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Author: Tinku Tara

Question-222783

Question Number 222783 by MathematicalUser2357 last updated on 07/Jul/25 $$\:\underbrace{ } \\ $$ Answered by AntonCWX8 last updated on 07/Jul/25 $${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{5}}{x}^{\mathrm{5}} −\frac{\mathrm{5}}{\mathrm{3}}{x}^{\mathrm{3}} +\mathrm{4}{x}+\mathrm{2} \\ $$$${f}'\left({x}\right)={x}^{\mathrm{4}}…

Simplify-cos214-i-sin146-cos10-i-sin10-cos66-i-sin246-

Question Number 222777 by hardmath last updated on 07/Jul/25 $$\mathrm{Simplify}: \\ $$$$\frac{\left(\mathrm{cos214}°\:+\:\boldsymbol{\mathrm{i}}\:\mathrm{sin146}°\right)\centerdot\left(\mathrm{cos10}°\:+\:\boldsymbol{\mathrm{i}}\:\mathrm{sin10}°\right)}{\left(\mathrm{cos66}°\:−\:\boldsymbol{\mathrm{i}}\:\mathrm{sin246}°\right)}\:=\:? \\ $$ Answered by Frix last updated on 07/Jul/25 $$\mathrm{cos}\:\mathrm{214}°\:+\mathrm{i}\:\mathrm{sin}\:\mathrm{146}°\:=\mathrm{e}^{\mathrm{i}\frac{\mathrm{73}\pi}{\mathrm{90}}} \\ $$$$\mathrm{cos}\:\mathrm{10}°\:+\mathrm{i}\:\mathrm{sin}\:\mathrm{10}°\:=\mathrm{e}^{\mathrm{i}\frac{\pi}{\mathrm{18}}} \\…

lim-x-0-2log-1-x-x-3x-2-x-1-2-x-3-

Question Number 222778 by Osefavour last updated on 07/Jul/25 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2log}\left(\mathrm{1}+\mathrm{x}\right)−\frac{\mathrm{x}\left(\mathrm{3x}+\mathrm{2}\right)}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }}{\mathrm{x}^{\mathrm{3}} } \\ $$ Answered by MrGaster last updated on 09/Jul/25 $$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{2}\left({x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}}…

Question-222779

Question Number 222779 by mnjuly1970 last updated on 07/Jul/25 Answered by gabthemathguy25 last updated on 08/Jul/25 $$\mathrm{this}\:\mathrm{is}\:\mathrm{actually}\:\mathrm{hard}.. \\ $$$$\mathrm{im}\:\mathrm{approximating}\:\mathrm{this}. \\ $$$$\approx−\mathrm{0}.\mathrm{0491713} \\ $$ Terms of…

Question-222800

Question Number 222800 by fantastic last updated on 07/Jul/25 Answered by mr W last updated on 07/Jul/25 $$\mathrm{2}×\frac{\pi{a}^{\mathrm{2}} }{\mathrm{6}}−\frac{\sqrt{\mathrm{3}}{a}^{\mathrm{2}} }{\mathrm{4}}=\left(\frac{\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\right){a}^{\mathrm{2}} \approx\mathrm{0}.\mathrm{614}{a}^{\mathrm{2}} \\ $$ Commented by…

Question-222798

Question Number 222798 by fantastic last updated on 07/Jul/25 Answered by mr W last updated on 07/Jul/25 $${r}=\frac{{a}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{4}} \\ $$$${area}\:{of}\:{semi}\:{circle}=\frac{\pi{r}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{3}\pi{a}^{\mathrm{2}} }{\mathrm{32}}\:\checkmark \\ $$ Terms…