Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 78894 by gopikrishnan last updated on 21/Jan/20

if U=x^3 +y^3 +6 sinz+11 then u_z (1,0,pi) is

$${if}\:{U}={x}^{\mathrm{3}} +{y}^{\mathrm{3}} +\mathrm{6}\:{sinz}+\mathrm{11}\:{then}\:{u}_{{z}} \left(\mathrm{1},\mathrm{0},{pi}\right)\:{is} \\ $$

Commented by mathmax by abdo last updated on 22/Jan/20

U(x,y,z) =x^3  +y^3  +6sinz +11 ⇒U(1,0,π)=1+0 +0 +11 =12

$${U}\left({x},{y},{z}\right)\:={x}^{\mathrm{3}} \:+{y}^{\mathrm{3}} \:+\mathrm{6}{sinz}\:+\mathrm{11}\:\Rightarrow{U}\left(\mathrm{1},\mathrm{0},\pi\right)=\mathrm{1}+\mathrm{0}\:+\mathrm{0}\:+\mathrm{11}\:=\mathrm{12} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com