Question Number 72161 by aliesam last updated on 25/Oct/19 Commented by aliesam last updated on 25/Oct/19 $${measure}\:{in}\:{radians}\:{of}\:{angle}\:{a}\: \\ $$$${respectively}\:{length}\:{of}\:{median}\: \\ $$$${from}\:{a} \\ $$ Commented by…
Question Number 6593 by Tawakalitu. last updated on 04/Jul/16 Commented by Yozzii last updated on 05/Jul/16 $${Given}\:{that}\:{p}\:{is}\:{prime}\:{and}\:{p}\leqslant\mathrm{100}, \\ $$$${show}\:{that}\:{there}\:{are}\:{solutions}\:\left({x},{y}\right),\: \\ $$$${x},{y}\in\mathbb{Z},\:{for}\:{which}\:{y}^{\mathrm{37}} \equiv{x}^{\mathrm{3}} +\mathrm{37}\:\left({mod}\:{p}\right). \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−…
Question Number 6487 by Temp last updated on 29/Jun/16 $$\mathrm{Prove}\:\mathrm{and}\:\mathrm{show}\:\mathrm{why}: \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{H}_{{k}} =\left({n}+\mathrm{1}\right)\left({H}_{{n}+\mathrm{1}} −\mathrm{1}\right) \\ $$$$\mathrm{Where}: \\ $$$${H}_{{k}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+…+\frac{\mathrm{1}}{{k}} \\ $$ Commented by…
Question Number 71984 by Henri Boucatchou last updated on 23/Oct/19 $$\:{x},\:{y}\:\in\:{I}\:\subset\:\mathbb{R}\:\:{prove}\:\:{that}\:\:\mid{cosx}\:−\:{siny}\mid\:\leqslant\:\mid{x}\:−\:{y}\mid \\ $$ Answered by mind is power last updated on 23/Oct/19 $$\mathrm{x}=\mathrm{y}=\mathrm{0}\:\mathrm{mistack} \\ $$$$\Rightarrow\mathrm{1}\leqslant\mathrm{0}…
Question Number 71925 by Henri Boucatchou last updated on 22/Oct/19 $$\:\boldsymbol{{Solve}}\:\:\sqrt{\mathrm{1}\:−\:\boldsymbol{{x}}^{\mathrm{2}} }\:=\:\boldsymbol{{x}}^{\mathrm{4}/\mathrm{3}} \\ $$ Answered by MJS last updated on 23/Oct/19 $${x}={t}^{\mathrm{3}/\mathrm{2}} \\ $$$$\sqrt{\mathrm{1}−{t}^{\mathrm{3}} }={t}^{\mathrm{2}}…
Question Number 71886 by aliesam last updated on 21/Oct/19 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 6340 by sanusihammed last updated on 24/Jun/16 Commented by nburiburu last updated on 24/Jun/16 $${seems}\:{it}\:{need}\:{more}\:{info}.\:{For}\:{the}\:{moment}\:{being}\:{could}\:{be}\:{that}\:{moving}\:{one}\:{unit}\:{from}\:{right}\:{to}\:{left}\:{adds}\:\mathrm{2}\:{units}\:{up}\:{and}\:{cost}\:\mathrm{1}\:{in}\:{the}\:{centre}. \\ $$$${However}\:{many}\:{other}\:{posibilities}\:{could}\:{explain}\:{the}\:{numbers}\:{in}\:{the}\:{diagram}. \\ $$ Commented by prakash jain…
Question Number 6311 by FilupSmith last updated on 23/Jun/16 $$\mathrm{According}\:\mathrm{to}\:\mathrm{WolframAlpha}: \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\prod}}\left(\mathrm{1}−{x}^{\left(−\mathrm{1}\right)^{{k}} } \right)=\left(\mathrm{1}−{x}\right)^{\lfloor\frac{{n}}{\mathrm{2}}\rfloor+\mathrm{1}} \left(\frac{{x}−\mathrm{1}}{{x}}\right)^{\lfloor\frac{{n}−\mathrm{1}}{\mathrm{2}}\rfloor+\mathrm{1}} \\ $$$$ \\ $$$$\mathrm{Can}\:\mathrm{anyone}\:\mathrm{work}\:\mathrm{out}\:\mathrm{how}? \\ $$ Commented by…
Question Number 6289 by sanusihammed last updated on 22/Jun/16 Commented by FilupSmith last updated on 22/Jun/16 $${a}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{5}}{\mathrm{6}}+…+\frac{\mathrm{2015}}{\mathrm{2016}} \\ $$$${b}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{4}}×\frac{\mathrm{4}}{\mathrm{6}}×…×\frac{\mathrm{2015}}{\mathrm{2016}} \\ $$$$\frac{{a}}{{b}}+\frac{{b}}{{a}}=?? \\ $$$$ \\ $$$${a}=\underset{{n}=\mathrm{1}}…
Question Number 71790 by jatin123 last updated on 20/Oct/19 Answered by $@ty@m123 last updated on 20/Oct/19 $$=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\mathrm{3}}{\mathrm{4}}×…..×\frac{\mathrm{98}}{\mathrm{99}}×\frac{\mathrm{99}}{\mathrm{100}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{100}}=\frac{\mathrm{1}}{\mathrm{50}} \\ $$ Commented by jatin123 last…