Question Number 186265 by mnjuly1970 last updated on 02/Feb/23 $$ \\ $$$$\:\:\:{f}\left({x}\right)=\:\sqrt{\:{x}\:−{a}}\:\:+\:\sqrt{\mathrm{3}{a}\:−{x}}\:\:\:{with}\:\left(\:{a}>\mathrm{0}\right) \\ $$$$\:\:\:\:{is}\:\:{given}\:.{If}\:\:,\:\:{f}_{\:{max}} \:.\:{f}_{\:{min}} \:=\:\sqrt{\mathrm{32}} \\ $$$$\:\:\:\:\:{find}\:\:,\:\:\:\:\:\:\:''\:\:\:{a}\:\:''\:\:=\:? \\ $$ Answered by mahdipoor last updated…
Question Number 55195 by behi83417@gmail.com last updated on 19/Feb/19 $$\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{ax}}+\sqrt{\mathrm{2}}\boldsymbol{\mathrm{b}}=\mathrm{0},\boldsymbol{\mathrm{has}}\:\mathrm{2}\:\boldsymbol{\mathrm{roots}}:\boldsymbol{\mathrm{c}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{d}},\boldsymbol{\mathrm{also}} \\ $$$$\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{cx}}+\sqrt{\mathrm{2}}\boldsymbol{\mathrm{d}}=\mathrm{0},\boldsymbol{\mathrm{has}}\:\mathrm{2}\:\boldsymbol{\mathrm{roots}}:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{b}}.\boldsymbol{\mathrm{such}} \\ $$$$\boldsymbol{\mathrm{that}}:\boldsymbol{\mathrm{a}},\:\boldsymbol{\mathrm{b}},\:\boldsymbol{\mathrm{c}},\:\boldsymbol{\mathrm{d}},\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{defferent}}\:\boldsymbol{\mathrm{non}}\:\boldsymbol{\mathrm{zero}}\: \\ $$$$\boldsymbol{\mathrm{numbers}}. \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{possible}}\:\boldsymbol{\mathrm{value}}\left(\boldsymbol{\mathrm{s}}\right)\:\boldsymbol{\mathrm{for}}:\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} +\boldsymbol{\mathrm{c}}^{\mathrm{2}} +\boldsymbol{\mathrm{d}}^{\mathrm{2}} . \\…
Question Number 186256 by Shrinava last updated on 02/Feb/23 $$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathbb{R}\:\left(\mathrm{m}\:,\:\mathrm{n}\right)\:\leqslant\:\mathbb{C}_{\boldsymbol{\mathrm{m}}+\boldsymbol{\mathrm{n}}} ^{\boldsymbol{\mathrm{m}}} \\ $$$$ \\ $$ Commented by mr W last updated on 02/Feb/23…
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Question Number 186253 by ajfour last updated on 02/Feb/23 Commented by ajfour last updated on 02/Feb/23 $${The}\:{curve}\:{is}\:{a}\:\:{cubic}\:\left(\:{y}={x}^{\mathrm{3}} −{x}\right), \\ $$$$\:{and}\:{that}\:\:\mathrm{0}<{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\:\centerdot \\ $$$${Let}\:{circle}\:{not}\:{necessarily}\:{be}\: \\ $$$${tangent}\:{to}\:{cubic}\:{curve}\:{at}\:\left({q},\:{c}\right). \\…
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Question Number 120711 by TITA last updated on 02/Nov/20 $${show}\:{that}\:\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}…=\frac{−\mathrm{1}}{\mathrm{8}} \\ $$ Answered by Dwaipayan Shikari last updated on 02/Nov/20 $${S}=\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+… \\ $$$${S}=\:\mathrm{1}+\left(\mathrm{2}+\mathrm{3}+\mathrm{4}\right)+\left(\mathrm{5}+\mathrm{6}+\mathrm{7}\right)+\left(\mathrm{8}+\mathrm{9}+\mathrm{10}\right)+… \\ $$$${S}=\mathrm{1}+\mathrm{9}+\mathrm{18}+\mathrm{27}+\mathrm{36}+..…
Question Number 186241 by a.lgnaoui last updated on 02/Feb/23 $${x}+\frac{\mathrm{1}}{{x}}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${x}? \\ $$ Commented by MJS_new last updated on 02/Feb/23 $$\mathrm{are}\:\mathrm{you}\:\mathrm{unable}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{a}\:\mathrm{simple}\:\mathrm{quadratic}? \\ $$ Commented…
Question Number 186232 by mustafazaheen last updated on 02/Feb/23 $$\mathrm{1}+\frac{{x}}{\mathrm{1}+\frac{{x}}{\mathrm{1}+\frac{{x}}{\centerdot_{\centerdot_{\centerdot} } }}}=\mathrm{5}\:\:\:\:\:\:\:\:\:{x}=? \\ $$ Answered by gungun last updated on 02/Feb/23 $$\mathrm{1}+\frac{{x}}{\mathrm{5}}=\mathrm{5} \\ $$$${x}=\mathrm{20} \\…
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