Question Number 133961 by Ahmed1hamouda last updated on 25/Feb/21 Answered by mathmax by abdo last updated on 27/Feb/21 $$\mathrm{y}^{'} \:+\mathrm{sinx}\:\mathrm{y}=−\mathrm{x}^{\mathrm{3}} \\ $$$$\mathrm{h}\rightarrow\mathrm{y}^{'} \:+\mathrm{sinxy}=\mathrm{0}\:\Rightarrow\mathrm{y}^{'} \:=−\mathrm{sinxy}\:\Rightarrow\frac{\mathrm{y}^{'} }{\mathrm{y}}=−\mathrm{sinx}\:\Rightarrow\mathrm{ln}\mid\mathrm{y}\mid=\mathrm{cosx}\:+\mathrm{c}\:\Rightarrow…
Question Number 2891 by 123456 last updated on 29/Nov/15 $$\mathrm{given} \\ $$$$\Gamma\left({z}\right)=\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:\frac{{n}!}{{z}\left({z}+\mathrm{1}\right)\centerdot\centerdot\centerdot\left({z}+{n}\right)}{n}^{{z}} \\ $$$$\mathrm{proof}\:\mathrm{that} \\ $$$$\left.{i}\right) \\ $$$$\Gamma\left({z}+\mathrm{1}\right)={z}\Gamma\left({z}\right) \\ $$$$\left.{ii}\right)\:\mathrm{wiertrass}\:\mathrm{definition}\:\mathrm{of}\:\mathrm{gamma}\:\mathrm{function} \\ $$$$\frac{\mathrm{1}}{\Gamma\left({z}\right)}={ze}^{{z}\gamma} \underset{{m}=\mathrm{1}} {\overset{+\infty}…
Question Number 133963 by bobhans last updated on 25/Feb/21 $$\mathcal{Y}\:=\:\int\:\frac{{dx}}{\:\sqrt[{\mathrm{6}}]{\mathrm{1}+{x}^{\mathrm{6}} }}? \\ $$ Answered by EDWIN88 last updated on 26/Feb/21 $$\mathbb{Y}=\:\int\:\frac{\mathrm{dx}}{\mathrm{x}\:\sqrt[{\mathrm{6}}]{\mathrm{1}+\mathrm{x}^{−\mathrm{6}} }}\:=\:\int\:\frac{\mathrm{x}^{\mathrm{6}} }{\mathrm{x}^{\mathrm{7}} \:\sqrt[{\mathrm{6}}]{\mathrm{1}+\mathrm{x}^{−\mathrm{6}} }}\:\mathrm{dx}\:…
Question Number 68425 by mr W last updated on 10/Sep/19 Commented by Prithwish sen last updated on 10/Sep/19 $$\mathrm{AB}=\mathrm{2x}\:\:\mathrm{AC}=\mathrm{x} \\ $$$$\mathrm{3x}×\mathrm{100}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\mathrm{2x}^{\mathrm{2}} ×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{2x}^{\mathrm{2}} −\mathrm{300x}=\mathrm{0}\:\Rightarrow\mathrm{x}=\mathrm{150}…
Question Number 2888 by prakash jain last updated on 29/Nov/15 $$\underset{\left({x},{y}\right)\rightarrow\left(\mathrm{0},\mathrm{0}\right)} {\mathrm{lim}}\:\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}=? \\ $$$$\mathrm{or} \\ $$$$\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\Re\left({z}\right)}{\mid{z}\mid} \\ $$ Answered by prakash jain…
Question Number 133957 by mathmax by abdo last updated on 25/Feb/21 $$\left.\mathrm{1}\right)\mathrm{decompose}\:\:\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{5}} \left(\mathrm{2x}−\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$\left.\mathrm{2}\right)\:\mathrm{find}\:\int_{\mathrm{1}} ^{\infty} \:\mathrm{F}\left(\mathrm{x}\right)\mathrm{dx} \\ $$ Answered by Olaf last updated…
Question Number 68422 by ajfour last updated on 10/Sep/19 Answered by ajfour last updated on 10/Sep/19 $${C}=\mathrm{10}{m}^{\mathrm{3}} {p}^{\mathrm{2}} +{b}\left({m}^{\mathrm{3}} +\mathrm{6}{m}^{\mathrm{2}} {p}+\mathrm{3}{mp}^{\mathrm{2}} \right) \\ $$$$\:+{c}\left(\mathrm{3}{m}^{\mathrm{2}} +\mathrm{6}{mp}+{p}^{\mathrm{2}}…
Question Number 68418 by necxxx last updated on 10/Sep/19 Commented by necxxx last updated on 10/Sep/19 $${The}\:{UDL}\:{is}\:\mathrm{2}.\mathrm{5}{KN}/{m}.\:{The}\:{first}\:{point} \\ $$$${load}\:{is}\:\mathrm{35}{KN}.\:{I}'{ve}\:{tried}\:{solving}\:{but}\:{I} \\ $$$${encountered}\:{some}\:{issues}\:{with}\:{the}\:{diagram}. \\ $$$${I}'{ll}\:{be}\:{really}\:{grateful}\:{to}\:{get}\:{this}\:{solved} \\ $$$${with}\:{the}\:{possible}\:{diagrams}.…
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Question Number 68414 by behi83417@gmail.com last updated on 10/Sep/19 $$\mathrm{Is}\:\mathrm{it}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{find}\:\mathrm{any}\:\mathrm{value}\:\mathrm{for} \\ $$$$\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}},\boldsymbol{\mathrm{c}}\:\mathrm{from}\:\mathrm{below}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equetions}? \\ $$$$\begin{cases}{\boldsymbol{\mathrm{sina}}+\boldsymbol{\mathrm{sinb}}=\boldsymbol{\mathrm{sinc}}}\\{\boldsymbol{\mathrm{cosa}}+\boldsymbol{\mathrm{cosb}}=\boldsymbol{\mathrm{cosc}}}\end{cases} \\ $$ Commented by kaivan.ahmadi last updated on 10/Sep/19 $$\begin{cases}{{sinacosa}+{sinbcosa}={sinccosa}}\\{−{sinacosa}−{sinacosb}=−{sinacosc}}\end{cases}\Rightarrow \\…