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z-3-7-6i-z-2-3-1-9i-z-2-7-9i-0-Resolve-the-equation-E-sachet-that-the-stop-image-one-any-solution-behoves-thru-the-righ-t-equation-y-x-

Question Number 158417 by LEKOUMA last updated on 03/Nov/21 $${z}^{\mathrm{3}} −\left(\mathrm{7}+\mathrm{6}{i}\right){z}^{\mathrm{2}} +\mathrm{3}\left(\mathrm{1}+\mathrm{9}{i}\right){z}+\mathrm{2}\left(\mathrm{7}−\mathrm{9}{i}\right)=\mathrm{0} \\ $$$${Resolve}\:{the}\:{equation}\:\left({E}\right)\:{sachet}\:{that}\: \\ $$$${the}\:{stop}\:{image}\:\:{one}\:{any}\:{solution}\:{behoves} \\ $$$${thru}\:{the}\:{righ}\:{t}\:{equation}\:{y}={x} \\ $$ Answered by MJS_new last updated…

Exercise-D-y-x-1-directed-by-u-1-1-his-normal-vector-is-v-2-2-1-Determinate-the-equation-of-such-as-t-v-S-D-S-2-Determinate-the-nature-and-caracteristics-of-th

Question Number 92862 by mathocean1 last updated on 09/May/20 $$\mathrm{Exercise} \\ $$$$\: \\ $$$$\left(\mathrm{D}\right):\mathrm{y}=\mathrm{x}−\mathrm{1}\:\:\mathrm{directed}\:\mathrm{by}\:\:\:\overset{\rightarrow} {\mathrm{u}}\left(\mathrm{1};\mathrm{1}\right). \\ $$$$\mathrm{his}\:\mathrm{normal}\:\mathrm{vector}\:\mathrm{is}\:\overset{\rightarrow} {\mathrm{v}}\left(\mathrm{2};−\mathrm{2}\right). \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Determinate}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\:\left(\Delta\right)\: \\ $$$$\mathrm{such}\:\mathrm{as}\:\mathrm{t}_{\overset{\rightarrow} {\mathrm{v}}} =\mathrm{S}_{\left(\mathrm{D}\right)} °\mathrm{S}_{\left(\Delta\right)}…

Any-proof-or-Idea-about-4-2-16-3-4-2-3-16-

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}} \\…

F-x-1-F-x-1-6-F-0-4-F-3-

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-158341

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-158325

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)>…