Question Number 191483 by Mastermind last updated on 24/Apr/23 $$\mathrm{Compute}\:\mathrm{the}\:\mathrm{min}−\mathrm{max}\:\mathrm{polynomial} \\ $$$$\mathrm{q}_{\mathrm{1}} ^{\ast} \left(\mathrm{x}\right)\:\mathrm{to}\:\mathrm{e}^{\mathrm{x}} \:\mathrm{on}\:\mathrm{interval}\:\left[−\mathrm{1},\:\mathrm{1}\right]. \\ $$$$ \\ $$$$\mathrm{Anyone}?? \\ $$ Terms of Service Privacy…
Question Number 125944 by Algoritm last updated on 15/Dec/20 Commented by JDamian last updated on 15/Dec/20 $${What}\:{is}\:{a}_{{n}} ? \\ $$ Commented by Algoritm last updated…
Question Number 191477 by Spillover last updated on 24/Apr/23 $$\mathrm{Prove}\:\mathrm{that} \\ $$$$ \\ $$$$\frac{\mathrm{tan}\:\mathrm{y}+\mathrm{sec}\:\mathrm{y}−\mathrm{1}}{\mathrm{tan}\:\mathrm{y}−\mathrm{sec}\:\mathrm{y}+\mathrm{1}}=\mathrm{tan}\:\mathrm{y}+\mathrm{sec}\:\mathrm{y} \\ $$$$ \\ $$ Answered by ARUNG_Brandon_MBU last updated on 24/Apr/23…
Question Number 60406 by Tawa1 last updated on 20/May/19 $$\mathrm{n}\:\in\:\mathbb{Z}^{+} ,\:\:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:\:\mathrm{x}^{−\mathrm{1}} \:\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of}\:\:\:\left(\mathrm{1}\:+\:\mathrm{x}\right)^{\mathrm{n}} \left(\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{n}} \\ $$ Answered by mr W last updated on 20/May/19 $$\left(\mathrm{1}+{x}\right)^{{n}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)^{{n}}…
Question Number 191473 by Spillover last updated on 24/Apr/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 125939 by Dwaipayan Shikari last updated on 15/Dec/20 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{cos}\mathrm{2}{x}−{tanx}.{cot}\left({tanx}\right)}{{sin}\mathrm{2}{x}−{tan}\left({tanx}\right){log}\left({cos}^{\mathrm{2}} {x}\right)}{dx} \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 191474 by Spillover last updated on 24/Apr/23 $$\mathrm{If}\:\mathrm{A}+\mathrm{B}+\mathrm{C}=\pi \\ $$$$\mathrm{Prove}\:\mathrm{that} \\ $$$$\mathrm{cos}\:\mathrm{2A}+\mathrm{cos}\:\mathrm{2B}+\mathrm{cos2C}+\mathrm{1}=−\mathrm{4cosAcos}\:\mathrm{Bcos}\:\mathrm{C} \\ $$$$ \\ $$ Commented by Tinku Tara last updated on…
Question Number 60398 by Kunal12588 last updated on 20/May/19 $${I}\:{am}\:{very}\:{sorry}\:{mr}\:{W}\:{sir}\:{for}\:{taking}\:{your}\: \\ $$$${valuable}\:{time} \\ $$$${thank}\:{you}\:{very}\:{much}\:{i}\:{got}\:{my}\:{mistake}. \\ $$$${I}\:{will}\:{work}\:{on}\:{my}\:{basics}\:{thank}\:{you} \\ $$$${and}\:{I}\:{am}\:{very}\:{sorry}. \\ $$ Commented by mr W last…
Question Number 125925 by zarminaawan last updated on 15/Dec/20 $$\frac{{d}^{\mathrm{3}} {y}}{{dx}^{\mathrm{3}} }−\mathrm{2}\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{4}\frac{{dy}}{{dx}}−\mathrm{8}{y}=\mathrm{0} \\ $$ Answered by liberty last updated on 15/Dec/20 $${HE}\:\equiv\:{z}^{\mathrm{3}} −\mathrm{2}{z}^{\mathrm{2}}…
Question Number 191457 by Mingma last updated on 24/Apr/23 Commented by Tinku Tara last updated on 24/Apr/23 $$\mathrm{The}\:\mathrm{user}\:\mathrm{is}\:\mathrm{blocked}\:\mathrm{from}\:\mathrm{making}\:\mathrm{further} \\ $$$$\mathrm{posts}. \\ $$ Commented by mr…