Question Number 142325 by loveineq last updated on 30/May/21 $$\mathrm{Given}\:\mathrm{that}\:{a}\:\geqslant\:\mathrm{1}\:\geqslant\:{b}\:>\:\mathrm{0}.\:\mathrm{Prove}\:\mathrm{the}\:\mathrm{followings}:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\left({a}−{b}\right)^{\mathrm{2}} \:\leqslant\:\left({a}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{1}−{b}\right)^{\mathrm{2}} \:\leqslant\:\left({a}−{b}\right)^{\mathrm{2}} \:\:\:\: \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}}\left({a}−{b}\right)^{\mathrm{3}} \:\leqslant\:\left({a}−\mathrm{1}\right)^{\mathrm{3}} +\left(\mathrm{1}−{b}\right)^{\mathrm{3}} \:\leqslant\:\left({a}−{b}\right)^{\mathrm{3}} \\ $$$$ \\ $$…
Question Number 76790 by john santu last updated on 30/Dec/19 $${calculate}\:\frac{{sin}\mathrm{72}^{{o}} +{sin}\mathrm{148}^{{o}} +{sin}\mathrm{140}^{{o}} }{{sin}\mathrm{36}^{{o}} ×{sin}\mathrm{74}^{{o}} ×{sin}\mathrm{70}^{{o}} }. \\ $$ Commented by benjo 1/2 santuyy last…
Question Number 11255 by 786786AM last updated on 18/Mar/17 $$\mathrm{If}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{4}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{an}\:\mathrm{A}.\mathrm{P}.,\:\mathrm{is}\:\mathrm{p},\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{8}\:\mathrm{terms}\:\mathrm{is}\:\mathrm{q}\:\mathrm{and}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first} \\ $$$$\mathrm{12}\:\mathrm{terms}\:\mathrm{is}\:\mathrm{r},\:\mathrm{express}\:\left(\mathrm{3p}+\mathrm{r}\right)\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{q}. \\ $$ Answered by ajfour last updated on 18/Mar/17 $$\mathrm{p}=\mathrm{2}\left(\mathrm{2a}+\mathrm{3d}\right) \\ $$$$\mathrm{q}=\mathrm{4}\left(\mathrm{2a}+\mathrm{7d}\right) \\…
Question Number 76791 by john santu last updated on 30/Dec/19 $${calculate}\:\frac{{sin}\mathrm{44}^{{o}} +{sin}\mathrm{66}^{{o}} +{sin}\mathrm{70}^{{o}} }{{cos}\mathrm{22}^{{o}} ×{cos}\mathrm{33}^{{o}} ×{cos}\mathrm{35}^{{o}} }. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 142326 by bramlexs22 last updated on 30/May/21 Answered by EDWIN88 last updated on 30/May/21 $$\:\:\frac{\mathrm{180}\left(\mathrm{n}−\mathrm{2}\right)}{\mathrm{n}}\:=\:\mathrm{4}\left(\frac{\cancel{\mathrm{180}}\:^{\mathrm{36}} }{\cancel{\mathrm{5}}}\:\right) \\ $$$$\:\frac{\mathrm{180n}−\mathrm{360}}{\mathrm{n}}\:=\:\mathrm{144}\:\Rightarrow\mathrm{180n}−\mathrm{360}=\mathrm{144n} \\ $$$$\Rightarrow\:\mathrm{36n}\:=\:\mathrm{360}\:;\:\mathrm{n}=\mathrm{10} \\ $$$$\: \\…
Question Number 11249 by uni last updated on 18/Mar/17 $${x}\in\left(\mathrm{0},\mathrm{2}\pi\right) \\ $$$$\mathrm{2cos}^{\mathrm{2}} {x}\:+{sinx}−\mathrm{1}=\mathrm{0}\:\Rightarrow\Sigma{x}=?\: \\ $$ Answered by ajfour last updated on 18/Mar/17 $$\mathrm{2}−\mathrm{2sin}\:^{\mathrm{2}} \mathrm{x}+\mathrm{sin}\:\mathrm{x}−\mathrm{1}=\mathrm{0} \\…
Question Number 76782 by mathmax by abdo last updated on 30/Dec/19 $${calculate}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{x}} \:\:\frac{{sin}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }{dx} \\ $$ Answered by mind is power last…
Question Number 76783 by mathmax by abdo last updated on 30/Dec/19 $${let}\:{f}\left({x}\right)={x}^{\mathrm{3}} \:\:\:\:,\mathrm{2}\pi\:{periodic}\:{odd}\:{developp}\:{f}\:{at}\:{fourier}\:{serie} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 76780 by mathmax by abdo last updated on 30/Dec/19 $${find}\:{A}=\int_{−\infty} ^{+\infty} \:{x}\:{e}^{−{x}^{\mathrm{2}} } {arctan}\left({x}−\frac{\mathrm{1}}{{x}}\right){dx} \\ $$ Commented by mathmax by abdo last updated…
Question Number 11245 by uni last updated on 18/Mar/17 $${cos}\mathrm{10}×{cos}\mathrm{20}×{cos}\mathrm{40}=? \\ $$ Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 18/Mar/17 $$\frac{\mathrm{2}{sin}\mathrm{10}.{cos}\mathrm{10}.{cos}\mathrm{20}.{cos}\mathrm{40}}{\mathrm{2}{sin}\mathrm{10}}= \\ $$$$\frac{{sin}\mathrm{20}.{cos}\mathrm{20}.{cos}\mathrm{40}}{\mathrm{2}{sin}\mathrm{10}}=\frac{\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{40}.{cos}\mathrm{40}}{\mathrm{2}{sin}\mathrm{10}}= \\ $$$$\frac{\frac{\mathrm{1}}{\mathrm{4}}{sin}\mathrm{80}}{\mathrm{2}{sin}\mathrm{10}}=\frac{\mathrm{1}}{\mathrm{8}}.\frac{{cos}\mathrm{10}}{{sin}\mathrm{10}}=\frac{\mathrm{1}}{\mathrm{8}}\mathrm{cot}\:\mathrm{10}\:\:\blacksquare \\…