Question Number 202009 by sonukgindia last updated on 18/Dec/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 202011 by sonukgindia last updated on 18/Dec/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 202016 by sonukgindia last updated on 18/Dec/23 Answered by MM42 last updated on 18/Dec/23 $${x}^{\mathrm{55}} −\mathrm{1}=\mathrm{253}×\mathrm{8}{k}\Rightarrow“{x}''\:{is}\:{odd} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{54}} +{x}^{\mathrm{53}} +…+\mathrm{1}\right)=\mathrm{253}×\mathrm{8}{k} \\ $$$${x}−\mathrm{1}=\mathrm{8}{k}'\Rightarrow{x}\overset{\mathrm{8}} {\equiv}\mathrm{1}…
Question Number 202017 by sonukgindia last updated on 18/Dec/23 Answered by Mathspace last updated on 18/Dec/23 $${f}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{sin}\left({nx}\right)}{\mathrm{2}^{{n}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\pi} {f}\left({x}\right){dx}=\int_{\mathrm{0}} ^{\pi}…
Question Number 201949 by sonukgindia last updated on 16/Dec/23 Answered by Frix last updated on 17/Dec/23 $${n}^{{k}} \equiv{n}\mathrm{mod2024};\:{n}\in\left\{\mathrm{529},\:\mathrm{737},\:\mathrm{760},\:\mathrm{1265},\:\mathrm{1288},\:\mathrm{1496}\right\} \\ $$$$\Rightarrow \\ $$$$\Sigma{n}^{{n}} \equiv\mathrm{3mod2024} \\ $$…
Question Number 201956 by sonukgindia last updated on 16/Dec/23 Commented by mahdipoor last updated on 16/Dec/23 $$\mathrm{5}^{\mathrm{2}^{\mathrm{117}} } −\mathrm{1}=\left(\mathrm{5}^{\mathrm{2}^{\mathrm{116}} } −\mathrm{1}\right)\left(\mathrm{5}^{\mathrm{2}^{\mathrm{116}} } +\mathrm{1}\right)= \\ $$$$\left(\mathrm{5}^{\mathrm{2}^{\mathrm{115}}…
Question Number 201936 by aurpeyz last updated on 15/Dec/23 Commented by aurpeyz last updated on 15/Dec/23 $${Using}\:{complex}\:{analysis}\:{method} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 201906 by sonukgindia last updated on 15/Dec/23 Answered by AST last updated on 15/Dec/23 $${sin}\left({x}\right)=\frac{{s}}{{h}}…\left({i}\right) \\ $$$$\left({ii}\right)…\frac{{sin}\left(\mathrm{2}{x}\right)}{{h}}=\frac{{sin}\left({x}\right)}{\mathrm{2}{s}}\Rightarrow\mathrm{2}{sin}\left({x}\right)=\frac{{sin}\left({x}\right)}{{sin}\left(\mathrm{2}{x}\right)} \\ $$$$\Rightarrow{sin}\left(\mathrm{2}{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{x}=\mathrm{15}° \\ $$ Answered by…
Question Number 201901 by sonukgindia last updated on 15/Dec/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 201902 by sonukgindia last updated on 15/Dec/23 Terms of Service Privacy Policy Contact: info@tinkutara.com