Question Number 89956 by swizanjere@gmail.com last updated on 20/Apr/20 $$\mathrm{simplify}\kappa\mathrm{giving}\kappa\mathrm{your}\kappa\mathrm{answer}\kappa\mathrm{in}\kappa\mathrm{index}\kappa\mathrm{form} \\ $$$$\sqrt{\frac{\mathrm{ac}^{\mathrm{2}} }{\mathrm{9a}^{\mathrm{2}} \mathrm{c}^{\mathrm{4}} }} \\ $$ Commented by john santu last updated on 20/Apr/20…
Question Number 89953 by swizanjere@gmail.com last updated on 20/Apr/20 $$\mathrm{solvethefollowingequation} \\ $$$$\mathrm{5}^{\mathrm{2x}+\mathrm{y}} =\mathrm{625and2}^{\mathrm{4x}\nmid\mathrm{2y}} =\frac{\mathrm{1}}{\mathrm{6}} \\ $$ Commented by jagoll last updated on 20/Apr/20 $$\mathrm{2x}+\mathrm{y}\:=\:\mathrm{4} \\…
Question Number 89936 by ~blr237~ last updated on 20/Apr/20 $${Prove}\:{that}\:\underset{{p}\geqslant\mathrm{1},{q}\geqslant\mathrm{1}} {\sum}\:\:\frac{\mathrm{1}}{{pq}\left({p}+{q}−\mathrm{1}\right)}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{3}}\: \\ $$ Answered by maths mind last updated on 20/Apr/20 $$\underset{{p},{q}\geqslant\mathrm{2}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 89934 by ~blr237~ last updated on 20/Apr/20 $$\left.{Let}\:{x}\in\right]\mathrm{0};\mathrm{1}\left[\:\:{Prove}\:{that}\right. \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}^{{n}} }{\mathrm{1}+{x}^{{n}} }\:+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−{x}\right)^{{n}} }{\mathrm{1}−{x}^{{n}} }\:=\:\mathrm{0} \\ $$ Terms of Service…
Question Number 89937 by ~blr237~ last updated on 20/Apr/20 $${Prove}\:{that}\:{for}\:{all}\:{complex}\:{such}\:{as}\:\mid{z}\mid<\mathrm{1}= \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{z}^{{n}} }{\left({z}^{{n}} −\mathrm{1}\right)^{\mathrm{2}} }\:+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{nz}^{{n}} }{{z}^{{n}} −\mathrm{1}}\:=\:\mathrm{0}\: \\ $$ Commented by…
Question Number 24387 by chernoaguero@gmail.com last updated on 17/Nov/17 $$\boldsymbol{\mathrm{Two}}\:\boldsymbol{\mathrm{particle}}\:\boldsymbol{\mathrm{move}}\:\boldsymbol{\mathrm{along}}\:\boldsymbol{\mathrm{an}}\:\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{axis}}. \\ $$$$\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{position}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{particle}}\:\mathrm{1}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{given}}; \\ $$$$\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{x}}=\mathrm{6}.\mathrm{00}\boldsymbol{\mathrm{t}}^{\mathrm{2}} +\mathrm{3}.\mathrm{00}\boldsymbol{\mathrm{t}}+\mathrm{2}.\mathrm{00}\left(\frac{\boldsymbol{\mathrm{m}}}{\boldsymbol{\mathrm{s}}}\right) \\ $$$$\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{acceleration}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{particle}}\:\mathrm{2}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{given}} \\ $$$$\boldsymbol{\mathrm{by}}\:\:\boldsymbol{\mathrm{a}}=−\mathrm{8}.