Question Number 143006 by mathdanisur last updated on 08/Jun/21 $$\mathrm{1}.\:{y}\frac{\partial{z}}{\partial{x}}\:+\:{z}\frac{\partial{z}}{\partial{y}}\:=\:\frac{{y}}{{x}} \\ $$$$\mathrm{2}.\:{x}^{\mathrm{2}} \frac{\partial{z}}{\partial{x}}\:−\:{xy}\frac{\partial{z}}{\partial{y}}\:+\:{y}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{3}.\:\begin{cases}{\frac{\partial{z}}{\partial{x}}\:=\:\frac{{z}}{{x}}}\\{\frac{\partial{z}}{\partial{y}}\:=\:\frac{\mathrm{2}{z}}{{y}}}\end{cases} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 77459 by jagoll last updated on 06/Jan/20 $$ \\ $$$$ \\ $$$${the}\:{greatest}\:{value}\:{of}\:{n}\:{so}\:{that}\: \\ $$$${only}\:{one}\:{value}\:{of}\:{k}\:{satisfies}\: \\ $$$$\frac{\mathrm{8}}{\mathrm{15}}<\frac{{n}}{{n}+{k}}<\frac{\mathrm{7}}{\mathrm{13}}\:{is}\: \\ $$ Answered by MJS last updated…
Question Number 77451 by BK last updated on 06/Jan/20 Commented by MJS last updated on 06/Jan/20 $$\mathrm{anyway}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\mathrm{stupid}.\:\mathrm{there}\:\mathrm{is}\:\mathrm{only} \\ $$$$\mathrm{one}\:\mathrm{power}\:\mathrm{of}\:\mathrm{7}\:\mathrm{in}\:\mathrm{50}! \\ $$ Answered by mr W…
Question Number 11914 by ahmet last updated on 04/Apr/17 $${Turevlenebilir}\:{bir}\:{f}\:{fonksiyonu}\:{icin} \\ $$$${f}\left({x}+{y}\right)={f}\left({x}\right)+{f}\left({y}\right)+\mathrm{2}{xy}\:{ve}\:{f}'\left(\mathrm{0}\right)=−\mathrm{3} \\ $$$${old}.{gore}\:{f}'\left(\mathrm{2}\right)=? \\ $$$${czm}\because\:\:{f}\left({x}+{y}\right)={f}\left({x}\right)+{f}\left({y}\right)+\mathrm{2}{xy} \\ $$$${y}\:{yi}\:{sabit}\:{kabul}\:{edersek} \\ $$$${f}'\left({x}+{y}\right)={f}'\left({x}\right)+\mathrm{2}{y} \\ $$$${x}=\mathrm{0},{y}=\mathrm{2}\:{icn} \\ $$$${f}'\left(\mathrm{2}\right)={f}'\left(\mathrm{0}\right)+\mathrm{2}.\mathrm{2} \\…
Question Number 142986 by mnjuly1970 last updated on 08/Jun/21 Commented by MJS_new last updated on 08/Jun/21 $$\mathrm{if}\:{y}={x}\:\mathrm{the}\:\mathrm{expression}\:\mathrm{is}\:\mathrm{constantly}\:\:\sqrt[{\mathrm{3}}]{\mathrm{4}} \\ $$ Commented by mnjuly1970 last updated on…
Question Number 11901 by ahmet last updated on 04/Apr/17 $$\frac{\mathrm{7}{cos}^{\mathrm{2}} {x}+{sin}^{\mathrm{2}} {x}−\mathrm{3}}{\mathrm{2}{cos}^{\mathrm{2}} {x}−{sin}^{\mathrm{2}} {x}}=? \\ $$$${czm}\because\:\:\frac{\mathrm{7}{cos}^{\mathrm{2}} {x}+\mathrm{1}−{cos}^{\mathrm{2}} {x}−\mathrm{3}}{\mathrm{2}{cos}^{\mathrm{2}} {x}−{sin}^{\mathrm{2}} {x}} \\ $$$$\frac{\mathrm{6}{cos}^{\mathrm{2}} {x}−\mathrm{2}}{\mathrm{2}{cos}^{\mathrm{2}} {x}−\left(\mathrm{1}−{cos}^{\mathrm{2}} {x}\right)}=\frac{\mathrm{6}{cos}^{\mathrm{2}}…
Question Number 142968 by bramlexs22 last updated on 08/Jun/21 $${Given}\:{p}<{x}<{q}\:{is}\:{solution}\:{set} \\ $$$${inequality}\:\mathrm{1}+\mathrm{2}^{{x}} +\mathrm{2}^{\mathrm{2}{x}} +\mathrm{2}^{\mathrm{3}{x}} +…>\mathrm{2} \\ $$$${for}\:{x}\neq\mathrm{1}.\:{find}\:{the}\:{value}\:{of}\: \\ $$$$\mathrm{5}{p}−\mathrm{3}{q}\:. \\ $$ Answered by mr W…
Question Number 77412 by BK last updated on 06/Jan/20 Commented by Tony Lin last updated on 06/Jan/20 $$\zeta\left({s}\right)=\frac{\mathrm{1}}{\mathrm{1}^{{s}} }+\frac{\mathrm{1}}{\mathrm{2}^{{s}} }+\frac{\mathrm{1}}{\mathrm{3}^{{s}} }+\centerdot\centerdot\centerdot+\frac{\mathrm{1}}{{n}^{{s}} } \\ $$$$\zeta\left(\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}}…
Question Number 11869 by Nayon last updated on 03/Apr/17 $${Why}\:{the}\:{expansion}\:{of}\:\:\left({a}+{b}\right)^{−{n}} \:{follows} \\ $$$${newton}'{s}\:{expansion}\:{rule}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com