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Author: Tinku Tara

solve-the-EDP-1-x-f-x-y-f-y-0-2-x-f-x-f-y-z-2-x-3-x-2-2-f-x-2-y-2-2-f-y-2-0-

Question Number 198355 by lapache last updated on 18/Oct/23 $${solve}\:{the}\:{EDP} \\ $$$$\mathrm{1}−\:\:\:{x}\frac{\partial{f}}{\partial{x}}−{y}\frac{\partial{f}}{\partial{y}}=\mathrm{0} \\ $$$$\mathrm{2}−\:\:\:{x}\frac{\partial{f}}{\partial{x}}−\frac{\partial{f}}{\partial{y}}=\frac{{z}^{\mathrm{2}} }{{x}} \\ $$$$ \\ $$$$\mathrm{3}−\:\:{x}^{\mathrm{2}} \frac{\partial^{\mathrm{2}} {f}}{\partial{x}^{\mathrm{2}} }−{y}^{\mathrm{2}} \frac{\partial^{\mathrm{2}} {f}}{\partial{y}^{\mathrm{2}} }=\mathrm{0}…

Question-198377

Question Number 198377 by sonukgindia last updated on 18/Oct/23 Answered by witcher3 last updated on 19/Oct/23 $$=\int_{\mathrm{0}} ^{\infty} \mathrm{xe}^{−\mathrm{4x}^{\mathrm{2}} } .\mathrm{xe}^{\frac{−\mathrm{9}}{\mathrm{x}^{\mathrm{2}} }} \mathrm{dx}\:\:\:\mathrm{IBP} \\ $$$$=\left[−\frac{\mathrm{e}^{−\mathrm{4x}^{\mathrm{2}}…

1-1-5-1-4-8-1-7-11-1-10-14-

Question Number 198372 by cortano12 last updated on 18/Oct/23 $$\:\:\:\frac{\mathrm{1}}{\mathrm{1}×\mathrm{5}}\:+\:\frac{\mathrm{1}}{\mathrm{4}×\mathrm{8}}\:+\:\frac{\mathrm{1}}{\mathrm{7}×\mathrm{11}}\:+\frac{\mathrm{1}}{\mathrm{10}×\mathrm{14}}\:+\:\ldots=? \\ $$ Commented by Frix last updated on 19/Oct/23 $$\frac{\mathrm{1}}{\mathrm{8}}+\frac{\sqrt{\mathrm{3}}\pi}{\mathrm{36}} \\ $$ Answered by mr…

Question-198339

Question Number 198339 by cherokeesay last updated on 18/Oct/23 Answered by witcher3 last updated on 18/Oct/23 $$\mathrm{didint}\:\mathrm{existe}\: \\ $$$$\left(\mathrm{x},\mathrm{y}\right)=\left(\frac{\mathrm{1}}{\mathrm{n}},\mathrm{2}−\frac{\mathrm{1}}{\mathrm{n}}\right);\mathrm{n}\geqslant\mathrm{1} \\ $$$$\mathrm{n}\rightarrow\infty \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }\right)\rightarrow\infty…

Of-men-that-attended-a-party-30-of-them-wore-coats-20-wore-ties-and-10-wore-hats-There-were-4-men-who-wore-coats-and-tie-or-tie-and-hat-or-coat-and-hat-14-men-wore-tie-only-with-no-coat-and-hat-

Question Number 198332 by necx122 last updated on 17/Oct/23 $${Of}\:{men}\:{that}\:{attended}\:{a}\:{party},\:\mathrm{30}\:{of} \\ $$$${them}\:{wore}\:{coats},\:\mathrm{20}\:{wore}\:{ties}\:{and}\:\mathrm{10} \\ $$$${wore}\:{hats}.\:{There}\:{were}\:\mathrm{4}\:{men}\:{who}\:{wore} \\ $$$${coats}\:{and}\:{tie},\:{or}\:{tie}\:{and}\:{hat}\:{or}\:{coat}\:{and} \\ $$$${hat}.\:\mathrm{14}\:{men}\:{wore}\:{tie}\:{only}\:{with}\:{no} \\ $$$${coat}\:{and}\:{hat}.\:{Find}\:{the}\:{number}\:{of}\:{men} \\ $$$${who}\:{wore}: \\ $$$$\left.{a}\right)\:{coat},\:{tie}\:{and}\:{hat} \\…

Question-198302

Question Number 198302 by sonukgindia last updated on 17/Oct/23 Answered by mr W last updated on 17/Oct/23 $${z}_{\mathrm{1}} =\mathrm{1}+{i}\sqrt{\mathrm{3}}=\mathrm{2}{e}^{{i}\left(\frac{\pi}{\mathrm{3}}\right)} \\ $$$${z}_{\mathrm{2}} =\mathrm{2}{e}^{{i}\left(\frac{\pi}{\mathrm{3}}+\frac{\mathrm{2}\pi}{\mathrm{3}}\right)} =\mathrm{2}{e}^{{i}\pi} =−\mathrm{2} \\…

Let-x-r-r-1-n-be-n-positive-real-numbers-Show-That-x-1-1-x-1-2-x-2-1-x-1-2-x-2-2-x-n-1-x-1-2-x-2-2-x-n-2-lt-n-

Question Number 198296 by York12 last updated on 17/Oct/23 $${Let}\:\left\{{x}_{{r}} \right\}_{{r}=\mathrm{1}} ^{{n}} {be}\:{n}\:{positive}\:{real}\:{numbers}\:{Show}\:{That}: \\ $$$$\frac{{x}_{\mathrm{1}} }{\mathrm{1}+{x}_{\mathrm{1}} ^{\mathrm{2}} }+\frac{{x}_{\mathrm{2}} }{\mathrm{1}+{x}_{\mathrm{1}} ^{\mathrm{2}} +{x}_{\mathrm{2}} ^{\mathrm{2}} }+…+\frac{{x}_{{n}} }{\mathrm{1}+{x}_{\mathrm{1}} ^{\mathrm{2}}…