\mathrm{00}\boldsymbol{\mathrm{t}}\left(\frac{\boldsymbol{\mathrm{m}}}{\boldsymbol{\mathrm{s}}^{\mathrm{2}} }\right)\:\boldsymbol{\mathrm{and}},\mathrm{at}\:\mathrm{t}=\mathrm{0},\boldsymbol{\mathrm{its}} \\ $$$$\boldsymbol{\mathrm{velocity}}\:\boldsymbol{\mathrm{is}}\:\mathrm{20}\:\left(\frac{\boldsymbol{\mathrm{m}}}{\boldsymbol{\mathrm{s}}}\right).\boldsymbol{\mathrm{when}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{velocities}} \\ $$$$\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{particles}}\:\boldsymbol{\mathrm{match}}, \\…
Question Number 24371 by Tinkutara last updated on 16/Nov/17 $$\mathrm{Two}\:\mathrm{blocks}\:\mathrm{are}\:\mathrm{moving}\:\mathrm{together}\:\mathrm{under} \\ $$$$\mathrm{the}\:\mathrm{action}\:\mathrm{of}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{horizontal} \\ $$$$\mathrm{external}\:\mathrm{force}\:{F}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{smaller}\:\mathrm{block}\:\mathrm{is} \\ $$$$\mathrm{at}\:\mathrm{rest}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{the}\:\mathrm{bigger}\:\mathrm{block} \\ $$$$\mathrm{due}\:\mathrm{to}\:\mathrm{the}\:\mathrm{friction}\:\mathrm{between}\:\mathrm{them},\:\mathrm{then} \\ $$$$\mathrm{the}\:\mathrm{normal}\:\mathrm{reaction}\:\mathrm{between}\:\mathrm{the}\:\mathrm{bigger} \\ $$$$\mathrm{block}\:\mathrm{and}\:\mathrm{floor}\:\mathrm{is} \\ $$ Commented…
Question Number 24369 by ajfour last updated on 16/Nov/17 Commented by ajfour last updated on 16/Nov/17 $${If}\:{a}\:{uniform}\:{rod}\:{of}\:{mass}\:\boldsymbol{{m}}, \\ $$$${length}\:\boldsymbol{{L}},\:{and}\:{hooked}\:{at}\:{end}\:{A} \\ $$$${to}\:{a}\:{string}\:{of}\:{length}\:\boldsymbol{{l}}\:{be} \\ $$$${released}\:{in}\:{horizontal}\:{position}, \\ $$$${then}\:{a}\:{little}\:{time}\:{later}\:{find}\:\boldsymbol{\phi}…
Question Number 155400 by TVTA last updated on 30/Sep/21 $${x}^{\mathrm{2}} +{x}+\mathrm{5}{xy}+\mathrm{6}{y}^{\mathrm{2}} +\mathrm{2}{y}−\mathrm{2}=\mathrm{0} \\ $$ Commented by MJS_new last updated on 30/Sep/21 $$\mathrm{hyperbola}\:\mathrm{with}\:\mathrm{center}\:\begin{bmatrix}{\mathrm{2}}\\{−\mathrm{1}}\end{bmatrix}\:\mathrm{and}\:\mathrm{axes} \\ $$$${a}=\mathrm{2}\sqrt{−\mathrm{7}+\mathrm{5}\sqrt{\mathrm{2}}} \\…
Question Number 24301 by Tinkutara last updated on 15/Nov/17 $$\mathrm{In}\:\mathrm{the}\:\mathrm{figure}\:\mathrm{shown}\:\mathrm{below},\:\mathrm{all}\:\mathrm{surfaces} \\ $$$$\mathrm{are}\:\mathrm{smooth},\:\mathrm{strings}\:\mathrm{and}\:\mathrm{pulley}\:\mathrm{are}\:\mathrm{ideal}. \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{wedge}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{with}\:\mathrm{acceleration} \\ $$$${a}\:\mathrm{towards}\:\mathrm{the}\:\mathrm{right},\:\mathrm{then}\:\mathrm{the}\:\mathrm{acceleration} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{block}\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{the}\:\mathrm{wedge} \\ $$$$\mathrm{at}\:\mathrm{that}\:\mathrm{instant}\:\mathrm{is} \\ $$ Commented by Tinkutara…