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Category: Arithmetic

Question-68203

Question Number 68203 by peter frank last updated on 07/Sep/19 Commented by Cmr 237 last updated on 07/Sep/19 $$\eth\mathrm{f}=\frac{\eth\mathrm{f}}{\eth\mathrm{x}}+\frac{\eth\mathrm{f}}{\eth\mathrm{y}}+\frac{\eth\mathrm{f}}{\eth\mathrm{z}} \\ $$$$\:\:\:\:=\mathrm{2xy}+\mathrm{cosh}\left(\mathrm{yz}\right)+\mathrm{x}^{\mathrm{2}} +\mathrm{xzsinh}\left(\mathrm{yz}\right)+\mathrm{xysinh}\left(\mathrm{yz}\right) \\ $$ Commented…

Question-68191

Question Number 68191 by peter frank last updated on 06/Sep/19 Answered by mind is power last updated on 06/Sep/19 $${sin}\left({a}−{b}\right)={sin}\left({a}\right){cos}\left({b}\right)−{cos}\left({a}\right){sin}\left({b}\right) \\ $$$$\frac{{sin}\left({a}−{b}\right)}{{cos}\left({a}\right){cos}\left({b}\right)}={tg}\left({a}\right)−{tg}\left({b}\right) \\ $$$$\Rightarrow\frac{{sin}\left({a}−{b}\right)}{{cos}\left({a}\right){cos}\left({b}\right)}={tg}\left({b}\right)\left({k}−\mathrm{1}\right) \\…

According-to-Wikipedia-and-WolframAlpha-the-sign-function-sgn-x-is-defined-as-sgn-x-x-x-x-x-for-x-0-and-satisfies-sgn-x-x-1-x-but-sgn-0-0-In-short-sgn-x-1-f

Question Number 2400 by Filup last updated on 19/Nov/15 $$\mathrm{According}\:\mathrm{to}\:\mathrm{Wikipedia}\:\mathrm{and}\:\mathrm{WolframAlpha}, \\ $$$$\mathrm{the}\:\mathrm{sign}\:\mathrm{function},\:\mathrm{sgn}\left({x}\right),\:\mathrm{is}\:\mathrm{defined}\:\mathrm{as}: \\ $$$$ \\ $$$$\mathrm{sgn}\left({x}\right)\equiv\frac{{x}}{\mid{x}\mid}=\frac{\mid{x}\mid}{{x}}\:\:\:\mathrm{for}\:{x}\neq\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{satisfies}: \\ $$$$\mathrm{sgn}\left({x}\right)=\sqrt{{x}}\sqrt{\frac{\mathrm{1}}{{x}}} \\ $$$$\boldsymbol{{but}} \\ $$$$\mathrm{sgn}\left(\mathrm{0}\right)=\mathrm{0} \\…

Is-there-a-way-to-evaluate-the-following-S-2-2-2-2-2-

Question Number 2275 by Filup last updated on 13/Nov/15 $$\mathrm{Is}\:\mathrm{there}\:\mathrm{a}\:\mathrm{way}\:\mathrm{to}\:\mathrm{evaluate}\:\mathrm{the}\:\mathrm{following}: \\ $$$$ \\ $$$${S}=\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+…}}}} \\ $$ Answered by Rasheed Soomro last updated on 13/Nov/15 $${S}=\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+…}}}}…