Question Number 157873 by HongKing last updated on 29/Oct/21 $$\mathrm{if}:\:\boldsymbol{\alpha}\:=\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{x}}}{\mathrm{1}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{x}}}\:\:\mathrm{e}^{-\boldsymbol{\pi\mathrm{y}}\left(\mathrm{1}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }\right)} \:\mathrm{dydx} \\ $$$$\mathrm{find}:\:\sqrt{\mathrm{19683}\boldsymbol{\alpha}^{\mathrm{6}} \:-\:\mathrm{94041}\boldsymbol{\alpha}^{\mathrm{4}} \:+\:\mathrm{105786}\boldsymbol{\alpha}^{\mathrm{2}} } \\ $$$$ \\…
Question Number 157871 by HongKing last updated on 29/Oct/21 $$\mathrm{if}\:\:\:\mathrm{x};\mathrm{y}>\mathrm{0}\:\:\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{2}} -\mathrm{x}+\mathrm{1}}\:+\:\frac{\mathrm{y}}{\mathrm{y}^{\mathrm{2}} -\mathrm{y}+\mathrm{1}}\:+\:\frac{\mathrm{xy}}{\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} -\mathrm{xy}+\mathrm{1}}\:\leqslant \\ $$$$\leqslant\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} -\mathrm{x}+\mathrm{1}}\:+\:\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{y}^{\mathrm{2}} -\mathrm{y}+\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} -\mathrm{xy}+\mathrm{1}} \\…
Question Number 92335 by jagoll last updated on 06/May/20 $$\sqrt{\left\{\mathrm{x}\right\}}\:=\:\mathrm{1}+\:\mathrm{ln}\left(\mathrm{x}\right)\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 92324 by I want to learn more last updated on 06/May/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{x}\:\:\mathrm{for}\:\mathrm{which}\:\:\:\:\underset{\mathrm{n}\:\:=\:\:\mathrm{0}} {\overset{\mathrm{n}\:\:=\:\:\infty} {\sum}}\:\mathrm{16}\left(\frac{\mathrm{3}}{\mathrm{4}}\mathrm{x}\:\:+\:\:\mathrm{1}\right)^{\mathrm{n}} \\ $$$$\left(\mathrm{a}\right)\:\:\:\mathrm{Is}\:\mathrm{convergent} \\ $$$$\left(\mathrm{b}\right)\:\:\:\mathrm{Is}\:\mathrm{equal}\:\mathrm{to}\:\:\mathrm{10}\frac{\mathrm{2}}{\mathrm{3}} \\ $$ Answered by mr…
Question Number 157855 by HongKing last updated on 28/Oct/21 Commented by Rasheed.Sindhi last updated on 30/Oct/21 $${x},{y},{z}\:{are}\:{measures}\:{of}\:{sides}\:{of}\:\bigtriangleup\mathrm{ABC}? \\ $$ Commented by HongKing last updated on…
Question Number 157851 by emanuelMcCarthy last updated on 28/Oct/21 Commented by otchereabdullai@gmail.com last updated on 29/Oct/21 $$\left.\mathrm{22}\right)\:\mathrm{Let}\:\mathrm{g}\left(\mathrm{x}\right)=−\mathrm{2x}^{\mathrm{3}} −\mathrm{3x}^{\mathrm{2}} +\mathrm{13bx}+\mathrm{20} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}+\mathrm{a}=\mathrm{0}\:\Rightarrow\mathrm{x}=−\mathrm{a} \\ $$$$\:\:\:\mathrm{so}\:\:\:\mathrm{f}\left(−\mathrm{a}\right)=−\mathrm{56} \\ $$$$−\mathrm{2}\left(−\mathrm{a}\right)^{\mathrm{3}}…
Question Number 157836 by zakirullah last updated on 28/Oct/21 $${find}\:{the}\:{indicated}\:{higher}\:{order}\:{derivative} \\ $$$${of}\:{the}\:{following}\:{function} \\ $$$${f}\left({x}\right)\:=\:\left({x}^{\mathrm{3}} +\mathrm{4}{x}−\mathrm{5}\right)^{\mathrm{4}} ,\:{f}\left({x}\right)^{{iv}} \\ $$ Answered by tounghoungko last updated on 29/Oct/21…
Question Number 157839 by gsk2684 last updated on 28/Oct/21 $${find}\:{the}\:{last}\:{four}\:{digits}\:{of}\: \\ $$$$\mathrm{11}^{\mathrm{15999}} ? \\ $$ Answered by Rasheed.Sindhi last updated on 29/Oct/21 $$\mathrm{11}^{\mathrm{15999}} \equiv{x}\left({mod}\:\mathrm{10}^{\mathrm{4}} \right)…
Question Number 157835 by HongKing last updated on 28/Oct/21 Answered by Rasheed.Sindhi last updated on 29/Oct/21 $$\overline {{abcd}}\::\:\:{a}<{b}<{c}<{d}\: \\ $$$$\underset{−} {\wedge\:{a}+{b}+{c}+{d}\equiv\mathrm{1}\left[\mathrm{3}\right]\wedge\:{a}+{b}+{c}+{d}\equiv\mathrm{5}\left[\mathrm{11}\right]} \\ $$$$\bullet\mathrm{1234}\leqslant\overline {{abcd}}\leqslant\mathrm{6789} \\…
Question Number 157832 by mathlove last updated on 28/Oct/21 Answered by mr W last updated on 28/Oct/21 $${x}^{{x}^{\frac{\mathrm{3}}{\mathrm{2}}} } ={x}^{{x}} {x}^{\frac{{x}}{\mathrm{2}}} \\ $$$${x}^{{x}^{\frac{\mathrm{3}}{\mathrm{2}}} −\frac{\mathrm{3}{x}}{\mathrm{2}}} =\mathrm{1}…