Question Number 158396 by zakirullah last updated on 03/Nov/21 $${Any}\:{proof}\:{or}\:{Idea}\:{about}; \\ $$$$\frac{\mathrm{4}}{\mathrm{2}}\boldsymbol{\div}\frac{\mathrm{16}}{\mathrm{3}}\:=\:\frac{\mathrm{4}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{16}} \\ $$ Answered by MJS_new last updated on 03/Nov/21 $$\mathrm{it}'\mathrm{s}\:\mathrm{just}\:\mathrm{the}\:\mathrm{definition} \\ $$$$\frac{{a}}{{b}}\boldsymbol{\div}\frac{{c}}{{d}}=\frac{{a}}{{b}}×\frac{{d}}{{c}} \\…
Question Number 158379 by Khalmohmmad last updated on 03/Nov/21 $${F}\left({x}+\mathrm{1}\right)−{F}\left({x}−\mathrm{1}\right)=\mathrm{6} \\ $$$${F}\left(\mathrm{0}\right)=\mathrm{4} \\ $$$${F}\left(\mathrm{3}\right)=? \\ $$ Commented by mr W last updated on 03/Nov/21 $${you}\:{can}\:{get}\:{F}\left({x}+\mathrm{2}\right)={F}\left({x}\right)+\mathrm{6}.…
Question Number 158341 by aliibrahim1 last updated on 02/Nov/21 Answered by puissant last updated on 03/Nov/21 $${That}\:{is}\:\varepsilon>\mathrm{0},\:{let}'{s}\:{seek}\:\alpha>\mathrm{0}\:/\:\forall{x}\in\mathbb{R}\:,\:\mathrm{0}<\mid{x}−\mathrm{2}\mid<\alpha\:\Rightarrow\:\mid{x}^{\mathrm{2}} −\mathrm{4}\mid<\varepsilon \\ $$$${for}\:{x}\in\mathbb{R}\:,\:\mid{x}^{\mathrm{2}} −\mathrm{4}\mid\:=\:\mid{x}−\mathrm{2}\mid.\mid{x}+\mathrm{2}\mid\:{we}\:{have}\:\mid{x}−\mathrm{2}\mid<\mathrm{2}\:\rightarrow\:\mathrm{0}<{x}+\mathrm{2}<\mathrm{6} \\ $$$$\Rightarrow\:\mid{x}^{\mathrm{2}} −\mathrm{4}\mid<\:\mathrm{6}\mid{x}−\mathrm{2}\mid\:;\:\mid{x}^{\mathrm{2}} −\mathrm{4}\mid<\varepsilon\:\rightarrow\:\mathrm{6}\mid{x}−\mathrm{2}\mid<\varepsilon\:\rightarrow\:\mid{x}−\mathrm{2}\mid<\frac{\varepsilon}{\mathrm{6}}…
Question Number 158325 by SANOGO last updated on 02/Nov/21 Answered by puissant last updated on 03/Nov/21 $${X}=\left({x},{y},{z}\right)\in\:{ker}\left({A}\right)\:\Leftrightarrow\:{AX}=\mathrm{0} \\ $$$$\Rightarrow\:\left\{\mathrm{4}{x}−\mathrm{2}{y}−\mathrm{2}{z}=\mathrm{0}\:;\:{x}−{z}=\mathrm{0}\:;\:\mathrm{3}{x}−\mathrm{2}{y}−{z}=\mathrm{0}\right. \\ $$$${x}={z}\:\Rightarrow\:\mathrm{2}{x}−\mathrm{2}{y}=\mathrm{0}\:\Rightarrow\:{x}={y}={z} \\ $$$${X}=\left({x},{y},{z}\right)=\left({x},{x},{x}\right)={x}\left(\mathrm{1},\mathrm{1},\mathrm{1}\right) \\ $$$${ker}\left({A}\right)=<\left(\mathrm{1},\mathrm{1},\mathrm{1}\right)>…
Question Number 158330 by mkam last updated on 02/Nov/21 Commented by mkam last updated on 02/Nov/21 $$????? \\ $$ Terms of Service Privacy Policy Contact:…
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Question Number 158301 by LEKOUMA last updated on 02/Nov/21 $${proven}\:{that}\: \\ $$$$\mathrm{1}^{\mathrm{0}} =\mathrm{1}\:{et}\:{que}\:\mathrm{0}!=\mathrm{1} \\ $$ Commented by Rasheed.Sindhi last updated on 02/Nov/21 $$\mathrm{1}=\frac{\mathrm{1}}{\mathrm{1}}=\frac{\mathrm{1}^{\mathrm{1}} }{\mathrm{1}^{\mathrm{1}} }=\mathrm{1}^{\mathrm{1}−\mathrm{1}}…
Question Number 27224 by mrW1 last updated on 03/Jan/18 Commented by mrW1 last updated on 03/Jan/18 $${The}\:{friction}\:{coefficient}\:{betweem}\:{the} \\ $$$${blocks}\:{is}\:\mu. \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{tension}\:{in}\:{the}\:{thread}. \\ $$$$\left.\mathrm{2}\right)\:{what}\:{is}\:{the}\:{inclination}\:{angle}\:{such} \\ $$$${that}\:{the}\:{tension}\:{in}\:{thread}\:{is}\:{maximum}?…
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Question Number 158285 by saly last updated on 02/Nov/21 Commented by saly last updated on 02/Nov/21 $$\:\:\:{Help}\:{me}? \\ $$ Terms of Service Privacy Policy Contact:…