Question Number 78820 by mathocean1 last updated on 20/Jan/20 $$\mathrm{please}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{fomula}\:\mathrm{to}\: \\ $$$$\mathrm{determinate}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{of}\: \\ $$$$\mathrm{bissectors}\:\mathrm{in}\:\mathrm{triangle}? \\ $$ Answered by MJS last updated on 21/Jan/20 $$\mathrm{bisector}\:\mathrm{between}\:\mathrm{2}\:\mathrm{lines} \\…
Question Number 144359 by mathdanisur last updated on 24/Jun/21 $$\int\:\frac{{x}^{\mathrm{4}} {e}^{{x}} \:{dx}}{\left({x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{3}} +\mathrm{12}{x}^{\mathrm{2}} +\mathrm{24}{x}+\mathrm{24}+\mathrm{72}{e}^{{x}} \right)^{\mathrm{2}} }\:=\:? \\ $$ Answered by mitica last updated on…
Question Number 78814 by M±th+et£s last updated on 20/Jan/20 Commented by mind is power last updated on 21/Jan/20 $$\mathrm{2}{nd}\:{Way} \\ $$$$\alpha^{\mathrm{2}} \beta^{\mathrm{2}} \gamma^{\mathrm{2}} \delta^{\mathrm{2}} +\alpha^{\mathrm{2}}…
Question Number 144340 by santoshachary1985 last updated on 24/Jun/21 $$ \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{2ab}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{2ab}−\mathrm{3}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{2ab}−\mathrm{3}=\mathrm{0} \\ $$$$\:\mathrm{3} \\ $$…
Question Number 144334 by mathdanisur last updated on 24/Jun/21 $${Reduct}\:{it}:\:\:\frac{\boldsymbol{{z}}^{\mathrm{8}} \:+\:\boldsymbol{{z}}\:+\:\mathrm{1}}{\boldsymbol{{z}}^{\mathrm{5}} \:+\:\boldsymbol{{z}}\:+\:\mathrm{1}} \\ $$ Answered by Olaf_Thorendsen last updated on 24/Jun/21 $$\mathrm{Z}\:=\:\frac{{z}^{\mathrm{8}} +{z}+\mathrm{1}}{{z}^{\mathrm{5}} +{z}+\mathrm{1}} \\…
Question Number 144300 by mathdanisur last updated on 24/Jun/21 Answered by Rasheed.Sindhi last updated on 25/Jun/21 $$\left({x}+\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} −{a}^{\mathrm{2}{n}+\mathrm{1}} ={x}\left({x}+\mathrm{1}\right){Q}\left({x}\right)+{ux}+{v} \\ $$$${For}\:{x}\in\mathbb{Z} \\ $$$${For}\:{x}=\mathrm{0} \\ $$$$\mathrm{1}−{a}^{\mathrm{2}{n}+\mathrm{1}}…
Question Number 78762 by ajfour last updated on 20/Jan/20 $$\mathrm{3}{acr}^{\mathrm{2}} \left(\mathrm{1}−{r}\right)+\mathrm{3}{apr}\left(\mathrm{1}−{r}\right)\left({pa}+{qb}\right) \\ $$$$+\mathrm{3}{bqr}\left(\mathrm{1}−{r}\right)\left({pa}+{qb}\right) \\ $$$$\:\:\:=\:\mathrm{3}\left(\mathrm{1}−{r}\right)^{\mathrm{2}} \left({pa}+{qb}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$${Find}\:{p},\:{q},\:{r}\:{such}\:{that}\:{the}\:{equation} \\ $$$${is}\:{satisfied}\:{for}\:{general}\:{any} \\ $$$${values}\:{of}\:{a},{b},{c}.\: \\…
Question Number 13223 by Tinkutara last updated on 16/May/17 $$\mathrm{If}\:{a},\:{b},\:{c}\:\mathrm{are}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{triangle}\:\mathrm{show}\:\mathrm{that} \\ $$$$\left(\mathrm{1}\:+\:\frac{{b}−{c}}{{a}}\right)^{{a}} \left(\mathrm{1}\:+\:\frac{{c}−{a}}{{b}}\right)^{{b}} \left(\mathrm{1}\:+\:\frac{{a}−{b}}{{c}}\right)^{{c}} \:<\:\mathrm{1} \\ $$ Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 17/May/17 $${if}:\:\:{a}={b}={c}\Rightarrow{RHS}=\mathrm{1}\nless{LHS}=\mathrm{1}…
Question Number 78755 by TawaTawa last updated on 20/Jan/20 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\:\:\:\:\mathrm{xy}\:+\:\mathrm{5x}\:+\:\mathrm{5y}\:\:=\:\:−\:\mathrm{25}\:\:\:\:\:\:…\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\mathrm{yz}\:+\:\mathrm{3y}\:+\:\mathrm{5z}\:\:=\:\:−\:\mathrm{15}\:\:\:\:\:\:…\:\left(\mathrm{ii}\right) \\ $$$$\:\:\:\:\mathrm{xz}\:+\:\mathrm{5z}\:+\:\mathrm{3x}\:\:=\:\:−\:\mathrm{15}\:\:\:\:\:\:…\:\left(\mathrm{iii}\right) \\ $$ Commented by Tony Lin last updated on…
Question Number 144264 by loveineq last updated on 24/Jun/21 $$\mathrm{Let}\:{a},{b}\:>\:\mathrm{0}\:\mathrm{and}\:\mathrm{2}{a}+{b}\:=\:\mathrm{3}.\:\mathrm{Prove}\:\mathrm{the}\:\mathrm{followings}:\:\:\:\:\:\:\:\:\:\: \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}}{{n}}{a}\left({b}+\mathrm{4}\right)+\mathrm{3}{b}^{\frac{\mathrm{1}}{{n}}} \:\leqslant\:\frac{\mathrm{10}+\mathrm{3}{n}}{{n}},\:\forall{n}\in\mathbb{N}^{+} \geqslant\mathrm{1}. \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{na}\left({b}+\mathrm{4}\right)+\mathrm{3}{b}^{{n}} \:\geqslant\:\mathrm{10}{n}+\mathrm{3},\:\forall{n}\in\mathbb{N}^{+} \geqslant\mathrm{2}. \\ $$$$ \\ $$ Terms of Service…