Question Number 98815 by M±th+et+s last updated on 16/Jun/20 $${what}\:{is}\:{heisenberg}\:{uncertainty}\:{principle}? \\ $$$$ \\ $$ Answered by Rio Michael last updated on 17/Jun/20 $$\mathrm{This}\:\mathrm{should}\:\mathrm{be}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{laws}\:\mathrm{of}\:\mathrm{Quantum} \\ $$$$\mathrm{mechanics}\:\mathrm{which}\:\mathrm{states}\:\mathrm{that}\:'\mathrm{it}\:\mathrm{is}\:\mathrm{impossible}\:\mathrm{to}\:\mathrm{know}…
Question Number 33271 by NECx last updated on 14/Apr/18 $${For}\:{a}\:{couple}\:{of}\:{weeks}\:{now}.{Two} \\ $$$${men}\:{have}\:{been}\:{missing}\:{here}\:. \\ $$$${Please}\:{come}\:{around}.{We}\:{so}\:{much} \\ $$$${enjoy}\:{everyone}'{s}\:{presence}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 33270 by NECx last updated on 14/Apr/18 $${For}\:{a}\:{couple}\:{of}\:{weeks}\:{now}.{Two} \\ $$$${men}\:{have}\:{been}\:{missing}\:{here}\:. \\ $$$${Please}\:{come}\:{around}.{We}\:{so}\:{much} \\ $$$${enjoy}\:{everyone}'{s}\:{presence}. \\ $$ Commented by Tinkutara last updated on 14/Apr/18…
Question Number 164310 by ZiYangLee last updated on 16/Jan/22 Answered by Rasheed.Sindhi last updated on 16/Jan/22 $${Let}\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{ab}}\:={c} \\ $$$${Obviously}\:{c}\geqslant{a}\:\&\:{c}\geqslant{b} \\ $$$${So}\:{the}\:{greater}\:{angle}\:{is}\:{opposite}\:{to}\:{c} \\ $$$${Now},\:{c}^{\mathrm{2}}…
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Question Number 164300 by otchereabdullai@gmail.com last updated on 16/Jan/22 $$\mathrm{Question} \\ $$$$\mathrm{a}.\mathrm{which}\:\mathrm{numbers}\:\mathrm{have}\:\mathrm{an}\:\mathrm{odd}\:\mathrm{number} \\ $$$$\:\:\:\:\:\mathrm{of}\:\mathrm{divisors} \\ $$$$\mathrm{b}.\:\:\mathrm{Is}\:\mathrm{there}\:\mathrm{a}\:\mathrm{number}\:\mathrm{with}\:\mathrm{exactly}\:\mathrm{13} \\ $$$$\:\:\:\:\:\:\mathrm{divisors} \\ $$$$\mathrm{c}.\:\:\mathrm{Generalize} \\ $$ Answered by mr…
Question Number 33218 by mondodotto@gmail.com last updated on 13/Apr/18 $$\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{angle}}\:\boldsymbol{\mathrm{bisector}} \\ $$$$\:\mathrm{3}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{{xy}}−\mathrm{5}\boldsymbol{{y}}^{\mathrm{2}} =\mathrm{0} \\ $$ Answered by MJS last updated on 13/Apr/18 $${y}_{\mathrm{1}} =−\frac{\mathrm{2}+\sqrt{\mathrm{19}}}{\mathrm{5}}{x}…
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Question Number 164266 by naka3546 last updated on 15/Jan/22 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\:\left({x}−\mathrm{2}\right)\:−\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{5}}\:\:=\:\:? \\ $$ Answered by ajfour last updated on 15/Jan/22 $${x}=\frac{\mathrm{1}}{{h}}\rightarrow\infty \\ $$$${L}=\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}−\mathrm{2}{h}−\left(\mathrm{1}+\mathrm{2}{h}−\mathrm{5}{h}^{\mathrm{2}}…