$QUADRATIC \: \: \: RESOLUTIONS \\ \\ \\ \\Function \: to \: model \: the \: height \: of \: the \\ Grasshopper \: above \: the \: ground \: is \: \\ given \: by \: \: - \\ \\ h(t) \: \: = \: \: - {t}^{2} \: + \: \frac{4}{3} t \: + \: \frac{1}{4} \\ \\ \\ When \: the \: grasshopper \: will \: be \: at \: \\ the \: ground \: , \: \: \\ \\ h(t) \: = \: 0 \\ \\ - {t}^{2} \: + \: \frac{4}{3} t \: + \: \frac{1}{4} \: = \: 0 \\ \\ - 12{t}^{2} \: + \: 16 t \: + 3 \: = \: 0 \\ \\ \: \: \: 12 {t}^{2} \: - \: 16t \: - \: 3 \: = \: 0 \\ \\ \: \: \: 12 {t}^{2} \: + \: 2t \: - \: 18t \: - 3 \: = \: 0 \\ \\ \: \: 2t \: (6t + 1) \: - 3 \: (6t + 1) \: = \: 0 \\ \\ \: (2t - 3) \: (6t + 1) \: \: = \: \: 0 \\ \\ Neglecting \: the \: negative \: value \: \\ as \: time \: cannot \: be \: negative \: \: ,\\ \\ We \: get \: \: - \: \\ \\ t \: \: = \: \: \frac{3}{2} \\ \\ \\ Hence \: \: , \: \: After \: \: || \: \: 1.5 \: seconds \: \: || \: , \: the \: \\ Grasshopper \: will \: land \: on \: the \: ground \: .$

3 0

3 years ago

Can you vive me please equivalent fraccions of 14/80 6/18 80/100 Please

-BARSIC- [3]

14/80 = 7/40
6/18 = 1/3
80/100 = 4/5

8 0

2 years ago

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