Question Number 184473 by Tawa11 last updated on 07/Jan/23 Choose 4 random points in a sphere to form a tetrahedron inside the sphere. What is…
Question Number 184478 by Shrinava last updated on 07/Jan/23 Answered by mr W last updated on 08/Jan/23 $$“\boldsymbol{{generating}}\:\boldsymbol{{function}}''\:\boldsymbol{{method}} \\ $$$$\left({x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…\right)^{\mathrm{4}} =\frac{{x}^{\mathrm{4}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} }={x}^{\mathrm{4}}…
Question Number 184472 by mr W last updated on 07/Jan/23 $${solve}\:{in}\:\mathbb{R}^{\mathrm{3}} \\ $$$$\begin{cases}{{x}+\frac{\mathrm{1}}{{y}}=\mathrm{3}}\\{{y}+\frac{\mathrm{1}}{{z}}=\mathrm{4}}\\{{z}+\frac{\mathrm{1}}{{x}}=\mathrm{5}}\end{cases} \\ $$ Answered by liuxinnan last updated on 07/Jan/23 $${actually}\:\mathbb{R}^{\mathrm{3}} =\mathbb{R} \\…
Question Number 118936 by mathocean1 last updated on 20/Oct/20 $${show}\:{by}\:{recurrence}\:{that}: \\ $$$${for}\:{n}\:\in\:\mathbb{N}^{\ast} ,\:\mathrm{2}^{\mathrm{6}{n}−\mathrm{5}} +\mathrm{3}^{\mathrm{2}{n}\:} \:{is}\:{divisible}\:{by} \\ $$$$\mathrm{11}. \\ $$ Answered by Dwaipayan Shikari last updated…
Question Number 184475 by Shrinava last updated on 07/Jan/23 $$\mathrm{If}\:\:\:\mathrm{A}\:=\:\mathrm{20}°\:\:\:\mathrm{and}\:\:\:\mathrm{B}\:=\:\mathrm{25}° \\ $$$$\mathrm{Find}\:\:\:\left(\mathrm{1}\:+\:\mathrm{tanA}\right)\left(\mathrm{1}\:+\:\mathrm{tanB}\right) \\ $$ Answered by cortano1 last updated on 07/Jan/23 $$\:\mathrm{tan}\:\mathrm{45}\:=\:\frac{\mathrm{tan}\:\mathrm{20}°+\mathrm{tan}\:\mathrm{25}°}{\mathrm{1}−\mathrm{tan}\:\mathrm{20}°\:\mathrm{tan}\:\mathrm{25}°} \\ $$$$\Rightarrow\mathrm{1}−\mathrm{tan}\:\mathrm{20}°\:\mathrm{tan}\:\mathrm{25}°=\mathrm{tan}\:\mathrm{20}°+\mathrm{tan}\:\mathrm{25}° \\…
Question Number 118937 by harckinwunmy last updated on 20/Oct/20 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{general}\:\mathrm{rule}\: \\ $$$$\mathrm{for}\:\mathrm{factorising}\:\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)^{\mathrm{n}} ? \\ $$ Answered by mathmax by abdo last updated on 20/Oct/20 $$\mathrm{i}\:\mathrm{think}\:\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\mathrm{developping}…
Question Number 118930 by Ar Brandon last updated on 20/Oct/20 $$\mathrm{Let}\:\left[{x}\right]\:\mathrm{denote}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{integer}\:\leqslant{x}.\:\mathrm{Then}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{ordered}\:\mathrm{pair}\:\left({x},\mathrm{y}\right),\:\mathrm{where}\:{x}\:\mathrm{and}\:\mathrm{y}\:\mathrm{are} \\ $$$$\mathrm{positive}\:\mathrm{integers}\:\mathrm{less}\:\mathrm{than}\:\mathrm{30}\:\mathrm{such}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\left[\frac{{x}}{\mathrm{2}}\right]+\left[\frac{\mathrm{2}{x}}{\mathrm{3}}\right]+\left[\frac{\mathrm{y}}{\mathrm{4}}\right]+\left[\frac{\mathrm{4y}}{\mathrm{5}}\right]=\frac{\mathrm{7}{x}}{\mathrm{6}}+\frac{\mathrm{21y}}{\mathrm{20}} \\ $$$$\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{4} \\ $$ Answered by…
Question Number 118907 by frc2crc last updated on 20/Oct/20 $${where}\:{can}\:{u}\:{fond}\:{a}\:{formula}\:{for} \\ $$$${denesting}\:{ramanujan}\:{type}\:{roots}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 184438 by Shrinava last updated on 06/Jan/23 $$\mathrm{If}\:\:\:\mathrm{a}\:−\:\frac{\mathrm{4}}{\:\sqrt{\mathrm{a}}}\:=\:\mathrm{17} \\ $$$$\mathrm{Find}\:\:\:\mathrm{a}\:−\:\mathrm{4}\:\sqrt{\mathrm{a}}\:=\:? \\ $$ Answered by Frix last updated on 06/Jan/23 $${a}−\frac{\mathrm{4}}{\:\sqrt{{a}}}=\mathrm{17} \\ $$$$\mathrm{Let}\:{t}=\sqrt{{a}}\:\Rightarrow\:{t}>\mathrm{0} \\…
Question Number 184434 by a.lgnaoui last updated on 06/Jan/23 $$\left(\mathrm{2}{a}−{b}\right){b}+\frac{\mathrm{5}}{\mathrm{3}{a}}−\mathrm{7}{b}^{\mathrm{2}} +\mathrm{3}{a}=\mathrm{0} \\ $$$${Evaluer}\:\:\boldsymbol{{b}}\:{en}\:{fonction}\:{de}\:\boldsymbol{{a}} \\ $$ Answered by Frix last updated on 06/Jan/23 $$\mathrm{Transform}\:\mathrm{to} \\ $$$${b}^{\mathrm{2}}…