Question Number 153574 by otchereabdullai@gmail.com last updated on 08/Sep/21 $$\left.\:\mathrm{1}\right)\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{a}\:\mathrm{spherical}\:\mathrm{ballon}\:\mathrm{is}\: \\ $$$$\mathrm{increasing}\:\mathrm{at}\:\mathrm{the}\:\mathrm{rate}\:\mathrm{of}\:\mathrm{3cms}^{−\mathrm{1}} \:.\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{rate}\:\mathrm{in}\:\mathrm{the}\:\mathrm{surface}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{ballon} \\ $$$$\mathrm{when}\:\mathrm{its}\:\mathrm{radius}\:\mathrm{is}\:\mathrm{10cm} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:\mathrm{evaluate}\:\int\frac{\mathrm{5}}{\left(\mathrm{3x}−\mathrm{2}\right)^{\mathrm{4}} }\mathrm{dx} \\ $$ Answered…
Question Number 153563 by ZiYangLee last updated on 08/Sep/21 $$\int\:\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{2}{x}}\:}\:{dx}\:=? \\ $$ Answered by puissant last updated on 08/Sep/21 $${u}=\sqrt{\mathrm{2}{x}}\:\rightarrow\:{u}^{\mathrm{2}} =\mathrm{2}{x}\:\rightarrow\:{x}=\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\:{dx}={udu} \\ $$$${K}=\int\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{2}{x}}}{dx}=\int\frac{{u}}{\mathrm{1}+{u}}{du} \\…
Question Number 88021 by naka3546 last updated on 07/Apr/20 Commented by MJS last updated on 07/Apr/20 $$\mathrm{I}\:\mathrm{get}\:\mathrm{4}\vee\mathrm{5}…\:\mathrm{not}\:\mathrm{sure}\:\mathrm{how}\:\mathrm{I}\:\mathrm{got}\:\mathrm{there}… \\ $$ Commented by naka3546 last updated on…
Question Number 22475 by mondodotto@gmail.com last updated on 19/Oct/17 Commented by Rasheed.Sindhi last updated on 19/Oct/17 $$\left(\mathrm{b}\right)\mathrm{9}^{\mathrm{8x}^{\mathrm{2}} } \\ $$$$\sqrt{\mathrm{9}^{\mathrm{16x}^{\mathrm{2}} } }=\left(\mathrm{9}^{\mathrm{16x}^{\mathrm{2}} } \right)^{\mathrm{1}/\mathrm{2}} =\mathrm{9}^{\mathrm{16x}^{\mathrm{2}}…
Question Number 153537 by SANOGO last updated on 08/Sep/21 Answered by puissant last updated on 08/Sep/21 $$\left.\mathrm{18}\right) \\ $$$${on}\:{remarque}\:{d}'{abord}\:{que} \\ $$$${x}^{{x}} =\mathrm{1}+{xlnx}+….+\frac{\left({xlnx}\right)^{{n}} }{{n}!}\left(\mathrm{1}+\varepsilon\left({x}\right)\right) \\ $$$$\varepsilon\left({x}\right)\rightarrow\mathrm{0}\:{quand}\:{x}\rightarrow\mathrm{0}…
Question Number 153512 by naka3546 last updated on 08/Sep/21 $${a},{b},{c}\:\:\in\:\:\mathbb{Z} \\ $$$$\mid{a}−{b}\mid^{\mathrm{3}} \:+\:\mid{b}−{c}\mid^{\mathrm{3}} \:=\:\mathrm{1} \\ $$$${Find}\:\:{the}\:\:{value}\:\:{of} \\ $$$$\:\:\:\:\:\:\mid{a}−{b}\mid\:+\:\mid{b}−{c}\mid\:+\:\mid{c}−{a}\mid \\ $$ Answered by MJS_new last updated…
Question Number 22434 by gopikrishnan005@gmail.com last updated on 18/Oct/17 $$\boldsymbol{\mathrm{I}}\mathrm{m}\left(\mathrm{2z}+\mathrm{1}\right)/\left(\mathrm{iz}−\mathrm{1}\right)=\mathrm{2} \\ $$ Answered by ajfour last updated on 18/Oct/17 $$\mathrm{2}{z}+\mathrm{1}=\mathrm{2}{iz}−\mathrm{2} \\ $$$$\mathrm{2}{x}+\mathrm{2}{iy}+\mathrm{1}=\mathrm{2}{ix}−\mathrm{2}{y}−\mathrm{2} \\ $$$$\Rightarrow\:\:\:\:\mathrm{2}{x}+\mathrm{2}{y}=−\mathrm{3}\:\:\:{and} \\…
Question Number 87966 by M±th+et£s last updated on 07/Apr/20 Answered by ajfour last updated on 07/Apr/20 $${y}={kx}^{\mathrm{2}} \\ $$$$\mathrm{8}=\mathrm{4}{k}\:\:\Rightarrow\:\:{k}=\mathrm{2}\:\:\Rightarrow\:{y}=\mathrm{2}{x}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{x}^{\mathrm{2}} =\mathrm{4}\left(\frac{\mathrm{2}+{a}}{\mathrm{8}}\right){y}\:=\:\frac{{y}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:{a}=\mathrm{2}\:. \\…
Question Number 87963 by M±th+et£s last updated on 07/Apr/20 Commented by M±th+et£s last updated on 07/Apr/20 $${sorry}\:\:\mathrm{9}{y}^{\mathrm{2}} −\mathrm{16}{x}^{\mathrm{2}} =\mathrm{144} \\ $$ Terms of Service Privacy…
Question Number 153488 by talminator2856791 last updated on 07/Sep/21 $$\:\mathrm{my}\:\mathrm{notifications}\:\mathrm{dont}\:\mathrm{work}. \\ $$$$\:\mathrm{am}\:\mathrm{i}\:\mathrm{the}\:\mathrm{only}\:\mathrm{one}\:\mathrm{with}\:\mathrm{this}\:\mathrm{problem}?\: \\ $$$$\:\mathrm{how}\:\mathrm{can}\:\mathrm{i}\:\mathrm{contact}\:\mathrm{tinku}\:\mathrm{tara}\:\mathrm{and}\: \\ $$$$\:\mathrm{do}\:\mathrm{they}\:\mathrm{still}\:\mathrm{update}\:\mathrm{the}\:\mathrm{app}?\: \\ $$ Commented by Tinku Tara last updated on…