Question Number 76171 by Master last updated on 24/Dec/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 76169 by Master last updated on 24/Dec/19 Answered by mr W last updated on 24/Dec/19 $${yx}+\mathrm{27}{y}=\mathrm{154} \\ $$$${xy}+\mathrm{27}{x}=\mathrm{30} \\ $$$$\mathrm{2}{xy}+\mathrm{27}\left({x}+{y}\right)=\mathrm{184}\:\:\:…\left({i}\right) \\ $$$${y}−{x}=\frac{\mathrm{124}}{\mathrm{27}} \\…
Question Number 141700 by BHOOPENDRA last updated on 22/May/21 Commented by BHOOPENDRA last updated on 22/May/21 $${Anyone}\:{please}\:{ans}. \\ $$ Commented by BHOOPENDRA last updated on…
Question Number 76167 by Master last updated on 24/Dec/19 Answered by behi83417@gmail.com last updated on 24/Dec/19 $$\Rightarrow\begin{cases}{\mathrm{3x}−\mathrm{x}^{\mathrm{2}} −\mathrm{3y}+\mathrm{xy}=\mathrm{1}−\mathrm{3x}−\mathrm{xy}+\mathrm{3x}^{\mathrm{2}} \mathrm{y}}\\{\mathrm{2x}−\mathrm{x}^{\mathrm{2}} +\mathrm{2y}−\mathrm{xy}=\mathrm{1}−\mathrm{2x}+\mathrm{xy}−\mathrm{2x}^{\mathrm{2}} \mathrm{y}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{\mathrm{x}−\mathrm{5y}+\mathrm{2xy}=−\mathrm{x}−\mathrm{2xy}+\mathrm{5x}^{\mathrm{2}} \mathrm{y}}\\{\mathrm{5x}−\mathrm{2x}^{\mathrm{2}} −\mathrm{y}=\mathrm{2}−\mathrm{5x}+\mathrm{x}^{\mathrm{2}}…
Question Number 141703 by SLVR last updated on 22/May/21 Commented by SLVR last updated on 22/May/21 $${kindly}\:{give}\:{how}\:{you}\:{wrote} \\ $$$${Mr}.{Olaf}.. \\ $$ Commented by SLVR last…
Question Number 10627 by pan123 last updated on 20/Feb/17 $$\mathrm{60}^{\mathrm{15}} ×\mathrm{30}^{\mathrm{8}} ×\mathrm{35}^{\mathrm{6}} ×\mathrm{60}^{\mathrm{15}} =? \\ $$ Answered by mrW1 last updated on 20/Feb/17 $$\mathrm{60}=\mathrm{2}^{\mathrm{2}} ×\mathrm{3}×\mathrm{5}…
Question Number 141699 by cesarL last updated on 22/May/21 $${If}\:{lim}_{{x}\rightarrow\mathrm{1}} =\frac{\sqrt[{{k}}]{{x}}−\mathrm{1}}{{x}−\mathrm{1}}={L}\neq\mathrm{0}\:\:{Find}\:{lim}_{{x}\rightarrow\mathrm{0}} \frac{\sqrt{{x}+\mathrm{1}}−\mathrm{1}}{\:\sqrt[{{k}}]{{x}+\mathrm{1}}−\mathrm{1}} \\ $$ Answered by iloveisrael last updated on 22/May/21 $${Given}\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt[{{k}\:}]{{x}}−\mathrm{1}}{{x}−\mathrm{1}}\:=\:{L}\:{equivalent} \\ $$$${to}\:\underset{{t}\rightarrow\mathrm{1}}…
Question Number 141693 by Fikret last updated on 22/May/21 $${n}\in{N}^{+} \\ $$$${a}_{\mathrm{5}} ={a}_{\mathrm{13}} =\mathrm{0} \\ $$$${b}_{{n}+\mathrm{1}} −{b}_{{n}} =\mathrm{2} \\ $$$${b}_{{n}} ={a}_{{n}+\mathrm{1}} −{a}_{{n}} \:\:\:\Rightarrow\:\:{a}_{\mathrm{1}} =? \\…
Question Number 141692 by ArielVyny last updated on 22/May/21 $${Daniel}\:{and}\:{Bruno}\:{are}\:{playing}\:{with}\:{perfect}\:{cube} \\ $$$${Daniel}\:{is}\:{the}\:{first}\:{player}\:{if}\:{he}\:{obtains}\:\mathrm{1}\:{or}\:\mathrm{2} \\ $$$${he}\:{wins}\:{the}\:{game}\:{and}\:{the}\:{party}\:{stopping} \\ $$$${or}\:{else}\:{Bruno}\:{plays}\:{and}\:{if}\:{he}\:{have}\:\left\{\mathrm{3}.\mathrm{4}.\mathrm{6}\right\}\:{Bruno}\:{won}\:{and}\:{the}\:{game}\:{stopping} \\ $$$${Determine}\:{the}\:{probability}\:{that}\:{Daniel}\:{winand}\:{the}\:{probability}\:{that}\:{Bruno}\:{win} \\ $$$$ \\ $$ Answered by MJS_new…
Question Number 10621 by ketto last updated on 20/Feb/17 Answered by mrW1 last updated on 20/Feb/17 $${x}\:{boys}\:{and}\:{y}\:{girls} \\ $$$${x}−{y}=\mathrm{10}\:\:\:\:\:\left({i}\right) \\ $$$${x}=\mathrm{2}\left({y}+\mathrm{1}\right)\:{or} \\ $$$${x}−\mathrm{2}{y}=\mathrm{2}\:\:\:\:\:\left({ii}\right) \\ $$$$…