Question Number 170091 by Beginner last updated on 16/May/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 170084 by Beginner last updated on 16/May/22 Commented by cortano1 last updated on 16/May/22 $$\:\:\:\mathrm{sin}\:{x}=\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{cos}\:{x}+\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\:\mathrm{sin}\:{x}\right) \\ $$$$\:\:\:\mathrm{sin}\:{x}=\:\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{cos}\:{x}\:+\frac{\mathrm{3}}{\mathrm{2}}\sqrt{\mathrm{3}}\:\mathrm{sin}\:{x} \\ $$$$\:\left(\frac{\mathrm{2}−\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)\mathrm{sin}\:{x}\:=\:\frac{\mathrm{3}}{\mathrm{2}}\:\mathrm{cos}\:{x} \\ $$$$\:\:\:\mathrm{tan}\:{x}\:=\:\frac{\mathrm{3}}{\mathrm{2}−\mathrm{3}\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{3}\left(\mathrm{2}+\mathrm{3}\sqrt{\mathrm{3}}\right)}{\mathrm{4}−\mathrm{27}} \\ $$$$\:\:\:\mathrm{tan}\:{x}\:=\:−\frac{\mathrm{6}+\mathrm{9}\sqrt{\mathrm{3}}}{\mathrm{23}}…
Question Number 104542 by mohammad17 last updated on 22/Jul/20 Answered by Dwaipayan Shikari last updated on 22/Jul/20 $$\left.\mathrm{2}\right) \\ $$$${tan}^{−\mathrm{1}} \left(\frac{{y}}{{x}}\right)={log}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right) \\ $$$$\frac{\frac{\mathrm{1}}{{x}}.\frac{{dy}}{{dx}}−\frac{{y}}{{x}^{\mathrm{2}}…
Question Number 170077 by MathsFan last updated on 16/May/22 $$ \\ $$How do I find for the accurate measure of sine from 1 to 30…
Question Number 104536 by Anindita last updated on 22/Jul/20 $$\boldsymbol{\mathrm{The}}\:\boldsymbol{\mathrm{sum}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{two}}\:\boldsymbol{\mathrm{numbers}}\:\boldsymbol{\mathrm{is}}\:\mathrm{384}\:\boldsymbol{\mathrm{and}} \\ $$$$\mathrm{t}\boldsymbol{\mathrm{heir}}\:\boldsymbol{\mathrm{GCD}}\:\boldsymbol{\mathrm{is}}\:\mathrm{48}.\:\boldsymbol{\mathrm{What}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{LCM}}\:\boldsymbol{\mathrm{of}} \\ $$$$\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{numbers}}\:? \\ $$ Commented by 1549442205PVT last updated on 22/Jul/20 $$\mathrm{Similar}\:\mathrm{to}\:\mathrm{Q104413} \\…
Question Number 104539 by Anindita last updated on 22/Jul/20 $$ \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}}} \\ $$ Answered by OlafThorendsen last updated on 22/Jul/20 $$=\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}}} \\ $$$$=\:\mathrm{1}−\frac{\mathrm{3}}{\mathrm{2}}\:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\…
Question Number 170054 by daus last updated on 15/May/22 Answered by greougoury555 last updated on 15/May/22 $$\:\begin{cases}{{a}=\mathrm{9}^{{x}} }\\{{b}=\mathrm{12}^{{x}} }\\{{a}+{b}=\mathrm{16}^{{x}} }\end{cases}\Rightarrow\mathrm{9}^{{x}} +\mathrm{12}^{{x}} \:=\:\mathrm{16}^{{x}} \\ $$$$\Rightarrow\left(\frac{\mathrm{9}}{\mathrm{12}}\right)^{{x}} +\mathrm{1}=\left(\frac{\mathrm{16}}{\mathrm{12}}\right)^{{x}}…
Question Number 170050 by DAVONG last updated on 14/May/22 $$\mathrm{Show}\:\mathrm{that},\forall\mathrm{a},\mathrm{b},\mathrm{c}\in\mathbb{R}\::\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} +\mathrm{12}\geqslant\mathrm{4}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right) \\ $$ Answered by mindispower last updated on 15/May/22 $$\mathrm{12}=\mathrm{4}+\mathrm{4}+\mathrm{4} \\ $$$$\mathrm{4}+{x}^{\mathrm{2}}…
Question Number 104509 by bobhans last updated on 22/Jul/20 $$\left(\mathrm{1}\right)\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:−\mathrm{7}\:\frac{{dy}}{{dx}}\:+\mathrm{12}{y}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:{y}'''−\mathrm{3}{y}''+\mathrm{4}{y}'−\mathrm{2}{y}=\:{e}^{{x}} −\mathrm{cos}\:{x} \\ $$$$\left(\mathrm{3}\right){y}'''+\mathrm{3}{y}''+\mathrm{5}{y}'+\mathrm{3}{y}=\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{cot}\:\mathrm{4}{x}−\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} } \\ $$ Answered by…
Question Number 104486 by Anindita last updated on 21/Jul/20 $$\boldsymbol{\mathrm{If}}\:\boldsymbol{\mathrm{A}}\:=\:\begin{bmatrix}{\mathrm{3}}&{−\mathrm{4}}&{\mathrm{2}}\\{−\mathrm{2}}&{\mathrm{1}}&{\mathrm{0}}\\{−\mathrm{1}}&{−\mathrm{1}}&{\mathrm{1}}\end{bmatrix}\boldsymbol{\mathrm{and}}\: \\ $$$$\boldsymbol{\mathrm{B}}\:=\:\begin{bmatrix}{\mathrm{1}}&{\mathrm{2}}&{−\mathrm{2}}\\{\mathrm{2}}&{\mathrm{5}}&{−\mathrm{4}}\\{\mathrm{3}}&{\mathrm{7}}&{−\mathrm{5}}\end{bmatrix}\boldsymbol{\mathrm{then}}\: \\ $$$$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{AB}}\:=\:\boldsymbol{\mathrm{BA}}\:=\:\boldsymbol{\mathrm{I}}_{\mathrm{3}} \:. \\ $$ Answered by OlafThorendsen last updated on 21/Jul/20 $$\mathrm{Just}\:\mathrm{calculate}\:\mathrm{B}^{−\mathrm{1}}…