Question Number 148915 by DELETED last updated on 01/Aug/21 Answered by DELETED last updated on 01/Aug/21 $$\frac{\mathrm{1}}{\mathrm{Rp}}\:=\:\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{1}}{\mathrm{8}}=\frac{\mathrm{2}}{\mathrm{8}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{Rp}=\mathrm{4}\:\Omega \\ $$$$ \\ $$$$\mathrm{V}_{\mathrm{p}} =\mathrm{V}_{\mathrm{rs1}} =\mathrm{2}\:\mathrm{Vol}…
Question Number 148914 by naka3546 last updated on 01/Aug/21 $${Solve}\:\:{the}\:\:{equation} \\ $$$$\:\:\:\mathrm{2}^{{x}} \:+\:{x}\:=\:\mathrm{11} \\ $$$${with}\:\:{Omega}\:\:{Function}\:. \\ $$ Commented by naka3546 last updated on 01/Aug/21 $${thank}\:{you},\:{sir}.…
Question Number 17828 by Abbas-Nahi last updated on 11/Jul/17 Answered by Tinkutara last updated on 11/Jul/17 $$\mathrm{sin}\:\left(\alpha\:+\:\beta\right)\:=\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta\:+\:\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta \\ $$$$\mathrm{sin}\:\left(\alpha\:−\:\beta\right)\:=\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta\:−\:\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta \\ $$$$\mathrm{sin}\:\left(\alpha\:+\:\beta\right)\:+\:\mathrm{sin}\:\left(\alpha\:−\:\beta\right)\:=\:\mathrm{2}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta \\ $$$$\mathrm{Let}\:\alpha\:+\:\beta\:=\:{x}\:\mathrm{and}\:\alpha\:−\:\beta\:=\:{y} \\ $$$$\mathrm{so}\:\mathrm{that}\:\alpha\:=\:\frac{{x}\:+\:{y}}{\mathrm{2}}\:\mathrm{and}\:\beta\:=\:\frac{{x}\:−\:{y}}{\mathrm{2}}.…
Question Number 148884 by 0731619 last updated on 01/Aug/21 Answered by aleks041103 last updated on 03/Aug/21 $$\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{{a}+{x}^{\mathrm{13}} }=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\mathrm{1}+\left(\frac{{x}}{{a}^{\mathrm{1}/\mathrm{13}} }\right)^{\mathrm{13}} }= \\…
Question Number 17805 by ibraheem160 last updated on 10/Jul/17 $${If}\:\theta={t}^{{n}} {e}^{−{r}^{\frac{\mathrm{2}}{{ut}}} } ,{find}\:{the}\:{value}\:{of} \\ $$$${n}\:{which}\:{will}\:{make}\:\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\:\frac{\partial}{\partial{r}}\left({r}^{\mathrm{2}} \frac{\partial\theta}{\left.\partial{r}\right)}\right. \\ $$$${equal}\:{to}\:\frac{\partial\theta}{\partial{t}} \\ $$ Terms of Service Privacy…
Question Number 17782 by Mr easymsn last updated on 10/Jul/17 $${let}\:{a},{b},{c},{x},{y}\:{and}\:{z}\:{be}\:{complex}\:{numbers} \\ $$$${such}\:{that}\:: \\ $$$${a}=\frac{{b}+{c}}{{x}−\mathrm{2}},\:{b}=\frac{{c}+{a}}{{y}−\mathrm{2}},\:{c}=\frac{{a}+{b}}{{z}−\mathrm{2}} \\ $$$${if}\:{xy}+{yz}+{zx}=\mathrm{1000}\:{and}\:{x}+{y}+{z}=\mathrm{2016}, \\ $$$${find}\:{the}\:{value}\:{of}\:{xyz} \\ $$ Commented by ajfour last…
Question Number 17779 by Mr easymsn last updated on 10/Jul/17 $${show}\:{that}\:\left\{{lo}\underset{{a}} {{g}ab}\right\}\left\{{lo}\underset{{b}} {{g}ab}\right\}={loga}\underset{{a}} {{b}}+{loga}\underset{{b}} {{b}} \\ $$ Answered by Tinkutara last updated on 11/Jul/17 $$\left(\mathrm{log}_{{a}}…
Question Number 83295 by Rio Michael last updated on 29/Feb/20 $$\mathrm{Expand}\:\mathrm{ln}\:\left(\mathrm{1}\:+\:\mathrm{sinh}\:{x}\right)\:\mathrm{as}\:\mathrm{a}\:\mathrm{series}\:\mathrm{in} \\ $$$$\mathrm{ascending}\:\mathrm{powers}\:\mathrm{of}\:{x}\:\mathrm{up}\:\mathrm{to}\:\mathrm{and}\:\mathrm{including} \\ $$$$\mathrm{the}\:\mathrm{term}\:\mathrm{in}\:{x}^{\mathrm{3}} \:.\:\mathrm{Hence}\:,\:\mathrm{show}\:\mathrm{that}\: \\ $$$$\:\left(\mathrm{1}\:+\:\mathrm{sinh}\:{x}\right)^{\frac{\mathrm{3}}{{x}}} \:\cong\:{e}^{\mathrm{2}} \left(\mathrm{1}\:−{x}\:+\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$ Terms of…
Question Number 148825 by DELETED last updated on 31/Jul/21 Answered by DELETED last updated on 31/Jul/21 $$\Sigma\mathrm{IR}+\Sigma\mathrm{E}=\mathrm{0} \\ $$$$\mathrm{i}_{\mathrm{1}} +\mathrm{i}_{\mathrm{2}} =\mathrm{i}_{\mathrm{3}} \:\rightarrow\mathrm{i}_{\mathrm{2}} \:=\mathrm{i}_{\mathrm{3}} −\mathrm{i}_{\mathrm{2}} \:……\left(\mathrm{1}\right)…
Question Number 148804 by saly last updated on 31/Jul/21 Commented by saly last updated on 31/Jul/21 $$\:\:\:{Construct}\:\:{a}\:{graph}\:\:{of}\:{a}\:{functoin}? \\ $$$$\:\:{help}\:{me}\:…… \\ $$ Answered by ArielVyny last…