Question Number 147791 by Tawa11 last updated on 23/Jul/21 $$\mathrm{If}\:\:\:\:\:\:\:\mathrm{a}\:\:+\:\:\mathrm{b}\:\:+\:\:\mathrm{c}\:\:+\:\:\mathrm{d}\:\:\:=\:\:\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{a}^{\mathrm{2}} \:\:+\:\:\mathrm{b}^{\mathrm{2}} \:\:+\:\:\mathrm{c}^{\mathrm{2}} \:\:+\:\:\mathrm{d}^{\mathrm{2}} \:\:=\:\:\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{a}^{\mathrm{3}} \:\:+\:\:\mathrm{b}^{\mathrm{3}} \:\:+\:\:\mathrm{c}^{\mathrm{3}} \:\:+\:\:\mathrm{d}^{\mathrm{3}} \:\:=\:\:\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{a}^{\mathrm{4}} \:\:+\:\:\mathrm{b}^{\mathrm{4}}…
Question Number 16699 by mrW1 last updated on 25/Jun/17 $$\mathrm{Related}\:\mathrm{to}\:\mathrm{Q16675}: \\ $$$$ \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{intersection}\:\mathrm{points} \\ $$$$\mathrm{of}\:\mathrm{graph}\:\mathrm{sin}\:\mathrm{x}=\frac{\mathrm{x}}{\mathrm{10}}. \\ $$$$ \\ $$$$\mathrm{Let}'\mathrm{s}\:\mathrm{see}\:\mathrm{sin}\:\mathrm{x}\:=\:\frac{\mathrm{x}}{\mathrm{n}}\:\mathrm{with}\:\mathrm{n}>\mathrm{1}. \\ $$$$\mathrm{For}\:\mathrm{n}\leqslant\mathrm{1}\:\mathrm{there}\:\mathrm{is}\:\mathrm{one}\:\mathrm{intersection}\:\mathrm{point}. \\ $$$$ \\…
Question Number 147737 by liberty last updated on 23/Jul/21 $$\:\underset{{n}\geqslant\mathrm{1}} {\sum}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…+\frac{\mathrm{1}}{{n}}\right)^{\mathrm{2}} =? \\ $$ Answered by ArielVyny last updated on 24/Jul/21 $$\left(\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}}\right)^{\mathrm{2}} \leqslant\Sigma\frac{\mathrm{1}}{{n}^{\mathrm{2}}…
Question Number 82191 by jagoll last updated on 19/Feb/20 $${find}\:{the}\:{solution} \\ $$$$\sqrt{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}\:}\:\:>\:\:{x}−\mathrm{2}\: \\ $$ Commented by arkanmath7@gmail.com last updated on 19/Feb/20 $${x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}\:\:\:>\:\:{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{4}…
Question Number 16632 by tawa tawa last updated on 24/Jun/17 $$\mathrm{e}^{\mathrm{xy}} \:+\:\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{sin}\left(\mathrm{x}\:+\:\mathrm{y}\right) \\ $$$$\mathrm{Express}\:\mathrm{the}\:\mathrm{variable}\:\mathrm{y}\:\mathrm{interms}\:\mathrm{of}\:\mathrm{x}\:\mathrm{only}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 147697 by mathdanisur last updated on 22/Jul/21 $${f}\left(\mathrm{0}\right)+{f}\left(\mathrm{1}\right)+{f}\left(\mathrm{2}\right)+…+{f}\left({n}\right)={n}!−{n}\centerdot{a} \\ $$$${f}\left(\mathrm{16}\right)=\mathrm{15}\centerdot\left(\mathrm{15}!−\mathrm{1}\right) \\ $$$${find}\:\:\boldsymbol{{a}}=? \\ $$ Answered by Olaf_Thorendsen last updated on 22/Jul/21 $$\underset{{k}=\mathrm{0}} {\overset{{n}}…
Question Number 147699 by mathdanisur last updated on 22/Jul/21 $$\boldsymbol{{a}}^{\mathrm{2}} \:=\:\mathrm{36}^{\mathrm{7}} \:+\:\mathrm{9}^{\boldsymbol{{b}}} \:+\:\mathrm{6}^{\mathrm{8}} \:\:;\:\:\boldsymbol{{a}}\in\mathbb{Z} \\ $$$${find}\:\:\boldsymbol{{b}}=? \\ $$ Answered by Olaf_Thorendsen last updated on 22/Jul/21…
Question Number 147689 by mathdanisur last updated on 22/Jul/21 $$\left(\mathrm{234}\right)_{\mathrm{5}} \:\centerdot\:\left(\mathrm{23}\right)_{\mathrm{5}} \:=\:\left({x}\right)_{\mathrm{5}} \:\:\Rightarrow\:\:\boldsymbol{{x}}=? \\ $$ Answered by Olaf_Thorendsen last updated on 22/Jul/21 $$\left(\mathrm{234}\right)_{\mathrm{5}} .\left(\mathrm{23}\right)_{\mathrm{5}} \\…
Question Number 82149 by TawaTawa last updated on 18/Feb/20 $$\mathrm{Evaluate}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{coefficient}\:\mathrm{of}\:\:\:\:\left(\mathrm{7}\:−\:\mathrm{5x}\right)^{−\:\mathrm{3}} \\ $$ Commented by TawaTawa last updated on 18/Feb/20 $$\mathrm{Please}\:\mathrm{working}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$ Answered by mr…
Question Number 147673 by mnjuly1970 last updated on 22/Jul/21 Answered by Rasheed.Sindhi last updated on 22/Jul/21 $$\sqrt[{\mathrm{3}}]{{x}}+\sqrt[{\mathrm{3}}]{{x}−\mathrm{2}}=\sqrt[{\mathrm{3}}]{{x}−\mathrm{1}} \\ $$$$\left(\sqrt[{\mathrm{3}}]{{x}}+\sqrt[{\mathrm{3}}]{{x}−\mathrm{2}}\right)^{\mathrm{3}} =\left(\sqrt[{\mathrm{3}}]{{x}−\mathrm{1}}\right)^{\mathrm{3}} \\ $$$${x}+{x}−\mathrm{2}+\mathrm{3}\sqrt[{\mathrm{3}}]{{x}}\sqrt[{\mathrm{3}}]{{x}−\mathrm{2}}\left(\sqrt[{\mathrm{3}}]{{x}}+\sqrt[{\mathrm{3}}]{{x}−\mathrm{2}}\right)={x}−\mathrm{1} \\ $$$$\mathrm{3}\sqrt[{\mathrm{3}}]{{x}}\sqrt[{\mathrm{3}}]{{x}−\mathrm{2}}\left(\sqrt[{\mathrm{3}}]{{x}−\mathrm{1}}\right)=−{x}+\mathrm{1} \\…