Question Number 160179 by cortano last updated on 25/Nov/21 $$\:\mathrm{5}^{\mathrm{2}\left({x}−\mathrm{2}\right)} \left(\mathrm{5}^{\mathrm{2}\left({x}−\mathrm{1}\right)} \right)^{\left({x}+\mathrm{1}\right)} \:>\:\mathrm{125}^{{x}−\mathrm{1}} \\ $$ Commented by yeti123 last updated on 25/Nov/21 $$\mathrm{5}^{\mathrm{2}\left({x}−\mathrm{2}\right)} \left(\mathrm{5}^{\mathrm{2}\left({x}−\mathrm{1}\right)} \right)^{\left({x}+\mathrm{1}\right)}…
Question Number 160175 by HongKing last updated on 25/Nov/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 160168 by HongKing last updated on 25/Nov/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 160169 by HongKing last updated on 25/Nov/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 160171 by HongKing last updated on 25/Nov/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 160170 by HongKing last updated on 25/Nov/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 160160 by cortano last updated on 25/Nov/21 $$\:\:\:\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\frac{\mathrm{1}}{\left(\mathrm{2}{k}\right)!\left(\mathrm{2}{n}−{k}\right)!}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 94603 by byaw last updated on 19/May/20 $$\mathrm{List}\:\mathrm{the}\:\mathrm{elements}\:\mathrm{in}\: \\ $$$${C}=\left\{{x}:{x}\:\mathrm{is}\:\mathrm{an}\:{x}^{\mathrm{2}} \leqslant\mathrm{4},\:\mathrm{integer}\right\} \\ $$ Commented by Rasheed.Sindhi last updated on 20/May/20 $${C}=\left\{−\mathrm{2},−\mathrm{1},\mathrm{0},\mathrm{1},\mathrm{2}\right\} \\ $$…
Question Number 94601 by I want to learn more last updated on 19/May/20 Answered by Rasheed.Sindhi last updated on 20/May/20 $$\alpha+\beta+\gamma=\mathrm{6}………………………\left({i}\right) \\ $$$$\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\gamma^{\mathrm{3}}…
Question Number 29049 by NECx last updated on 03/Feb/18 $${Prove}\:{that} \\ $$$$\:\:\:\:\:{e}^{{i}\pi} +\mathrm{1}=\mathrm{0} \\ $$ Answered by Penguin last updated on 03/Feb/18 $${e}^{{i}\theta} =\mathrm{cos}\left(\theta\right)+{i}\mathrm{sin}\left(\theta\right) \\…