Question Number 62591 by hovea cw last updated on 23/Jun/19 $$\mathrm{If}\:\:\mathrm{2}^{{x}} =\:\mathrm{3}^{{y}} =\:\mathrm{6}^{{z}} ,\:\mathrm{then}\:\frac{\mathrm{1}}{{x}}\:+\:\frac{\mathrm{1}}{{y}}\:+\:\frac{\mathrm{1}}{{z}}\:=\:\_\_\_\_. \\ $$ Commented by mr W last updated on 23/Jun/19 $${could}\:{be}\:{any}\:{value}\:{in}\:\left(−\infty,\infty\right)\:{except}\:{zero}.…
Question Number 62388 by hovea cw last updated on 20/Jun/19 $$\mathrm{If}\:\mathrm{tan}\:\theta=\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{and}\:\mathrm{tan}\:\phi=\frac{\mathrm{1}}{\mathrm{3}},\:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\:\theta\:+\:\phi\:\:\:\mathrm{is} \\ $$ Commented by Tony Lin last updated on 20/Jun/19 $${tan}\left(\theta+\emptyset\right)=\frac{{tan}\theta+{tan}\emptyset}{\mathrm{1}−{tan}\theta{tan}\phi}=\mathrm{1} \\…
Question Number 193426 by 073 last updated on 13/Jun/23 $$\int\sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{3}} +\mathrm{1}}\mathrm{dx}=? \\ $$$$\mathrm{solution}? \\ $$ Answered by witcher3 last updated on 15/Jun/23 $$\mathrm{x}^{\mathrm{3}} =\mathrm{t} \\…
Question Number 193414 by mokys last updated on 13/Jun/23 Answered by AST last updated on 13/Jun/23 $${Z}\:{and}\:{Z}_{\mathrm{2}} \:{in}\:{the}\:{closed}\:{unit}\:{disk}\Rightarrow\mid{Z}\mid,\mid{Z}_{\mathrm{2}} \mid\leqslant\mathrm{1} \\ $$$$\mid{Z}_{\mathrm{1}} −{Z}_{\mathrm{2}} \mid\geqslant\mathrm{1}\Rightarrow\left({Z}_{\mathrm{1}} −{Z}_{\mathrm{2}} \right)\left(\overset{−}…
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Question Number 62086 by Cypher1207 last updated on 15/Jun/19 $$\mathrm{The}\:\mathrm{coefficient}\:\mathrm{of}\:{x}^{{m}} \:\mathrm{in} \\ $$$$\left(\mathrm{1}+{x}\right)^{{p}} +\left(\mathrm{1}+{x}\right)^{{p}+\mathrm{1}} +…+\left(\mathrm{1}+{x}\right)^{{n}} ,\:{p}\leqslant\:{m}\leqslant\:{n} \\ $$$$\mathrm{is}\: \\ $$ Answered by mr W last…
Question Number 61948 by subhankar10 last updated on 12/Jun/19 $$\mathrm{The}\:\mathrm{vectors}\:\boldsymbol{\mathrm{a}},\:\boldsymbol{\mathrm{b}},\:\boldsymbol{\mathrm{c}}\:\mathrm{are}\:\mathrm{equal}\:\mathrm{in}\:\mathrm{length} \\ $$$$\mathrm{and}\:\mathrm{taken}\:\mathrm{pairwise},\:\mathrm{they}\:\mathrm{make}\:\mathrm{equal} \\ $$$$\mathrm{angles}.\:\mathrm{If}\:\:\boldsymbol{\mathrm{a}}=\boldsymbol{\mathrm{i}}+\boldsymbol{\mathrm{j}},\:\:\boldsymbol{\mathrm{b}}=\boldsymbol{\mathrm{j}}+\boldsymbol{\mathrm{k}}\:\:\mathrm{and}\:\:\boldsymbol{\mathrm{c}}\:\mathrm{makes} \\ $$$$\mathrm{an}\:\mathrm{obtuse}\:\mathrm{angle}\:\mathrm{with}\:{X}−\mathrm{axis},\:\mathrm{then}\:\boldsymbol{\mathrm{c}}= \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 61614 by Tajaddin last updated on 05/Jun/19 $$\underset{−\mathrm{1}} {\overset{\mathrm{1}} {\int}}\:\frac{{d}}{{dx}}\:\left(\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{{x}}\right){dx}\:= \\ $$ Commented by maxmathsup by imad last updated on 05/Jun/19 $${we}\:{have}\:\frac{{d}}{{dx}}\left({arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right)\:=\frac{−\mathrm{1}}{{x}^{\mathrm{2}}…
Question Number 60461 by ashutosh last updated on 21/May/19 $$\mathrm{Two}\:\mathrm{cogged}\:\mathrm{wheels},\:\mathrm{of}\:\mathrm{which}\:\mathrm{one}\:\mathrm{has} \\ $$$$\mathrm{16}\:\mathrm{cogs}\:\mathrm{and}\:\mathrm{other}\:\mathrm{has}\:\mathrm{27},\:\mathrm{work}\:\mathrm{into}\: \\ $$$$\mathrm{each}\:\mathrm{other}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{latter}\:\mathrm{turns}\:\mathrm{80}\:\mathrm{times}\:\mathrm{in}\: \\ $$$$\mathrm{three}\:\mathrm{quarters}\:\mathrm{of}\:\mathrm{a}\:\mathrm{minute},\:\mathrm{how}\:\mathrm{often} \\ $$$$\mathrm{does}\:\mathrm{the}\:\mathrm{other}\:\mathrm{turn}\:\mathrm{in}\:\mathrm{8}\:\mathrm{seconds}? \\ $$ Answered by MJS last updated…
Question Number 59976 by soufiane last updated on 16/May/19 $$\underset{\:\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{{f}\left({x}\right)}{{f}\left({x}\right)+{f}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}\:{dx}\:= \\ $$ Answered by tanmay last updated on 16/May/19 $${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{f}\left({x}\right)}{{f}\left({x}\right)+{f}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}{dx} \\…