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y-x-2-17x-56-x-2-y-min-

Question Number 122379 by bounhome last updated on 16/Nov/20 $${y}=\frac{{x}^{\mathrm{2}} +\mathrm{17}{x}−\mathrm{56}}{{x}−\mathrm{2}} \\ $$$${y}_{{min}} =…? \\ $$ Answered by MJS_new last updated on 16/Nov/20 $${y}'=\frac{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{22}}{\left({x}−\mathrm{2}\right)^{\mathrm{2}}…

Question-122350

Question Number 122350 by shaker last updated on 16/Nov/20 Answered by liberty last updated on 16/Nov/20 $$\:\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\frac{{x}^{{a}} −{a}^{{x}} }{{a}^{{x}} −{a}^{{a}} }\:=\:\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:\frac{{ax}^{{a}−\mathrm{1}} −{a}^{{x}} .\mathrm{ln}\:{a}}{{a}^{{x}}…

Question-187857

Question Number 187857 by thean last updated on 23/Feb/23 Answered by a.lgnaoui last updated on 23/Feb/23 $$\left.{a}\right){z}^{\mathrm{2}} =−{i}=\mathrm{cos}\:\left(−\frac{\pi}{\mathrm{2}}\right)+{i}\mathrm{sin}\:\left(−\frac{\pi}{\mathrm{2}}\right) \\ $$$${z}=\left[\mathrm{cos}\:\left(−\frac{\pi}{\mathrm{2}}\right)+{i}\mathrm{sin}\:\left(−\frac{\pi}{\mathrm{2}}\right)\right]^{\frac{\mathrm{1}}{\mathrm{2}}} = \\ $$$$\mathrm{cos}\:\left(−\frac{\pi}{\mathrm{4}}\right)+{i}\mathrm{sin}\:\left(−\frac{\pi}{\mathrm{4}}\right)= \\ $$$$\:\:\:\:{arg}\left({z}\right)=\left(−\frac{\pi}{\mathrm{4}};\frac{\mathrm{3}\pi}{\mathrm{4}}\right)…

Given-that-1-1-Prove-that-without-using-the-exact-value-of-1-1-Thank-you-

Question Number 56772 by Hassen_Timol last updated on 23/Mar/19 $$\mathrm{Given}\:\mathrm{that}\:: \\ $$$$\:\:\:\:\frac{\mathrm{1}}{\:\Phi\:}\:\:=\:\:\frac{\Phi}{\:\mathrm{1}\:+\:\Phi\:} \\ $$$$ \\ $$$$\mathrm{Prove}\:\mathrm{that}\::\:{without}\:{using}\:{the}\:{exact} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{value}\:{of}\:\Phi… \\ $$$$\:\:\:\:\:\frac{\:\mathrm{1}\:}{\:\Phi\:}\:\:=\:\:\Phi\:−\:\mathrm{1} \\ $$$$ \\ $$$$ \\…

Question-122295

Question Number 122295 by ZiYangLee last updated on 15/Nov/20 Answered by som(math1967) last updated on 15/Nov/20 $$\mathrm{tan}\left(\mathrm{A}+\mathrm{B}\right) \\ $$$$=\frac{\mathrm{sin}\left(\mathrm{A}+\mathrm{B}\right)}{\mathrm{cos}\left(\mathrm{A}+\mathrm{B}\right)}=\frac{\mathrm{sinAcosB}+\mathrm{cosAsinB}}{\mathrm{cosAcosB}−\mathrm{sinASinB}} \\ $$$$=\frac{\frac{\mathrm{sinAcosB}+\mathrm{cosAsinB}}{\mathrm{cosAcosB}}}{\frac{\mathrm{cosAcosB}−\mathrm{sinAsinB}}{\mathrm{cosAcosB}}} \\ $$$$=\frac{\mathrm{tanA}+\mathrm{tanB}}{\mathrm{1}−\mathrm{tanAtanB}} \\ $$$$\mathrm{now}…