Question Number 191310 by Mingma last updated on 22/Apr/23 Answered by HeferH last updated on 23/Apr/23 Commented by HeferH last updated on 23/Apr/23 $$\mathrm{say}:\:\mathrm{AC}\:=\:\mathrm{b},\:\mathrm{OC}\:=\:\mathrm{a},\:\mathrm{AB}\:=\:\mathrm{m} \\…
Question Number 191305 by Mingma last updated on 22/Apr/23 Commented by mr W last updated on 22/Apr/23 $${there}\:{is}\:{no}\:{unique}\:{solution}.\:{but}\:{there} \\ $$$${is}\:{a}\:{r}_{{min}} . \\ $$ Answered by…
Question Number 191304 by Mingma last updated on 22/Apr/23 Answered by cortano12 last updated on 22/Apr/23 $$\:\mathrm{x}=\mathrm{t}^{\mathrm{2}} −\mathrm{2}\: \\ $$$$\:\left(\mathrm{t}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{3}} −\mathrm{3}\left(\mathrm{t}^{\mathrm{2}} −\mathrm{2}\right)−\mathrm{t}=\mathrm{0} \\ $$$$\:\left(\mathrm{t}−\mathrm{2}\right)\left(\mathrm{2t}+\mathrm{1}−\sqrt{\mathrm{5}}\right)\left(\mathrm{2t}+\mathrm{1}+\sqrt{\mathrm{5}}\right)\left(\frac{\mathrm{1}}{\mathrm{4}}\mathrm{t}^{\mathrm{3}}…
Question Number 60234 by aliesam last updated on 19/May/19 Answered by ajfour last updated on 19/May/19 $$\mathrm{z}=\mathrm{x}−\mathrm{c} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 191307 by cortano12 last updated on 22/Apr/23 Answered by HeferH last updated on 22/Apr/23 $$\mathrm{60}°\:+\:\mathrm{90}°\:+\:?\:=\:\mathrm{180}° \\ $$$$\:?\:=\:\mathrm{30}° \\ $$ Commented by AROUNAMoussa last…
Question Number 125768 by Don08q last updated on 13/Dec/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{2}×\mathrm{2}\:\mathrm{matrix}\:{A}\:\mathrm{such}\:\mathrm{that} \\ $$$$\:\:\begin{pmatrix}{−\mathrm{4}}&{\:\:\:\:\:\mathrm{0}}\\{\:\:\:\:\mathrm{0}}&{−\mathrm{4}}\end{pmatrix}\:\:\:−\:\:{A}\:\:=\:\:{A}^{−\mathrm{1}} \begin{pmatrix}{\mathrm{3}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{3}}\end{pmatrix} \\ $$ Answered by 676597498 last updated on 13/Dec/20 $${A}=\begin{pmatrix}{{a}}&{{b}}\\{{c}}&{{d}}\end{pmatrix}\: \\ $$$${det}\left({A}\right)={ad}−{cd}…
Question Number 191301 by Mingma last updated on 22/Apr/23 Answered by a.lgnaoui last updated on 23/Apr/23 $$\mathrm{a};\mathrm{b};\mathrm{c}>\mathrm{0}\:\:\:\Rightarrow\left(\mathrm{a}+\mathrm{b}+\mathrm{c}>\mathrm{0}\:\:\:\mathrm{et}\:\mathrm{abc}>\mathrm{0}\right) \\ $$$$\mathrm{poxons}\:\:\mathrm{x}=\mathrm{abc}\:\:\:\:\:\mathrm{y}=\mathrm{a}+\mathrm{b}+\mathrm{c}\:\:\:\mathrm{xy}=\mathrm{1} \\ $$$$\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}=\boldsymbol{\mathrm{y}}−\boldsymbol{\mathrm{c}}\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{c}}=\boldsymbol{\mathrm{y}}−\boldsymbol{\mathrm{b}} \\ $$$$\:\:\left(\boldsymbol{\mathrm{y}}−\boldsymbol{\mathrm{c}}\right)\left(\boldsymbol{\mathrm{y}}−\boldsymbol{\mathrm{b}}\right)\:=\boldsymbol{\mathrm{y}}^{\mathrm{2}} −\left(\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\right)\boldsymbol{\mathrm{y}}+\boldsymbol{\mathrm{bc}} \\…
Question Number 191300 by Mingma last updated on 22/Apr/23 Answered by Tinku Tara last updated on 22/Apr/23 $$\left({a}^{\mathrm{2}} −\mathrm{3}\right)\left({a}^{\mathrm{2}} −\mathrm{2}\right)\left({a}^{\mathrm{2}} −\mathrm{1}\right)\left({a}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} +\mathrm{1}\right)\left({a}^{\mathrm{2}} +\mathrm{2}\right)\left({a}^{\mathrm{2}} +\mathrm{3}\right)=\mathrm{8}!…
Question Number 191303 by Mingma last updated on 22/Apr/23 Answered by mahdipoor last updated on 22/Apr/23 $$\int{f}\left({x}\right)=\int{xf}^{'} \left({x}\right)−\sqrt{\mathrm{2}{x}−{x}^{\mathrm{2}} }=\left[{xf}\left({x}\right)−\int{f}\left({x}\right)\right]−\int\sqrt{\mathrm{2}{x}−{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\:\mathrm{1}} {f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left[{xf}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 191302 by Mingma last updated on 22/Apr/23 Answered by a.lgnaoui last updated on 22/Apr/23 $$\boldsymbol{\mathrm{S}}_{\mathrm{1}} =\boldsymbol{\mathrm{S}}\left(\boldsymbol{\mathrm{OAB}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{AB}×\mathrm{h}_{\mathrm{1}} \:\:\:\:=\frac{\mathrm{CD}×\mathrm{h}_{\mathrm{1}} }{\mathrm{4}}\:\:\:\left(\mathrm{h}_{\mathrm{1}} =\mathrm{OH}_{\mathrm{1}} \right) \\ $$$$\:\:\:\frac{\mathrm{CD}×\mathrm{h}_{\mathrm{1}} }{\mathrm{4}}=\mathrm{8}\:\:\:\Rightarrow\boldsymbol{\mathrm{CD}}.\boldsymbol{\mathrm{h}}_{\mathrm{1}}…