Question Number 95416 by john santu last updated on 25/May/20 $$\mathrm{It}\:\mathrm{takes}\:\mathrm{12}\:\mathrm{hours}\:\mathrm{to}\:\mathrm{fill}\:\mathrm{a}\:\mathrm{swimming}\: \\ $$$$\mathrm{pool}\:\mathrm{using}\:\mathrm{2}\:\mathrm{pipes}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{larger}\: \\ $$$$\mathrm{pipe}\:\mathrm{used}\:,\:\mathrm{for}\:\mathrm{4}\:\mathrm{hours}\:\mathrm{and}\:\mathrm{the}\: \\ $$$$\mathrm{small}\:\mathrm{pipe}\:\mathrm{for}\:\mathrm{9}\:\mathrm{hours},\:\mathrm{only}\:\mathrm{half} \\ $$$$\mathrm{the}\:\mathrm{pool}\:\mathrm{is}\:\mathrm{filled}.\:\mathrm{How}\:\mathrm{long}\:\mathrm{would}\: \\ $$$$\mathrm{it}\:\mathrm{take}\:\mathrm{for}\:\mathrm{each}\:\mathrm{pipe}\:\mathrm{alone}\:\mathrm{to}\: \\ $$$$\mathrm{fill}\:\mathrm{the}\:\mathrm{pool}? \\ $$…
Question Number 160950 by geron last updated on 09/Dec/21 Answered by mindispower last updated on 09/Dec/21 $${y}=\frac{\mathrm{2}{x}}{\mathrm{1}−{x}^{\mathrm{2}} },{x}=\frac{\mathrm{2}{z}}{\mathrm{1}−{z}^{\mathrm{2}} },{z}=\frac{\mathrm{2}{y}}{\mathrm{1}−{y}^{\mathrm{2}} } \\ $$$${just}\:{x}={tg}\left({a}\right),{y}={tg}\left({b}\right),{z}={tg}\left({c}\right) \\ $$$${withe}\:{tg}\left(\mathrm{2}{t}\right)=\frac{\mathrm{2}{tg}\left({t}\right)}{\mathrm{1}−{tg}^{\mathrm{2}} \left({t}\right)}…
Question Number 95394 by I want to learn more last updated on 24/May/20 $$\mathrm{Solve}:\:\:\:\mathrm{x}\:\:+\:\:\mathrm{y}\:\:=\:\:\mathrm{3}\:\:\:\:\:\:….\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{y}} \:\:+\:\:\mathrm{y}^{\mathrm{x}} \:\:=\:\:\mathrm{6}\:\:\:\:…..\:\:\left(\mathrm{ii}\right) \\ $$ Commented by mr W last…
Question Number 160921 by HongKing last updated on 09/Dec/21 $$\mathrm{Find}: \\ $$$$\boldsymbol{\Omega}\:=\underset{\:\mathrm{1}} {\overset{\:\mathrm{21}} {\int}}\:\frac{\mathrm{dx}}{\boldsymbol{\mathrm{e}}^{\left[\mathrm{2}\boldsymbol{\mathrm{x}}\:+\:\frac{\mathrm{1}}{\mathrm{4}}\right]} }\:\:\:;\:\:\:\left[\ast\right]-\mathrm{GIF} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 160923 by cortano last updated on 09/Dec/21 $$\:\:\begin{cases}{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2021}} =\:\mathrm{z}}\\{\left(\mathrm{x}+\mathrm{z}\right)^{\mathrm{2021}} \:=\:\mathrm{y}}\\{\left(\mathrm{y}+\mathrm{z}\right)^{\mathrm{2021}} \:=\:\mathrm{x}}\end{cases} \\ $$$$\:\:\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=… \\ $$ Commented by Rasheed.Sindhi last updated on 09/Dec/21 $$\:\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)…
Question Number 160922 by KONE last updated on 09/Dec/21 $${monter}\:{que}\:\left(\mathrm{3}+\sqrt{\mathrm{5}}\right)^{{n}} +\left(\mathrm{3}−\sqrt{\mathrm{5}}\right)^{{n}\:} \\ $$$${est}\:{divisible}\:{par}\:\mathrm{2}^{{n}} \\ $$$${beson}\:{d}'{aide}\:{svp} \\ $$ Answered by mindispower last updated on 09/Dec/21 $${Starte}\:{withe}…
Question Number 160910 by geron last updated on 09/Dec/21 Answered by blackmamba last updated on 09/Dec/21 $$\:\:\:\:\:\:\sqrt[{\mathrm{3}}]{{a}^{\mathrm{2}} }+\sqrt[{\mathrm{3}}]{{b}^{\mathrm{2}} }+\mathrm{1}+\mathrm{2}\sqrt[{\mathrm{3}}]{{ab}}−\mathrm{2}\sqrt[{\mathrm{3}}]{{a}}−\mathrm{2}\sqrt[{\mathrm{3}}]{{b}}\:=\:\mathrm{49}+\mathrm{20}\sqrt[{\mathrm{3}}]{\mathrm{6}} \\ $$$$ \\ $$ Commented by…
Question Number 29837 by abdo imad last updated on 12/Feb/18 $${let}\:{give}\:\:{T}_{{n}} \left({x}\right)={cos}\left({n}\:{arcosx}\right)\:{with}\:{x}\in\left[−\mathrm{1},\mathrm{1}\right] \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:{T}_{{n}} \:{is}\:{a}\:{polynomial}\:{and}\:{T}_{{n}} \in{Z}\left[{x}\right] \\ $$$$\left.\mathrm{2}\right){calculate}\:{T}_{\mathrm{1}} ,\:{T}_{\mathrm{2}} ,\:{T}_{\mathrm{3}} ,{and}\:{T}_{\mathrm{4}} \\ $$$$\left.\mathrm{3}\right)\:{prove}\:{that}\:{T}_{{n}+\mathrm{2}} \left({x}\right)=\mathrm{2}{x}\:{T}_{{n}+\mathrm{1}} \left({x}\right)−{T}_{{n}}…
Question Number 95373 by Don08q last updated on 24/May/20 $$\: \\ $$$$\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{m}\:\mathrm{for}\:\mathrm{which}\:\mathrm{the}\:\mathrm{roots} \\ $$$$\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:{x}^{\mathrm{3}} \:+\:\mathrm{6}{x}^{\mathrm{2}} \:+\:\mathrm{11}{x}\:+{m}\:=\:\mathrm{0} \\ $$$$\:\mathrm{form}\:\mathrm{a}\:\mathrm{linear}\:\mathrm{sequence}. \\ $$$$ \\ $$ Commented by Rasheed.Sindhi…
Question Number 29832 by abdo imad last updated on 12/Feb/18 $${p}\:{is}\:{a}\:{polynomial}\:{having}\:{n}\:{roots}\:{x}_{{i}} \:\:{with}\:{x}_{{i}} \neq{x}_{{j}} \:{for}\:{i}\neq{j} \\ $$$${prove}\:{that}\:\sum_{{i}=\mathrm{1}} ^{{n}} \:\:\frac{{p}^{''} \left({x}_{{i}} \right)}{{p}^{'} \left({x}_{{i}} \right)}=\mathrm{0} \\ $$ Terms…