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Author: Tinku Tara

an-object-placed-20cm-from-a-converging-lens-forms-a-magnified-clear-image-on-a-screen-when-the-lens-is-moved-20cm-towards-the-screen-a-smaller-clear-image-is-formed-on-the-screen-calculate-the-fo

Question Number 133860 by aurpeyz last updated on 24/Feb/21 $${an}\:{object}\:{placed}\:\mathrm{20}{cm}\:{from}\:{a}\:{converging} \\ $$$${lens}\:{forms}\:{a}\:{magnified}\:{clear}\:{image} \\ $$$${on}\:{a}\:{screen}.\:{when}\:{the}\:{lens}\:{is}\:{moved} \\ $$$$\mathrm{20}{cm}\:{towards}\:{the}\:{screen}.\:{a}\:{smaller} \\ $$$${clear}\:{image}\:{is}\:{formed}\:{on}\:{the}\:{screen}.\: \\ $$$${calculate}\:{the}\:{forcal}\:{length}\:{of}\:{the} \\ $$$${lens}. \\ $$$$\left({a}\right)\:\mathrm{1}.\mathrm{33} \\…

y-1-2x-7-3-

Question Number 68327 by 9102176137086 last updated on 08/Sep/19 $${y}=\left(\mathrm{1}−\mathrm{2}{x}^{−\mathrm{7}} \right)^{\mathrm{3}} \\ $$ Answered by $@ty@m123 last updated on 09/Sep/19 $${Please}\:{do}\:{not}\:{spoil}\:{the}\:{flow}\:{of}\:{this}\:{forum} \\ $$$${with}\:{a}\:{flood}\:{of}\:{similar}\:{questions}. \\ $$$${See}\:{solved}\:{examples}\:{in}\:{your}…

advanced-calculus-prove-that-0-1-ln-2-1-x-x-dx-2-3-1-x-t-0-1-ln-2-t-1-t-dt-0-1-n-0-ln-

Question Number 133857 by mnjuly1970 last updated on 24/Feb/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:…..#{advanced}\:\:\:\:……………\:\:\:{calculus}#….. \\ $$$$\:\:\:\:{prove}\:\:{that}\::::\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{{x}}{dx}\overset{?} {=}\mathrm{2}\zeta\left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\overset{\mathrm{1}−{x}={t}} {=}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left({t}\right)}{\mathrm{1}−{t}}{dt}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \underset{{n}=\mathrm{0}} {\overset{\infty}…

U-0-1-U-1-2-U-n-2-3-2-U-n-1-1-2-U-n-Determinate-the-smallest-integer-n-0-such-that-n-n-0-we-have-U-n-3-10-4-

Question Number 133856 by mathocean1 last updated on 24/Feb/21 $$ \\ $$$$\begin{cases}{{U}_{\mathrm{0}} =\mathrm{1}}\\{{U}_{\mathrm{1}} =\mathrm{2}}\\{\:{U}_{{n}+\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{2}}{U}_{{n}+\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}{U}_{{n}} }\end{cases} \\ $$$${Determinate}\:{the}\:{smallest}\:{integer} \\ $$$${n}_{\mathrm{0}} \:{such}\:{that}\:\forall\:{n}\geqslant{n}_{\mathrm{0}} \:{we}\:{have}\:\mid{U}_{{n}} −\mathrm{3}\mid\leqslant\mathrm{10}^{−\mathrm{4}} \\…

Question-133859

Question Number 133859 by mnjuly1970 last updated on 24/Feb/21 Answered by Ñï= last updated on 24/Feb/21 $$\underset{{n}=−\infty} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\left({x}+{n}\pi\right)^{\mathrm{2}} } \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{\mathrm{1}}{\left({x}+{n}\pi\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({x}−{n}\pi\right)^{\mathrm{2}}…

call-lim-n-H-n-ln-n-proof-that-is-finite-and-0-1-

Question Number 2783 by 123456 last updated on 27/Nov/15 $$\mathrm{call}\:\gamma:=\underset{{n}\rightarrow+\infty} {\mathrm{lim}H}_{{n}} −\mathrm{ln}\:{n} \\ $$$$\mathrm{proof}\:\mathrm{that}\:\gamma\:\mathrm{is}\:\mathrm{finite}\:\mathrm{and}\:\gamma\in\left(\mathrm{0},\mathrm{1}\right) \\ $$ Commented by Filup last updated on 27/Nov/15 $$\mathrm{I}\:\mathrm{am}\:\mathrm{curious}\:\mathrm{as}\:\mathrm{to}\:\mathrm{how}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{these} \\…

4sin-3x-e-4x-4-

Question Number 68316 by 9102176137086 last updated on 08/Sep/19 $$\int\left(\mathrm{4sin}\:\mathrm{3}{x}+\frac{{e}^{\mathrm{4}{x}} }{\mathrm{4}}\right) \\ $$ Commented by mathmax by abdo last updated on 08/Sep/19 $$=\mathrm{4}\int\:{sin}\left(\mathrm{3}{x}\right){dx}\:+\frac{\mathrm{1}}{\mathrm{4}}\int\:{e}^{\mathrm{4}{x}} {dx}\:+{c} \\…

x-3-2x-4-x-

Question Number 68315 by 9102176137086 last updated on 08/Sep/19 $$\int\left(\frac{{x}^{−\mathrm{3}} +\mathrm{2}{x}−\mathrm{4}}{{x}}\right) \\ $$ Commented by mathmax by abdo last updated on 10/Sep/19 $$=\int\:\left({x}^{−\mathrm{4}} \:+\mathrm{2}\:−\frac{\mathrm{4}}{{x}}\right){dx}\:=\frac{\mathrm{1}}{−\mathrm{4}+\mathrm{1}}{x}^{−\mathrm{4}+\mathrm{1}} \:+\mathrm{2}{x}−\mathrm{4}{ln}\mid{x}\mid\:+{c}…