Question Number 5914 by sanusihammed last updated on 05/Jun/16 $${A}\:{certain}\:{telephone}\:{company}\:{can}\:{get}\:\mathrm{10000}\:{subscribers}\:\:\:{t}. \\ $$$${yearly}\:{rate}\:{at}\:\mathrm{50}\:{naira}\:{each}\:.\:{it}\:{will}\:{get}\:\mathrm{1000}\:{more}\:{subscribers} \\ $$$${for}\:{each}\:\mathrm{1}.\mathrm{00}\:{naira}\:{decrease}\:{in}\:{the}\:{rate}.\:{what}\:{rate}\:{will}\:{yield}\: \\ $$$${the}\:{maximum}\:{yearly}\:{income}\:{and}\:{what}\:{will}\:{this}\:{income}\:{be}\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 5913 by sanusihammed last updated on 05/Jun/16 $${Suppose}\:{a}\:{drug}\:{manufacturer}\:{can}\:{sell}\:\:{x}\:\:{bottles}\:{of}\:{drug}\:{at} \\ $$$${price}\:{P}\:\:=\:\mathrm{200}\:−\:\mathrm{0}.\mathrm{01}{x}\:{kobo}\:{and}\:{that}\:{it}\:{cost}\:\:{y}\:=\:\mathrm{50}{x}\:+\:\mathrm{20000} \\ $$$${kobo}\:{to}\:{produce}\:{x}\:{bottles}\:{of}\:{drug}.\:{What}\:{is}\:{the}\:{production}\:{level}\: \\ $$$${for}\:{maximum}\:{profit}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 71432 by mezihloic last updated on 15/Oct/19 $${is}\:\mathrm{1}\frac{\mathrm{5}}{\mathrm{3}\:}\:{an}\:{example}\:{of}\:{a}\:{mixed}\:{fraction} \\ $$ Commented by Prithwish sen last updated on 15/Oct/19 $$\boldsymbol{\mathrm{It}}\:\boldsymbol{\mathrm{will}}\:\boldsymbol{\mathrm{be}}\:\mathrm{2}\frac{\mathrm{2}}{\mathrm{3}} \\ $$ Commented by…
Question Number 136950 by BHOOPENDRA last updated on 28/Mar/21 Answered by Olaf last updated on 28/Mar/21 $$\overset{\wedge} {{f}}\:^{{c}} \left(\nu\right)\:=\:\int_{−\infty} ^{+\infty} {f}\left({s}\right)\mathrm{cos}\left(\mathrm{2}\pi\nu{s}\right){ds}\:=\:\mathrm{Re}\overset{\wedge} {{f}}\left(\nu\right) \\ $$$$\mathcal{F}\left(\frac{\mathrm{1}}{{s}}\right)\:=\:−{i}\pi\mathrm{sign}\left(\nu\right) \\…
Question Number 5859 by wanderer last updated on 02/Jun/16 $${what}\:{is}\:{the}\:{intutive}\:{understanding}\:{of}\: \\ $$$${eigenvalues}\:{and}\:{vectors}.{with}\:{practical}\: \\ $$$${examples}. \\ $$ Commented by FilupSmith last updated on 02/Jun/16 $$\mathrm{A}\:\mathrm{vector}\:\mathrm{is}\:\mathrm{a}\:\mathrm{direction}\:\mathrm{and}\:\mathrm{magnitude}. \\…
Question Number 136930 by I want to learn more last updated on 27/Mar/21 Commented by I want to learn more last updated on 27/Mar/21 Commented…
Question Number 5857 by FilupSmith last updated on 02/Jun/16 $$\mathrm{tan}\:\theta\:=\:\frac{{b}}{{a}} \\ $$$$\mathrm{For}\:\mathrm{what}\:\mathrm{range}/\mathrm{values}\:\mathrm{of}\:\theta\:\mathrm{gives} \\ $$$${a}\geqslant{b}? \\ $$ Answered by 123456 last updated on 02/Jun/16 $${a}\geqslant{b} \\…
Question Number 136899 by BHOOPENDRA last updated on 27/Mar/21 Answered by Olaf last updated on 28/Mar/21 $$\mathrm{3}. \\ $$$${f}\left({x}\right)\:=\:{x}^{\mathrm{2}} ,\:−\mathrm{2}\leqslant{x}\leqslant\mathrm{2} \\ $$$${a}_{\mathrm{0}} \left({f}\right)\:=\:\frac{\mathrm{1}}{\mathrm{T}}\int_{−\frac{\mathrm{T}}{\mathrm{2}}} ^{+\frac{\mathrm{T}}{\mathrm{2}}} {f}\left({x}\right){dx}…
Question Number 136892 by Rayan1997 last updated on 27/Mar/21 $$\int\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }{dy} \\ $$$$ \\ $$ Answered by Olaf last updated on 27/Mar/21 $$\mathrm{F}\left({x}\right)\:=\:\int\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}}…
Question Number 136885 by BHOOPENDRA last updated on 27/Mar/21 Answered by bramlexs22 last updated on 27/Mar/21 $$\lambda^{\mathrm{3}} −\left(\mathrm{trace}\:\mathrm{A}\right)\lambda^{\mathrm{2}} +\:\begin{pmatrix}{\mathrm{minor}\:\mathrm{of}\:\mathrm{the}\:\mathrm{terms}}\\{\mathrm{on}\:\mathrm{the}\:\mathrm{leading}\:\mathrm{diag}\:\mathrm{A}\:}\end{pmatrix}\lambda−\mathrm{det}\left(\mathrm{A}\right)=\mathrm{0} \\ $$$$\lambda^{\mathrm{3}} −\mathrm{3}\lambda^{\mathrm{2}} +\left(\begin{vmatrix}{\mathrm{1}\:\:\mathrm{2}}\\{\mathrm{2}\:\:\mathrm{1}}\end{vmatrix}+\begin{vmatrix}{\mathrm{1}\:\:\:\mathrm{0}}\\{\mathrm{1}\:\:\:\mathrm{1}}\end{vmatrix}+\begin{vmatrix}{\:\:\mathrm{1}\:\:\:\:\:\mathrm{2}}\\{−\mathrm{1}\:\:\:\mathrm{1}}\end{vmatrix}\right)\lambda−\mathrm{3}\:=\mathrm{0} \\ $$$$\lambda^{\mathrm{3}}…