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Author: Tinku Tara

Question-204395

Question Number 204395 by Thierrybadouana last updated on 15/Feb/24 Answered by Lindemann last updated on 20/Feb/24 $$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }×\frac{{x}^{\mathrm{2}} }{{sin}^{\mathrm{2}} \left({x}\right)}\right)\:=\:\mathrm{1} \\ $$ Terms…

Question-204389

Question Number 204389 by Ngarmadji last updated on 15/Feb/24 Answered by Frix last updated on 15/Feb/24 $$\frac{\mathrm{2}{x}}{\mathrm{3}}−\frac{\mathrm{2}{x}}{\mathrm{5}}=\mathrm{1}\:\:\frac{{x}}{\mathrm{3}}−\frac{{x}}{\mathrm{5}}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\mathrm{5}{x}−\mathrm{3}{x}=\frac{\mathrm{15}}{\mathrm{2}}\:\:{x}=\frac{\mathrm{15}}{\mathrm{4}} \\ $$$$\frac{\mathrm{3}{x}}{\mathrm{5}}−\frac{\mathrm{2}{x}}{\mathrm{15}}+\frac{{x}}{\mathrm{3}}=−\frac{\mathrm{7}}{\mathrm{15}}\:\:\frac{\mathrm{9}{x}−\mathrm{2}{x}+\mathrm{5}{x}}{\mathrm{15}}=−\frac{\mathrm{7}}{\mathrm{15}}\:\:…\:{x}=−\frac{\mathrm{7}}{\mathrm{12}} \\ $$ Terms of Service Privacy…

Question-204384

Question Number 204384 by Abdullahrussell last updated on 15/Feb/24 Commented by Frix last updated on 15/Feb/24 $$\mathrm{35}^{\mathrm{36}} =\mathrm{10}{m}+\mathrm{5} \\ $$$$\mathrm{36}^{\mathrm{35}} =\mathrm{10}{n}+\mathrm{6} \\ $$$$\mathrm{4}^{\mathrm{10}{m}+\mathrm{5}} =\mathrm{100}{k}+\mathrm{24} \\…

3-7-11-7-11-15-11-15-19-39-43-47-

Question Number 204348 by BaliramKumar last updated on 14/Feb/24 $$\mathrm{3}×\mathrm{7}×\mathrm{11}\:+\:\mathrm{7}×\mathrm{11}×\mathrm{15}\:+\:\mathrm{11}×\mathrm{15}×\mathrm{19}\:+\:……….+\:\:\mathrm{39}×\mathrm{43}×\mathrm{47}\:=\:?\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$ Answered by MM42 last updated on 14/Feb/24 $$\underset{\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\:\left(\mathrm{4}{n}−\mathrm{1}\right)\left(\mathrm{4}{n}+\mathrm{3}\right)\left(\mathrm{4}{n}+\mathrm{7}\right) \\ $$$$=\underset{\mathrm{1}} {\overset{\mathrm{10}}…

Question-204350

Question Number 204350 by universe last updated on 14/Feb/24 Answered by mr W last updated on 14/Feb/24 $${P}\left({a}\right)={a}^{\mathrm{4}} +{a}^{\mathrm{2}} {b}+{ac}+{d}=\mathrm{1}\:\:\:\:…\left({i}\right) \\ $$$${P}\left({b}\right)={ab}^{\mathrm{3}} +{b}^{\mathrm{3}} +{bc}+{d}=−\mathrm{1}\:\:\:…\left({ii}\right) \\…