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Question-27909

Question Number 27909 by mondodotto@gmail.com last updated on 16/Jan/18 Commented by abdo imad last updated on 16/Jan/18 $${we}\:{have}\:\frac{{d}}{{dx}}{ln}\left(\mathrm{1}−{x}\right)=\:\frac{−\mathrm{1}}{\mathrm{1}−{x}}=−\sum_{{n}=\mathrm{0}} ^{\propto} \:{x}^{{n}} \\ $$$$\Rightarrow\:{ln}\left(\mathrm{1}−{x}\right)=\:−\sum_{{n}=\mathrm{0}} ^{\propto} \:\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:=−\sum_{{n}=\mathrm{1}}…

To-tinkutara-I-have-forget-my-old-password-and-has-to-create-a-new-one-Is-there-any-way-to-retrive-my-old-account-

Question Number 93442 by PRITHWISH SEN 2 last updated on 13/May/20 $$\mathrm{To}\:\mathrm{tinkutara} \\ $$$$\mathrm{I}\:\mathrm{have}\:\mathrm{forget}\:\mathrm{my}\:\mathrm{old}\:\mathrm{password}\:\mathrm{and}\:\mathrm{has}\:\mathrm{to}\:\mathrm{create}\:\mathrm{a} \\ $$$$\mathrm{new}\:\mathrm{one}.\:\mathrm{Is}\:\mathrm{there}\:\mathrm{any}\:\mathrm{way}\:\mathrm{to}\:\mathrm{retrive}\:\mathrm{my}\:\mathrm{old}\:\mathrm{account}. \\ $$ Commented by i jagooll last updated on…

Question-93434

Question Number 93434 by naka3546 last updated on 13/May/20 Commented by naka3546 last updated on 13/May/20 $${AT}^{\:\mathrm{2}} +{BT}^{\:\mathrm{2}} \:=\:\mathrm{3312} \\ $$$${CT}^{\:\mathrm{2}} +{DT}^{\:\mathrm{2}} \:=\:\mathrm{3308} \\ $$$${Radius}\:\:{of}\:\:{circle}\:\:{is}\:…\:?…

New-version-of-the-app-with-usability-enhacements-is-now-available-on-playstore-Pleasd-update-app-Comment-on-this-post-for-problem-feedback-suggestions-

Question Number 93371 by Tinku Tara last updated on 12/May/20 $$\mathrm{New}\:\mathrm{version}\:\mathrm{of}\:\mathrm{the}\:\mathrm{app}\:\mathrm{with}\:\mathrm{usability} \\ $$$$\mathrm{enhacements}\:\mathrm{is}\:\mathrm{now}\:\mathrm{available}\:\mathrm{on} \\ $$$$\mathrm{playstore}. \\ $$$$\mathrm{Pleasd}\:\mathrm{update}\:\mathrm{app}.\: \\ $$$$\mathrm{Comment}\:\mathrm{on}\:\mathrm{this}\:\mathrm{post}\:\mathrm{for} \\ $$$$\mathrm{problem}/\mathrm{feedback}\:\mathrm{suggestions}. \\ $$ Commented by…

1-2-3-4-

Question Number 158899 by ehab last updated on 10/Nov/21 $$\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{4}}= \\ $$ Answered by EbrimaDanjo last updated on 10/Nov/21 $$\mathrm{the}\:\mathrm{LCM}\:\mathrm{of}\:\mathrm{denominators}\:=\mathrm{4} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\:\frac{\mathrm{2}\:+\:\mathrm{3}}{\mathrm{4}}\:\:\:\Rightarrow\:\frac{\mathrm{5}}{\mathrm{4}}\:=\mathrm{1}\frac{\mathrm{1}}{\mathrm{4}} \\…