Question Number 179328 by bekkiyor last updated on 28/Oct/22 $${y}\left({x}\right)={cos}\mathrm{2}{x},\:\:\left(\frac{{d}}{{dx}}{y}\left({x}\right)\right)^{{n}} =\:? \\ $$ Answered by pablo1234523 last updated on 28/Oct/22 $${y}=\mathrm{cos}\:\mathrm{2}{x} \\ $$$$\frac{{dy}}{{dx}}=−\mathrm{2sin}\:\mathrm{2}{x} \\ $$$$\left(\frac{{dy}}{{dx}}\right)^{{n}}…
Question Number 179327 by greougoury555 last updated on 28/Oct/22 $$\:{Find}\:{polynomial}\:{u},{v}\:\in{Q}\left[{x}\right]\:{such} \\ $$$$\:\:{that}\:\left({x}^{\mathrm{4}} −\mathrm{1}\right){u}\left({x}\right)+\left({x}^{\mathrm{7}} −\mathrm{1}\right){v}\left({x}\right)=\left({x}−\mathrm{1}\right) \\ $$ Answered by manxsol last updated on 28/Oct/22 $$\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{3}} +{x}^{\mathrm{2}}…
Question Number 179302 by mnjuly1970 last updated on 28/Oct/22 Commented by Frix last updated on 28/Oct/22 $${p}\left({x}\right){p}\left(\frac{\mathrm{1}}{{x}}\right)={p}\left({x}\right)+{p}\left(\frac{\mathrm{1}}{{x}}\right) \\ $$$$\Rightarrow \\ $$$${p}\left({x}\right)=\frac{{p}\left(\frac{\mathrm{1}}{{x}}\right)}{{p}\left(\frac{\mathrm{1}}{{x}}\right)−\mathrm{1}} \\ $$$${p}\left({x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{polynomial}\:\Rightarrow \\ $$$${p}\left({x}\right)={c}_{\mathrm{0}}…
Question Number 113767 by Khanacademy last updated on 15/Sep/20 $$ \\ $$$$ \\ $$ Commented by Khanacademy last updated on 15/Sep/20 $$\left(\boldsymbol{{AB}}//\boldsymbol{{CD}}\right)\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\boldsymbol{{E}}\:\epsilon\boldsymbol{{BC}} \\…
Question Number 48224 by gunawan last updated on 21/Nov/18 $${e}^{{z}} =\mathrm{1}−\sqrt{\mathrm{3}}{i} \\ $$$${z}=.. \\ $$ Answered by Smail last updated on 21/Nov/18 $${z}={ln}\left(\mathrm{1}−\sqrt{\mathrm{3}}{i}\right)={ln}\left(\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\right) \\ $$$$={ln}\mathrm{2}+{ln}\left({e}^{−{i}\frac{\pi}{\mathrm{3}}}…
Question Number 48225 by gunawan last updated on 21/Nov/18 $$\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{exp}\left(\frac{\mathrm{2}+\pi\mathrm{i}}{\mathrm{4}}\right)=\sqrt{\frac{{e}}{\mathrm{2}}}\left(\mathrm{1}+{i}\right) \\ $$$$\mathrm{cos}\:\left({z}_{\mathrm{1}} +{z}_{\mathrm{2}} \right)=\mathrm{cos}\:{z}_{\mathrm{1}} \mathrm{cos}\:{z}_{\mathrm{2}} −\mathrm{sin}\:{z}_{\mathrm{1}} \mathrm{sin}\:{z}_{\mathrm{2}} \\ $$ Answered by Smail last…
Question Number 48222 by gunawan last updated on 21/Nov/18 $${f}\left({x}\right)=\underset{{i}=\mathrm{0}} {\overset{{n}} {\Sigma}}{a}_{{i}} {x}^{{i}} ={a}_{{n}} {x}^{{n}} +{a}_{{n}−\mathrm{1}} {x}^{{n}−\mathrm{1}} +{a}_{{n}−\mathrm{2}} {x}^{{n}−\mathrm{2}} +\ldots+{a}_{\mathrm{2}} {x}^{\mathrm{2}} +{a}_{\mathrm{1}} {x}+{a}_{\mathrm{0}} \\ $$$${f}^{−\mathrm{1}}…
Question Number 179284 by Agnibhoo98 last updated on 27/Oct/22 $$\mathrm{Solve}\:\mathrm{for}\:\boldsymbol{{x}} \\ $$$$\frac{{a}}{{ax}\:−\:\mathrm{1}}\:+\:\frac{{b}}{{bx}\:−\:\mathrm{1}}\:=\:{a}\:+\:{b}\:\left[{x}\:\neq\:\frac{\mathrm{1}}{{a}},\:\frac{\mathrm{1}}{{b}}\right] \\ $$ Answered by FelipeLz last updated on 27/Oct/22 $$\frac{{a}\left({bx}−\mathrm{1}\right)+{b}\left({ax}−\mathrm{1}\right)}{\left({ax}−\mathrm{1}\right)\left({bx}−\mathrm{1}\right)}\:=\:{a}+{b} \\ $$$$\frac{\mathrm{2}{abx}−\left({a}+{b}\right)}{{abx}^{\mathrm{2}} −\left({a}+{b}\right){x}+\mathrm{1}}\:=\:{a}+{b}…
Question Number 179282 by Shrinava last updated on 27/Oct/22 $$\mathrm{m}^{\mathrm{2}} \:=\:\mathrm{n}\:+\:\mathrm{2} \\ $$$$\mathrm{n}^{\mathrm{2}} \:=\:\mathrm{m}\:+\:\mathrm{2} \\ $$$$\mathrm{4mn}\:−\:\mathrm{m}^{\mathrm{3}} \:−\:\mathrm{n}^{\mathrm{3}} \:=\:?\:\:\:\left(\mathrm{m}\neq\mathrm{n}\right) \\ $$ Answered by Acem last updated…
Question Number 48204 by ajfour last updated on 20/Nov/18 $$\mathrm{2}\left({x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}\right)\left({y}^{\mathrm{4}} −\mathrm{3}{y}^{\mathrm{2}} +\mathrm{4}\right)=\mathrm{7} \\ $$$${Find}\:\left({x},{y}\right)\:. \\ $$ Answered by ajfour last updated on 20/Nov/18…