Question Number 60021 by Tawa1 last updated on 17/May/19 Answered by tanmay last updated on 17/May/19 $${dW}=\overset{\rightarrow} {{F}}.{d}\overset{\rightarrow} {{r}} \\ $$$$\:\:\:\:\:\:\:\:\:={Fdrco}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }\frac{{q}_{\mathrm{1}} {q}_{\mathrm{2}}…
Question Number 60022 by meme last updated on 17/May/19 $${construct}\:{M}^{'} \:{z}^{'} =\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{z}+\mid{z}\mid}{\mathrm{3}}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 191069 by uchihayahia last updated on 17/Apr/23 $$ \\ $$$$\:{both}\:{has}\:{same}\:{number}\:{of}\:{degrees},\:{edges} \\ $$$$\:{and}\:{vertices},\:{i}\:{think}\:{the}\:{graph}\:{below}\:{are}\:{not} \\ $$$$\:{isomorhic}\:{hos}\:{do}\:{i}\:{prove}\:{it}? \\ $$$$ \\ $$ Commented by uchihayahia last updated…
Question Number 125490 by Dwaipayan Shikari last updated on 11/Dec/20 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{3}} −\mathrm{2}}{\left({x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}} }\sqrt{{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +\mathrm{1}}\:{dx} \\ $$ Terms of Service Privacy Policy…
Question Number 191009 by pascal889 last updated on 16/Apr/23 Answered by Frix last updated on 16/Apr/23 $$\int\frac{\mathrm{3}{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{4}}{dt}=\frac{\mathrm{3}{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{4}}\int{dt}=\frac{\mathrm{3}{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{4}}{t}+{C} \\ $$ Commented by ARUNG_Brandon_MBU…
Question Number 59935 by maxmathsup by imad last updated on 16/May/19 $${sir}\:{malwan}\:{you}\:{must}\:{revise}\:\:{analytical}\:{function}\:{and}\:{complex}\:{analysis}… \\ $$ Commented by malwaan last updated on 16/May/19 $$\mathrm{O}.\mathrm{K}.\:{Sir} \\ $$ Terms…
Question Number 190972 by pascal889 last updated on 15/Apr/23 Answered by Frix last updated on 15/Apr/23 $$\frac{\mathrm{1}}{\mathrm{2}−\sqrt{\mathrm{3}}}=\mathrm{2}+\sqrt{\mathrm{3}}\wedge\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{3}}}=\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$${a}=\mathrm{2}\wedge{b}=\sqrt{\mathrm{3}} \\ $$$$\left({a}+{b}\right)^{\mathrm{2}} +\left({a}−{b}\right)^{\mathrm{2}} =\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)=\mathrm{14}…
Question Number 190961 by pascal889 last updated on 15/Apr/23 Answered by cortano12 last updated on 16/Apr/23 $$\:\Rightarrow\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{3x}^{\mathrm{2}} +\mathrm{8}\right)=\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{5x}+\mathrm{10}\right) \\ $$$$\:\Rightarrow\mathrm{3x}^{\mathrm{2}} −\mathrm{5x}−\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{3x}+\mathrm{1}\right)\left(\mathrm{x}−\mathrm{2}\right)=\mathrm{0}\:…
Question Number 190960 by pascal889 last updated on 15/Apr/23 Commented by Rasheed.Sindhi last updated on 15/Apr/23 $$\sqrt{{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{28}}\:+\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{8}}\:=\sqrt{\mathrm{6}{x}^{\mathrm{2}} −\mathrm{11}{x}−\mathrm{52}}\:\:\:\:\:? \\ $$ Commented by pascal889…