Question Number 68122 by TawaTawa last updated on 05/Sep/19 Commented by TawaTawa last updated on 05/Sep/19 $$\mathrm{Please}\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{the}\:\mathrm{workings}\:\mathrm{here}.\:\:\mathrm{Help}\:\mathrm{me}\:\mathrm{explain}\:\mathrm{please} \\ $$ Answered by mind is power last…
Question Number 133653 by greg_ed last updated on 23/Feb/21 $$\boldsymbol{\mathrm{hi}},\:\boldsymbol{\mathrm{everybody}}\:! \\ $$$$\boldsymbol{\mathrm{how}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\pi}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{irrational}}\:\boldsymbol{\mathrm{number}}\:??? \\ $$ Answered by Dwaipayan Shikari last updated on 23/Feb/21 $$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{9}}−\frac{\mathrm{1}}{\mathrm{11}}+…=\frac{\pi}{\mathrm{4}} \\ $$$${As}\:{It}\:{is}\:{an}\:{Infinte}\:{series}\:{so}\:{it}\:{can}\:{never}\:{be}\:{rational}…
Question Number 2548 by Filup last updated on 22/Nov/15 $$\mathrm{For}\:\mathrm{a}\:\mathrm{function}\:{y}={f}\left({x}\right), \\ $$$$\mathrm{inflection}\:\mathrm{points}/\mathrm{stationary}\:\mathrm{points}\:\mathrm{are} \\ $$$$\mathrm{when}\:\:\frac{{df}}{{dx}}=\mathrm{0}. \\ $$$$ \\ $$$$\mathrm{For}\:\mathrm{a}\:\mathrm{function}\:{z}={f}\left({x},\:{y}\right),\:\mathrm{can}\:\mathrm{you}\:\mathrm{find} \\ $$$$\mathrm{these}\:\mathrm{points}\:\mathrm{through}\:\mathrm{a}\:\mathrm{similar}\:\mathrm{method}? \\ $$$$ \\ $$$$\mathrm{Is}\:\mathrm{it}\:\mathrm{something}\:\mathrm{like}\:\frac{\partial{f}}{\partial{x}}=\mathrm{0}\:\mathrm{and}\:\frac{\partial{f}}{\partial{y}}=\mathrm{0}? \\…
Question Number 2545 by Rasheed Soomro last updated on 22/Nov/15 $$\mathcal{P}{rove}\:{that}\:\mathrm{3}^{\mathrm{3}{n}} −\mathrm{26}{n}−\mathrm{1}\:{is}\:{divisible}\:{by}\:\mathrm{676}, \\ $$$${where}\:{n}\in\mathbb{Z}^{+} \\ $$ Commented by Yozzi last updated on 22/Nov/15 $${Use}\:{induction}. \\…
Question Number 2544 by Rasheed Soomro last updated on 22/Nov/15 $$\mathcal{C}{an}\:{you}\:\mathcal{G}{eneralize}\:{the}\:{following}? \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{3}+…+{n}=\frac{\mathrm{1}}{\mathrm{2}}\left({n}\right)\left({n}+\mathrm{1}\right) \\ $$$$\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +…+{n}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{6}}\left({n}\right)\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right. \\ $$$$\mathrm{1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +\mathrm{3}^{\mathrm{3}} +…+{n}^{\mathrm{3}} =\left[\frac{\mathrm{1}}{\mathrm{2}}\left({n}\right)\left({n}+\mathrm{1}\right)\right]^{\mathrm{2}}…
Question Number 133612 by benjo_mathlover last updated on 23/Feb/21 $$\mathrm{If}\:\begin{cases}{\mathrm{x}+\mathrm{y}+\mathrm{2z}=\mathrm{k}}\\{\mathrm{x}+\mathrm{2y}+\mathrm{z}=\mathrm{k}}\\{\mathrm{2x}+\mathrm{y}+\mathrm{z}=\mathrm{k}}\end{cases}\:;\:\mathrm{k}\neq\:\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} \:=? \\ $$$$ \\ $$ Commented by mr W last updated…
Question Number 68063 by TawaTawa last updated on 04/Sep/19 Answered by MJS last updated on 04/Sep/19 $$\mathrm{tricky}\:\mathrm{but}\:\mathrm{easy} \\ $$$$\left(\mathrm{1}\right)\:\:{x}^{\mathrm{3}} −\mathrm{3}{y}^{\mathrm{2}} {x}={a}+\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:\:{y}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} {y}={a}…
Question Number 133568 by bemath last updated on 23/Feb/21 $$\mathrm{Proof}\:\mathrm{the}\:\mathrm{series}\:\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{2}}{\mathrm{9}+\mathrm{2n}\left(\mathrm{ln}\:\mathrm{n}\right)^{\mathrm{2}} } \\ $$$$\mathrm{convergent} \\ $$$$ \\ $$ Answered by EDWIN88 last updated on…
Question Number 133542 by help last updated on 22/Feb/21 Commented by help last updated on 22/Feb/21 $${x}^{\mathrm{3}{n}+\mathrm{2}} +{x}+\mathrm{1}\:\:{is}\:{divisible}\:{by}\:{x}^{\mathrm{2}} +{x}+\mathrm{1} \\ $$$${correct}\:{question} \\ $$ Terms of…
Question Number 67997 by ajfour last updated on 03/Sep/19 $$\mathrm{5}{y}^{\mathrm{2}} +\mathrm{2}{axy}+{b}=\mathrm{0} \\ $$$${ay}^{\mathrm{2}} +\mathrm{2}{bx}+\mathrm{5}{c}=\mathrm{0} \\ $$$$\left(\mathrm{5}{x}+\mathrm{3}{a}\right){y}^{\mathrm{2}} +\left(\mathrm{4}{ax}^{\mathrm{2}} \right){y}−{bx}−\mathrm{5}{c}=\mathrm{0} \\ $$$$\mathrm{5}{y}^{\mathrm{2}} −{x}\left(\mathrm{5}{x}+\mathrm{2}{a}\right){y}−{ax}^{\mathrm{3}} −\mathrm{3}{b}=\mathrm{0} \\ $$$${Please}\:{solve}\:{simultaneously} \\…