Question Number 160409 by quvonnn last updated on 29/Nov/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 160408 by quvonnn last updated on 29/Nov/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 94874 by i jagooll last updated on 21/May/20 Commented by hknkrc46 last updated on 21/May/20 $$\sqrt{\mathrm{2}−\mathrm{7x}}+\mathrm{2x}=\mathrm{0} \\ $$$$\bigstar\:\sqrt[{\mathrm{2n}}]{\mathrm{f}\left(\mathrm{x}\right)}\geqslant\mathrm{0}\:\:\wedge\:\mathrm{f}\left(\mathrm{x}\right)\geqslant\mathrm{0}\:;\:\mathrm{n}\in\mathbb{Z}^{+} \\ $$$$\bigstar\:\mathrm{n}=\mathrm{1}\:\Rightarrow\:\sqrt[{\mathrm{2}}]{\mathrm{f}\left(\mathrm{x}\right)}=\sqrt{\mathrm{f}\left(\mathrm{x}\right)} \\ $$$$\Rightarrow\:\sqrt{\mathrm{2}−\mathrm{7x}}=−\mathrm{2x} \\…
Question Number 160405 by HongKing last updated on 29/Nov/21 $$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{2}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{2}}{\mathrm{1}+\frac{\mathrm{1}}{….}}}}}\:\:\:\Rightarrow\:\:\mathrm{x}^{\mathrm{2}} \:=\:? \\ $$ Commented by quvonnn last updated on 29/Nov/21 $$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{2}}{{x}}}={x} \\ $$$$\left({x}+\mathrm{2}\right)+{x}={x}\left({x}+\mathrm{2}\right) \\ $$$$\mathrm{2}{x}+\mathrm{2}={x}^{\mathrm{2}}…
Question Number 94868 by O Predador last updated on 21/May/20 Commented by john santu last updated on 21/May/20 $$\mathrm{n}^{\mathrm{2}} \:=\:\mathrm{2}^{\mathrm{2n}} \:\Rightarrow\:\mathrm{n}\:=\:\mathrm{2}^{\mathrm{n}} \: \\ $$$$\mathrm{ln}\left(\mathrm{n}\right)\:=\:\mathrm{n}\:\mathrm{ln}\left(\mathrm{2}\right)\: \\…
Question Number 160398 by HongKing last updated on 29/Nov/21 $$\mathrm{Find}: \\ $$$$\boldsymbol{\Omega}\:=\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\frac{\mathrm{log}\left(\mathrm{x}\:+\:\mathrm{1}\right)}{\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{1}}\:\mathrm{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 160391 by akolade last updated on 29/Nov/21 Answered by FongXD last updated on 29/Nov/21 $$\bullet\:\left(\mathrm{x}^{\mathrm{x}} \right)'=\left(\mathrm{e}^{\mathrm{xlnx}} \right)'=\mathrm{e}^{\mathrm{xlnx}} \left(\mathrm{lnx}+\mathrm{1}\right)=\mathrm{x}^{\mathrm{x}} \left(\mathrm{lnx}+\mathrm{1}\right) \\ $$$$\bullet\:\left(\mathrm{x}^{\mathrm{x}^{\mathrm{x}} } \right)'=\left(\mathrm{e}^{\mathrm{x}^{\mathrm{x}}…
Question Number 160373 by HongKing last updated on 28/Nov/21 $$\mathrm{Compare}\:\mathrm{it}: \\ $$$$\frac{\mathrm{1}\:-\:\mathrm{sin}\:\left(\mathrm{10}°\right)}{\mathrm{cos}\:\left(\mathrm{10}°\right)}\:\:\:\:\:\mathrm{and}\:\:\:\:\:\mathrm{1} \\ $$ Answered by ghimisi last updated on 28/Nov/21 $$\frac{{sin}\mathrm{90}−{sin}\mathrm{10}}{{cos}\mathrm{10}}=\frac{\mathrm{2}{sin}\mathrm{40}{cos}\mathrm{50}}{{sin}\mathrm{80}}= \\ $$$$\frac{\mathrm{2}{sin}\mathrm{40}{sin}\mathrm{40}}{\mathrm{2}{sin}\mathrm{40}{cos}\mathrm{40}}={tg}\mathrm{40}<{tg}\mathrm{45}=\mathrm{1} \\…
Question Number 160375 by HongKing last updated on 28/Nov/21 Commented by ghimisi last updated on 28/Nov/21 $$\frac{\mathrm{1}}{\mathrm{2}{b}+\mathrm{1}}\:{or}\:\frac{\mathrm{1}}{\mathrm{4}{b}+\mathrm{1}} \\ $$$$ \\ $$ Commented by HongKing last…
Question Number 94835 by Jidda28 last updated on 21/May/20 $$\mathrm{proof}\:\mathrm{or}\:\mathrm{disproof}\:\mathrm{that}\:\mathrm{if}\:\mathrm{a}\:\mathrm{quotient} \\ $$$$\mathrm{group}\:\frac{\mathrm{G}}{\mathrm{H}}\:\mathrm{is}\:\mathrm{abelian}\:\mathrm{then}\:\mathrm{G}\:\mathrm{must}\:\mathrm{be}\:\mathrm{abelian}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com