Question Number 132119 by Hamid52 last updated on 11/Feb/21 $$\begin{cases}{ {x}+ {y}= }\\{ {x}− {y}= }\end{cases} \\ $$$$ \\ $$ Answered by Raxreedoroid last updated…
Question Number 132116 by aurpeyz last updated on 11/Feb/21 $${find}\:{the}\:{magnitude}\:{and}\:{direction}\:{of} \\ $$$${the}\:{vector}\:{r}=\mathrm{3}{i}−\mathrm{4}{j}\:{to}\:{the}\:{nearest}\: \\ $$$${degree} \\ $$$$\left({a}\right)\:\mathrm{7}{N}\:\mathrm{143}^{\mathrm{0}} \:\left({b}\right)\:\mathrm{5}{N}\:\mathrm{143}^{\mathrm{0}} \:\left({c}\right)\:\mathrm{5}{N}\:\mathrm{127}^{\mathrm{0}} \\ $$$$\left({d}\right)\:\mathrm{7}{N}\:\mathrm{127}^{\mathrm{0}} \\ $$ Answered by Olaf…
Question Number 132107 by weltr last updated on 11/Feb/21 $$\left({arcsin}\mid\mathrm{ln}^{\mathrm{3}} \left({cot}\mathrm{4}\:−\:{x}\right)\mid\:+\:\mathrm{4}!\right)^{\frac{\mathrm{1}}{\mathrm{19}}} \:=\:\mathrm{5} \\ $$ Answered by Olaf last updated on 11/Feb/21 $$\left(\mathrm{arcsin}\mid\mathrm{ln}^{\mathrm{3}} \left(\mathrm{cot4}−{x}\right)\mid+\mathrm{4}!\right)^{\frac{\mathrm{1}}{\mathrm{19}}} \:=\:\mathrm{5} \\…
Question Number 66472 by hmamarques1994@gmail.com last updated on 15/Aug/19 $$\begin{cases}{\sqrt[{\sqrt{\mathrm{6}}}]{\boldsymbol{\mathrm{x}}}+\sqrt[{\sqrt{\mathrm{5}}}]{\boldsymbol{\mathrm{y}}}=\mathrm{11}}\\{\frac{\sqrt[{\sqrt{\mathrm{5}}}]{\boldsymbol{\mathrm{y}}}}{\:\sqrt[{\sqrt{\mathrm{6}}}]{\boldsymbol{\mathrm{x}}}}=\mathrm{1}\frac{\mathrm{1}}{\mathrm{5}}}\end{cases} \\ $$$$\: \\ $$$$\:\boldsymbol{\mathrm{Qual}}\:\:\acute {\boldsymbol{\mathrm{e}}}\:\:\boldsymbol{\mathrm{o}}\:\:\boldsymbol{\mathrm{par}}\:\:\boldsymbol{\mathrm{ordenado}}\:\:\boldsymbol{\mathrm{na}}\:\:\boldsymbol{\mathrm{forma}}\:\:\boldsymbol{\mathrm{a}}^{\sqrt{\boldsymbol{\mathrm{p}}}} \:\:\boldsymbol{\mathrm{e}}\:\:\boldsymbol{\mathrm{b}}^{\sqrt{\boldsymbol{\mathrm{q}}}} \\ $$$$\:\boldsymbol{\mathrm{que}}\:\:\boldsymbol{\mathrm{satisfaz}}\:\:\boldsymbol{\mathrm{o}}\:\:\boldsymbol{\mathrm{sistema}}\:\:\boldsymbol{\mathrm{como}}\:\:\boldsymbol{\mathrm{possivel}}\:\:\boldsymbol{\mathrm{e}}\:\:\boldsymbol{\mathrm{determinado}}? \\ $$ Answered by MJS last updated…
Question Number 132098 by rs4089 last updated on 11/Feb/21 Answered by Ar Brandon last updated on 11/Feb/21 $$\mathrm{S}_{\mathrm{n}} =\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{n}}{\mathrm{n}^{\mathrm{2}} +\mathrm{kn}+\mathrm{k}^{\mathrm{2}} } \\ $$$$\underset{\mathrm{n}\rightarrow\infty}…
Question Number 1024 by brahim last updated on 19/May/15 $$\mathrm{4}{x}+\mathrm{5}=\mathrm{6}−\mathrm{4}\left(\mathrm{1}−{x}\right) \\ $$ Answered by prakash jain last updated on 19/May/15 $$\mathrm{4}{x}+\mathrm{5}=\mathrm{6}−\mathrm{4}+\mathrm{4}{x} \\ $$$$\mathrm{Equation}\:\mathrm{has}\:\mathrm{no}\:\mathrm{solution}. \\ $$…
Question Number 1012 by rpatle69@gmail.com last updated on 13/May/15 $${find}\:{the}\:{equation}\:{of}\:{circle}\:{whose} \\ $$$${parametric}\:{form}\:{is}\:{given}\:{by}\: \\ $$$${x}=\mathrm{3cos}\:\theta+\mathrm{5}\:{and}\:{y}=\:\mathrm{3sin}\:\theta−\mathrm{7}\:{and} \\ $$$${second}\:{part}\:{is}\:{x}=\mathrm{4cos}\:\theta−\mathrm{3}\:{and} \\ $$$${y}=\mathrm{4sin}\:\theta+\mathrm{4}.\:{find}\:{centre}\:{and}\: \\ $$$${radius}\:{of}\:{above}\:{circle}. \\ $$$$ \\ $$$$ \\…
Question Number 66544 by naka3546 last updated on 17/Aug/19 $${Find}\:\:{a},\:{b},\:{c}\:\:{which}\:\:{fulfill}\:\:\: \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{{x}\left({a}\:+\:{b}\:\mathrm{cos}\:{x}\right)\:−\:{c}\:\mathrm{sin}\:{x}}{{x}^{\mathrm{5}} }\:\:=\:\:\mathrm{1} \\ $$ Answered by Tanmay chaudhury last updated on 17/Aug/19 $$\underset{{x}\rightarrow\mathrm{0}}…
Question Number 66543 by hmamarques1994@gmail.com last updated on 17/Aug/19 $$\:\mathrm{3}^{\boldsymbol{{x}}} =\mathrm{3}\boldsymbol{{x}} \\ $$$$\: \\ $$$$\:\boldsymbol{{x}}=? \\ $$ Commented by gunawan last updated on 17/Aug/19 $${x}=\mathrm{1}…
Question Number 1000 by rpatle69@gmail.com last updated on 13/May/15 $$\int{e}^{{x}\:\:} {sinx}\:{dx}=?\:\:{plz}\:{give}\:{me}\:{answer}\:{soon}. \\ $$ Commented by rpatle69@gmail.com last updated on 13/May/15 $${please}\:{soon}\:{guys} \\ $$ Answered by…