Question Number 146300 by akolade last updated on 12/Jul/21 Answered by nimnim last updated on 12/Jul/21 $${cosB}=\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{ac}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left({n}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +\left({n}^{\mathrm{2}} −\mathrm{2}{n}\right)^{\mathrm{2}}…
Question Number 15179 by Joel577 last updated on 08/Jun/17 $$\mathrm{If}\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{4}{y}\:=\:\mathrm{0}\: \\ $$$${y}\left(\mathrm{0}\right)\:=\:\mathrm{1}\:\mathrm{and}\:{y}\left(\frac{\pi}{\mathrm{6}}\right)\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{How}\:\mathrm{to}\:\mathrm{find}\:{y}\left({x}\right)\:? \\ $$ Answered by sma3l2996 last updated on 08/Jun/17…
Question Number 146164 by mnjuly1970 last updated on 11/Jul/21 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{calulate}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{S}\::\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\mathrm{H}_{\frac{{n}}{\mathrm{2}}\:} }{\:\mathrm{2}^{\:{n}} }\:=? \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:…….{m}.{n}. \\ $$ Answered…
Question Number 146063 by mnjuly1970 last updated on 10/Jul/21 $$ \\ $$$$\:\:\:\:{if}\:\:{g}\left({x}\right)=\frac{{x}^{\:\mathrm{2}} −{x}}{\mathrm{2}{x}−\mathrm{1}}\:\:\:,\:{D}_{{g}} =\:\left[\mathrm{1}\:,\:\infty\right) \\ $$$$\:\:\:\:,\:{lim}_{{x}\rightarrow\infty} \frac{{g}^{\:−\mathrm{1}} \left({x}\right)}{{ax}\:+\:{b}}\:=\:{b}−{a}\:\:\left({a}\:<\mathrm{0}\:\right) \\ $$$$\:\:{then}\:{find}\:\:{the}\:{value}\:{of}\:{Max}\:\left({b}\:\right) \\ $$$$\:\: \\ $$$$\:\:{D}_{\:{g}} \:=\:{Domain}\:…
Question Number 146054 by mnjuly1970 last updated on 10/Jul/21 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{I}\::=\:\int_{\mathrm{0}} ^{\:\:\infty} {e}^{\:−{x}} \:.\:\mathrm{J}_{\frac{\mathrm{1}}{\mathrm{2}}} \:\left({x}\:\right)\:{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{J}_{{v}\:} \:\left({x}\:\right)\:=\:{x}^{\:{v}} \:\underset{{n}=\mathrm{0}} {\overset{\:\infty} {\sum}}\frac{\left(−\:\mathrm{1}\:\right)^{\:{n}} \:{x}^{\:\mathrm{2}{n}} }{\mathrm{2}^{\:{n}\:+\:{v}} \:{n}\:!\:\Gamma\:\left(\:{n}\:+\:{v}\:+\mathrm{1}\:\right)}…
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Question Number 145701 by alcohol last updated on 07/Jul/21 Commented by MJS_new last updated on 07/Jul/21 $$\mathrm{1}. \\ $$$$\mathrm{square}−\mathrm{2}×\left(\mathrm{square}−\mathrm{quarter}\:\mathrm{circle}\right)= \\ $$$$=−\mathrm{square}+\mathrm{2}×\mathrm{quarter}\:\mathrm{circle}= \\ $$$$=\mathrm{half}\:\mathrm{circle}−\mathrm{square}=\mathrm{4}^{\mathrm{2}} \pi/\mathrm{2}−\mathrm{4}^{\mathrm{2}} =…
Question Number 145620 by Study last updated on 06/Jul/21 $$\frac{{d}}{{dx}}\left(\sqrt[{\mathrm{3}}]{{x}+\sqrt[{\mathrm{3}}]{{x}+\sqrt[{\mathrm{3}}]{{x}+…}}}\right)=? \\ $$ Answered by Olaf_Thorendsen last updated on 06/Jul/21 $${y}\:=\:{f}\left({x}\right)\:=\:\sqrt[{\mathrm{3}}]{{x}+{f}\left({x}\right)}\:=\:\sqrt[{\mathrm{3}}]{{x}+{y}} \\ $$$${y}^{\mathrm{3}} \:=\:{x}+{y} \\ $$$$\mathrm{3}{y}^{\mathrm{2}}…
Question Number 145602 by mnjuly1970 last updated on 06/Jul/21 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..\mathrm{Advanced}\:………\mathrm{Calculus}….. \\ $$$$\:\:\:\mathrm{Q}::\:\:\:\:\:\:\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\::: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\begin{array}{|c|c|}{\:{i}\:::\:\:\:\boldsymbol{\phi}\::=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{Ln}\:\left(\:\Gamma\:\left(\:\mathrm{2}\:+\:{x}\:\right)\:\right){dx}\:=\:?\:\:\:\:}\\{\:{ii}\:::\:\:\:\Omega\::=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\mathrm{1}}{\:{n}\:\left(\:\mathrm{2}{n}\:+\:\mathrm{3}\:\right)}\:=\:?}\\\hline\end{array} \\ $$$$ \\…
Question Number 14513 by tawa tawa last updated on 01/Jun/17 $$\mathrm{If}\:\:\mathrm{y}^{\mathrm{2}} \left(\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}} \right)\:=\:\mathrm{1}\:−\:\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{Show}\:\mathrm{that},\:\:\:\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{1}\:−\:\mathrm{y}^{\mathrm{4}} }{\mathrm{1}\:−\:\mathrm{x}^{\mathrm{4}} } \\ $$ Answered by mrW1 last updated…