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Question Number 88606 by TawaTawa1 last updated on 11/Apr/20

Commented by jagoll last updated on 12/Apr/20

$$\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{line}$$$$\mathrm{y}=\mathrm{mx}\Rightarrow 4\mathrm{t}-{\mathrm{t}}^2=\mathrm{m}.\mathrm{t}$$$$\mathrm{m}=4-\mathrm{t}\Rightarrow \mathrm{y}=(4-\mathrm{t})\mathrm{x}$$$$\mathrm{line}\mathrm{and}\mathrm{parabolic}\mathrm{intersect}$$$$\Rightarrow 4\mathrm{x}-{\mathrm{x}}^2=4\mathrm{x}-\mathrm{tx}$$$${\mathrm{x}}^2-\mathrm{tx}=0,\mathrm{\Delta }={\mathrm{t}}^2$$$$\mathrm{area}=\frac{{\mathrm{t}}^2\sqrt{{\mathrm{t}}^2}}{6.1}=\frac{9}{2}\Rightarrow {\mathrm{t}}^3=27$$$$\mathrm{t}=3$$

Commented by TawaTawa1 last updated on 12/Apr/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by mahdi last updated on 11/Apr/20

$${\int }_0^{\mathrm{t}}(4\mathrm{x}-{\mathrm{x}}^2)\mathrm{dx}-\frac{\mathrm{t}(4\mathrm{t}-{\mathrm{t}}^2)}{2}=\frac{9}{2}$$$${[{2\mathrm{x}}^2-\frac{{\mathrm{x}}^3}{3}]}_0^{\mathrm{t}}-\frac{{4\mathrm{t}}^2-{\mathrm{t}}^3}{2}=\frac{{\mathrm{t}}^3}{6}=\frac{9}{2}\Rightarrow \mathrm{t}=3$$

Commented by TawaTawa1 last updated on 11/Apr/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by mr W last updated on 11/Apr/20

$$\frac{2t}{3}[\frac{4t}{2}-\frac{t^2}{4}-\frac{1}{2}(4t-t^2)]=\frac{9}{2}$$$$t^3=27$$$$\Rightarrow t=3$$

Commented by TawaTawa1 last updated on 11/Apr/20

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by TawaTawa1 last updated on 11/Apr/20

$$\mathrm{Sir},\mathrm{you}\mathrm{are}\mathrm{solving}\mathrm{like}\mathrm{a}\mathrm{professional}\mathrm{for}\mathrm{me}.$$

Commented by mr W last updated on 12/Apr/20

$$seeQ88758$$

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