Question Number 60308 by Tawa1 last updated on 19/May/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 60307 by Tawa1 last updated on 19/May/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 60256 by rahul 19 last updated on 19/May/19 Answered by mr W last updated on 19/May/19 $$\mathrm{2}{x}+\mathrm{4}{y}=\mathrm{2}\left({x}+{y}\right)+\mathrm{2}{y}\leqslant\mathrm{8}+\mathrm{2}{y} \\ $$$$\leqslant\mathrm{8}+\mathrm{2}\left(\mathrm{4}−{x}\right)=\mathrm{16}−\mathrm{2}{x}\leqslant\mathrm{16}\:! \\ $$$$ \\ $$$${z}=\mathrm{2}{x}+\mathrm{5}{y}=\mathrm{2}\left({x}+{y}\right)+\mathrm{3}{y}\leqslant\mathrm{8}+\mathrm{3}{y}…
Question Number 60257 by rahul 19 last updated on 19/May/19 $${The}\:{maximum}\:{value}\:{of}\:{Z}=\mathrm{4}{x}+\mathrm{2}{y} \\ $$$${subject}\:{to}\:{constraints}\:\mathrm{2}{x}+\mathrm{3}{y}\leqslant\mathrm{18}\:, \\ $$$${x}+{y}\geqslant\mathrm{10}\:{and}\:{x},{y}\geqslant\mathrm{0}\:{is}\:? \\ $$ Commented by rahul 19 last updated on 19/May/19…
Question Number 60247 by pete last updated on 19/May/19 $$\mathrm{A}\:\mathrm{uniform}\:\mathrm{pole}\:\mathrm{PQ},\:\mathrm{30}\:\mathrm{m}\:\mathrm{long}\:\mathrm{and}\:\mathrm{of}\:\mathrm{mass} \\ $$$$\mathrm{4}\:\mathrm{kg}\:\mathrm{is}\:\mathrm{carried}\:\mathrm{by}\:\mathrm{a}\:\mathrm{boy}\:\mathrm{at}\:\mathrm{P}\:\mathrm{and}\:\mathrm{a}\:\mathrm{man}\: \\ $$$$\mathrm{8}\:\mathrm{m}\:\mathrm{away}\:\mathrm{from}\:\mathrm{Q}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{from} \\ $$$$\mathrm{P}\:\mathrm{where}\:\mathrm{a}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{20}\:\mathrm{kg}\:\mathrm{should}\:\mathrm{be}\:\mathrm{attached} \\ $$$$\mathrm{so}\:\mathrm{that}\:\mathrm{the}\:\mathrm{man}'\mathrm{s}\:\mathrm{support}\:\mathrm{is}\:\mathrm{twice}\:\mathrm{that} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{boy},\:\mathrm{if}\:\mathrm{the}\:\mathrm{system}\:\mathrm{is}\:\mathrm{in}\:\mathrm{equilibrium} \\ $$$$\left[\mathrm{Take}\:\mathrm{g}=\mathrm{10ms}^{−\mathrm{2}} \right] \\ $$…
Question Number 125781 by Dwaipayan Shikari last updated on 13/Dec/20 $$\int_{\mathrm{0}} ^{\infty} \frac{\sqrt{{x}}}{\mathrm{1}−{x}^{\mathrm{2}} }.\frac{\mathrm{1}}{{e}^{\mathrm{2}\pi{x}} −\mathrm{1}}{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 60227 by Tawa1 last updated on 19/May/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 60219 by ANTARES VY last updated on 19/May/19 $$\boldsymbol{\mathrm{C}}=\frac{\mathrm{2}\boldsymbol{\pi\varepsilon\varepsilon}_{\mathrm{0}} \boldsymbol{\mathrm{L}}}{\boldsymbol{\mathrm{ln}}\left(\frac{\boldsymbol{\mathrm{R}}_{\mathrm{2}} }{\boldsymbol{\mathrm{R}}_{\mathrm{1}} }\right)}. \\ $$$$\boldsymbol{\mathrm{prove}}. \\ $$ Commented by ANTARES VY last updated on…
Question Number 60202 by meme last updated on 18/May/19 $${construct}\:{the}\:{point}\:{M}^{'} =\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\left({z}+\mid{z}\mid\right)}{\mathrm{2}}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 125708 by Dwaipayan Shikari last updated on 13/Dec/20 $$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}−{tanhx}}{\:\sqrt[{\mathrm{5}}]{{tanhx}}}{dx} \\ $$ Answered by mindispower last updated on 13/Dec/20 $${e}^{{x}} ={t} \\…