Category: Others

Question Number 170492 by 2407 last updated on 25/May/22 Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 39416 by Rio Mike last updated on 06/Jul/18 $${Given}\:{that}\:{f}\left({x}\right)\:{is}\:{a}\:{cubic}\:{function} \\ $$$${and}\:{f}\left({x}\right)\:=\:{x}^{\mathrm{3}} \:\:−\:\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\:+\:\mathrm{5}{x}\:−\:\mathrm{7} \\ $$$$\left.{a}\right)\:{find}\:{one}\:{factor}\:{of}\:{f}\left({x}\right) \\ $$$$\left.{b}\right)\:{find}\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:{for}\:{f}\left({x}\right) \\ $$$$\left.{c}\right)\:{hence}\:{Evaluate}\:\:{y}\:=\:\int_{\mathrm{0}} ^{\infty} {f}\left({x}\right).…

Question Number 170485 by Mastermind last updated on 24/May/22 Answered by Rasheed.Sindhi last updated on 25/May/22 $$\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:\right)^{{x}} +\left(\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\:\right)^{{x}} =\mathrm{2}\:;\:{x}=? \\ $$$$\:\:\:\:\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:\right)^{{x}} }=\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:\right)^{{x}} }\centerdot\frac{\left(\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\:\right)^{{x}} }{\left(\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\:\right)^{{x}} }…

Question Number 170480 by 2407 last updated on 24/May/22 Commented by JDamian last updated on 24/May/22 $$\Sigma\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)}\:\:{is}\:{a}\:{telescoping}\:{series} \\ $$$$\Sigma\frac{\mathrm{1}}{\mathrm{3}^{{k}+\mathrm{1}} }\:\:\:{is}\:{a}\:{GP} \\ $$$${Then}\:{I}\:{guess}\:{you}\:{can}\:{solve}\:{by}\:\:{yourself} \\ $$ Commented…

Question Number 39395 by Rio Mike last updated on 06/Jul/18 $${Given}\:{that}\:\theta\:{is}\:{an}\:{obtuse}\: \\ $$$${angle}\:{find}\:{tan}\:\theta\:{if} \\ $$$${cos}\:\theta\:=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$ \\ $$ Commented by MrW3 last updated on…