Question Number 134319 by aurpeyz last updated on 02/Mar/21 $${the}\:{radii}\:{of}\:{curvatures}\:{of}\:{front}\:{and} \\ $$$${rear}\:{surfaces}\:{of}\:{a}\:{thin}\:{convex}\:{lens}\: \\ $$$${are}\:\mathrm{30}{cm}\:{and}\:\mathrm{12}{cm}\:{respectively}.\: \\ $$$${what}\:{is}\:{its}\:{focal}\:{length}\:{when}\:{placed} \\ $$$${inside}\:{water}?\:\left({take}\:{the}\:{refractive}\right. \\ $$$${indices}\:{of}\:{glass}\:{and}\:{water}\:{to}\:{be}\:\mathrm{1}.\mathrm{52} \\ $$$${and}\:\mathrm{1}.\mathrm{33}\:{respectively} \\ $$ Commented…
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Question Number 3239 by Rasheed Soomro last updated on 08/Dec/15 $$\mathcal{I}\:{don}'{t}\:{know}\:{the}\:{value}\:{of}\:\:{Log}\left(−\mathrm{1}\right)\:{but}\:{I}\:{calculate} \\ $$$${it}\:{in}\:{the}\:{following}\:{way}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\left(−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\:\:\:\:\:\Rightarrow\:\:{Log}\left(−\mathrm{1}\right)^{\mathrm{2}} ={Log}\left(\mathrm{1}\right) \\ $$$$\:\:\:\:\:\Rightarrow\:\:\mathrm{2}×{Log}\left(−\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\Rightarrow\:{Log}\left(−\mathrm{1}\right)=\frac{\mathrm{0}}{\mathrm{2}}=\mathrm{0} \\ $$$${Am}\:\mathcal{I}\:{correct}?\:\mathcal{I}{f}\:{no},{why}?…
Question Number 68775 by rajesh4661kumar@gmail.com last updated on 15/Sep/19 Answered by $@ty@m123 last updated on 15/Sep/19 $${Let}\:{a}\:{be}\:\:{the}\:{first}\:{term}\:{and}\:{d}\:{be}\:{the} \\ $$$${common}\:{difference}\:{of}\:{AP}. \\ $$$${a}+\left({p}−\mathrm{1}\right){d}={A}\:…\left(\mathrm{1}\right) \\ $$$${a}+\left({q}−\mathrm{1}\right){d}={AR}\:\:….\left(\mathrm{2}\right) \\ $$$${a}+\left({r}−\mathrm{1}\right){d}={AR}^{\mathrm{2}}…
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Question Number 68768 by aliesam last updated on 15/Sep/19 $$\sqrt[{\mathrm{3}}]{\mathrm{2}{x}−\mathrm{1}}\:+\sqrt[{\mathrm{3}}]{{x}−\mathrm{1}}\:=\:\mathrm{1} \\ $$ Commented by kaivan.ahmadi last updated on 15/Sep/19 $${t}={x}−\mathrm{1}\Rightarrow{x}={t}+\mathrm{1}\Rightarrow \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{2}{t}+\mathrm{1}}+\sqrt[{\mathrm{3}}]{{t}}=\mathrm{1}\Rightarrow\sqrt[{\mathrm{3}}]{\mathrm{2}{t}+\mathrm{1}}=\mathrm{1}−\sqrt[{\mathrm{3}}]{{t}}\Rightarrow \\ $$$$\mathrm{2}{t}+\mathrm{1}=\mathrm{1}−\mathrm{3}\sqrt[{\mathrm{3}}]{{t}}+\mathrm{3}\sqrt[{\mathrm{3}}]{{t}^{\mathrm{2}} }−{t}\Rightarrow\mathrm{3}{t}=\mathrm{3}\sqrt[{\mathrm{3}}]{{t}}\left(\sqrt[{\mathrm{3}}]{{t}}−\mathrm{1}\right)\Rightarrow…
Question Number 134301 by bramlexs22 last updated on 02/Mar/21 $$\Omega\:=\:\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{4}} }\:{dx} \\ $$ Answered by EDWIN88 last updated on 02/Mar/21 $$\mathrm{replace}\:\mathrm{x}\:\mathrm{by}\:\frac{\mathrm{1}}{\mathrm{x}}\:\mathrm{yields}\:…
Question Number 68767 by aliesam last updated on 15/Sep/19 Commented by ~ À ® @ 237 ~ last updated on 15/Sep/19 $${Let}\:{named}\:{it}\:\:{f}_{{n}} \left({x}\right) \\ $$$${Let}\:{named}\:\:\forall\:{a}>\mathrm{0}\:\:{g}\left({a},{x}\right)=\int\:\frac{{dx}}{{a}+{x}^{\mathrm{2}}…
Question Number 134303 by bramlexs22 last updated on 02/Mar/21 $$\mathcal{F}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{16}\:\mathrm{arctan}\:\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx} \\ $$ Answered by Ñï= last updated on 02/Mar/21 $$\mathcal{F}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{16}\:\mathrm{arctan}\:\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}}…
Question Number 134302 by bramlexs22 last updated on 02/Mar/21 $$\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{ln}\:{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\mathrm{dx} \\ $$ Answered by Dwaipayan Shikari last updated on 02/Mar/21 $${I}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}}…