Question Number 33317 by mondodotto@gmail.com last updated on 14/Apr/18 $$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{radius}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{circle}} \\ $$$$\boldsymbol{\mathrm{wich}}\:\boldsymbol{\mathrm{inscribes}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{equalateral}}\:\boldsymbol{\mathrm{triangle}} \\ $$$$\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{perimeter}}\:\boldsymbol{\mathrm{of}}\:\mathrm{24}\boldsymbol{\mathrm{cm}} \\ $$ Answered by MJS last updated on 14/Apr/18 $$\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{sure}\:\mathrm{if}\:\mathrm{you}\:\mathrm{mean}\:\mathrm{the} \\…
Question Number 164376 by muneer0o0 last updated on 16/Jan/22 Answered by alephzero last updated on 16/Jan/22 $$\int\frac{\mathrm{d}{x}}{\:\sqrt{{x}^{\mathrm{3}} +\mathrm{64}}}\:=\:? \\ $$$${x}^{\mathrm{3}} \:=\:\left(\sqrt{{x}^{\mathrm{3}} }\right)^{\mathrm{2}} \\ $$$$\int\frac{\mathrm{d}{u}}{\:\sqrt{{u}^{\mathrm{2}} +{a}}}\:=\:\mathrm{ln}\:\mid{u}+\sqrt{{u}^{\mathrm{2}}…
Question Number 98843 by HamraboyevFarruxjon last updated on 16/Jun/20 $$\boldsymbol{{I}}'\boldsymbol{{ve}}\:\boldsymbol{{got}}\:\boldsymbol{{a}}\:\boldsymbol{{question}}: \\ $$$$\boldsymbol{{What}}\:\boldsymbol{{differences}}\:\mathrm{2}.\mathrm{083}\:\boldsymbol{{and}}\:\mathrm{2}.\mathrm{084}\:\boldsymbol{{versions}}\:\boldsymbol{{with}}? \\ $$ Commented by Tinku Tara last updated on 16/Jun/20 $$\mathrm{No}\:\mathrm{major}\:\mathrm{difference}.\:\mathrm{Some}\:\mathrm{devices} \\ $$$$\mathrm{are}\:\mathrm{giving}\:\mathrm{problem}\:\mathrm{while}\:\mathrm{opening}…
Question Number 98815 by M±th+et+s last updated on 16/Jun/20 $${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 $${Let}\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{ab}}\:={c} \\ $$$${Obviously}\:{c}\geqslant{a}\:\&\:{c}\geqslant{b} \\ $$$${So}\:{the}\:{greater}\:{angle}\:{is}\:{opposite}\:{to}\:{c} \\ $$$${Now},\:{c}^{\mathrm{2}}…
Question Number 164311 by saly last updated on 16/Jan/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 164300 by otchereabdullai@gmail.com last updated on 16/Jan/22 $$\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}…