Question Number 140784 by ajfour last updated on 12/May/21 $${Let}\:{the}\:{i}-{j}\:{plane}\:{be}\:{the}\:{complex} \\ $$$$\:{plane},\:{with}\:{basic}\:{operations} \\ $$$$\:\:{ij}=−{i} \\ $$$$\:\:{ji}=−{j} \\ $$$$\:\:{i}^{\mathrm{2}} =−\mathrm{1} \\ $$$$\:\:{j}^{\mathrm{2}} =−\mathrm{1} \\ $$$${z}={r}+{xi}+{yj}\:\:\:\:{w}={s}+{pi}+{qj} \\…
Question Number 9698 by rahulrai441992@gmail.com last updated on 26/Dec/16 $${a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \\ $$ Answered by geovane10math last updated on 26/Dec/16 $${a}^{\mathrm{2}} \:+\:{ab}\:+\:{b}^{\mathrm{2}} \:=\:{a}^{\mathrm{2}} \:+\:\mathrm{2}{ab}\:+\:{b}^{\mathrm{2}} \:−\:{ab}\:=…
Question Number 75218 by 21042004 last updated on 08/Dec/19 $$\mathrm{Simplify}:\:\left({x}^{\mathrm{4}} −\mathrm{3}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{12}{x}\right)\::\:\left({x}^{\mathrm{2}} +\mathrm{4}\right) \\ $$ Commented by JDamian last updated on 08/Dec/19 $${That}\:{is}\:{not}\:{a}\:{equation} \\…
Question Number 140751 by peter frank last updated on 12/May/21 Answered by Ar Brandon last updated on 12/May/21 $$\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{n}} =\underset{\mathrm{r}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\overset{\mathrm{n}} {\:}\mathrm{C}_{\mathrm{r}} \mathrm{x}^{\mathrm{r}} \\…
Question Number 75176 by 21042004 last updated on 08/Dec/19 $$\mathrm{P}=\sqrt{\mathrm{25}{x}−\mathrm{50}}−\mathrm{14}\sqrt{\frac{{x}−\mathrm{2}}{\mathrm{4}}}+\sqrt{\mathrm{9}{x}−\mathrm{18}},\:\:{x}\geqslant\mathrm{2} \\ $$$$\left.\mathrm{a}\right)\:\mathrm{Simplify}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\left.\mathrm{b}\right)\:\mathrm{Find}\:\mathrm{x}\:\mathrm{if}\:\mathrm{P}=\mathrm{3} \\ $$ Answered by Kunal12588 last updated on 08/Dec/19 $${P}=\sqrt{\mathrm{25}\left({x}−\mathrm{2}\right)}−\mathrm{14}\frac{\sqrt{{x}−\mathrm{2}}}{\:\sqrt{\mathrm{4}}}+\sqrt{\mathrm{9}\left({x}−\mathrm{2}\right)} \\…
Question Number 140709 by mathdanisur last updated on 11/May/21 $$\frac{{z}^{\mathrm{999}} −\mathrm{1}}{\left({z}^{\mathrm{2}} +\mathrm{1}\right)\centerdot\left({z}^{\mathrm{2}} +{z}+\mathrm{1}\right)} \\ $$$${find}\:{the}\:{sum}\:{of}\:{the}\:{coefficientes}\:{of} \\ $$$${the}\:{polynomial}\:{obtained}\:{bh}\:{dividing} \\ $$$${the}\:{expression}\:{by}\:{the}\:{polynomial}… \\ $$ Commented by mr W…
Question Number 75158 by peter frank last updated on 08/Dec/19 Answered by mr W last updated on 08/Dec/19 $${let}\:{u}=\sqrt{\frac{\mathrm{7}{ga}}{\mathrm{2}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{mu}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} +{mga}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$$\Rightarrow{v}^{\mathrm{2}}…
Question Number 9617 by tawakalitu last updated on 20/Dec/16 Commented by sou1618 last updated on 21/Dec/16 $${m}=\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}×\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}}×\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}}}×… \\ $$$$\mathrm{put}\:{a}_{{n}} :\:{a}_{\mathrm{1}} =\sqrt{\frac{\mathrm{1}}{\mathrm{2}}},{a}_{{n}+\mathrm{1}} =\sqrt{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{a}_{{n}} } \\ $$$${m}={a}_{\mathrm{1}}…
Question Number 140669 by john_santu last updated on 11/May/21 $$\:{x}^{\mathrm{3}} +\mathrm{1}\:=\:\mathrm{2}\:\sqrt[{\mathrm{3}\:}]{\mathrm{2}{x}−\mathrm{1}} \\ $$ Answered by bemath last updated on 11/May/21 $$\:\frac{\mathrm{x}^{\mathrm{3}} +\mathrm{1}}{\mathrm{2}}\:=\:\sqrt[{\mathrm{3}\:}]{\mathrm{2x}−\mathrm{1}} \\ $$$$\:\mathrm{set}\:\mathrm{g}\left(\mathrm{x}\right)=\:\sqrt[{\mathrm{3}\:}]{\mathrm{2x}−\mathrm{1}}\:\Rightarrow\frac{\mathrm{x}^{\mathrm{3}} +\mathrm{1}}{\mathrm{2}}\:=\:\mathrm{g}^{−\mathrm{1}}…
Question Number 75136 by chess1 last updated on 07/Dec/19 Commented by MJS last updated on 07/Dec/19 $$\frac{\mathrm{1}}{\mathrm{10}} \\ $$ Commented by mathmax by abdo last…