Question Number 12620 by @ANTARES_VY last updated on 27/Apr/17 $$\boldsymbol{\alpha\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{bx}}+\boldsymbol{\mathrm{c}}. \\ $$$$\boldsymbol{\mathrm{y}}\left(\mathrm{1}\right)=\boldsymbol{\alpha}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}=\mathrm{3}\:\:\boldsymbol{\mathrm{max}} \\ $$$$\boldsymbol{\mathrm{y}}\left(−\mathrm{1}\right)=\boldsymbol{\alpha}−\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}=\mathrm{0}\:\:\boldsymbol{\mathrm{min}}. \\ $$$$\boldsymbol{\mathrm{y}}\left(\mathrm{5}\right)=\mathrm{25}\boldsymbol{\alpha}+\mathrm{5}\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}=????? \\ $$ Commented by 9(/’D-B 3HE1H last updated…
Question Number 12619 by @ANTARES_VY last updated on 27/Apr/17 $$\boldsymbol{\alpha\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{bx}}+\boldsymbol{\mathrm{c}}. \\ $$$$\boldsymbol{\mathrm{y}}\left(\mathrm{8}\right)=\mathrm{64}\boldsymbol{\alpha}+\mathrm{8}\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}=\mathrm{0}.\:\:\boldsymbol{\mathrm{max}} \\ $$$$\boldsymbol{\mathrm{y}}\left(\mathrm{6}\right)=\mathrm{36}\boldsymbol{\alpha}+\mathrm{6}\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}=−\mathrm{12}\:\:\boldsymbol{\mathrm{min}} \\ $$$$\sqrt{\boldsymbol{\alpha}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}}=??? \\ $$ Commented by prakash jain last updated…
Question Number 143688 by mohammad17 last updated on 17/Jun/21 Commented by mohammad17 last updated on 17/Jun/21 $${help}\:{me}\:{sir} \\ $$ Commented by mohammad17 last updated on…
Question Number 78153 by ajfour last updated on 14/Jan/20 Commented by ajfour last updated on 14/Jan/20 $${Find}\:{h}\:{in}\:{terms}\:{of}\:{c}. \\ $$ Answered by mr W last updated…
Question Number 143684 by Huy last updated on 17/Jun/21 $$\mathrm{x}^{\mathrm{3}} +\mathrm{x}−\mathrm{1}=^{\mathrm{3}} \sqrt{\mathrm{2x}^{\mathrm{3}} +\mathrm{11}}+\sqrt{\mathrm{5x}^{\mathrm{2}} +\mathrm{16}} \\ $$$$\mathrm{Find}\:\mathrm{x}\in\mathbb{R} \\ $$ Answered by TheHoneyCat last updated on 17/Jun/21…
Question Number 12613 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 26/Apr/17 Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 28/Apr/17 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 12612 by frank ntulah last updated on 26/Apr/17 $$\mathrm{Use}\:\mathrm{Newtown}\:\mathrm{Raphson}\:\mathrm{method}\:\mathrm{to}\:\mathrm{find}\:\mathrm{aproximate} \\ $$$$\mathrm{value}\:\mathrm{of}\:\:\boldsymbol{{X}}=\sqrt{\left(\frac{\mathrm{7}}{\boldsymbol{{x}}+\mathrm{1}}\right)}\:,\mathrm{starting}\:\mathrm{with}\:\boldsymbol{{x}}_{\mathrm{0}} =\mathrm{2}. \\ $$$$\boldsymbol{{perform}}\:\mathrm{4}\:\boldsymbol{{iteration}}\:\boldsymbol{{and}}\:\boldsymbol{{all}}\:\boldsymbol{{iteration}}\: \\ $$$$\boldsymbol{{should}}\:\boldsymbol{{be}}\:\boldsymbol{{presented}}\:\boldsymbol{{in}}\:\mathrm{4}\:\boldsymbol{{decimal}}\:\boldsymbol{{places}} \\ $$ Answered by mrW1 last updated…
Question Number 12611 by tawa last updated on 26/Apr/17 $$\mathrm{A}\:\mathrm{drum}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{100g}\:\mathrm{is}\:\mathrm{rolled}\:\mathrm{into}\:\mathrm{the}\:\mathrm{deck}\:\mathrm{of}\:\mathrm{a}\:\mathrm{lorry}\:\mathrm{1}.\mathrm{5m}\:\mathrm{above}\:\mathrm{a}\: \\ $$$$\mathrm{horizontal}\:\mathrm{floor}\:\mathrm{using}\:\mathrm{a}\:\mathrm{plank}\:\mathrm{4m}\:\mathrm{long}.\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{workdone}\:\mathrm{against} \\ $$$$\mathrm{gravity}\:\mathrm{during}\:\mathrm{the}\:\mathrm{process}.\:\left(\mathrm{g}\:=\:\mathrm{10m}/\mathrm{s}^{\mathrm{2}} \right). \\ $$ Answered by mrW1 last updated on 27/Apr/17 $${W}={mgh}=\mathrm{0}.\mathrm{1}×\mathrm{10}×\mathrm{1}.\mathrm{5}=\mathrm{1}.\mathrm{5}\:{J}…
Question Number 143680 by mathlove last updated on 17/Jun/21 $$\mathrm{tan}\:\mathrm{76}=\mathrm{4} \\ $$$$\mathrm{sin}\:^{\mathrm{2}} \mathrm{14}=? \\ $$ Commented by mr W last updated on 17/Jun/21 $$\mathrm{tan}\:\mathrm{14}=\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{76}}=\frac{\mathrm{1}}{{a}} \\…
Question Number 78147 by aliesam last updated on 14/Jan/20 Terms of Service Privacy Policy Contact: info@tinkutara.com