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Question-170488

Question Number 170488 by leicianocosta last updated on 25/May/22 Answered by FelipeLz last updated on 25/May/22 $$\left.\mathrm{1}\right)\: \\ $$$$\:\:\:\:{a}\left({t}\right)\:=\:\frac{{d}}{{dt}}\left[{v}\left({t}\right)\right]\:=\:\frac{{d}}{{dt}}\left[\mathrm{2}+\mathrm{3}{t}+\mathrm{5}{t}^{\mathrm{2}} \right]\:=\:\mathrm{3}+\mathrm{10}{t} \\ $$$$\:\:\:\:{a}\left(\mathrm{5}\right)\:=\:\mathrm{53}\:{m}\centerdot{s}^{−\mathrm{2}} \\ $$$$\left.\mathrm{2}\right) \\…

Given-that-log-4-y-1-log-4-x-y-m-and-log-2-y-1-log-2-x-m-1-show-that-y-2-1-8-m-

Question Number 170468 by MathsFan last updated on 24/May/22 $$\:\mathrm{G}{iven}\:{that}\:{log}_{\mathrm{4}} \left({y}−\mathrm{1}\right)+{log}_{\mathrm{4}} \left(\frac{{x}}{{y}}\right)={m} \\ $$$$\:{and}\:{log}_{\mathrm{2}} \left({y}+\mathrm{1}\right)−{log}_{\mathrm{2}} {x}={m}−\mathrm{1}, \\ $$$$\:{show}\:{that}\:{y}^{\mathrm{2}} =\mathrm{1}−\mathrm{8}^{{m}} \\ $$ Answered by cortano1 last…

Question-104905

Question Number 104905 by mohammad17 last updated on 24/Jul/20 Answered by OlafThorendsen last updated on 24/Jul/20 $${x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}\:=\:\mathrm{0} \\ $$$$\frac{\mathrm{1}−{x}^{\mathrm{5}} }{\mathrm{1}−{x}}\:=\:\mathrm{0} \\ $$$${x}^{\mathrm{5}}…