Question Number 226778 by Spillover last updated on 16/Dec/25 $${Solve}\:{the}\:{following}\:{D}.{E} \\ $$$$\left({a}\right)\:\frac{{dy}}{{dx}}+\mathrm{2}{y}={xy}^{\mathrm{2}} \\ $$$$\left({b}\right)\:\frac{{dy}}{{dx}}+\mathrm{3}\frac{{y}}{{x}}=\mathrm{2}{x}^{\mathrm{4}} {y}^{\mathrm{4}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 226779 by Spillover last updated on 14/Dec/25 $${By}\:{using}\:{De}\:{Moivres}\:{theorm} \\ $$$${simplify} \\ $$$$\left({a}\right)\frac{\left(\mathrm{cos}\:\frac{\pi}{\mathrm{2}}−{i}\mathrm{sin}\:\frac{\pi}{\mathrm{2}}\right)\left(\mathrm{cos}\:\frac{\pi}{\mathrm{3}}+{i}\mathrm{sin}\:\frac{\pi}{\mathrm{3}}\right)}{\mathrm{cos}\:\frac{\pi}{\mathrm{3}}−{i}\mathrm{sin}\:\frac{\pi}{\mathrm{3}}} \\ $$$$\left({b}\right)\frac{\mathrm{cos}\:\frac{\pi}{\mathrm{8}}+{i}\mathrm{sin}\:\frac{\pi}{\mathrm{8}}}{\mathrm{cos}\:\frac{\pi}{\mathrm{6}}+{i}\mathrm{sin}\:\frac{\pi}{\mathrm{6}}} \\ $$ Answered by Frix last updated on 14/Dec/25…
Question Number 226775 by aba_math last updated on 14/Dec/25 $${Prove}\:{that}\:\left({a}−{b}\right)\left({a}−{c}\right)\left({a}−{d}\right)\left({b}−{c}\right)\left({b}−{d}\right)\left({c}−{d}\right)\:{divisible}\:{by}\:\mathrm{12} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 226785 by fantastic2 last updated on 14/Dec/25 Commented by fantastic2 last updated on 15/Dec/25 $${there}\:{are}\:{three}\:{small}\:{balls}\:{of}\:{mass}\:{m}_{\mathrm{1},} {m}_{\mathrm{2}} \:{and}\:{m}_{\mathrm{3}} \\ $$$${all}\:{are}\:{hanging}\:{from}\:{a}\:{point}\:{O}\:{by}\:{a}\:{string} \\ $$$${length}\:{l}\:. \\ $$$${what}\:{charge}\:{should}\:{be}\:{given}\:{to}\:{them}\:{so}…
Question Number 226770 by gregori last updated on 14/Dec/25 Answered by TonyCWX last updated on 14/Dec/25 $$\mathrm{Stewart}'\mathrm{s}\:\mathrm{Theorem}: \\ $$$${b}^{\mathrm{2}} {x}+{b}^{\mathrm{2}} {y}=\left({x}+{y}\right)\left({z}^{\mathrm{2}} −{xy}\right) \\ $$$${b}^{\mathrm{2}} \left({x}+{y}\right)=\left({x}+{y}\right)\left({z}^{\mathrm{2}}…
Question Number 226771 by Spillover last updated on 14/Dec/25 Answered by Frix last updated on 14/Dec/25 $${a}\neq\pm\mathrm{1} \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{say}\:{a}>\mathrm{0}\:\mathrm{because} \\ $$$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{d}\theta}{\mathrm{1}−\mathrm{2}{a}\mathrm{cos}\:\theta\:+{a}^{\mathrm{2}} }=\underset{\mathrm{0}} {\overset{\pi}…
Question Number 226766 by mr W last updated on 13/Dec/25 Answered by mahdipoor last updated on 13/Dec/25 $$\mathrm{if}\:\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{ax}+\mathrm{b}\:\Rightarrow \\ $$$$\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{2x}+\mathrm{1}\:\mathrm{or}\:−\mathrm{2x}−\mathrm{3} \\ $$$$\mathrm{but}\:\mathrm{its}\:\mathrm{only}\:\mathrm{answer}? \\ $$$$\mathrm{all}\:\mathrm{function}\:\mathrm{can}\:\mathrm{show}\:\mathrm{as}\:\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{i}=\mathrm{0}} {\overset{\mathrm{m}}…
Question Number 226755 by Hanuda354 last updated on 13/Dec/25 Answered by TonyCWX last updated on 13/Dec/25 $${A}_{{Green}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{8}^{\mathrm{2}} \right)\left(\frac{\pi}{\mathrm{3}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{8}^{\mathrm{2}} \right)\left(\frac{\pi}{\mathrm{3}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\pi\left(\mathrm{4}^{\mathrm{2}} \right)=\frac{\mathrm{40}}{\mathrm{3}}\pi−\mathrm{16}\sqrt{\mathrm{3}} \\ $$$${A}_{{Blue}\:} =\:\left(\frac{\mathrm{12}−\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{2}\pi}{\mathrm{12}}\right)\left(\mathrm{8}^{\mathrm{2}} \right)=\mathrm{64}−\mathrm{16}\sqrt{\mathrm{3}}−\frac{\mathrm{32}}{\mathrm{3}}\pi…
Question Number 226732 by Spillover last updated on 12/Dec/25 Answered by TonyCWX last updated on 12/Dec/25 $$\mathrm{Characteristic}\:\mathrm{Equation}: \\ $$$$\lambda^{\mathrm{2}} +\mathrm{2}\lambda−\mathrm{8}=\mathrm{0}\:\Rightarrow\:\lambda_{\mathrm{1}} =−\mathrm{4}\:\mathrm{and}\:\lambda_{\mathrm{2}} =\mathrm{2} \\ $$$$ \\…
Question Number 226745 by fantastic2 last updated on 12/Dec/25 $${if}\:\frac{{x}}{{lm}−{n}^{\mathrm{2}} }=\frac{{y}}{{mn}−{l}^{\mathrm{2}} }=\frac{{z}}{{nl}−{m}^{\mathrm{2}} } \\ $$$${then}\:{show}\:{lx}+{my}+{nz}=\mathrm{0} \\ $$$$ \\ $$ Commented by fantastic2 last updated on…