Question Number 226780 by Spillover last updated on 14/Dec/25 $${By}\:{using}\:{concept}\:{of}\:{complex} \\ $$$${number} \\ $$$${show}\:{that} \\ $$$$\mathrm{tan}\:\mathrm{5}\theta=\frac{\mathrm{tan}\:^{\mathrm{5}} \theta−\mathrm{10tan}\:^{\mathrm{3}} \theta+\mathrm{5tan}\:\theta}{\mathrm{5tan}\:^{\mathrm{4}} \theta−\mathrm{10tan}\:^{\mathrm{2}} \theta+\mathrm{1}} \\ $$ Answered by Frix…
Question Number 226776 by Spillover last updated on 14/Dec/25 $${Approximate}\:\int_{\mathrm{0}} ^{\mathrm{1}} {xe}^{{x}^{\mathrm{2}} } {dx}\:{with}\:\mathrm{6}\:{ordinates}. \\ $$$${Use}\:{both}\:{rules}\:{Simpsons}\:{and} \\ $$$${Trapozoidal}\:{rules},{hence}\:{evaluate}\:{and} \\ $$$${calculate}\:{the}\:{percentage}\:{error} \\ $$$${commetted}\:{for}\:{each}\:{case}.{Give}\:{comments} \\ $$$$ \\…
Question Number 226777 by Spillover last updated on 16/Dec/25 $${Show}\:{that} \\ $$$$\:\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\mathrm{2}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{−\mathrm{1}} {dx}=\frac{\mathrm{4}−\pi}{\mathrm{4}} \\ $$$${Hence}\:{by}\:{using}\:{Simpson}^{'} {s} \\ $$$${rule}\:{find}\:{the}\:{value}\:\:{of}\:\pi\:{with}\: \\ $$$${eleven}\:{ordinates}. \\…
Question Number 226778 by Spillover last updated on 16/Dec/25 $${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}}…