Question Number 51250 by Tawa1 last updated on 25/Dec/18 $$\mathrm{Given}\:\mathrm{that}\:\:\:\mathrm{z}_{\mathrm{1}} \:=\:\mathrm{R}_{\mathrm{1}} \:+\:\mathrm{R}\:+\:\mathrm{j}\omega\mathrm{L}\:;\:\:\:\mathrm{z}_{\mathrm{2}} \:=\:\mathrm{R}_{\mathrm{2}} \:;\:\:\mathrm{z}_{\mathrm{3}} \:=\:\frac{\mathrm{1}}{\mathrm{j}\omega\mathrm{C}_{\mathrm{3}} } \\ $$$$\mathrm{and}\:\:\mathrm{z}_{\mathrm{4}} \:=\:\mathrm{R}_{\mathrm{4}} \:+\:\frac{\mathrm{1}}{\mathrm{j}\omega\mathrm{C}_{\mathrm{4}} }\:\:\mathrm{and}\:\mathrm{also}\:\mathrm{that}\:\:\:\mathrm{z}_{\mathrm{1}} \mathrm{z}_{\mathrm{3}} \:\:=\:\:\mathrm{z}_{\mathrm{2}} \mathrm{z}_{\mathrm{4}} \:,\:\:\:\mathrm{express}\:…
Question Number 51248 by Tawa1 last updated on 25/Dec/18 $$\mathrm{If}\:\:\:\:\:\frac{\mathrm{R}_{\mathrm{1}} \:+\:\mathrm{j}\omega\mathrm{L}}{\mathrm{R}_{\mathrm{3}} }\:\:=\:\:\frac{\mathrm{R}_{\mathrm{2}} }{\mathrm{R}_{\mathrm{4}} \:−\:\mathrm{j}\:\frac{\mathrm{1}}{\omega\mathrm{C}}}\:\:,\:\:\:\mathrm{where}\:\:\mathrm{R}_{\mathrm{1}} ,\:\mathrm{R}_{\mathrm{2}} ,\:\mathrm{R}_{\mathrm{3}} ,\:\mathrm{R}_{\mathrm{4}} ,\:\omega,\:\mathrm{L}\:\mathrm{and}\:\mathrm{C} \\ $$$$\mathrm{are}\:\mathrm{real}\:,\:\:\mathrm{show}\:\mathrm{that}\:\:\:\:\mathrm{L}\:=\:\frac{\mathrm{C}\:\mathrm{R}_{\mathrm{2}} \mathrm{R}_{\mathrm{3}} }{\omega^{\mathrm{2}} \mathrm{C}^{\mathrm{2}} \mathrm{R}_{\mathrm{4}} ^{\mathrm{2}}…
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Question Number 182253 by Tawa11 last updated on 06/Dec/22 Commented by Tawa11 last updated on 06/Dec/22 $$\mathrm{Sirs},\:\mathrm{see}\:\mathrm{whag}\:\mathrm{I}\:\mathrm{did}. \\ $$$$\:\:\mathrm{h}\:\:=\:\:\mathrm{5}\:\:+\:\:\mathrm{30t}\:\:−\:\:\mathrm{5t}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\mathrm{h}'\:\:\:=\:\:\:\mathrm{30}\:\:\:−\:\:\:\mathrm{10t} \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{10t}\:\:\:=\:\:\:\mathrm{30} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{t}\:\:\:=\:\:\:\frac{\mathrm{30}}{\mathrm{10}}…
Question Number 116685 by mathocean1 last updated on 05/Oct/20 $$\mathrm{Given}\:\mathrm{the}\:\mathrm{equality}: \\ $$$$\mathrm{1}+\mathrm{3}+\mathrm{5}+…+\left(\mathrm{2p}+\mathrm{1}\right)=\left(\mathrm{p}+\mathrm{1}\right)^{\mathrm{2}\:} \:\:\:\mathrm{p}\:\in\:\mathbb{N}^{\ast} \\ $$$$ \\ $$$$\mathrm{Show}\:\mathrm{this}\:\mathrm{equality}\:\mathrm{is}\:\mathrm{true}\:\mathrm{when} \\ $$$$\mathrm{we}\:\mathrm{replace}\:\mathrm{p}\:\mathrm{by}\:\mathrm{p}+\mathrm{1} \\ $$ Answered by TANMAY PANACEA…
Question Number 116610 by moh175 last updated on 05/Oct/20 $$ \\ $$$$\:\:\:\:\:{solve}\:: \\ $$$$\:\:\:\:\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}{h}} \:+\:\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{{h}} \:=\:\mathrm{1} \\ $$ Commented by Dwaipayan Shikari last updated on…
Question Number 182139 by Mastermind last updated on 04/Dec/22 $$\mathrm{Find}\:\mathrm{constant}\:\mathrm{a},\:\mathrm{b},\:\mathrm{so}\:\mathrm{that} \\ $$$$\mathrm{y}\left(\mathrm{t}\right)=\left(\mathrm{t}+\mathrm{3}\right)\mathrm{e}^{\mathrm{2t}} \:\mathrm{is}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{IVP} \\ $$$$\mathrm{y}^{'} =\mathrm{ay}+\mathrm{e}^{\mathrm{2t}} ,\:\:\:\:\:\:\:\:\:\:\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{b} \\ $$$$ \\ $$$$. \\ $$ Commented by…
Question Number 182069 by Mastermind last updated on 03/Dec/22 $$\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{3xy}=\mathrm{5x} \\ $$$$ \\ $$$$\mathrm{solve} \\ $$ Commented by CElcedricjunior last updated on 04/Dec/22 $$\left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}}…