Question Number 169382 by infinityaction last updated on 29/Apr/22 $$\:\:\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\:\left[\boldsymbol{\mathrm{v}}\right]\:\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{v}}\:\boldsymbol{\mathrm{den}{o}\mathrm{tes}}\:\boldsymbol{\mathrm{maximum}}\:\: \\ $$$$\:\:\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:+\:\boldsymbol{\mathrm{y}}^{\mathrm{2}} \:,\:\boldsymbol{\mathrm{where}}\:\left(\boldsymbol{\mathrm{x}}+\mathrm{5}\right)^{\mathrm{2}} \:+\:\left(\boldsymbol{\mathrm{y}}−\mathrm{12}\right)^{\mathrm{2}} \:=\:\mathrm{14} \\ $$$$\:\:\:\:\left(\boldsymbol{\mathrm{hint}}\:\left[\bullet\right]\:\boldsymbol{\mathrm{repersent}}\:\boldsymbol{\mathrm{greatest}}\:\boldsymbol{\mathrm{integer}}\:\boldsymbol{\mathrm{function}}\:\boldsymbol{\mathrm{of}}\:“\:\bullet''\right) \\ $$$$ \\ $$$$\: \\ $$ Commented…
Question Number 169356 by Shrinava last updated on 29/Apr/22 $$\mathrm{1}.\:\mathrm{y}\:=\:\mathrm{arcsin}\left(\mathrm{sinx}\right)\:\Rightarrow\:\mathrm{y}^{'} =? \\ $$$$\mathrm{2}.\:\mathrm{y}\:=\:\mathrm{sin}\:\sqrt{\mathrm{x}\:+\:\mathrm{1}}\:\Rightarrow\:\mathrm{y}^{'} =? \\ $$$$\mathrm{3}.\:\mathrm{y}\:=\:\mathrm{ln}^{\mathrm{5}} \:\mathrm{sinx}\:\Rightarrow\:\mathrm{y}^{'} =? \\ $$$$\mathrm{4}.\:\mathrm{y}\:=\:\mathrm{cos}\left(\mathrm{2x}\:+\:\mathrm{3}\right)\:\Rightarrow\:\mathrm{y}^{'} =? \\ $$ Answered by…
Question Number 169354 by Shrinava last updated on 29/Apr/22 $$\mathrm{1}.\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\mathrm{1}\:+\:\mathrm{x}} \\ $$$$\mathrm{2}.\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\left(\mathrm{6}\:-\:\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx} \\ $$$$\mathrm{3}.\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{cosx}}{\mathrm{1}\:+\:\mathrm{sinx}}\:\mathrm{dx} \\ $$ Answered by…
Question Number 169349 by Shrinava last updated on 29/Apr/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 103804 by ajfour last updated on 17/Jul/20 $$\:\:\:\boldsymbol{{S}}{olve}\:{for}\:\boldsymbol{{r}} \\ $$$$\:\:\frac{\boldsymbol{{h}}−\boldsymbol{{p}}^{\mathrm{2}} }{\boldsymbol{{p}}−\boldsymbol{{r}}}=\frac{\mathrm{1}}{\mathrm{2}\boldsymbol{{p}}}\:\:\:\:,\:\:\:\:\frac{\boldsymbol{{h}}−\boldsymbol{{q}}}{\boldsymbol{{q}}^{\mathrm{2}} −\boldsymbol{{r}}}=\:\mathrm{2}\boldsymbol{{q}} \\ $$$$\left(\boldsymbol{{h}}−\boldsymbol{{p}}^{\mathrm{2}} \right)^{\mathrm{2}} +\left(\boldsymbol{{p}}−\boldsymbol{{r}}\right)^{\mathrm{2}} =\boldsymbol{{r}}^{\mathrm{2}} \\ $$$$\:\left(\boldsymbol{{h}}−\boldsymbol{{q}}\right)^{\mathrm{2}} +\left(\boldsymbol{{q}}^{\mathrm{2}} −\boldsymbol{{r}}\right)^{\mathrm{2}} =\boldsymbol{{r}}^{\mathrm{2}} \\…
Question Number 169304 by mathlove last updated on 28/Apr/22 $$\mathrm{3}^{{x}} =\mathrm{4} \\ $$$$\mathrm{4}^{{y}} =\mathrm{12}\:\:\:\:\:\overset{{faind}\:{value}\:{of}\:\left(\frac{{x}+\mathrm{1}}{\mathrm{2}{xy}}\right)=?} {\:} \\ $$ Commented by infinityaction last updated on 28/Apr/22 $$\:\:\:\:\:\:\:\mathrm{3}^{{x}}…
Question Number 169291 by infinityaction last updated on 28/Apr/22 $$ \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{x}^{{x}^{.^{.^{{x}} } } } −{x}!}{{x}!^{{x}!} −\mathrm{1}}\:=\:?? \\ $$$$ \\ $$ Commented by infinityaction…
Question Number 169280 by mathlove last updated on 28/Apr/22 $${a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{25}{a}^{\mathrm{2}} {b}^{\mathrm{2}} =\mathrm{0} \\ $$$${b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{36}{b}^{\mathrm{2}} {c}^{\mathrm{2}} =\mathrm{0} \\ $$$${a}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{49}{a}^{\mathrm{2}} {c}^{\mathrm{2}}…
Question Number 103736 by I want to learn more last updated on 17/Jul/20 $$\mathrm{Solve}:\:\:\:\:\:\:\mathrm{a}^{\mathrm{2}} \:\:+\:\:\mathrm{c}^{\mathrm{2}} \:\:=\:\:\:\mathrm{196}\:\:\:\:\:\:…\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}^{\mathrm{2}} \:\:+\:\:\left(\mathrm{c}\:\:−\:\:\mathrm{a}\right)^{\mathrm{2}} \:\:=\:\:\mathrm{169}\:\:\:\:\:…\:\left(\mathrm{ii}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{c}^{\mathrm{2}} \:\:+\:\:\left(\mathrm{b}\:\:−\:\:\mathrm{c}\right)^{\mathrm{2}} \:\:=\:\:\mathrm{225}\:\:\:\:\:…\:\left(\mathrm{iii}\right) \\…
Question Number 169271 by Shrinava last updated on 27/Apr/22 Answered by mr W last updated on 27/Apr/22 $${a}={r}\:\mathrm{cot}\:\frac{{B}}{\mathrm{2}}+{r}\:\mathrm{cot}\:\frac{{C}}{\mathrm{2}} \\ $$$$\frac{{a}}{{r}}=\mathrm{cot}\:\frac{{B}}{\mathrm{2}}+\mathrm{cot}\:\frac{{C}}{\mathrm{2}}\geqslant\mathrm{2}\left(\mathrm{cot}\:\frac{{B}}{\mathrm{2}}\:\mathrm{cot}\:\frac{{C}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\left(\frac{{a}}{{r}}\right)^{\mathrm{6}} \geqslant\mathrm{2}^{\mathrm{6}} ×\left(\mathrm{cot}\:\frac{{B}}{\mathrm{2}}\:\mathrm{cot}\:\frac{{C}}{\mathrm{2}}\right)^{\mathrm{3}} \\…