Question Number 195751 by sonukgindia last updated on 09/Aug/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 195750 by sonukgindia last updated on 09/Aug/23 Commented by mr W last updated on 09/Aug/23 $$?=\mathrm{3}×\left(\mathrm{8}+\mathrm{8}\right)−\mathrm{8}−\mathrm{15}=\mathrm{25} \\ $$ Commented by SajaRashki last updated…
Question Number 195746 by Humble last updated on 09/Aug/23 Answered by mr W last updated on 09/Aug/23 $$\rho={x}\rho_{\mathrm{2}} +\left(\mathrm{1}−{x}\right)\rho_{\mathrm{1}} \\ $$$$\Rightarrow{x}=\frac{\rho−\rho_{\mathrm{1}} }{\rho_{\mathrm{2}} −\rho_{\mathrm{1}} }=\frac{\mathrm{0}.\mathrm{917}−\mathrm{0}.\mathrm{876}}{\mathrm{1}.\mathrm{049}−\mathrm{0}.\mathrm{876}}\approx\mathrm{23}.\mathrm{7\%} \\…
Question Number 195708 by mokys last updated on 08/Aug/23 $$\boldsymbol{{Prove}}\:\boldsymbol{{that}}\::\:\boldsymbol{{log}}_{\left(\sqrt{\boldsymbol{{a}}}\:−\:\boldsymbol{{b}}\right)} \left(\sqrt{\boldsymbol{{a}}}\:+\boldsymbol{{b}}\right)\:=\:−\mathrm{1} \\ $$ Commented by mr W last updated on 08/Aug/23 $${you}\:{can}\:{not}\:{prove}\:{things}\:{like}\:{x}+{y}=\mathrm{2}. \\ $$ Commented…
Question Number 195704 by sonukgindia last updated on 08/Aug/23 Answered by mr W last updated on 08/Aug/23 $${A}=\left({a}+{b}\right)^{\mathrm{2}} \\ $$$${A}_{\mathrm{1}} =\frac{\left({a}+{b}\right){b}}{\mathrm{2}} \\ $$$${A}_{\mathrm{2}} =\left(\frac{{b}}{\:\sqrt{\left({a}+{b}\right)^{\mathrm{2}} +{b}^{\mathrm{2}}…
Question Number 195707 by sonukgindia last updated on 08/Aug/23 Answered by mr W last updated on 08/Aug/23 $${y}''={y}'\frac{{d}\left({y}'\right)}{{dy}} \\ $$$${p}={y}' \\ $$$${p}\frac{{dp}}{{dy}}=\frac{\mathrm{1}}{{y}^{\mathrm{3}} } \\ $$$${pdp}=\frac{{dy}}{{y}^{\mathrm{3}}…
Question Number 195702 by Humble last updated on 08/Aug/23 Answered by mr W last updated on 08/Aug/23 $$\rho_{\mathrm{1}} {gh}_{\mathrm{1}} =\rho_{\mathrm{2}} {gh}_{\mathrm{2}} \\ $$$$\Rightarrow\rho_{\mathrm{2}} =\frac{{h}_{\mathrm{1}} \rho_{\mathrm{1}}…
Question Number 195697 by sonukgindia last updated on 08/Aug/23 Answered by MM42 last updated on 08/Aug/23 $${if}\:\:{mean}\:\:\:{lim}_{{x}\rightarrow\infty} \:\sqrt[{{x}}]{{x}\:}\:\:{then} \\ $$$${A}={lim}_{{x}\rightarrow\infty} \:\sqrt[{{x}}]{{x}} \\ $$$$\Rightarrow{lnA}={lim}_{{x}\rightarrow\infty} \:\frac{{lnx}}{{x}}\:=\mathrm{0}\Rightarrow{A}=\mathrm{1} \\…
Question Number 195725 by sonukgindia last updated on 08/Aug/23 Commented by mr W last updated on 13/Aug/23 $$\Rightarrow{Q}\mathrm{195931} \\ $$ Terms of Service Privacy Policy…
Question Number 195721 by Humble last updated on 08/Aug/23 Answered by mr W last updated on 08/Aug/23 $${volume}\:{of}\:{copper}\:{V}_{{cu}} =\frac{\mathrm{25}}{\mathrm{9}.\mathrm{8}×\mathrm{8}.\mathrm{93}}=\mathrm{0}.\mathrm{286}\:{ml} \\ $$$${total}\:{volume}\:{V}=\frac{\mathrm{25}−\mathrm{20}}{\mathrm{9}.\mathrm{8}×\mathrm{1}.\mathrm{0}}=\mathrm{0}.\mathrm{510}\:{ml} \\ $$$${volume}\:{of}\:{bubble}\:{V}_{{b}} =\mathrm{0}.\mathrm{510}−\mathrm{0}.\mathrm{286}=\mathrm{0}.\mathrm{224}\:{ml} \\…