Question Number 54700 by peter frank last updated on 09/Feb/19 $$\mathrm{F}{i}\mathrm{nd}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{to}\:\mathrm{infinity} \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\:}\mathrm{cos}\:\theta+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\:\mathrm{2}\theta−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{cos}\:\mathrm{3}\theta+………. \\ $$$$ \\ $$ Commented by maxmathsup by imad last updated on…
Question Number 185743 by Rupesh123 last updated on 26/Jan/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 119928 by peter frank last updated on 28/Oct/20 Answered by bramlexs22 last updated on 28/Oct/20 $${mx}\:=\:{y}−\left(\frac{\mathrm{3}{m}^{\mathrm{2}} +\mathrm{4}}{\mathrm{4}{m}}\right) \\ $$$$\:{x}\:=\:\frac{{y}}{{m}}−\left(\frac{\mathrm{3}{m}^{\mathrm{2}} +\mathrm{4}}{\mathrm{4}{m}^{\mathrm{2}} }\right)\:\leftarrow{substitute} \\ $$$${to}\:{quadratic}\:{function}\:{give}…
Question Number 54322 by peter frank last updated on 02/Feb/19 Answered by kaivan.ahmadi last updated on 02/Feb/19 $$\mathrm{ax}+\mathrm{by}=\mathrm{c}\:\mathrm{and}\:\mathrm{bx}−\mathrm{ay}=\mathrm{d}\:\mathrm{are}\:\mathrm{perpendicular} \\ $$$$\mathrm{abx}^{\mathrm{2}} +\left(\mathrm{b}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} \right)\mathrm{xy}−\mathrm{aby}^{\mathrm{2}} =\mathrm{cd} \\…
Question Number 185386 by Rupesh123 last updated on 20/Jan/23 Commented by floor(10²Eta[1]) last updated on 21/Jan/23 $$\mathrm{a}^{\mathrm{2}} +\mathrm{2ab}+\mathrm{b}^{\mathrm{2}} +\mathrm{1990a}+\mathrm{1990b}+\mathrm{36}= \\ $$$$=\mathrm{a}^{\mathrm{2}} +\mathrm{1990a}+\mathrm{36}+\mathrm{b}^{\mathrm{2}} +\mathrm{1990b}+\mathrm{36} \\ $$$$\Rightarrow\mathrm{2ab}=\mathrm{36}\Rightarrow\mathrm{ab}=\mathrm{18}…
Question Number 185269 by Rupesh123 last updated on 19/Jan/23 Answered by aleks041103 last updated on 19/Jan/23 $${p}=\sqrt{{n}−\mathrm{8}}+\sqrt{{n}+\mathrm{8}}\Rightarrow{n}\geqslant\mathrm{8} \\ $$$${p}^{\mathrm{2}} =\mathrm{2}{n}−\mathrm{2}\sqrt{{n}^{\mathrm{2}} −\mathrm{64}} \\ $$$$\Rightarrow\sqrt{\mathrm{4}\left({n}^{\mathrm{2}} −\mathrm{64}\right)}=\mathrm{2}{n}−{p}^{\mathrm{2}} \\…
Question Number 185270 by Mingma last updated on 19/Jan/23 Answered by mr W last updated on 20/Jan/23 $$\mathrm{2}{a}+\mathrm{1}\leqslant{d}\leqslant\mathrm{9}\:\Rightarrow{a}\leqslant\mathrm{4}\:\cap\:{a}\neq\mathrm{0} \\ $$$$\mathrm{2}{d}=\mathrm{10}+{a} \\ $$$${a}=\mathrm{4}:\:{d}=\mathrm{7}\:\Rightarrow\mathrm{2}×\mathrm{4}+\mathrm{1}=\mathrm{9}\:>\:\mathrm{7}\:\Rightarrow{rejected} \\ $$$${a}=\mathrm{2}:\:{d}=\mathrm{6}\:\Rightarrow\mathrm{2}×\mathrm{2}+\mathrm{1}=\mathrm{5}\:<\mathrm{6}\:\Rightarrow{ok} \\…
Question Number 185229 by Mingma last updated on 18/Jan/23 Answered by Rasheed.Sindhi last updated on 21/Jan/23 $${A}_{{S}} =\frac{{n}}{\mathrm{2}}\left[\mathrm{2}{c}+\mathrm{2}\left({n}−\mathrm{1}\right)\right] \\ $$$${G}_{{S}} =\frac{{c}\left(\mathrm{1}−\mathrm{2}^{{n}} \right)}{\mathrm{1}−\mathrm{2}} \\ $$$$\mathrm{2}\centerdot{A}_{{S}} ={G}_{{S}}…
Question Number 185167 by Rupesh123 last updated on 18/Jan/23 Answered by aba last updated on 18/Jan/23 $$\mathrm{P}=\left(\mathrm{1}+\mathrm{2}+…+\mathrm{9}\right)\mathrm{N}=\mathrm{45N} \\ $$$$\mathrm{if}\:\mathrm{N}\:\mathrm{is}\:\mathrm{an}\:\mathrm{even}\:\mathrm{integer},\:\mathrm{then}\:\mathrm{P}=\mathrm{45N}\:\mathrm{will}\:\mathrm{end}\:\mathrm{in}\:\mathrm{0} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{not}\:\mathrm{possible}\:\mathrm{as}\:\mathrm{all}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{are}\:\mathrm{same} \\ $$$$\mathrm{if}\:\mathrm{N}\:\mathrm{is}\:\mathrm{an}\:\mathrm{odd}\:\mathrm{integer},\:\mathrm{then}\:\mathrm{P}=\mathrm{45N}\:\mathrm{will}\:\mathrm{end}\:\mathrm{in}\:\mathrm{5} \\ $$$$\mathrm{P}=\mathrm{45N}=\mathrm{555}…\mathrm{55}…
Question Number 119575 by mathocean1 last updated on 25/Oct/20 $$\mathrm{a}\:\in\:\mathbb{N}.\:\mathrm{a}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{3}. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Show}\:\mathrm{that}\:\mathrm{a}^{\mathrm{3}} \equiv−\mathrm{1}\left[\mathrm{9}\right]\:\mathrm{or}\:\mathrm{a}^{\mathrm{3}} \equiv\mathrm{1}\left[\mathrm{9}\right]. \\ $$$$\left.\mathrm{2}\right)\:\mathrm{Given}\:\mathrm{a};\:\mathrm{b};\:\mathrm{c}\:\in\:\mathbb{Z}. \\ $$$$\left.\mathrm{Deduct}\:\mathrm{from}\:\mathrm{1}\right)\:\mathrm{that}\:\mathrm{if}\:\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} \equiv\mathrm{0}\left[\mathrm{9}\right]\:,\:\mathrm{then}\:\: \\ $$$$\mathrm{one}\:\mathrm{of}\:\mathrm{integers}\:\mathrm{a};\:\mathrm{b};\:\mathrm{c}\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{3}. \\ $$…