Question Number 131084 by greg_ed last updated on 01/Feb/21 $$\mathrm{let}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{in}\:\mathbb{R}\::\: \\ $$$${x}^{\mathrm{4}} −{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0}. \\ $$ Answered by liberty last updated on 01/Feb/21 $$\mathrm{x}_{\mathrm{1}}…
Question Number 17 by user1 last updated on 25/Jan/15 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{regression}\:\mathrm{coefficient}\:{b}_{{xy}} \:\mathrm{between} \\ $$$${x}\:\mathrm{and}\:{y}\:\mathrm{for}\:\mathrm{the}\:\mathrm{following}\:\mathrm{data}: \\ $$$$\Sigma{x}=\mathrm{24},\:\Sigma{y}=\mathrm{44},\:\Sigma{xy}=\mathrm{306},\:\Sigma{x}^{\mathrm{2}} =\mathrm{164}, \\ $$$$\Sigma{y}^{\mathrm{2}} =\mathrm{574}\:\mathrm{and}\:{n}=\mathrm{4}. \\ $$ Answered by user1 last…
Question Number 131075 by benjo_mathlover last updated on 01/Feb/21 $$\mathrm{If}\:\mathrm{a}−\mathrm{3}=−\mathrm{b}−\mathrm{4}=−\mathrm{c}−\mathrm{5}=\mathrm{d}+\mathrm{6}=\mathrm{e}+\mathrm{7}= \\ $$$$\mathrm{a}−\mathrm{b}−\mathrm{c}+\mathrm{d}+\mathrm{e}+\mathrm{8}\:\mathrm{then}\:\mathrm{a}−\mathrm{b}−\mathrm{c}+\mathrm{d}+\mathrm{e}\:=? \\ $$ Answered by EDWIN88 last updated on 01/Feb/21 $${a}\:={k}+\mathrm{3}\:;\:{b}=−\mathrm{4}−{k}\:;\:{c}=−{k}−\mathrm{5} \\ $$$${d}={k}−\mathrm{6}\:;\:{e}={k}−\mathrm{7} \\…
Question Number 144083 by mathdanisur last updated on 21/Jun/21 $${if}\:\:{a};{b};{c}>\mathrm{0}\:\:{then}: \\ $$$$\frac{\mathrm{9}{abc}}{{a}^{\mathrm{3}} \:+\:{b}^{\mathrm{3}} \:+\:{c}^{\mathrm{3}} }\:+\:\frac{{a}^{\mathrm{2}} }{{bc}}\:+\:\frac{{b}^{\mathrm{2}} }{{ca}}\:+\:\frac{{c}^{\mathrm{2}} }{{ab}}\:\geqslant\:\mathrm{6} \\ $$ Answered by mitica last updated…
Question Number 13002 by Joel577 last updated on 10/May/17 $$\mathrm{3}^{\left({x}\:−\:\mathrm{3}\right)\left({x}\:−\:{y}\:−\:\mathrm{2}\right)} \:=\:\mathrm{1} \\ $$$$\mathrm{5}^{\left({x}^{\mathrm{2}} \:−\:\mathrm{2}{xy}\:+\:{y}^{\mathrm{2}} \:+\:{x}\:−\:{y}\:−\:\mathrm{3}/\mathrm{2}\right)} \:=\:\sqrt{\mathrm{5}} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x}\:\mathrm{and}\:{y} \\ $$ Answered by 433 last updated…
Question Number 144049 by physicstutes last updated on 20/Jun/21 $$\:\mathrm{Given}\:\mathrm{the}\:\mathrm{equation}\:\:\mathrm{1000}\:=\:\mathrm{2000}\left(\frac{\mathrm{1}−\left(\mathrm{1}+{t}\right)^{−{n}} }{{t}}\right) \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{t}. \\ $$ Answered by Olaf_Thorendsen last updated on 21/Jun/21 $$\mathrm{1000}\:=\:\mathrm{2000}\left(\frac{\mathrm{1}−\left(\mathrm{1}+{t}\right)^{−{n}} }{{t}}\right) \\…
Question Number 78503 by jagoll last updated on 18/Jan/20 $${if}\:{x},{y}\:>\mathrm{1}\: \\ $$$${prove}\:\frac{{x}^{\mathrm{2}} }{{y}−\mathrm{1}}+\frac{{y}^{\mathrm{2}} }{{x}−\mathrm{1}}\geqslant\mathrm{8} \\ $$ Answered by ~blr237~ last updated on 18/Jan/20 $$\:\mathrm{We}\:\mathrm{know}\:\mathrm{that}\:\mathrm{for}\:\mathrm{all}\:\mathrm{a},\mathrm{b}\in\mathbb{R}\: \\…
Question Number 144031 by mathdanisur last updated on 20/Jun/21 Answered by mindispower last updated on 20/Jun/21 $$\frac{\mathrm{1}}{{sin}\left(\mathrm{2}^{{k}} \right)}+\frac{\mathrm{1}}{{tg}\left(\mathrm{2}^{{k}} \right)}=\frac{\mathrm{1}+{cos}\left(\mathrm{2}{k}\right)}{{sin}\left(\mathrm{2}{k}\right)}=\frac{\mathrm{1}}{{tg}\left(\mathrm{2}^{{k}−\mathrm{1}} \right)} \\ $$$$\frac{\mathrm{1}}{{sin}\left(\mathrm{2}^{{k}} \right)}=\frac{\mathrm{1}}{{tg}\left(\mathrm{2}^{{k}−\mathrm{1}} \right)}−\frac{\mathrm{1}}{{tg}\left(\mathrm{2}^{{k}} \right)}…
Question Number 144021 by mathdanisur last updated on 20/Jun/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 78465 by TawaTawa last updated on 17/Jan/20 Answered by mind is power last updated on 17/Jan/20 $$\zeta\left(\mathrm{z}\right)=\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{z}} } \\ $$$$\mathrm{Re}\left(\zeta\left(\mathrm{z}\right)\right)=\mathrm{Re}\left\{\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{Re}\left(\mathrm{z}\right)+\mathrm{iIm}\left(\mathrm{z}\right)} }\right\}…