Someone please help me with this algebra

Mathematics

1 answer:

labwork [276]3 years ago

3 0

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

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Vera_Pavlovna [14]

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points }\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&k&(~ x &,& y~) 
% (c,d)
&l&(~ 0.6 &,& -5.6~)
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{ x_2 + x_1}{2}\quad ,\quad \cfrac{ y_2 + y_1}{2} \right)$

$\bf \left( \cfrac{0.6+x}{2}~~,~~\cfrac{-5.6+y}{2} \right)~=~\stackrel{midpoint}{(4.8~,~-7.8)}\quad 
\begin{cases}
\cfrac{0.6+x}{2}=4.8\\\\
0.6+x=9.6\\
\boxed{x=9}\\
---------\\
\cfrac{-5.6+y}{2}=-7.8\\\\
-5.6+y=-15.6\\
\boxed{y=10}
\end{cases}$

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