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Prove-arctan1-arctan2-arctan3-pi-

Question Number 151943 by Huy last updated on 24/Aug/21 $$\mathrm{Prove}\:\mathrm{arctan1}+\mathrm{arctan2}+\mathrm{arctan3}=\pi \\ $$ Commented by puissant last updated on 24/Aug/21 $${x}={arctan}\mathrm{1}\:,\:{y}={arctan}\mathrm{2}\:,\:{z}={arctan}\mathrm{3} \\ $$$${p}={x}+{y} \\ $$$$\Rightarrow\:{tan}\left({p}\right)={tan}\left({x}+{y}\right)=\frac{{tanx}+{tany}}{\mathrm{1}−{tanxtany}} \\…

Question-151876

Question Number 151876 by tabata last updated on 23/Aug/21 Answered by Olaf_Thorendsen last updated on 23/Aug/21 $$\mathrm{A}\:=\:\mathrm{0}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{5}}+\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{5}}{\mathrm{7}}+… \\ $$$$\mathrm{A}\:=\:\frac{\mathrm{0}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{5}}+\frac{\mathrm{4}}{\mathrm{6}}+\frac{\mathrm{5}}{\mathrm{7}}+… \\ $$$$\mathrm{A}\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}}{{n}+\mathrm{2}}\:\:\:\mathrm{diverges} \\ $$…

4x-y-11-2y-5x-5-x-2z-9-z-y-x-

Question Number 20798 by ANTARES_VY last updated on 03/Sep/17 $$\begin{cases}{\mathrm{4}\boldsymbol{{x}}−\boldsymbol{{y}}=\mathrm{11}}\\{\mathrm{2}\boldsymbol{{y}}+\mathrm{5}\boldsymbol{{x}}=\mathrm{5}}\\{\boldsymbol{{x}}−\mathrm{2}\boldsymbol{{z}}=\mathrm{9}}\end{cases}\:\:\:\:\boldsymbol{{z}}+\boldsymbol{{y}}−\boldsymbol{{x}}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question-20744

Question Number 20744 by mondodotto@gmail.com last updated on 02/Sep/17 Answered by $@ty@m last updated on 02/Sep/17 $${ATQ}, \\ $$$${Let}\:{the}\:{equation}\:{of}\:{the}\:{st}.\:{line}\:{be} \\ $$$${y}={mx}+{c}\:−−\left(\mathrm{1}\right) \\ $$$${It}\:{passes}\:{through}\:\left(\mathrm{4},−\mathrm{2}\right) \\ $$$$\therefore\:−\mathrm{2}=\mathrm{4}{m}+{c}…

1-Let-R-be-a-relation-on-a-set-A-1-2-3-4-5-6-defined-by-R-a-b-a-b-9-then-find-A-R-and-R-1-B-domain-and-range-of-R-and-R-1-C-is-R-R-1-

Question Number 151805 by abdurehime last updated on 23/Aug/21 $$\mathrm{1}.\mathrm{Let}\:\mathrm{R}\:\mathrm{be}\:\mathrm{a}\:\mathrm{relation}\:\mathrm{on}\:\mathrm{a}\:\mathrm{set}\: \\ $$$$\mathrm{A}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6}\right\}\:\mathrm{defined}\:\mathrm{by}\: \\ $$$$\mathrm{R}\left\{\left(\mathrm{a},\mathrm{b}\right):\mathrm{a}+\mathrm{b}\leqslant\mathrm{9}\:\mathrm{then}\:\mathrm{find}\right. \\ $$$$\mathrm{A}.\:\:\mathrm{R}\:\mathrm{and}\:\mathrm{R}^{−\mathrm{1}} \\ $$$$\mathrm{B}.\:\:\mathrm{domain}\:\mathrm{and}\:\mathrm{range}\:\mathrm{of}\:\mathrm{R}\:\mathrm{and}\:\mathrm{R}^{−\mathrm{1}} \\ $$$$\mathrm{C}.\mathrm{is}\:\mathrm{R}=\mathrm{R}^{−\mathrm{1}} ?? \\ $$$$ \\ $$…