Question Number 21264 by AKBAR ALU HUSAIN last updated on 18/Sep/17 $$\mathrm{2x}+\mathrm{3y}+\mathrm{2z}=\mathrm{2} \\ $$ Answered by mrW1 last updated on 18/Sep/17 $$\mathrm{assumed}\:\mathrm{x},\mathrm{y},\mathrm{z}\in\mathbb{Z} \\ $$$$\mathrm{2x}+\mathrm{3y}=\mathrm{u} \\…
Question Number 152335 by Olaf_Thorendsen last updated on 27/Aug/21 $$\mathrm{La}\:\mathrm{maison}\:\mathrm{d}'\mathrm{en}\:\mathrm{face} \\ $$$$ \\ $$$$\mathrm{Un}\:\mathrm{facteur}\:\mathrm{sonne}\:\grave {\mathrm{a}}\:\mathrm{une}\:\mathrm{porte}\:\mathrm{pour} \\ $$$$\mathrm{d}\acute {\mathrm{e}livrer}\:\mathrm{un}\:\mathrm{recommand}\acute {\mathrm{e}}.\:\mathrm{Un}\:\mathrm{p}\grave {\mathrm{e}re}\:\mathrm{de} \\ $$$$\mathrm{famille}\:\mathrm{ouvre}.\:\mathrm{Il}\:\mathrm{dit}\:\mathrm{au}\:\mathrm{facteur}\:: \\ $$$$ \\…
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Question Number 152333 by Rankut last updated on 27/Aug/21 $${please}\:{how}\:{can}\:{i}\:{use}\:{this}\:{app}\:{to} \\ $$$${create}\:{tables}\:{for}\:{my}\:{data}\:{please} \\ $$ Commented by Rankut last updated on 28/Aug/21 $${thank}\:{you}\:{sir} \\ $$ Terms…
Question Number 152329 by tabata last updated on 27/Aug/21 $${prove}\:{the}\:{function}\:{is}\:{dont}\:{continuous}\: \\ $$$${uniformly}\:{f}\left({z}\right)={x}^{\mathrm{2}} −{iy}^{\mathrm{2}} \\ $$ Commented by tabata last updated on 27/Aug/21 $$?????? \\ $$…
Question Number 21253 by gopikrishnan005@gmail.com last updated on 17/Sep/17 $${the}\:{hcf}\:{of}\:{two}\:{numbers}\:{is}\:\mathrm{75}\:{and}\:{their}\:{lcm}\:{is}\:\mathrm{3375}\:{if}\:{one}\:{of}\:{the}\:{numbers}\:{is}\:\mathrm{675}\:{find}\:{the}\:{other}\:{number} \\ $$ Answered by Tinkutara last updated on 17/Sep/17 $$\mathrm{Product}\:\mathrm{of}\:\mathrm{two}\:\mathrm{numbers}\:=\:\mathrm{HCF}×\mathrm{LCM} \\ $$$$\mathrm{675}×{x}=\mathrm{75}×\mathrm{3375} \\ $$$${x}=\mathrm{375} \\…
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Question Number 152299 by saly last updated on 27/Aug/21 Commented by saly last updated on 27/Aug/21 $$\:\:\: \\ $$$$\:{Do}\:{you}\:\:{help}\:{me}\:? \\ $$$$ \\ $$ Commented by…
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Question Number 152247 by mathocean1 last updated on 26/Aug/21 $${show}\:{that}\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:\frac{\frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}}\:−\mathrm{1}}{{x}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Answered by mr W last updated on 26/Aug/21 $$\mathrm{ln}\:\left(\mathrm{1}+{x}\right)={x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−\frac{{x}^{\mathrm{4}} }{\mathrm{4}}+……