Question Number 46998 by OTCHRRE ABDULLAI last updated on 03/Nov/18 $${In}\:{physics}\:{how}\:{do}\:{we}\:{find}\: \\ $$$${average}\:{half}-{life}? \\ $$$$ \\ $$$${please}\:{i}\:{need}\:{help} \\ $$ Answered by peter frank last updated…
Question Number 46996 by Raj Singh last updated on 03/Nov/18 Answered by tanmay.chaudhury50@gmail.com last updated on 03/Nov/18 $${A}\:{perform}\:\frac{\mathrm{1}}{\mathrm{15}}\:{portion}\:{of}\:{work}\:{in}\:\mathrm{1}\:{day} \\ $$$${B}\:{perform}\:\frac{\mathrm{1}}{\mathrm{20}}{portion}\:{of}\:{work}\:{in}\:\mathrm{1}\:{day} \\ $$$${both}\:{perform}\:\left(\frac{\mathrm{1}}{\mathrm{15}}+\frac{\mathrm{1}}{\mathrm{20}}=\frac{\mathrm{4}+\mathrm{3}}{\mathrm{60}}=\frac{\mathrm{7}}{\mathrm{60}}{portion}\:{in}\:\mathrm{1}\:{day}\right. \\ $$$${both}\:{perform}\:\mathrm{6}×\frac{\mathrm{7}}{\mathrm{60}}=\frac{\mathrm{7}}{\mathrm{10}}{portion}\:{in}\:\mathrm{6}\:{day} \\…
Question Number 112454 by bemath last updated on 08/Sep/20 $$\left(\mathrm{1}\right)\:\mathrm{find}\:\mathrm{the}\:\mathrm{locus}\:\mid\mathrm{z}−\mathrm{z}_{\mathrm{1}} \mid\:=\:\mathrm{2}\:\mathrm{meets} \\ $$$$\mathrm{the}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{axis} \\ $$$$\left(\mathrm{2}\right)\mathrm{On}\:\mathrm{a}\:\mathrm{single}\:\mathrm{Argand}\:\mathrm{diagram},\:\mathrm{sketch} \\ $$$$\mathrm{the}\:\mathrm{loci}\:\rightarrow\begin{cases}{\mid\mathrm{z}−\mathrm{z}_{\mathrm{1}} \mid=\mathrm{2}}\\{\mathrm{arg}\left(\mathrm{z}−\mathrm{z}_{\mathrm{2}} \right)=\frac{\pi}{\mathrm{4}}}\end{cases} \\ $$ Commented by MJS_new last…
Question Number 112430 by Dwaipayan Shikari last updated on 07/Sep/20 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\frac{\mathrm{1}}{\mathrm{1}−{x}^{{n}} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 177957 by infinityaction last updated on 11/Oct/22 $$\mathrm{let}\:\:\mathrm{a}\:>\:\mathrm{0}\:\mathrm{find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{infinite}\: \\ $$$$\mathrm{series} \\ $$$$\:\mathrm{1}\:+\:\frac{\left(\mathrm{loga}\right)^{\mathrm{2}} \:}{\mathrm{2}!}\:+\:\frac{\left(\mathrm{loga}\right)^{\mathrm{4}} \:}{\mathrm{4}!}\:+\:\frac{\left(\mathrm{loga}\right)^{\mathrm{6}} \:}{\mathrm{6}!}… \\ $$ Answered by mr W last updated…
Question Number 177922 by Lekhraj last updated on 11/Oct/22 Answered by mr W last updated on 11/Oct/22 $$\left({i}\right) \\ $$$${W}={mgs}\:\mathrm{sin}\:\theta+\frac{{m}\left({v}_{\mathrm{2}} ^{\mathrm{2}} −{v}_{\mathrm{1}} ^{\mathrm{2}} \right)}{\mathrm{2}} \\…
Question Number 46805 by 786786AM last updated on 31/Oct/18 $$\mathrm{In}\:\mathrm{an}\:\mathrm{A}.\mathrm{p}.,\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{first}\:\mathrm{n}\:\mathrm{terms}\:\mathrm{is}\:\mathrm{P}\:,\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{next}\:\mathrm{n}\:\mathrm{terms}\:\mathrm{is}\:\mathrm{Q}\:\mathrm{and} \\ $$$$\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{further}\:\mathrm{next}\:\mathrm{n}\:\mathrm{terms}\:\mathrm{is}\:\mathrm{R}.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{P},\:\mathrm{Q},\:\mathrm{R}\:\mathrm{is}\:\mathrm{an}\:\mathrm{A}.\mathrm{P}. \\ $$$$ \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 31/Oct/18 $${first}\:{term}\:{is}\:{a}\:\:{and}\:{common}\:{difference}\:{is}\:{d} \\…
Question Number 177870 by infinityaction last updated on 10/Oct/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 177859 by neinhaltsieger369 last updated on 10/Oct/22 Answered by a.lgnaoui last updated on 10/Oct/22 $${anser}\:{ix}\left({c}\right)\:\:\:\:\frac{\mathrm{3}}{\mathrm{7}} \\ $$$$\partial{W}=\frac{\partial{W}}{\partial{x}}\partial{x}+\frac{\partial{W}}{\partial{y}}\partial{y}+\frac{\partial{W}}{\partial{z}} \\ $$$$=\frac{\mathrm{1}}{\:\left(\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \:}\:\right)^{\mathrm{2}} }\left({x}+{y}+{z}\right)=\frac{{x}+{y}+{z}}{\:{x}^{\mathrm{2}}…
Question Number 177835 by infinityaction last updated on 09/Oct/22 Answered by mr W last updated on 09/Oct/22 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{x}^{{n}−\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}}…