Question Number 74284 by arthur.kangdani@gmail.com last updated on 21/Nov/19 Commented by mr W last updated on 21/Nov/19 $${a}_{\mathrm{2}} =\mathrm{4} \\ $$$${a}_{\mathrm{21}} =\mathrm{99} \\ $$$$\Sigma=\frac{\left(\mathrm{4}+\mathrm{99}\right)×\mathrm{20}}{\mathrm{2}}=\mathrm{1030} \\…
Question Number 139778 by chiamaka last updated on 01/May/21 $${prove}\:{that}\:{the}\:{absolute}\:{valje}\:{of}\:{z}\mathrm{1}+{z}\mathrm{2}<={absolute}\:{value}\:{of}\:{z}\mathrm{1}+{absolute}\:{value}\:{of}\:{z}\mathrm{2} \\ $$ Answered by mr W last updated on 01/May/21 Commented by mr W last…
Question Number 8704 by vuckintv last updated on 22/Oct/16 $${Solving}\:{for}\:{A}. \\ $$$${U}\left({z}\right)\:=\:{U}_{{b}} +\frac{\mathrm{2}{A}}{{h}+\mathrm{1}}\left(\rho×{g}×{sin}\left(\alpha\right)\right)^{{n}} \left[{H}^{{n}+\mathrm{1}} −\left({H}−{Z}\right)^{{n}+\mathrm{1}} \right] \\ $$ Answered by Rasheed Soomro last updated on…
Question Number 8695 by vuckintv last updated on 22/Oct/16 $${solving}\:{for}\:{B}? \\ $$$${U}_{{b}} \:=\:{U}_{{s}} \:−\:\frac{\mathrm{2}}{{n}+\mathrm{1}}\left(\frac{\rho×{g}×{sin}\alpha}{{B}}\right)^{{n}} {H}^{{n}+\mathrm{1}} \\ $$ Answered by FilupSmith last updated on 22/Oct/16 $${U}_{{b}}…
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Question Number 74214 by rajesh4661kumar@gmail.com last updated on 20/Nov/19 Answered by mr W last updated on 20/Nov/19 $$\mathrm{9744}{k}+\mathrm{4500}\:{with}\:{k}=\mathrm{0},\:\mathrm{1},\:\mathrm{2},\:… \\ $$$${the}\:{smallest}\:{number}\:{is}\:\mathrm{4500} \\ $$ Commented by rajesh4661kumar@gmail.com…
Question Number 8674 by tawakalitu last updated on 20/Oct/16 Answered by sandy_suhendra last updated on 21/Oct/16 $$\mathrm{when}\:\mathrm{series}\:: \\ $$$$\mathrm{C}_{\mathrm{1}} =\mathrm{0}.\mathrm{10}\:\mu\mathrm{F} \\ $$$$\mathrm{C}_{\mathrm{2}} =\mathrm{0}.\mathrm{20}\:\mu\mathrm{F} \\ $$$$\frac{\mathrm{1}}{\mathrm{C}_{\mathrm{series}}…
Question Number 74211 by rajesh4661kumar@gmail.com last updated on 20/Nov/19 Commented by $@ty@m123 last updated on 20/Nov/19 $${Question}\:{not}\:{clear}. \\ $$$${Pl}.\:{type}\:{the}\:{question}\:{using}\:{this}\:{app}. \\ $$ Terms of Service Privacy…