Question Number 125882 by harckinwunmy last updated on 14/Dec/20 $$\mathrm{suppose}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{set}\:\mathrm{of}\left\{\mathrm{3}>\mid\mathrm{c}+\mathrm{1}\mid\&\mid\mathrm{d}−\mathrm{1}\mid<\mathrm{10}\right\} \\ $$$$\mathrm{is}\:\mathrm{x}<\mathrm{c}+\mathrm{d}<\mathrm{y},\:\mathrm{then}\:\left(\mathrm{x},\mathrm{y}\right)=? \\ $$ Commented by talminator2856791 last updated on 15/Dec/20 $$\:\left(−\mathrm{12};\:\mathrm{11}\right)? \\ $$ Commented…
Question Number 60347 by ANTARES VY last updated on 20/May/19 Commented by ANTARES VY last updated on 20/May/19 $$\boldsymbol{\mathrm{calculate}} \\ $$ Answered by mr W…
Question Number 60346 by necx1 last updated on 20/May/19 $$\int{x}\mathrm{sec}\:^{\mathrm{3}} {xdx} \\ $$$${please}\:{help} \\ $$ Commented by kaivan.ahmadi last updated on 20/May/19 $${first}\:\int{sec}^{\mathrm{3}} {x}\:{dx}=\left(\frac{{tgx}}{{cosx}}+{ln}\mid{secx}+{tgx}\mid\right)/\mathrm{2} \\…
Question Number 125881 by aurpeyz last updated on 14/Dec/20 $${find}\:{the}\:{greatest}\:{coeeficient}\:{of} \\ $$$$\left(\mathrm{1}+\frac{{x}}{\mathrm{2}}\right)^{−\mathrm{2}} \\ $$$${i}\:{have}\:{applied}\:{Q}.\mathrm{125697}\:{but}\:{i}\:{got}\:{two} \\ $$$${negative}\:{values}\:{of}\:{n}\:{after}\:{i}\:{have} \\ $$$${made}\:{k}=\mathrm{2}{n}\:{for}\:{positive}\:{coefficient}. \\ $$$$ \\ $$$${I}\:{believe}\:{it}\:{should}\:{have}\:{a}\:{greatest} \\ $$$${coeeficient}\:{since}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{k}} {decreases}\:{as}…
Question Number 125876 by bramlexs22 last updated on 14/Dec/20 Commented by liberty last updated on 15/Dec/20 $${W}\:=\:\underset{{n}\rightarrow\infty} {\mathrm{Lim}}\:\underset{{i}\:=\:\mathrm{1}} {\overset{{n}} {\sum}}\rho{A}\left({x}_{{i}} ^{\ast} \right)\Delta{x}\:\:\:\:\:\:\: \\ $$$${W}\:=\:\underset{{n}\rightarrow\infty} {\mathrm{Lim}}\:\underset{{i}\:=\:\mathrm{1}}…
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Question Number 191409 by null last updated on 23/Apr/23 Answered by senestro last updated on 24/Apr/23 $$−\frac{{x}}{\mathrm{6}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{\mathrm{2}} }+\frac{{x}}{\mathrm{18}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)}+\frac{\mathrm{1}}{\mathrm{9}}\mathrm{ln}\:\left(\frac{\left(\mathrm{1}+{x}\right)^{\mathrm{3}} }{\mathrm{1}+{x}^{\mathrm{3}} }\right)+\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+{C} \\ $$ Terms…
Question Number 60337 by arcana last updated on 20/May/19 $$\mathrm{15},\mathrm{25},\mathrm{42},…? \\ $$ Commented by MJS last updated on 20/May/19 $$\mathrm{only}\:\mathrm{one}\:\mathrm{of}\:\infty\:\mathrm{solutions}: \\ $$$${ax}^{\mathrm{2}} +{bx}+{c}−{y}=\mathrm{0} \\ $$$${a}+{b}+{c}−\mathrm{15}=\mathrm{0}…
Question Number 191410 by otchereabdullai last updated on 23/Apr/23 $$\:{Three}\:{villages}\:{A},\:{B}\:{and}\:{C}\:{are}\:{on}\:{a}\: \\ $$$$\:{straight}\:{road}\:{and}\:{B}\:{is}\:{the}\:{mid}-{way} \\ $$$${between}\:{A}\:{and}\:{C}.\:{A}\:{motor}\:{cyclist} \\ $$$${moving}\:{with}\:{a}\:{uniform}\:{acceleration} \\ $$$${passes}\:{A},\:{B}\:{and}\:{C}.\:{The}\:{speeds}\:{with}\: \\ $$$${which}\:{the}\:{motorcyclist}\:{passes}\:{A}\:{and}\: \\ $$$${C}\:{are}\:\mathrm{20}{ms}^{−\mathrm{1}} \:{and}\:\mathrm{40}{ms}^{−} \:{respectively}. \\…
Question Number 125873 by Mathgreat last updated on 14/Dec/20 Answered by MJS_new last updated on 15/Dec/20 $${t}=\mathrm{sin}\:\mathrm{2}{x}\:=\mathrm{2sin}\:{x}\:\mathrm{cos}\:{x}\:=\mathrm{2sin}\:{x}\:\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{x}}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{sin}\:{x}\:=\begin{cases}{\mathrm{0}\leqslant{x}\leqslant\frac{\pi}{\mathrm{4}};\:\frac{\sqrt{\mathrm{1}+{t}}}{\mathrm{2}}−\frac{\sqrt{\mathrm{1}−{t}}}{\mathrm{2}}}\\{\frac{\pi}{\mathrm{4}}\leqslant{x}\leqslant\frac{\pi}{\mathrm{2}};\:\frac{\sqrt{\mathrm{1}+{t}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{1}−{t}}}{\mathrm{2}}}\end{cases} \\ $$$$\rightarrow\:{dx}=\begin{cases}{\frac{{dt}}{\mathrm{2}\sqrt{\mathrm{1}+{t}}\sqrt{\mathrm{1}−{t}}}}\\{−\frac{{dt}}{\mathrm{2}\sqrt{\mathrm{1}+{t}}\sqrt{\mathrm{1}−{t}}}}\end{cases} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}}…