Question Number 159111 by mathocean1 last updated on 13/Nov/21 $${Given}\:\overset{\rightarrow} {{A}}=\mathrm{5}{t}^{\mathrm{2}} \overset{\rightarrow} {{i}}+{t}\overset{\rightarrow} {{j}}−{t}^{\mathrm{3}} \overset{\rightarrow} {{k}}\:{and} \\ $$$$\overset{\rightarrow} {{B}}={sin}\left({t}\right)\overset{\rightarrow} {{i}}−{cos}\left({t}\right)\overset{\rightarrow} {{j}}. \\ $$$${Calculate}\:\frac{{d}\left(\overset{\rightarrow} {{A}}.\overset{\rightarrow} {{B}}\right)}{{dx}}\:;\:\frac{{d}\left(\overset{\rightarrow}…
Question Number 159099 by akolade last updated on 12/Nov/21 Answered by Rasheed.Sindhi last updated on 13/Nov/21 $${Let}\:{h}_{\mathrm{1}} \:{is}\:{the}\:{height}\:{of}\:{tower}\:{above}\: \\ $$$${kim}'{s}\:{position}\:{and}\:{h}_{\mathrm{2}} \:{the}\:{height}\:{below} \\ $$$${her}\:{position}. \\ $$$$\mathcal{T}{he}\:{height}\:{of}\:{tower}\:{h}={h}_{\mathrm{1}}…
Question Number 159079 by LEKOUMA last updated on 12/Nov/21 $${Prove}\:{that}\:\:\mathrm{2}+\sqrt{\mathrm{3}}\:,\:\frac{\sqrt{\mathrm{3}}−\mathrm{2}}{\mathrm{3}}\:\:{and}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}−\mathrm{5}}\: \\ $$$${are}\:{the}\:{number}\:{irrational} \\ $$ Answered by mindispower last updated on 13/Nov/21 $$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\in{Q}\Leftrightarrow\sqrt{\mathrm{3}}\in{Q} \\ $$$$\sqrt{\mathrm{3}}=\frac{{a}}{{b}},{a}\wedge{b}=\mathrm{1} \\…
Question Number 93534 by naka3546 last updated on 13/May/20 $${A}^{\mathrm{2}} \:\:=\:\:\begin{pmatrix}{\mathrm{7}}&{\mathrm{3}}\\{\mathrm{9}}&{\mathrm{4}}\end{pmatrix}\:\:\:\Rightarrow\:\:{A}\:=\:\begin{pmatrix}{{a}}&{{b}}\\{{c}}&{{d}}\end{pmatrix} \\ $$$${Find}\:\:{the}\:\:{all}\:\:{of}\:\:\:{different}\:\:{matrices}\:\:{A}\:\: \\ $$$$\left({i}\right)\:.\:{If}\:\:{a},\:{b},\:{c},\:{d}\:\in\:\mathbb{Z}\:\:\: \\ $$$$\left({ii}\right)\:.\:{If}\:\:{a},\:{b},\:{c},\:{d}\:\in\:\mathbb{R}^{+} \: \\ $$ Answered by prakash jain last…
Question Number 93530 by Shakhzod last updated on 13/May/20 $$\left(\mathrm{4}{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \right){dx}+\left(\mathrm{4}{y}^{\mathrm{2}} +\mathrm{3}{xy}+{x}^{\mathrm{2}} \right){dy}=\mathrm{0} \\ $$ Commented by Shakhzod last updated on 13/May/20 $$\left.{S}\left.{orry}.\right)\right) \\…
Question Number 93513 by Shakhzod last updated on 13/May/20 $${y}^{''} +{y}^{'} −\mathrm{2}{y}=\mathrm{0} \\ $$ Answered by i jagooll last updated on 13/May/20 $$\mathrm{homogenous}\:\mathrm{solution}\: \\ $$$$\lambda^{\mathrm{2}}…
Question Number 93512 by Shakhzod last updated on 13/May/20 $$\left({xy}+\mathrm{sin}\:{y}\right){dx}+\left(\mathrm{0}.\mathrm{5}{x}^{\mathrm{2}} +{x}\mathrm{cos}\:{y}\right){dy}={o} \\ $$ Commented by i jagooll last updated on 14/May/20 $$\mathrm{sin}\:\mathrm{ydx}+\mathrm{xcos}\:\mathrm{ydy}\:=\:−\mathrm{0}.\mathrm{5x}^{\mathrm{2}} \mathrm{dy}−\mathrm{xydx} \\ $$$$\mathrm{d}\left(\mathrm{xsin}\:\mathrm{y}\right)=\:−\:\left(\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}^{\mathrm{2}}…
Question Number 93510 by Shakhzod last updated on 13/May/20 $${y}^{'} −{y}.\mathrm{tan}\:{x}+{y}^{\mathrm{2}} \mathrm{cos}\:{x}=\mathrm{0} \\ $$ Commented by Shakhzod last updated on 13/May/20 $${Friends}\:{help}\:{me}\:{please}. \\ $$ Terms…
Question Number 93509 by Shakhzod last updated on 13/May/20 $$\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{5}{n}+\mathrm{6}}\: \\ $$ Commented by Shakhzod last updated on 13/May/20 $${Help}\:{me}\:{who}\:{can}\:{solve}\:{it}. \\ $$…
Question Number 159029 by otchereabdullai@gmail.com last updated on 12/Nov/21 $$\:\mathrm{In}\:\mathrm{a}\:\mathrm{box}\:\mathrm{of}\:\mathrm{marbles},\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{red} \\ $$$$\mathrm{marble}\:\mathrm{to}\:\mathrm{total}\:\mathrm{marble}\:\mathrm{is}\:\mathrm{2}\::\mathrm{5},\:\mathrm{the}\: \\ $$$$\mathrm{ratio}\:\mathrm{of}\:\mathrm{green}\:\mathrm{marble}\:\mathrm{to}\:\mathrm{the}\:\mathrm{total}\: \\ $$$$\mathrm{marble}\:\mathrm{is}\:\mathrm{3}:\mathrm{10}\:.\:\mathrm{If}\:\mathrm{the}\:\mathrm{marbles}\:\mathrm{that}\:\mathrm{are} \\ $$$$\mathrm{neither}\:\mathrm{red}\:\mathrm{nor}\:\mathrm{green}\:\mathrm{are}\:\mathrm{blue},\:\mathrm{how}\: \\ $$$$\mathrm{many}\:\mathrm{marbles}\:\mathrm{are}\:\mathrm{in}\:\mathrm{the}\:\mathrm{box}\:\mathrm{if}\:\mathrm{there} \\ $$$$\mathrm{are}\:\mathrm{40}\:\mathrm{marbles}\:\mathrm{in}\:\mathrm{the}\:\mathrm{box} \\ $$ Commented…