Question Number 128083 by AgnibhoMukhopadhyay last updated on 04/Jan/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{8}\:−\:{i}}{\mathrm{3}\:−\:\mathrm{2}{i}} \\ $$$$\:\mathrm{If}\:\mathrm{the}\:\mathrm{expression}\:\mathrm{above}\:\mathrm{is}\:\mathrm{rewritten}\: \\ $$$$\:\mathrm{in}\:\mathrm{the}\:\mathrm{form}\:{a}\:+\:{bi},\:\mathrm{where}\:{a}\:\mathrm{and}\:{b}\:\mathrm{are} \\ $$$$\:\mathrm{real}\:\mathrm{numbers},\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{a}? \\ $$$$\:\mathrm{A}.\:\mathrm{2} \\ $$$$\:\mathrm{B}.\:\frac{\mathrm{8}}{\mathrm{3}} \\ $$$$\:\mathrm{C}.\:\mathrm{3} \\ $$$$\:\mathrm{D}.\:\frac{\mathrm{11}}{\mathrm{3}} \\…
Question Number 128080 by mohammad17 last updated on 04/Jan/21 Answered by mathmax by abdo last updated on 04/Jan/21 $$\mathrm{z}^{\mathrm{4}} −\mathrm{1}=\sqrt{\mathrm{3}}\mathrm{i}\:\Rightarrow\mathrm{z}^{\mathrm{4}} \:=\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}\:=\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \:\mathrm{let}\:\mathrm{z}\:=\mathrm{re}^{\mathrm{i}\theta} \: \\ $$$$\mathrm{z}^{\mathrm{4}}…
Question Number 128079 by liberty last updated on 04/Jan/21 $$\Omega\:=\:\int\:\mathrm{ln}\:\left(\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} }\:\right)\mathrm{dx} \\ $$ Answered by bemath last updated on 04/Jan/21 $$\Omega=\mathrm{x}\:\mathrm{ln}\:\left(\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} }\:\right)−\int\:\frac{\mathrm{x}\left(\mathrm{1}+\frac{\mathrm{x}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}}…
Question Number 62542 by mr W last updated on 22/Jun/19 Commented by mr W last updated on 22/Jun/19 $${the}\:{distances}\:{of}\:{a}\:{point}\:{M}\:{to}\:{the} \\ $$$${vertexes}\:{of}\:{a}\:{triangle}\:{are}\:{p},{q},{r}. \\ $$$${find}\:{the}\:{side}\:{lengthes}\:{and}\:{hence}\:{the} \\ $$$${perimeter}\:{of}\:{the}\:{triangle}\:{with}\:{the}…
Question Number 62539 by aliesam last updated on 22/Jun/19 $$\mathrm{li}\underset{\mathrm{x}\rightarrow\infty} {\mathrm{m}}\frac{\mathrm{x}^{\mathrm{3}} +\mathrm{sin}\left(\mathrm{2x}\right)−\mathrm{2sin}\left(\mathrm{x}\right)}{\mathrm{arctan}\left(\mathrm{x}^{\mathrm{3}} \right)−\left(\mathrm{arctan}\left(\mathrm{x}\right)\right)^{\mathrm{3}} } \\ $$ Answered by tanmay last updated on 22/Jun/19 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}^{\mathrm{3}}…
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Question Number 128072 by benjo_mathlover last updated on 04/Jan/21 $$\:\frac{\mathrm{dy}}{\mathrm{dx}}\:+\:\mathrm{y}.\mathrm{sin}\:\mathrm{x}\:=\:\mathrm{y}^{\mathrm{2}} .\mathrm{sin}\:\mathrm{2x} \\ $$ Answered by bemath last updated on 04/Jan/21 Terms of Service Privacy Policy…
Question Number 128073 by I want to learn more last updated on 04/Jan/21 $$\int\:\frac{\mathrm{sin}\left(\mathrm{2x}\right)}{\left(\mathrm{1}\:\:−\:\:\mathrm{x}\right)^{\mathrm{3}} }\:\:\mathrm{dx} \\ $$ Answered by mindispower last updated on 04/Jan/21 $$\Leftrightarrow{Im}\int\frac{{e}^{\mathrm{2}{ix}}…
Question Number 62534 by peter frank last updated on 22/Jun/19 Answered by tanmay last updated on 22/Jun/19 $${bi}\:{metal}\:{changed}\:{to}\:{a}\:{shape}\:{concave}.. \\ $$$$\:\:{as}\:\:\:\smallsmile\:{and}\:{touch}\:{upper}\:.\:{point}\:{to}\:{activate} \\ $$$${A}\:…{so}\:{A}\:{is}\:{cooler} \\ $$$${and}\:{B}\:{is}\:{heater}…{i}\:{think}\:{so} \\…
Question Number 128068 by benjo_mathlover last updated on 04/Jan/21 $$\:\mathrm{Given}\:\mathrm{25x}^{\mathrm{2}} −\mathrm{30x}+\mathrm{7}=\mathrm{0}\:\mathrm{has}\:\mathrm{the} \\ $$$$\mathrm{roots}\:\mathrm{are}\:\mathrm{cos}\:\alpha\:\mathrm{and}\:\mathrm{cos}\:\beta.\:\mathrm{If}\:\mathrm{cos}\:\alpha−\mathrm{cos}\:\beta>\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{tan}\:\left(\frac{\alpha+\beta}{\mathrm{2}}\right).\mathrm{tan}\:\left(\frac{\alpha−\beta}{\mathrm{2}}\right)\:=? \\ $$ Answered by bemath last updated on 04/Jan/21 $$\:\mathrm{let}\:\frac{\alpha+\beta}{\mathrm{2}}=\mathrm{x}\:\wedge\:\frac{\alpha−\beta}{\mathrm{2}}=\mathrm{y}…