Question Number 4539 by love math last updated on 05/Feb/16 $${Two}\:{tractor},\:{working}\:{together},\: \\ $$$${have}\:{plowed}\:{the}\:{field}\:{for}\:\mathrm{48}\:{hours}. \\ $$$${If}\:{half}\:{of}\:{the}\:{field}\:{is}\:{plowed}\:{by}\: \\ $$$${one}\:{of}\:{them}\:{and}\:{then}\:{the}\:{remaining} \\ $$$${half}\:{to}\:{the}\:{other},\:{the}\:{work}\:{would} \\ $$$${be}\:{carried}\:{out}\:\mathrm{100}\:{hours}.\:{How}\:{many} \\ $$$${hours}\:{to}\:{plow}\:{the}\:{field}\:{if}\:{every}\:{tractor} \\ $$$${working}\:{separately}?…
Question Number 4524 by love math last updated on 05/Feb/16 $$\mathrm{3}^{\mathrm{3}{x}} +{x}^{\mathrm{2}} −\mathrm{9}>\mathrm{0} \\ $$ Commented by Yozzii last updated on 05/Feb/16 $$\mathrm{27}^{{x}} >\mathrm{9}−{x}^{\mathrm{2}} \:\:\:\left(\ast\right)…
Question Number 70040 by Shamim last updated on 30/Sep/19 $$\mathrm{If},\:\mathrm{a}^{\mathrm{2}} \mathrm{b}^{\mathrm{2}} \mathrm{c}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{c}^{\mathrm{3}} }\right)=\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} \:\mathrm{than} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{Successive}\:\mathrm{Proportional}. \\ $$ Commented by…
Question Number 4502 by #na last updated on 02/Feb/16 $$\frac{\mathrm{4}}{{x}−\mathrm{1}}−\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{2}}{{x}+\mathrm{1}}+\frac{\mathrm{6}}{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$ Answered by Rasheed Soomro last updated on 02/Feb/16 $$\frac{\mathrm{4}}{{x}−\mathrm{1}}−\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{2}}{{x}+\mathrm{1}}−\frac{\mathrm{6}}{\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)} \\…
Question Number 4492 by RasheedSindhi last updated on 01/Feb/16 $$ \\ $$$${By}\:{murging}\:{three}\:{sequences} \\ $$$$\:{a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,…{a}_{{n}} \:,\:{b}_{\mathrm{1}} ,{b}_{\mathrm{2}} ,…,{b}_{{n}} \:\&\:{c}_{\mathrm{1}} ,{c}_{\mathrm{2}} ,…,{c}_{{n}} \\ $$$${a}\:{new}\:{sequence}\: \\…
Question Number 4485 by Rasheed Soomro last updated on 31/Jan/16 $${By}\:\:{murging}\:\:{two}\:{sequences} \\ $$$$\:\:\:\:{a}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,…{a}_{{n}} \:\:\:{and}\:\:{b}_{\mathrm{1}} ,{b}_{\mathrm{2}} ,…,{b}_{{n}} \\ $$$${a}\:{new}\:{sequence}\: \\ $$$$\:\:\:\:{a}_{\mathrm{1}} ,{b}_{\mathrm{1}} ,{a}_{\mathrm{2}} ,{b}_{\mathrm{2}}…
Question Number 135536 by islamo last updated on 13/Mar/21 $$\left(\boldsymbol{{u}}^{\boldsymbol{{v}}} \right)'=? \\ $$$${u}\:,{v}\:\:\:{tow}\:{fonctions} \\ $$ Answered by Olaf last updated on 13/Mar/21 $${u}^{{v}} \:=\:{e}^{\mathrm{ln}\left({u}^{{v}} \right)}…
Question Number 69995 by behi83417@gmail.com last updated on 29/Sep/19 $$\:\:\mathrm{1}.\begin{cases}{\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} }+\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{y}}^{\mathrm{2}} }=\boldsymbol{\mathrm{a}}}\\{\sqrt{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}}+\sqrt{\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{y}}}=\boldsymbol{\mathrm{b}}}\end{cases}\:\:\:\:\:\left[\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}}\in\boldsymbol{\mathrm{R}}^{+} \right] \\ $$$$ \\ $$$$\:\:\:\mathrm{2}.\begin{cases}{\sqrt{\sqrt{\mathrm{x}}+\mathrm{y}}+\sqrt{\mathrm{x}+\sqrt{\mathrm{y}}}=\boldsymbol{\mathrm{a}}}\\{\sqrt{\sqrt{\mathrm{x}}−\mathrm{y}}+\sqrt{\mathrm{x}−\sqrt{\mathrm{y}}}=\boldsymbol{\mathrm{b}}}\end{cases} \\ $$$$\: \\ $$$$\:\:\:\mathrm{3}.\begin{cases}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} =\left(\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}\right)\boldsymbol{\mathrm{xy}}}\\{\boldsymbol{\mathrm{x}}^{\mathrm{3}}…
Question Number 135532 by mr W last updated on 13/Mar/21 Commented by mr W last updated on 13/Mar/21 $${an}\:{old}\:{question} \\ $$ Answered by Dwaipayan Shikari…
Question Number 4446 by Rasheed Soomro last updated on 28/Jan/16 $$\mathrm{Define}\:\mathrm{2}^{\sqrt{\mathrm{3}}} \:\mathrm{and}\:\mathrm{3}^{\sqrt{\mathrm{2}}} \:. \\ $$ Commented by Yozzii last updated on 29/Jan/16 $${Dedekind}\:{cuts}\:{of}\:{the}\:{real}\:{line}\:{might} \\ $$$${define}\:{each}\:{of}\:{those}\:{numbers}.\:…