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calculate-lim-x-0-sin-sin-sinx-1-cos-x-2-x-3-

Question Number 141932 by mathmax by abdo last updated on 24/May/21 $$\mathrm{calculate}\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{sin}\left(\mathrm{sin}\left(\mathrm{sinx}\right)\right)+\mathrm{1}−\mathrm{cos}\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{x}^{\mathrm{3}} } \\ $$ Answered by TheSupreme last updated on 24/May/21 $${sin}\left({sin}\left({sin}\left({x}\right)\right)\right)+\mathrm{1}−{cos}\left({x}^{\mathrm{2}}…

Given-that-a-3i-4j-5k-and-b-2i-2j-3k-and-c-6i-7j-8k-find-3a-2b-3c-

Question Number 10862 by Saham last updated on 28/Feb/17 $$\mathrm{Given}\:\mathrm{that}:\:\:\hat {\mathrm{a}}\:=\:\mathrm{3i}\:+\:\mathrm{4j}\:+\:\mathrm{5k}\:\:\mathrm{and}\:\:\hat {\mathrm{b}}\:=\:\mathrm{2i}\:+\:\mathrm{2j}\:+\:\mathrm{3k}\:\:\mathrm{and}\:\:\:\hat {\mathrm{c}}\:=\:\mathrm{6i}\:−\:\mathrm{7j}\:−\:\mathrm{8k}. \\ $$$$\mathrm{find} \\ $$$$\mathrm{3}\hat {\mathrm{a}}\:+\:\mathrm{2}\hat {\mathrm{b}}\:−\:\mathrm{3}\hat {\mathrm{c}} \\ $$ Commented by Zainal…

Determine-if-the-numbers-1-5-8-are-in-the-range-of-the-fuctions-f-x-2x-if-2-x-lt-2-3-if-x-2-

Question Number 141935 by cesarL last updated on 24/May/21 $${Determine}\:{if}\:{the}\:{numbers}\:\mathrm{1},\:\mathrm{5},\:\mathrm{8}\: \\ $$$${are}\:{in}\:{the}\:{range}\:{of}\:{the}\:{fuctions} \\ $$$$ \\ $$$${f}\left({x}\right)=\begin{cases}{\mathrm{2}{x}\:\:\:\:\:\:{if}\:\:−\mathrm{2}\leqslant{x}<\mathrm{2}}\\{\mathrm{3}\:\:\:\:\:\:\:\:\:{if}\:\:\:\:{x}=\mathrm{2}}\end{cases} \\ $$$$ \\ $$ Terms of Service Privacy Policy…

Find-all-the-solution-that-fulfilled-the-equation-below-1-1-x-x-1-1-1-2013-2013-

Question Number 10856 by Joel576 last updated on 27/Feb/17 $$\mathrm{Find}\:\mathrm{all}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{that}\:\mathrm{fulfilled}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{below} \\ $$$$\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{{x}}\right)^{{x}\:+\:\mathrm{1}} \:=\:\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{2013}}\right)^{\mathrm{2013}} \\ $$ Answered by DrDaveR last updated on 12/Mar/17 $${Just}\:−\mathrm{2014}.\:{There}\:{can}\:{be}\:{no}\:{positve}\:{solution}.\:{The}\:{function}\:{is} \\ $$$${monotinically}\:{decreasing}\:\:{so}\:{there}\:{could}\:{be}\:{one}\:{negative}\:…

3-1-2-3-4-2-3-4-5-3-4-5-2016-2014-2015-2016-

Question Number 10855 by Joel576 last updated on 27/Feb/17 $$\frac{\mathrm{3}}{\mathrm{1}!+\mathrm{2}!+\mathrm{3}!}\:+\:\frac{\mathrm{4}}{\mathrm{2}!+\mathrm{3}!+\mathrm{4}!}\:+\:\frac{\mathrm{5}}{\mathrm{3}!+\mathrm{4}!+\mathrm{5}!}\:+\:…\:+\:\frac{\mathrm{2016}}{\mathrm{2014}!+\mathrm{2015}!+\mathrm{2016}!}\:=\:? \\ $$ Answered by nume1114 last updated on 28/Feb/17 $$\:\:\:\:\frac{\mathrm{3}}{\mathrm{1}!+\mathrm{2}!+\mathrm{3}!}+\frac{\mathrm{4}}{\mathrm{2}!+\mathrm{3}!+\mathrm{4}!}+…+\frac{\mathrm{2016}}{\mathrm{2014}!+\mathrm{2015}!+\mathrm{2016}!} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\mathrm{2014}} {\sum}}\frac{{n}+\mathrm{2}}{{n}!+\left({n}+\mathrm{1}\right)!+\left({n}+\mathrm{2}\right)!} \\…