Question Number 175953 by rexford last updated on 10/Sep/22 Answered by Ar Brandon last updated on 10/Sep/22 $$\theta=\mathrm{arctan}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{n}+\mathrm{1}}\right)=\mathrm{arctan}\left(\frac{\left({n}+\mathrm{1}\right)−{n}}{\mathrm{1}+{n}\left({n}+\mathrm{1}\right)}\right) \\ $$$$\mathrm{tan}\theta=\frac{\left({n}+\mathrm{1}\right)−{n}}{\mathrm{1}+{n}\left({n}+\mathrm{1}\right)}=\frac{\mathrm{tan}\left(\mathrm{arctan}\left({n}+\mathrm{1}\right)\right)−\mathrm{tan}\left(\mathrm{arctan}\left({n}\right)\right)}{\mathrm{1}+\mathrm{tan}\left(\mathrm{arctan}\left({n}+\mathrm{1}\right)\right)\mathrm{tan}\left(\mathrm{arctan}\left({n}\right)\right)} \\ $$$$\Rightarrow\mathrm{tan}\theta=\mathrm{tan}\left(\left(\mathrm{arctan}\left({n}+\mathrm{1}\right)−\mathrm{arctan}\left({n}\right)\right)\right. \\ $$$$\Rightarrow\theta=\mathrm{arctan}\left({n}+\mathrm{1}\right)−\mathrm{arctan}\left({n}\right)…
Question Number 175954 by rexford last updated on 10/Sep/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 175925 by Mastermind last updated on 09/Sep/22 $$\mathrm{1}^{\mathrm{x}} +\mathrm{6}^{\mathrm{x}} +\mathrm{8}^{\mathrm{x}} =\mathrm{9}^{\mathrm{x}} \\ $$$$\mathrm{find}\:\mathrm{x}\:? \\ $$$$ \\ $$$$\mathrm{Mastermind} \\ $$ Commented by Ar Brandon…
Question Number 110385 by Study last updated on 28/Aug/20 Commented by Study last updated on 28/Aug/20 $${hrlp}\:{mr} \\ $$ Commented by Dwaipayan Shikari last updated…
Question Number 175888 by Mastermind last updated on 08/Sep/22 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equation} \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }{\mathrm{y}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)} \\ $$ Answered by mahdipoor last updated on 08/Sep/22 $$\Rightarrow\int\frac{{y}}{\mathrm{1}+{y}^{\mathrm{2}} }{dy}=\int\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}}…
Question Number 175746 by Mastermind last updated on 06/Sep/22 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equation} \\ $$$$\mathrm{2}\left(\mathrm{2xy}+\mathrm{4y}−\mathrm{3}\right)\mathrm{dx}+\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} \mathrm{dy}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{Mastermind} \\ $$ Answered by floor(10²Eta[1]) last updated on…
Question Number 175655 by Mastermind last updated on 04/Sep/22 Answered by a.lgnaoui last updated on 06/Sep/22 $$\mathrm{26} \\ $$ Commented by a.lgnaoui last updated on…
Question Number 175624 by infinityaction last updated on 04/Sep/22 $$\mathrm{let}\:\mathrm{p}\left(\mathrm{x}\right)\:=\:\mathrm{x}^{\mathrm{6}} +\mathrm{ax}^{\mathrm{5}} +\mathrm{bx}^{\mathrm{4}} +\mathrm{cx}^{\mathrm{3}} +\mathrm{dx}^{\mathrm{2}} +\mathrm{ex}+\mathrm{f} \\ $$$$\:\:\mathrm{be}\:\mathrm{a}\:\mathrm{polynomial}\:\mathrm{function}\:\mathrm{such} \\ $$$$\:\:\mathrm{that}\:\:\mathrm{p}\left(\mathrm{1}\right)\:=\:\mathrm{1}\:;\:\mathrm{p}\left(\mathrm{2}\right)\:=\:\mathrm{2}\:;\:\:\mathrm{p}\left(\mathrm{3}\right)\:=\:\mathrm{3} \\ $$$$\:\:\mathrm{p}\left(\mathrm{4}\right)\:=\:\mathrm{4}\:;\:\mathrm{p}\left(\mathrm{5}\right)\:=\:\mathrm{5}\:;\:\mathrm{p}\left(\mathrm{6}\right)\:=\:\mathrm{6}\:\:\mathrm{then} \\ $$$$\:\:\mathrm{find}\:\:\mathrm{p}\left(\mathrm{7}\right)\:=\:? \\ $$…
Question Number 44546 by Rk7547055934@gmail.com last updated on 01/Oct/18 $$ \\ $$ Answered by $@ty@m last updated on 01/Oct/18 $$… \\ $$ Terms of Service…
Question Number 109922 by Ar Brandon last updated on 26/Aug/20 Answered by 1549442205PVT last updated on 26/Aug/20 $$\mathrm{3x}^{\mathrm{2}} +\mathrm{12xy}+\mathrm{6y}=\mathrm{0}\Leftrightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{4xy}+\mathrm{2y}=\mathrm{0}\left(\mathrm{1}\right) \\ $$$$\left.\mathrm{i}\right)\mathrm{Case}\:\mathrm{x}=\mathrm{0}\Rightarrow\mathrm{y}=\mathrm{0}\:\mathrm{substituting}\:\mathrm{into}\:\mathrm{first}\:\mathrm{eqn}. \\ $$$$\mathrm{we}\:\mathrm{see}\:\mathrm{it}\:\mathrm{satisfy},\mathrm{so} \\…