Question Number 110706 by Aina Samuel Temidayo last updated on 30/Aug/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 45166 by peter frank last updated on 09/Oct/18 Answered by MrW3 last updated on 10/Oct/18 $${N}={number}\:{of}\:{students} \\ $$$${when}\:{the}\:{students}\:{are}\:{divided}\:{into} \\ $$$$\mathrm{12}\:{groups},\:{the}\:{last}\:{group}\:{has}\:\mathrm{5}\:{students}, \\ $$$${i}.{e}.\:{the}\:{other}\:\mathrm{11}\:{groups}\:{have}\:{x}\:{students} \\…
Question Number 110704 by Aina Samuel Temidayo last updated on 30/Aug/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\frac{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}{\mathrm{n}}+\frac{\mathrm{n}}{\mathrm{n}^{\mathrm{2}} +\mathrm{1}},\mathrm{n}>\mathrm{0} \\ $$ Answered by mr W last updated on…
Question Number 110675 by Aina Samuel Temidayo last updated on 30/Aug/20 $$\mathrm{Two}\:\mathrm{polynomials}\:\mathrm{P}\:\mathrm{and}\:\mathrm{Q}\:\mathrm{satisfy} \\ $$$$\mathrm{P}\left(−\mathrm{2x}+\mathrm{Q}\left(\mathrm{x}\right)\right)=\mathrm{Q}\left(−\mathrm{2x}+\mathrm{P}\left(\mathrm{x}\right)\right). \\ $$$$\mathrm{Given}\:\mathrm{that}\:\mathrm{Q}\left(\mathrm{x}\right)=\mathrm{x}^{\mathrm{2}} −\mathrm{4}\:\mathrm{and} \\ $$$$\mathrm{P}\left(\mathrm{x}\right)=\mathrm{ax}+\mathrm{b}.\:\mathrm{Find}\:\mathrm{2a}+\mathrm{b}. \\ $$ Answered by bemath last…
Question Number 110673 by Aina Samuel Temidayo last updated on 30/Aug/20 $$ \\ $$$$\mathrm{Let}\:\mathrm{k}\:\mathrm{be}\:\mathrm{a}\:\mathrm{real}\:\mathrm{number}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{inequality}\:\sqrt{\mathrm{x}−\mathrm{3}}+\sqrt{\mathrm{6}−\mathrm{x}}\geqslant\mathrm{k}\:\mathrm{has}\:\mathrm{a} \\ $$$$\mathrm{solution}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{k}. \\ $$ Answered by Her_Majesty…
Question Number 176206 by mathlove last updated on 15/Sep/22 Answered by TheHoneyCat last updated on 18/Sep/22 $$\mathrm{a}=\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{1}^{\mathrm{1}} .\left(\left(\mathrm{2}^{\mathrm{1}} \right)^{\mathrm{1}} \right)^{\mathrm{2}} =\mathrm{1}.\left(\mathrm{2}^{\mathrm{1}} \right)^{\mathrm{2}} =\mathrm{2}^{\mathrm{2}}…
Question Number 176189 by Linton last updated on 14/Sep/22 $${solve}\:{for}\:{x} \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{x}+\mathrm{1}} =\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{6}}\right)^{\mathrm{6}} \\ $$$${show}\:{working} \\ $$ Answered by BaliramKumar last updated on 14/Sep/22 $$\left(\frac{{x}+\mathrm{1}}{{x}}\right)^{{x}+\mathrm{1}}…
Question Number 176159 by peter frank last updated on 14/Sep/22 Answered by adhigenz last updated on 14/Sep/22 $$\mathrm{Let}\:{x}\:=\:\mathrm{cos}\:\alpha\:+\:{i}.\mathrm{sin}\:\alpha\:=\:{e}^{{i}.\alpha} \:\mathrm{and}\:{y}\:=\:\mathrm{cos}\:\beta\:+\:{i}.\mathrm{sin}\:\beta\:=\:{e}^{{i}.\beta} \\ $$$$\left(\mathrm{a}\right).\:{x}^{{m}} {y}^{{n}} \:+\:\frac{\mathrm{1}}{{x}^{{m}} {y}^{{n}} }\:=\:{e}^{{i}.{m}\alpha}…
Question Number 110620 by peter frank last updated on 29/Aug/20 Answered by Dwaipayan Shikari last updated on 29/Aug/20 $${f}\left({x}\right)={tan}^{−\mathrm{1}} {x} \\ $$$${f}'\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} \\…
Question Number 176155 by adhigenz last updated on 14/Sep/22 $$\mathrm{Find}\:\mathrm{how}\:\mathrm{many}\:\mathrm{distinct}\:\mathrm{integers}\:\mathrm{are}\:\mathrm{there}\:\mathrm{in}\:\mathrm{this}\:\mathrm{sequence}: \\ $$$$\lfloor\frac{\mathrm{1}^{\mathrm{2}} +\mathrm{1}}{\mathrm{100}}\rfloor,\:\lfloor\frac{\mathrm{2}^{\mathrm{2}} +\mathrm{2}}{\mathrm{100}}\rfloor,\:\lfloor\frac{\mathrm{3}^{\mathrm{2}} +\mathrm{3}}{\mathrm{100}}\rfloor,\:…,\:\lfloor\frac{\mathrm{100}^{\mathrm{2}} +\mathrm{100}}{\mathrm{100}}\rfloor \\ $$$$\mathrm{where}\:\lfloor{x}\rfloor\:\mathrm{is}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{integer}\:\mathrm{that}\:\mathrm{is}\:\mathrm{less}\:\mathrm{than}\:\mathrm{or}\:\mathrm{equal}\:\mathrm{to}\:{x} \\ $$ Commented by Rasheed.Sindhi last updated…