Question Number 65859 by ajfour last updated on 05/Aug/19 $${x}^{\mathrm{4}} +\mathrm{5}{x}^{\mathrm{2}} +\mathrm{20}{x}+\mathrm{104}=\mathrm{0} \\ $$$${solve}\:{for}\:{x}. \\ $$ Answered by ajfour last updated on 05/Aug/19 $${a}=\mathrm{5},\:{b}=\mathrm{20},\:{c}=\mathrm{104} \\…
Question Number 65853 by ajfour last updated on 05/Aug/19 $${x}^{\mathrm{4}} −\mathrm{23}{x}^{\mathrm{2}} +\mathrm{18}{x}+\mathrm{40}=\mathrm{0} \\ $$$${solve}\:{for}\:{x}. \\ $$ Answered by ajfour last updated on 05/Aug/19 $${generally}\:\:{if}\:{x}^{\mathrm{4}} +{ax}^{\mathrm{2}}…
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Question Number 131374 by shaker last updated on 04/Feb/21 Answered by MJS_new last updated on 04/Feb/21 $${n}!\approx\left(\frac{{n}}{\mathrm{e}}\right)^{{n}} \sqrt{\mathrm{2}\pi{n}} \\ $$$$\left(\frac{\left(\mathrm{4}{n}\right)!}{\left(\mathrm{3}{n}\right)!}\right)^{\mathrm{1}/{n}} \approx\left(\frac{\left(\frac{\mathrm{4}{n}}{\mathrm{e}}\right)^{\mathrm{4}{n}} \sqrt{\mathrm{8}\pi{n}}}{\left(\frac{\mathrm{3}{n}}{\mathrm{e}}\right)^{\mathrm{3}{n}} \sqrt{\mathrm{6}\pi{n}}}\right)^{\mathrm{1}/{n}} =\frac{\left(\frac{\mathrm{4}{n}}{\mathrm{e}}\right)^{\mathrm{4}} }{\left(\frac{\mathrm{3}{n}}{\mathrm{e}}\right)^{\mathrm{3}}…
Question Number 65830 by ajfour last updated on 04/Aug/19 Commented by ajfour last updated on 04/Aug/19 $${x}^{\mathrm{4}} +{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}} +{cx}+{d}=\mathrm{0} \\ $$$${let}\:{roots}\:{be}\:−{p},−{q},−{r},−{s}. \\ $$$${Find}\:{p},{q},{r},{s}\:{in}\:{terms}\:{of}\:{a},{b},{c},{d}. \\…
Question Number 131363 by shaker last updated on 04/Feb/21 Commented by MJS_new last updated on 04/Feb/21 $$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{we}\:\mathrm{can}\:\mathrm{exactly}\:\mathrm{solve}\:\mathrm{this}.\:\mathrm{we} \\ $$$$\mathrm{need}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{1}\:\mathrm{and}\:\mathrm{in}\:\mathrm{the}\:\mathrm{next} \\ $$$$\mathrm{step}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{a}\:\mathrm{polynome}\:\mathrm{of}\:\mathrm{degree}\:\mathrm{4}\:\mathrm{if} \\ $$$$\mathrm{we}\:\mathrm{are}\:\mathrm{smart}…\:\mathrm{otherwise}\:\mathrm{degree}\:\mathrm{6} \\…
Question Number 131343 by physicstutes last updated on 03/Feb/21 $$\mathrm{A}\:\mathrm{mapping}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{as}\:{G}\rightarrow{S}\:\mathrm{where}\:\left({G},×\right)\:\mathrm{and}\:\left({S},+\right), \\ $$$$\:\mathrm{show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{mapping}\:{f}\left({x}\right)\:=\:\mathrm{ln}\:{x}\:\mathrm{is}\:\mathrm{an}\:\mathrm{isomophism}. \\ $$ Answered by mindispower last updated on 04/Feb/21 $${isomlrphisme}\:{is}\:{bijection}\:\:{morphisme} \\ $$$${without}\:{knowing}\:{G}\:{and}\:{S}\:{we}\:{can}\:{say}\:{just} \\…
Question Number 258 by mchawla last updated on 25/Jan/15 $$\mathrm{If}\:\mathrm{log}_{\mathrm{3}} \mathrm{2},\:\mathrm{log}_{\mathrm{3}} \left(\mathrm{2}^{{x}} −\mathrm{5}\right)\:\mathrm{and}\:\mathrm{log}_{\mathrm{3}} \left(\mathrm{2}^{{x}} −\frac{\mathrm{7}}{\mathrm{2}}\right)\:\mathrm{are} \\ $$$$\mathrm{in}\:\mathrm{arithmetic}\:\mathrm{progression}\:\mathrm{then}\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x}. \\ $$ Answered by 123456 last…
Question Number 248 by sushmitak last updated on 25/Jan/15 $$\mathrm{If}\:{f}\left({x}\right)=\mathrm{sin}^{\mathrm{2}} {x}+\mathrm{sin}^{\mathrm{2}} \left({x}+\pi/\mathrm{3}\right)+\mathrm{cos}\:{x}\:\mathrm{cos}\:\left({x}+\pi/\mathrm{3}\right) \\ $$$$\mathrm{and}\:\:{g}\left(\mathrm{5}/\mathrm{4}\right)=\mathrm{1}\:\mathrm{then}\:\:{gof}\left({x}\right)=? \\ $$ Answered by 123456 last updated on 17/Dec/14 $${f}\left({x}\right)=\mathrm{sin}^{\mathrm{2}} {x}+\mathrm{sin}^{\mathrm{2}}…
Question Number 131309 by shaker last updated on 03/Feb/21 Answered by Ar Brandon last updated on 03/Feb/21 $$\mathrm{z}^{\mathrm{5}} +\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{i}=\mathrm{0}\:\Rightarrow\mathrm{z}^{\mathrm{5}} +\mathrm{e}^{\frac{\pi}{\mathrm{4}}\mathrm{i}} =\mathrm{0} \\ $$$$\mathrm{z}^{\mathrm{5}} =−\mathrm{e}^{\frac{\pi}{\mathrm{4}}\mathrm{i}} =\mathrm{e}^{\frac{\pi}{\mathrm{4}}\mathrm{i}+\left(\mathrm{2k}+\mathrm{1}\right)\pi\mathrm{i}}…