Question Number 74109 by FCB last updated on 19/Nov/19 Commented by mr W last updated on 19/Nov/19 $${for}\:{a}_{{k}} >\mathrm{0}\:{and}\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{k}} =\mathrm{1} \\ $$$${when}\:{a}_{{k}} =\frac{\mathrm{1}}{{n}},…
Question Number 139642 by ajfour last updated on 30/Apr/21 $${If}\:\:\mathrm{0}<{c}^{\mathrm{2}} <\frac{\mathrm{4}}{\mathrm{27}}\:\:,\:{and} \\ $$$${m}\left\{\mathrm{4}{c}^{\mathrm{2}} −{m}\left(\mathrm{1}+{m}\right)^{\mathrm{2}} \right\} \\ $$$$\:\:\:\:\:=\left\{{m}\left(\mathrm{1}+{m}\right)^{\mathrm{2}} −\mathrm{3}{c}^{\mathrm{2}} \right\}^{\mathrm{2}} \:\:{then} \\ $$$${find}\:{real}\:{values}\:{of}\:{m}\:{in}\:{terms} \\ $$$${of}\:{c}^{\mathrm{2}} .…
Question Number 139641 by EnterUsername last updated on 30/Apr/21 $$\mathrm{Let}\:{a},\:{b}\:\mathrm{be}\:\mathrm{non}-\mathrm{zero}\:\mathrm{complex}\:\mathrm{numbers}\:\mathrm{and}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} \:\mathrm{be} \\ $$$$\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:{z}^{\mathrm{2}} +{az}+{b}=\mathrm{0}.\:\mathrm{If}\:\mathrm{there}\:\mathrm{exists} \\ $$$$\lambda\geqslant\mathrm{4}\:\mathrm{such}\:\mathrm{that}\:{a}^{\mathrm{2}} =\lambda{b},\:\mathrm{then}\:\mathrm{the}\:\mathrm{points}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} \:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{origin} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{form}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle} \\…
Question Number 8560 by FilupSmith last updated on 16/Oct/16 $$\frac{{n}}{\mathrm{2}^{{k}} }={m},\:\:\:\:\:\:\:{n},{m},{k}\in\mathbb{Z} \\ $$$$\mathrm{For}\:\mathrm{any}\:\mathrm{integer}\:{n},\:\mathrm{how}\:\mathrm{many}\:\mathrm{times} \\ $$$$\mathrm{can}\:\mathrm{you}\:\mathrm{divide}\:\mathrm{it}\:\mathrm{by}\:\mathrm{2},\:\mathrm{such}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{result}\:\mathrm{is}\:\mathrm{always}\:\mathrm{a}\:\mathrm{whole}\:\mathrm{number}. \\ $$$$\: \\ $$$$\mathrm{e}.\mathrm{g}.\:\mathrm{120} \\ $$$$\mathrm{160}/\mathrm{2}=\mathrm{80} \\ $$$$\mathrm{80}/\mathrm{2}=\mathrm{40}…
Question Number 139600 by rexford last updated on 29/Apr/21 Answered by mr W last updated on 29/Apr/21 $$\mathrm{2}^{{x}} =\mathrm{4}{x} \\ $$$${e}^{{x}\mathrm{ln}\:\mathrm{2}} =\mathrm{4}{x} \\ $$$${xe}^{−{x}\mathrm{ln}\:\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}…
Question Number 74063 by TawaTawa last updated on 18/Nov/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 139590 by EnterUsername last updated on 29/Apr/21 $$\mathrm{If}\:{a},\:{b}\:\mathrm{and}\:{c}\:\mathrm{are}\:\mathrm{integers}\:\mathrm{not}\:\mathrm{all}\:\mathrm{simultaneously}\:\mathrm{equal} \\ $$$$\mathrm{and}\:{w}\neq\mathrm{1}\:\mathrm{is}\:\mathrm{a}\:\mathrm{cube}\:\mathrm{root}\:\mathrm{of}\:\mathrm{unity},\:\mathrm{then}\:\mathrm{the}\:\mathrm{minimum} \\ $$$$\mathrm{value}\:\mathrm{of}\:\mid{a}+{bw}+{cw}^{\mathrm{2}} \mid\:\mathrm{is}\: \\ $$$$\left(\mathrm{A}\right)\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:\sqrt{\mathrm{3}}/\mathrm{2}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{1}/\mathrm{2} \\ $$ Answered by MJS_new last updated on…
Question Number 139587 by mathlove last updated on 29/Apr/21 Answered by qaz last updated on 29/Apr/21 $${f}\left(\mathrm{2}\right)=\int_{\mathrm{1}} ^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{3}{xdx}+{f}\left(\mathrm{1}\right)=\left(\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{2}}{x}^{\mathrm{2}} \right)\mid_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{5}=\frac{\mathrm{71}}{\mathrm{6}} \\…
Question Number 139577 by bemath last updated on 29/Apr/21 $$\:\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{number}\frac{\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}+\frac{\left(\mathrm{y}−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}}+\frac{\left(\mathrm{z}−\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{6}}+\mathrm{3}=\mid\mathrm{x}−\mathrm{1}\mid+\mid\mathrm{y}−\mathrm{2}\mid+\mid\mathrm{z}−\mathrm{3}\mid\: \\ $$ Answered by mr W last updated on 29/Apr/21 $${a}=\mid{x}−\mathrm{1}\mid\geqslant\mathrm{0},\:{b}=\mid{y}−\mathrm{2}\mid\geqslant\mathrm{0},\:{c}=\mid{z}−\mathrm{3}\mid\geqslant\mathrm{0} \\…
Question Number 74024 by ajfour last updated on 18/Nov/19 $$\begin{cases}{{h}^{\mathrm{2}} +{y}^{\mathrm{2}} +\left({k}−{z}\right)^{\mathrm{2}} ={s}^{\mathrm{2}} }\\{{a}^{\mathrm{2}} +\left({b}−{y}\right)^{\mathrm{2}} +{z}^{\mathrm{2}} ={s}^{\mathrm{2}} }\\{{ah}+{y}\left({y}−{b}\right)+{z}\left({z}−{k}\right)=\mathrm{0}}\\{\frac{{h}+{a}}{\mathrm{2}}+{yz}−\left({b}−{y}\right)\left({k}−{z}\right)=\mathrm{1}}\\{{b}+{a}\left({k}−{z}\right)+{hz}=\mathrm{1}}\\{{k}+{h}\left({b}−{y}\right)+{ay}=\mathrm{1}}\end{cases} \\ $$$${Find}\:\:{s}_{{min}} \:{or}\:{at}\:{least}\:{express} \\ $$$$\:{s}={f}\left({y}\right)\:{or}\:{g}\left({z}\right). \\ $$…