Question Number 44900 by peter frank last updated on 06/Oct/18 Answered by MJS last updated on 06/Oct/18 $$\mathrm{1}\:\mathrm{large}\:\mathrm{pump}\:\mathrm{needs}\:{x}\:\mathrm{hours}\:\mathrm{to}\:\mathrm{fill}\:\mathrm{1}\:\mathrm{pool} \\ $$$$\mathrm{this}\:\mathrm{means}\:\mathrm{it}\:\mathrm{fills}\:\frac{\mathrm{1}}{{x}}\:\mathrm{pool}\:\mathrm{per}\:\mathrm{hour} \\ $$$$\mathrm{similar}\:\mathrm{1}\:\mathrm{small}\:\mathrm{pump}\:\mathrm{fills}\:\frac{\mathrm{1}}{{y}}\:\mathrm{pool}\:\mathrm{per}\:\mathrm{hour} \\ $$$$\mathrm{2}\:\mathrm{large}\:\mathrm{pumps}\:+\:\mathrm{1}\:\mathrm{small}\:\mathrm{pump}\:\mathrm{need}\:\mathrm{4}\:\mathrm{hours} \\…
Question Number 110410 by mathdave last updated on 28/Aug/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 110403 by mohammad17 last updated on 28/Aug/20 $${if}\:{W}\:{represents}\:{the}\:{Runesky}\:{determinant}\:{of}\:{the}\:{two} \\ $$$${independent}\:{solutions}\:{linearly}\:\left({y}_{\mathrm{1}} ,{y}_{\mathrm{2}} \right){of}\:{the}\:{equation}\:{y}^{''} +{p}\left({x}\right){y}^{'} +{Q}\left({x}\right)=\mathrm{0}\:{then}\:{demonstrate}\:{that}\:{W}\:{satisfies}\:{the}\:{differential}\:{equation}\:\left({W}^{\:'} +{p}\left({x}\right){W}=\mathrm{0}\right)\:{and}\:{solve}\:{this}\:{equation}\:{to}\:{qet}\:{W}\:? \\ $$$$ \\ $$$${help}\:{me}\:{sir}\:{please} \\ $$ Commented by…
Question Number 110395 by mathdave last updated on 28/Aug/20 $${prove}\:{to}\:{ealier}\:{problem}\:{of}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tanh}^{−\mathrm{1}} \left(\sqrt[{\mathrm{4}}]{{x}}\right)\mathrm{tanh}^{−\mathrm{1}} \left(\sqrt[{\mathrm{4}}]{{y}}\right)}{{x}\sqrt{{y}}}{dxdy}=\pi^{\mathrm{2}} \\ $$$${solution}\: \\ $$$${let} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}}…
Question Number 110367 by mathdave last updated on 28/Aug/20 Answered by Dwaipayan Shikari last updated on 28/Aug/20 $$\left.\mathrm{1}−\left({log}\left(−\mathrm{1}\right)\right)\right)^{\mathrm{2}} −\mathrm{4}{ilog}\left({i}\right) \\ $$$$\mathrm{1}−\left(\pi{i}\right)^{\mathrm{2}} −\mathrm{4}{i}\left(\frac{\pi{i}}{\mathrm{2}}\right) \\ $$$$\left(\pi+\mathrm{1}\right)^{\mathrm{2}} \\…
Question Number 110365 by mathdave last updated on 28/Aug/20 Answered by Rasheed.Sindhi last updated on 29/Aug/20 $${wxyz}=\mathrm{120}\:,\:{wxz}−{y}=\mathrm{26}, \\ $$$${xyz}−{w}=\mathrm{58}\:\:,\:\:{wz}+{xy}=\mathrm{22} \\ $$$$\frac{{wxyz}}{{y}}−{y}=\mathrm{26}\Rightarrow\frac{\mathrm{120}}{{y}}−{y}=\mathrm{26}….\left({ii}\right) \\ $$$$\frac{{wxyz}}{{w}}−{w}=\mathrm{58}\Rightarrow\frac{\mathrm{120}}{{w}}−{w}=\mathrm{58}…\left({iii}\right) \\ $$$$\frac{{wxyz}}{{xy}}+{xy}=\mathrm{22}\Rightarrow\frac{\mathrm{120}}{{xy}}+{xy}=\mathrm{22}…\left({iv}\right)…
Question Number 110350 by bobhans last updated on 28/Aug/20 $$\:\:\:\frac{{bob}}{{hans}} \\ $$$$\left(\mathrm{1}\right)\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\left({x}−\frac{\mathrm{9}}{{x}}\right)}{\mathrm{tan}\:\left({x}−\mathrm{3}\right)\mathrm{cos}\:\left(\frac{\mathrm{9}}{{x}}−{x}\right)}= \\ $$$$\left(\mathrm{2}\right)\left({x}\:\mathrm{tan}^{−\mathrm{1}} \left({y}\right)\right){dx}\:+\left(\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{1}+{y}^{\mathrm{2}} \right)}\right).\:{dy}\:=\mathrm{0}\: \\ $$ Answered by john santu last…
Question Number 44811 by jasno91 last updated on 05/Oct/18 Commented by math khazana by abdo last updated on 05/Oct/18 $${x}+\frac{\mathrm{1}}{{x}}=\mathrm{4}\:\Rightarrow\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\mathrm{16}\:\Rightarrow{x}^{\mathrm{2}} \:+\mathrm{2}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:=\mathrm{16}\:\Rightarrow \\ $$$${x}^{\mathrm{2}}…
Question Number 44812 by jasno91 last updated on 05/Oct/18 Commented by math khazana by abdo last updated on 05/Oct/18 $${x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} =\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{xy}\:=\mathrm{12}^{\mathrm{2}} \:−\mathrm{28}\:=\mathrm{144}−\mathrm{28} \\…
Question Number 44809 by jasno91 last updated on 05/Oct/18 Commented by Joel578 last updated on 05/Oct/18 $$\mathrm{Just}\:\mathrm{substitute}\:{x}\:\mathrm{with}\:\mathrm{12} \\ $$$$\mathrm{9}\left(\mathrm{12}^{\mathrm{2}} \right)\:+\:\mathrm{24}\left(\mathrm{12}\right)\:+\:\mathrm{16} \\ $$ Terms of Service…