Question Number 110810 by mohammad17 last updated on 30/Aug/20 $${m}^{\mathrm{4}} +\mathrm{2}{m}^{\mathrm{3}} +\mathrm{6}{m}^{\mathrm{2}} +\mathrm{2}{m}+\mathrm{5}=\mathrm{0} \\ $$$${find}\:{all}\:{roots}\:{of}\:{m}? \\ $$ Commented by mohammad17 last updated on 30/Aug/20 $${sir}\:{can}\:{you}\:{exactly}\:{the}\:{solution}\:{steb}\:{by}\:{steb}…
Question Number 110798 by princeshah703 last updated on 30/Aug/20 $${Use}\:{Laplace}\:{transform}\:{to}\:{solve} \\ $$$$\partial{u}/\partial{x}\:+\partial{u}/\partial{t}={x} \\ $$$${x}>\mathrm{0},{t}>\mathrm{0} \\ $$$${u}\left(\mathrm{0},{t}\right)=\mathrm{0},{u}\left({x},\mathrm{0}\right)=\mathrm{0},{t}>\mathrm{0},{x}>\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 110797 by ZiYangLee last updated on 30/Aug/20 $$\mathrm{If}\:\overset{\rightarrow} {{p}}=\begin{pmatrix}{{a}}\\{{b}}\end{pmatrix}\:\mathrm{and}\:\overset{\rightarrow} {{q}}=\begin{pmatrix}{{c}}\\{{d}}\end{pmatrix}, \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{area}\:\mathrm{bounded}\:\mathrm{by}\: \\ $$$$\overset{\rightarrow} {{p}},\overset{\rightarrow} {{q}}\:\mathrm{and}\:\overset{\rightarrow} {{p}}−\overset{\rightarrow\:} {{q}}\mathrm{is}\:\frac{\left({ad}−{bc}\right)}{\mathrm{2}}. \\ $$$$ \\ $$$$\mathrm{Hints}:\:\mathrm{Use}\:\mathrm{cosine}\:\mathrm{rule}\:\mathrm{and}\:\mathrm{sine}\:\mathrm{rule} \\…
Question Number 176316 by Matica last updated on 16/Sep/22 $$\:\:{xy}+{xz}=\mathrm{255} \\ $$$$\:{xz}−{yz}=\mathrm{224}. \\ $$$${find}\:{x}\:{y}\:{and}\:{z} \\ $$ Answered by Frix last updated on 16/Sep/22 $${x}\in\mathbb{R}\backslash\left\{\mathrm{0}\right\} \\…
Question Number 176318 by Matica last updated on 16/Sep/22 Commented by Rasheed.Sindhi last updated on 16/Sep/22 $${find}\:{area}\:{of}\:{quadrilateral}\:\mathrm{ABCD} \\ $$ Commented by mr W last updated…
Question Number 110779 by ZiYangLee last updated on 30/Aug/20 $$\mathrm{If}\:\mathrm{0}\leqslant{x}\leqslant\frac{\pi}{\mathrm{2}}, \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\frac{\mathrm{2}}{\pi}{x}\leqslant\mathrm{sin}\:{x}\leqslant{x} \\ $$$$\mathrm{without}\:\mathrm{graphical}\:\mathrm{method}. \\ $$ Answered by Her_Majesty last updated on 30/Aug/20 $${f}_{\mathrm{1}} \left({x}\right)=\frac{\mathrm{2}{x}}{\pi}…
Question Number 45239 by mondodotto@gmail.com last updated on 10/Oct/18 $$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{range}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{following}} \\ $$$$\left(\boldsymbol{\mathrm{i}}\right)\:\boldsymbol{\mathrm{f}}\left(\boldsymbol{{x}}\right)=\boldsymbol{\mathrm{arcsin}}\left(\boldsymbol{{x}}\right) \\ $$$$\left(\boldsymbol{\mathrm{ii}}\right)\boldsymbol{\mathrm{f}}\left(\boldsymbol{{x}}\right)=\boldsymbol{\mathrm{arccos}}\left(\boldsymbol{{x}}\right) \\ $$$$\left(\boldsymbol{\mathrm{iii}}\right)\boldsymbol{\mathrm{f}}\left(\boldsymbol{{x}}\right)=\boldsymbol{\mathrm{arctan}}\left(\boldsymbol{{x}}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 45225 by mondodotto@gmail.com last updated on 10/Oct/18 $$\mathrm{solve}\:\mathrm{for}\:{x} \\ $$$$\mathrm{9}^{{x}+\mathrm{1}} +\mathrm{3}^{\mathrm{2}{x}+\mathrm{1}} =\mathrm{36} \\ $$ Answered by malwaan last updated on 10/Oct/18 $$\mathrm{9}.\mathrm{9}^{\mathrm{x}} +\mathrm{3}.\mathrm{9}^{\mathrm{x}}…
Question Number 176293 by mokys last updated on 15/Sep/22 $${find}\:{n}^{{th}} \:{formolla}\:{of}\:<\mathrm{0},\mathrm{4},\mathrm{9},\mathrm{16},\mathrm{25}> \\ $$ Answered by Frix last updated on 16/Sep/22 $$\mathrm{for}\:<\mathrm{0},\:\mathrm{4},\:\mathrm{9},\:\mathrm{16},\:\mathrm{25}>\:\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{polynomial} \\ $$$${a}_{{n}} =−\frac{\left({n}−\mathrm{1}\right)\left({n}^{\mathrm{3}} −\mathrm{13}{n}^{\mathrm{2}}…
Question Number 176292 by mokys last updated on 15/Sep/22 Answered by floor(10²Eta[1]) last updated on 16/Sep/22 $$\varphi\left(\mathrm{9}\right)=\varphi\left(\mathrm{3}^{\mathrm{2}} \right)=\mathrm{3}^{\mathrm{2}} −\mathrm{3}=\mathrm{6}\Rightarrow\mathrm{a}^{\mathrm{6}} \equiv\mathrm{1}\left(\mathrm{mod9}\right),\:\mathrm{gcd}\left(\mathrm{a},\mathrm{9}\right)=\mathrm{1} \\ $$$$\mathrm{n}=\mathrm{6k}+\mathrm{r},\:\mathrm{k}\in\mathbb{Z},\:\mathrm{0}\leqslant\mathrm{r}\leqslant\mathrm{5} \\ $$$$\Rightarrow\mathrm{4}^{\mathrm{n}} +\mathrm{6n}−\mathrm{1}=\mathrm{4}^{\mathrm{6k}+\mathrm{r}}…