $$ Commented by mr W last updated on 22/Nov/22 $${all}\:{answers}\:{given}\:{are}\:{wrong}!…
Question Number 181041 by mnjuly1970 last updated on 20/Nov/22 $$ \\ $$$$\:\:\:\mathrm{I}{f}\:\:\:\:{f}\left({x}\right)=\:{x}\:−\:\lfloor\:\frac{{x}−\mathrm{1}}{\mathrm{3}}\:\rfloor\:,\:\:{g}\left(\:{x}\right)=\:\mathrm{2}^{\:{x}} \\ $$$$\:\:\:\:\:\:\:{then}\:\:,\:\:\:\:\:\:{R}_{\:{gof}} \:=\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 181020 by mr W last updated on 20/Nov/22 $${solve}\:{for}\:{x}>\mathrm{0} \\ $$$$\int_{\mathrm{0}} ^{{x}} \lfloor{t}\rfloor^{\mathrm{2}} {dt}=\mathrm{2}\left({x}−\mathrm{1}\right) \\ $$$$ \\ $$$$\left({Q}\mathrm{180780}\:{reposted}\right) \\ $$ Answered by mr…
Question Number 115484 by bemath last updated on 26/Sep/20 $${find}\:{the}\:{value}\:{of}\: \\ $$$$\frac{\mathrm{111}}{\mathrm{1}+\mathrm{1}+\mathrm{1}}\:,\:\frac{\mathrm{222}}{\mathrm{2}+\mathrm{2}+\mathrm{2}}\:,\:\frac{\mathrm{333}}{\mathrm{3}+\mathrm{3}+\mathrm{3}},\:\frac{\mathrm{444}}{\mathrm{4}+\mathrm{4}+\mathrm{4}} \\ $$$$\frac{\mathrm{555}}{\mathrm{5}+\mathrm{5}+\mathrm{5}},\:\frac{\mathrm{666}}{\mathrm{6}+\mathrm{6}+\mathrm{6}}\:,\:\frac{\mathrm{777}}{\mathrm{7}+\mathrm{7}+\mathrm{7}}\:,\:\frac{\mathrm{888}}{\mathrm{8}+\mathrm{8}+\mathrm{8}} \\ $$$$\frac{\mathrm{999}}{\mathrm{9}+\mathrm{9}+\mathrm{9}} \\ $$ Commented by Dwaipayan Shikari last updated on…
Question Number 181007 by mr W last updated on 20/Nov/22 $$\left[{Question}\:{from}\:{Frix}\:{sir}\right] \\ $$$$…\:\mathrm{I}\:\mathrm{once}\:\mathrm{played}\:\mathrm{the}\:\mathrm{game}\:\mathrm{Tetris}.\: \\ $$$$\mathrm{You}\:\mathrm{got}\:\mathrm{points}\:\mathrm{for}\:\mathrm{each}\:\mathrm{filled}\:\mathrm{line},\: \\ $$$$\mathrm{depending}\:\mathrm{on}\:\mathrm{how}\:\mathrm{many}\:\mathrm{lines}\:\mathrm{you}\: \\ $$$$\mathrm{filled}\:\mathrm{at}\:\mathrm{once}: \\ $$$$\mathrm{1}\:\mathrm{line}\:=\:\mathrm{1}\:\mathrm{point} \\ $$$$\mathrm{2}\:\mathrm{lines}\:=\:\mathrm{3}\:\mathrm{points} \\ $$$$\mathrm{3}\:\mathrm{lines}\:=\:\mathrm{6}\:\mathrm{points}…
Question Number 180977 by mr W last updated on 19/Nov/22 $${what}\:{is}\:{larger},\:\mathrm{3}^{\mathrm{100}} +\mathrm{4}^{\mathrm{100}} \:{or}\:\:\mathrm{5}^{\mathrm{100}} ? \\ $$ Answered by MJS_new last updated on 19/Nov/22 $$\mathrm{2}×\mathrm{4}^{{x}} \overset{?}…
Question Number 115436 by frc2crc last updated on 25/Sep/20 $$\sqrt{\sqrt[{\mathrm{3}}]{{a}}+\sqrt[{\mathrm{3}}]{{b}}}=\frac{\mathrm{1}}{\:\sqrt{{b}−{a}×{s}^{\mathrm{3}} }}\left(\frac{−{s}^{\mathrm{2}} ×\sqrt[{\mathrm{3}}]{{a}^{\mathrm{2}} }}{\mathrm{2}}+{s}\sqrt[{\mathrm{3}}]{{ab}}+\sqrt[{\mathrm{3}}]{{b}^{\mathrm{2}} }\right) \\ $$$${what}\:{is}\:{s}? \\ $$ Answered by MJS_new last updated on 25/Sep/20…