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Question-195874

Question Number 195874 by Humble last updated on 12/Aug/23 Answered by mr W last updated on 12/Aug/23 $${x}={v}_{\mathrm{0}} \mathrm{sin}\:\theta\:{t}+\frac{{g}\:\mathrm{cos}\:\alpha}{\mathrm{2}}\:{t}^{\mathrm{2}} \\ $$$${z}={v}_{\mathrm{0}} \mathrm{cos}\:\theta{t}−\frac{{g}\:\mathrm{sin}\:\alpha}{\mathrm{2}}\:{t}^{\mathrm{2}} \\ $$$$\left.{a}\right) \\…

Question-195838

Question Number 195838 by sonukgindia last updated on 11/Aug/23 Answered by sniper237 last updated on 11/Aug/23 $${Area}=\int_{{u}} ^{{v}} {f}\left({x}\right){dx}\overset{{t}={lnx}} {=}\int_{{a}} ^{{b}} \:\frac{{dt}}{\:\sqrt{\left({t}−{a}\right)\left({b}−{t}\right)}}\: \\ $$$$\:\:\:\overset{{s}=\frac{{a}+{b}}{\mathrm{2}}−{t}} {=}\int_{−{c}}…