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Question-195707

Question Number 195707 by sonukgindia last updated on 08/Aug/23 Answered by mr W last updated on 08/Aug/23 $${y}''={y}'\frac{{d}\left({y}'\right)}{{dy}} \\ $$$${p}={y}' \\ $$$${p}\frac{{dp}}{{dy}}=\frac{\mathrm{1}}{{y}^{\mathrm{3}} } \\ $$$${pdp}=\frac{{dy}}{{y}^{\mathrm{3}}…

Question-195697

Question Number 195697 by sonukgindia last updated on 08/Aug/23 Answered by MM42 last updated on 08/Aug/23 $${if}\:\:{mean}\:\:\:{lim}_{{x}\rightarrow\infty} \:\sqrt[{{x}}]{{x}\:}\:\:{then} \\ $$$${A}={lim}_{{x}\rightarrow\infty} \:\sqrt[{{x}}]{{x}} \\ $$$$\Rightarrow{lnA}={lim}_{{x}\rightarrow\infty} \:\frac{{lnx}}{{x}}\:=\mathrm{0}\Rightarrow{A}=\mathrm{1} \\…

Question-195721

Question Number 195721 by Humble last updated on 08/Aug/23 Answered by mr W last updated on 08/Aug/23 $${volume}\:{of}\:{copper}\:{V}_{{cu}} =\frac{\mathrm{25}}{\mathrm{9}.\mathrm{8}×\mathrm{8}.\mathrm{93}}=\mathrm{0}.\mathrm{286}\:{ml} \\ $$$${total}\:{volume}\:{V}=\frac{\mathrm{25}−\mathrm{20}}{\mathrm{9}.\mathrm{8}×\mathrm{1}.\mathrm{0}}=\mathrm{0}.\mathrm{510}\:{ml} \\ $$$${volume}\:{of}\:{bubble}\:{V}_{{b}} =\mathrm{0}.\mathrm{510}−\mathrm{0}.\mathrm{286}=\mathrm{0}.\mathrm{224}\:{ml} \\…

0-4-x-5-x-5-dx-

Question Number 195674 by Rodier97 last updated on 07/Aug/23 $$\:\:\:\int_{\mathrm{0}} ^{\mathrm{4}} \:\frac{{x}!}{\mathrm{5}!\left({x}−\mathrm{5}\right)!}\:{dx}\:=\:??? \\ $$ Commented by mr W last updated on 07/Aug/23 $${what}\:{do}\:{you}\:{mean}\:{with}\:\begin{pmatrix}{{x}}\\{\mathrm{5}}\end{pmatrix}\:? \\ $$…

Question-195681

Question Number 195681 by sonukgindia last updated on 07/Aug/23 Answered by Frix last updated on 07/Aug/23 $$=\frac{\mathrm{2}×\mathrm{3}×…×\mathrm{98}}{\mathrm{4}×\mathrm{5}×…×\mathrm{100}}=\mathrm{6}×\frac{\mathrm{98}!}{\mathrm{100}!}=\frac{\mathrm{6}}{\mathrm{99}×\mathrm{100}}=\frac{\mathrm{1}}{\mathrm{1650}} \\ $$ Answered by MM42 last updated on…

2-2-

Question Number 195627 by Boazmbura last updated on 06/Aug/23 $$\mathrm{2}+\mathrm{2} \\ $$ Commented by Frix last updated on 06/Aug/23 $${x}=\mathrm{2}+\mathrm{2}\:\Rightarrow\:\mathrm{3}.\mathrm{97}<{x}<\mathrm{4}.\mathrm{015} \\ $$ Terms of Service…