Question Number 8578 by tawakalitu last updated on 16/Oct/16 Answered by ridwan balatif last updated on 17/Oct/16 Commented by tawakalitu last updated on 17/Oct/16 $$\mathrm{Thanks}\:\mathrm{so}\:\mathrm{much}\:\mathrm{sir}.\:\mathrm{for}\:\mathrm{your}\:\mathrm{help}.\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}…
Question Number 74112 by TawaTawa last updated on 19/Nov/19 Answered by mind is power last updated on 19/Nov/19 $${let}\:{l}={side}\:{of}\:{squar}\Rightarrow{l}^{\mathrm{2}} +\frac{{l}}{\mathrm{2}}\left({x}+{z}+{y}\right)=\frac{\left({z}+{l}+{y}\right)}{\mathrm{2}}\left({l}+{x}\right) \\ $$$$\Leftrightarrow\mathrm{2}{l}^{\mathrm{2}} +{l}\left({x}+{y}+{z}\right)={l}\left({z}+{y}+{l}\right)+{x}\left({z}+{l}+{y}\right) \\ $$$$\Rightarrow\mathrm{2}{l}^{\mathrm{2}}…
Question Number 139645 by otchereabdullai@gmail.com last updated on 30/Apr/21 Commented by mr W last updated on 30/Apr/21 $${OD}={AO}×\mathrm{sin}\:\angle{OAB}=\mathrm{52}×\frac{\mathrm{5}}{\mathrm{13}}=\mathrm{20} \\ $$$${BD}=\sqrt{\mathrm{25}^{\mathrm{2}} −\mathrm{20}^{\mathrm{2}} }=\mathrm{15} \\ $$$${BC}=\mathrm{2}×{BD}=\mathrm{2}×\mathrm{15}=\mathrm{30} \\…
Question Number 139644 by mathdanisur last updated on 30/Apr/21 $${a};{b};{c}\in\mathbb{R},\:\mid{ax}^{\mathrm{5}} +{bx}^{\mathrm{3}} +{cx}\mid\leqslant\mathrm{1},\:\forall\mid{x}\mid\leqslant\mathrm{1} \\ $$$${proof}:\:\mid{a}\mid\leqslant\mathrm{16},\:\mid{b}\mid\leqslant\mathrm{20},\:\mid{c}\mid\leqslant\mathrm{5} \\ $$ Answered by ajfour last updated on 30/Apr/21 $${let}\:\:\mid{x}\mid={r} \\…
Question Number 74111 by FCB last updated on 19/Nov/19 Commented by MJS last updated on 20/Nov/19 $$\mathrm{no}\:“\mathrm{beautiful}''\:\mathrm{solution} \\ $$ Answered by mind is power last…
Question Number 139647 by LUFFY last updated on 30/Apr/21 Commented by LUFFY last updated on 30/Apr/21 $${which}\:{one}\:{is}\:{correct}??????\left({B}\:{or}\:{W}\right) \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 74109 by FCB last updated on 19/Nov/19 Commented by mr W last updated on 19/Nov/19 $${for}\:{a}_{{k}} >\mathrm{0}\:{and}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{k}} =\mathrm{1} \\ $$$${when}\:{a}_{{k}} =\frac{\mathrm{1}}{{n}},…
Question Number 139641 by EnterUsername last updated on 30/Apr/21 $$\mathrm{Let}\:{a},\:{b}\:\mathrm{be}\:\mathrm{non}-\mathrm{zero}\:\mathrm{complex}\:\mathrm{numbers}\:\mathrm{and}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} \:\mathrm{be} \\ $$$$\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:{z}^{\mathrm{2}} +{az}+{b}=\mathrm{0}.\:\mathrm{If}\:\mathrm{there}\:\mathrm{exists} \\ $$$$\lambda\geqslant\mathrm{4}\:\mathrm{such}\:\mathrm{that}\:{a}^{\mathrm{2}} =\lambda{b},\:\mathrm{then}\:\mathrm{the}\:\mathrm{points}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} \:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{origin} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{form}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle} \\…
Question Number 8571 by tawakalitu last updated on 16/Oct/16 Commented by sandy_suhendra last updated on 16/Oct/16 $$\mathrm{Is}\:\mathrm{the}\:\mathrm{potensial}\:\mathrm{of}\:\mathrm{supply}\:\mathrm{1}.\mathrm{0}\:\mathrm{volt}? \\ $$ Commented by tawakalitu last updated on…
Question Number 139642 by ajfour last updated on 30/Apr/21 $${If}\:\:\mathrm{0}<{c}^{\mathrm{2}} <\frac{\mathrm{4}}{\mathrm{27}}\:\:,\:{and} \\ $$$${m}\left\{\mathrm{4}{c}^{\mathrm{2}} −{m}\left(\mathrm{1}+{m}\right)^{\mathrm{2}} \right\} \\ $$$$\:\:\:\:\:=\left\{{m}\left(\mathrm{1}+{m}\right)^{\mathrm{2}} −\mathrm{3}{c}^{\mathrm{2}} \right\}^{\mathrm{2}} \:\:{then} \\ $$$${find}\:{real}\:{values}\:{of}\:{m}\:{in}\:{terms} \\ $$$${of}\:{c}^{\mathrm{2}} .…